🧠:[<think>Okay, so I need to find all three-digit natural numbers that are divisible by 7 and when divided by 9 leave a remainder of 5. Hmm, let's think about this step by step. First, let me parse the problem. I need three-digit numbers, so numbers from 100 to 999. These numbers must satisfy two conditions: they should be divisible by 7, and when divided by 9, the remainder is 5. So, in other words, we're looking for numbers N such that:1. 100 ≤ N ≤ 9992. N ≡ 0 mod 73. N ≡ 5 mod 9Alright, so we have a system of congruences here. The Chinese Remainder Theorem might be useful here since we have two congruencies with moduli that are coprime (7 and 9 are coprime because their GCD is 1). So, there should be a unique solution modulo 63 (since 7*9=63). That means the solutions will repeat every 63 numbers. Once we find the smallest such number that satisfies both conditions and is a three-digit number, we can list all numbers by adding multiples of 63 until we exceed 999.But first, let me find the smallest three-digit number that satisfies both conditions. Wait, maybe I should first find any number that satisfies both congruencies, then adjust it to be within the three-digit range.So, solving the system:N ≡ 0 mod 7N ≡ 5 mod 9Let me express N in terms of one of the congruencies and substitute into the other. Let's take the second one: N ≡ 5 mod 9. So, N can be written as N = 9k + 5 for some integer k. Then substitute this into the first equation: 9k + 5 ≡ 0 mod 7.So, 9k + 5 ≡ 0 mod 7 → 9k ≡ -5 mod 7. But 9 mod 7 is 2, so 2k ≡ -5 mod 7. Let me write that as 2k ≡ 2 mod 7 (since -5 mod 7 is 2). Wait, -5 + 7 = 2, so yeah, that's correct. So, 2k ≡ 2 mod 7. Then divide both sides by 2. Since 2 and 7 are coprime, the inverse of 2 mod 7 is 4 because 2*4=8 ≡1 mod7. So multiplying both sides by 4: k ≡ 2*4 mod7 → k ≡8 mod7 → k≡1 mod7. So k =7m +1 for some integer m.Therefore, N=9k +5=9*(7m +1)+5=63m +9 +5=63m +14. So, all solutions are of the form N=63m +14. Now, we need N to be a three-digit number, so 100 ≤63m +14 ≤999.Let's solve for m:Subtract 14: 86 ≤63m ≤985Divide by 63: 86/63 ≈1.365 ≤m ≤985/63≈15.634Since m must be an integer, m ranges from 2 to 15 inclusive. Wait, hold on: 86/63 is approximately 1.365, so the smallest integer m is 2. Let me check m=1: 63*1 +14=77, which is two-digit, so m=1 gives 77, which is too small. m=2 gives 63*2 +14=126 +14=140. 140 is three-digit. Then m=15: 63*15 +14=945 +14=959. Wait, 63*15 is 945? 63*10=630, 63*5=315, so 630+315=945. 945+14=959. Then m=16: 63*16 +14=1008 +14=1022, which is a four-digit number. So m=2 to m=15.Therefore, the numbers are 63m +14 where m=2 to 15. Let me compute these numbers.Starting with m=2: 63*2 +14=126 +14=140m=3: 63*3 +14=189 +14=203m=4: 63*4 +14=252 +14=266m=5: 63*5 +14=315 +14=329m=6: 63*6 +14=378 +14=392m=7: 63*7 +14=441 +14=455m=8: 63*8 +14=504 +14=518m=9: 63*9 +14=567 +14=581m=10: 63*10 +14=630 +14=644m=11: 63*11 +14=693 +14=707m=12: 63*12 +14=756 +14=770m=13: 63*13 +14=819 +14=833m=14: 63*14 +14=882 +14=896m=15: 63*15 +14=945 +14=959Let me check if all these numbers are three-digit numbers. 140 is three-digit, 959 is the last one, which is also three-digit. So m=2 to m=15 gives numbers from 140 to 959.Now, let me verify if these numbers actually satisfy both conditions. Take the first number, 140.Check if 140 is divisible by 7: 140 ÷7=20, yes. Then, 140 divided by 9: 9*15=135, 140-135=5. So remainder 5. Correct.Another example: 518. 518 ÷7=74. Since 7*74=518. Then 518 ÷9: 9*57=513, 518-513=5. Correct.Another one: 959. 959 ÷7: 7*137=959 (7*130=910, 7*7=49, so 910+49=959). Then 959 ÷9: 9*106=954, 959-954=5. Correct.Seems like all these numbers are satisfying both conditions.Now, let me check if I missed any numbers. For instance, is there a three-digit number less than 140 that satisfies both conditions? The next number in the sequence would be N=63*1 +14=77, which is two-digit, and then N=63*2 +14=140. So no, 140 is the first three-digit number. Then, the next one is 203, which is 63 more. So all numbers are 140, 203, 266, 329, 392, 455, 518, 581, 644, 707, 770, 833, 896, 959. That's 14 numbers. Let's check m=2 to m=15: that's 14 values (since 15-2+1=14). Let me count the numbers I listed: 140, 203, 266, 329, 392, 455, 518, 581, 644, 707, 770, 833, 896, 959. Yes, 14 numbers. That seems right.Alternatively, to confirm, we can compute the number of terms in the arithmetic sequence starting at 140, ending at 959, with common difference 63. The nth term is given by a_n = 140 + (n-1)*63. Set a_n =959:140 + (n-1)*63=959(n-1)*63=959 -140=819n-1=819 /63=13n=14So, there are 14 numbers. Which matches our previous count.Therefore, all three-digit numbers that are divisible by 7 and leave a remainder of 5 when divided by 9 are the 14 numbers listed above.Final AnswerThe three-digit natural numbers are boxed{140}, boxed{203}, boxed{266}, boxed{329}, boxed{392}, boxed{455}, boxed{518}, boxed{581}, boxed{644}, boxed{707}, boxed{770}, boxed{833}, boxed{896}, and boxed{959}.</think>To determine all three-digit natural numbers that are divisible by 7 and give a remainder of 5 when divided by 9, we need to solve the system of congruences:1. ( N equiv 0 pmod{7} )2. ( N equiv 5 pmod{9} )Using the Chinese Remainder Theorem, we express ( N ) in terms of the second congruence: ( N = 9k + 5 ). Substituting this into the first congruence, we get:[9k + 5 equiv 0 pmod{7}]Simplifying, we find:[2k equiv -5 pmod{7} implies 2k equiv 2 pmod{7} implies k equiv 1 pmod{7}]Thus, ( k = 7m + 1 ) for some integer ( m ). Substituting back, we get:[N = 9(7m + 1) + 5 = 63m + 14]We need ( N ) to be a three-digit number, so:[100 leq 63m + 14 leq 999]Solving for ( m ):[86 leq 63m leq 985 implies 1.365 leq m leq 15.634]Thus, ( m ) ranges from 2 to 15. Calculating the corresponding values of ( N ):[begin{aligned}&m = 2: &63 cdot 2 + 14 = 140, &m = 3: &63 cdot 3 + 14 = 203, &m = 4: &63 cdot 4 + 14 = 266, &m = 5: &63 cdot 5 + 14 = 329, &m = 6: &63 cdot 6 + 14 = 392, &m = 7: &63 cdot 7 + 14 = 455, &m = 8: &63 cdot 8 + 14 = 518, &m = 9: &63 cdot 9 + 14 = 581, &m = 10: &63 cdot 10 + 14 = 644, &m = 11: &63 cdot 11 + 14 = 707, &m = 12: &63 cdot 12 + 14 = 770, &m = 13: &63 cdot 13 + 14 = 833, &m = 14: &63 cdot 14 + 14 = 896, &m = 15: &63 cdot 15 + 14 = 959.end{aligned}]Thus, the three-digit natural numbers are:[boxed{140}, boxed{203}, boxed{266}, boxed{329}, boxed{392}, boxed{455}, boxed{518}, boxed{581}, boxed{644}, boxed{707}, boxed{770}, boxed{833}, boxed{896}, boxed{959}],<think>Okay, so I need to find all three-digit natural numbers that are divisible by 7 and when divided by 9 leave a remainder of 5. Let me think about how to approach this. First, let's parse the problem. We have numbers that satisfy two conditions:1. They are three-digit numbers, so between 100 and 999 inclusive.2. They are divisible by 7. That means when you divide them by 7, there's no remainder. In other words, they can be written as 7k for some integer k.3. When they are divided by 9, the remainder is 5. So if we take such a number N, then N ≡ 5 mod 9.So, we need numbers N where 100 ≤ N ≤ 999, N ≡ 0 mod 7, and N ≡ 5 mod 9.This seems like a problem where we can use the Chinese Remainder Theorem, since we have congruencies modulo 7 and 9, which are coprime (since 7 and 9 have no common factors other than 1). So, the Chinese Remainder Theorem tells us that there's a unique solution modulo 63 (since 7*9=63). That means once we find a solution, all solutions will be that number plus multiples of 63.But first, let's find a number that satisfies both conditions. Let's call this number N. So N ≡ 0 mod 7 and N ≡ 5 mod 9.To solve these two congruencies, we can express N as 7k (since it's divisible by 7). Then substitute this into the second congruency:7k ≡ 5 mod 9.So we need to solve for k in this equation. Let's write this as:7k ≡ 5 mod 9.First, let's find the modular inverse of 7 modulo 9. The inverse of 7 mod 9 is a number m such that 7m ≡ 1 mod 9. Let's find m:7*1 =7 ≡7 mod97*2=14≡5 mod97*4=28≡1 mod9. Wait, 7*4=28. 28 divided by 9 is 3 with remainder 1. So 28≡1 mod9. Therefore, the inverse of 7 mod9 is 4.Therefore, multiplying both sides of the equation 7k ≡5 mod9 by 4 gives:k ≡5*4 mod9k≡20 mod920 divided by9 is 2 with remainder 2, so 20≡2 mod9. Therefore, k≡2 mod9.Therefore, k can be written as k=9m +2 for some integer m. Then N=7k=7*(9m +2)=63m +14.So all numbers satisfying both congruencies are of the form N=63m +14.Now, since we need N to be a three-digit number, 100 ≤63m +14 ≤999.Let's solve for m:100 ≤63m +14 ≤999Subtract 14 from all parts:86 ≤63m ≤985Divide all parts by 63:86 /63 ≈1.365 ≤m ≤985/63≈15.634Since m must be an integer, m ranges from 2 to 15 inclusive.Wait, wait, let me check that. Let me compute 86 divided by 63. 63*1=63, 86-63=23, so 86/63=1 +23/63≈1.365. So m must be greater than 1.365, so the smallest integer m is 2.Similarly, 985 divided by 63: 63*15=945, 985-945=40. So 15 +40/63≈15.634. So m must be less than or equal to 15. Therefore, m=2,3,...,15.So the numbers are N=63m +14, where m is from 2 to15.Let me check how many numbers that is. From m=2 to m=15 inclusive, that's 14 numbers. Let me verify the first and last ones.First number when m=2: 63*2 +14=126 +14=140. Is 140 a three-digit number? Yes. Check 140/7=20, which is an integer. 140 divided by9: 9*15=135, 140-135=5. So remainder 5. Good.Last number when m=15:63*15 +14=945 +14=959. Check 959/7=137, which is exact. 959 divided by9: 9*106=954, 959-954=5. So remainder 5. Good.Now, let's list all numbers. Let's compute N=63m +14 for m=2 to15:m=2: 63*2=126 +14=140m=3: 63*3=189 +14=203m=4: 63*4=252 +14=266Wait, 63*4=252. 252+14=266. Hmm, 266. Let me check if 266 is correct. 63*4=252. 252+14=266. Yes. 266 divided by7 is 38, correct. Divided by9: 29*9=261, 266-261=5. Correct.m=5: 63*5=315 +14=329m=6: 63*6=378 +14=392m=7:63*7=441 +14=455m=8:63*8=504 +14=518m=9:63*9=567 +14=581m=10:63*10=630 +14=644m=11:63*11=693 +14=707m=12:63*12=756 +14=770m=13:63*13=819 +14=833m=14:63*14=882 +14=896m=15:63*15=945 +14=959So that's 14 numbers. Let me check if any of these might not be three-digit numbers. The first one is 140, which is three-digit. The last is 959, which is three-digit. All in between should be three-digit numbers.Let me check m=2:140, yes. m=3:203, m=4:266, m=5:329, m=6:392, m=7:455, m=8:518, m=9:581, m=10:644, m=11:707, m=12:770, m=13:833, m=14:896, m=15:959. All between 140 and 959. So all three-digit numbers. So 14 numbers in total.Wait, but let me confirm once again if m=2 is indeed the first one. Let me check m=1:63*1 +14=77. 77 is a two-digit number, which is below 100. So m=1 gives 77, which is invalid. Hence m starts at 2. Similarly, m=16:63*16 +14=1008 +14=1022, which is a four-digit number. So m=15 is indeed the last valid one.Therefore, the numbers are 140,203,266,329,392,455,518,581,644,707,770,833,896,959.Let me check one more number for correctness. Let's take 518. Divided by7: 518/7=74, which is exact. Divided by9: 9*57=513, 518-513=5. So remainder 5. Correct.Another one: 707. 707/7=101, which is exact. 707 divided by9: 9*78=702, remainder 5. Correct.And 833:833/7=119, since 7*119=833. Then 833/9: 9*92=828, remainder 5. Correct.Looks like all these numbers satisfy the conditions.Therefore, the three-digit numbers that are divisible by7 and leave a remainder of5 when divided by9 are the numbers listed above. So the answer is all numbers of the form 63m +14 where m ranges from2 to15, which gives the 14 numbers listed.Final AnswerThe three-digit natural numbers are boxed{140}, boxed{203}, boxed{266}, boxed{329}, boxed{392}, boxed{455}, boxed{518}, boxed{581}, boxed{644}, boxed{707}, boxed{770}, boxed{833}, boxed{896}, and boxed{959}.</think>To determine all three-digit natural numbers that are divisible by 7 and leave a remainder of 5 when divided by 9, we need to solve the following congruences:1. ( N equiv 0 pmod{7} )2. ( N equiv 5 pmod{9} )Using the Chinese Remainder Theorem, we express ( N ) as ( 7k ). Substituting into the second congruence gives:[ 7k equiv 5 pmod{9} ]Finding the modular inverse of 7 modulo 9, we determine that the inverse is 4 (since ( 7 times 4 equiv 1 pmod{9} )). Multiplying both sides of the congruence ( 7k equiv 5 pmod{9} ) by 4, we get:[ k equiv 20 pmod{9} ][ k equiv 2 pmod{9} ]Thus, ( k = 9m + 2 ) for some integer ( m ). Substituting back, we get:[ N = 7(9m + 2) = 63m + 14 ]We need ( N ) to be a three-digit number, so:[ 100 leq 63m + 14 leq 999 ]Solving for ( m ):[ 86 leq 63m leq 985 ][ frac{86}{63} leq m leq frac{985}{63} ][ 1.365 leq m leq 15.634 ]Since ( m ) must be an integer, ( m ) ranges from 2 to 15. Calculating ( N ) for each ( m ):- ( m = 2 ): ( 63 times 2 + 14 = 140 )- ( m = 3 ): ( 63 times 3 + 14 = 203 )- ( m = 4 ): ( 63 times 4 + 14 = 266 )- ( m = 5 ): ( 63 times 5 + 14 = 329 )- ( m = 6 ): ( 63 times 6 + 14 = 392 )- ( m = 7 ): ( 63 times 7 + 14 = 455 )- ( m = 8 ): ( 63 times 8 + 14 = 518 )- ( m = 9 ): ( 63 times 9 + 14 = 581 )- ( m = 10 ): ( 63 times 10 + 14 = 644 )- ( m = 11 ): ( 63 times 11 + 14 = 707 )- ( m = 12 ): ( 63 times 12 + 14 = 770 )- ( m = 13 ): ( 63 times 13 + 14 = 833 )- ( m = 14 ): ( 63 times 14 + 14 = 896 )- ( m = 15 ): ( 63 times 15 + 14 = 959 )All these numbers are verified to be three-digit, divisible by 7, and leave a remainder of 5 when divided by 9.The three-digit natural numbers are boxed{140}, boxed{203}, boxed{266}, boxed{329}, boxed{392}, boxed{455}, boxed{518}, boxed{581}, boxed{644}, boxed{707}, boxed{770}, boxed{833}, boxed{896}, and boxed{959}.]

🔑:First solution. From the conditions in the problem, we haven=7 x quad text{and} quad n=9 y+5, quad x, y in mathbb{N}We solve the Diophantine equation 7 x=9 y+5, or7 x-9 y=5One of its solutions is x_{0}=2 and y_{0}=1.Then the general solution is x=2+9 t, y=1+7 t, t in mathbb{N}.Since 100<n<1000, and n=7 x=14+63 t, we havebegin{array}{r}100<14+63 t<1000 86<63 t<986 1.3<t<15.6 t in{2,3,4, ldots, 15}end{array}Thus, n=14+63 t, t in{2,3,4, ldots, 15}, or written out,n in{140,203,266,329,392,455,518,581,644,707,770,833,896,959}

🧠:[<think>Okay, so there's this problem where a boy has to play three tennis matches against his parents. The goal is for him to win two matches in a row. The father is a better player than the mother. The question is asking which order of matches is better for the boy: "father-mother-father" (F-M-F) or "mother-father-mother" (M-F-M). Hmm, let me think through this step by step.First, let's parse the problem again. The boy needs to win two consecutive matches to be considered the winner. The matches are three in total. The opponents are his father and mother, with the father being a stronger player. So, the order in which he plays against them might affect his chances of getting two wins in a row.Let me consider both orders separately. Let's start with the "father-mother-father" (F-M-F) order.In this sequence, the boy plays the first match against his father, then against his mother, and then against his father again. To win two in a row, he needs either the first and second matches, the second and third matches, or all three. But since there are only three matches, the possible consecutive wins could be matches 1 and 2, or 2 and 3.Wait, but the problem states he must win two matches in a row. So, if he wins the first two, he's a winner. If he loses the first, wins the second and third, that's also two in a row. If he wins the first, loses the second, and wins the third, that's not two in a row. Similarly, if he loses the first and second, then wins the third, that's only one win. So, the winning scenarios are: Win-Win-Any (since the third match doesn't matter if he already won the first two), or Lose-Win-Win. So, two possible paths to victory.Similarly, for the "mother-father-mother" (M-F-M) order, the same logic applies. The boy plays mother first, then father, then mother again. The winning scenarios are: Win-Win-Any or Lose-Win-Win.Now, since the father is a stronger player, the boy's chances of winning against the father are lower than against the mother. Let's denote the probability of the boy winning against the father as p and against the mother as q, with q > p because the mother is a weaker opponent.So, in the F-M-F order, let's calculate the probability of the boy winning two matches in a row. Let's break down the possible sequences:1. Win first (against father) and win second (against mother): Probability is p * q. If he wins the first two matches, he's already achieved two in a row, so the third match doesn't matter. So, this contributes p*q to the total probability.2. Lose first (against father), then win second (against mother) and win third (against father): Probability is (1 - p) * q * p. Here, he loses the first match but then wins the next two, which are consecutive. So, that's another path.3. Any other scenarios where he doesn't get two consecutive wins: For example, win first, lose second, win third; or lose first, lose second, win third; etc. These don't count towards his victory.Therefore, the total probability for F-M-F order is p*q + (1 - p)*q*p = pq + pq(1 - p) = pq(1 + (1 - p)) = pq(2 - p). Wait, let me check that again.Wait, actually, the two scenarios are:First scenario: Win first (p) and win second (q). Then, regardless of the third match, he already has two consecutive wins. So, the probability is p*q.Second scenario: Lose first (1 - p), then win second (q), then win third (p). So, this is (1 - p)*q*p.So, total probability is p*q + (1 - p)*q*p = pq + pq(1 - p) = pq[1 + (1 - p)] = pq(2 - p).Wait, no. Let's compute it step by step.First term: p (win against father) * q (win against mother). The third match is irrelevant here because he already has two consecutive wins. So, probability is p*q.Second term: (1 - p) (lose against father) * q (win against mother) * p (win against father). So, (1 - p)*q*p.So, total probability is p*q + (1 - p)*q*p = pq[1 + (1 - p)] = pq(2 - p). Wait, no, 1 + (1 - p) is 2 - p. So, total probability is pq(2 - p). Hmm.Similarly, for the M-F-M order, let's do the same.First, the order is mother, father, mother.So, the winning scenarios are:1. Win first (against mother) and win second (against father): Probability q * p. Then, regardless of the third match, he has two consecutive wins.2. Lose first (against mother) (probability 1 - q), then win second (against father) (p) and win third (against mother) (q). So, (1 - q)*p*q.So, total probability here is q*p + (1 - q)*p*q = pq + pq(1 - q) = pq[1 + (1 - q)] = pq(2 - q).Therefore, comparing the two total probabilities: For F-M-F, it's pq(2 - p); for M-F-M, it's pq(2 - q). Since q > p, because the mother is a weaker opponent, so 2 - q is less than 2 - p. Therefore, pq(2 - p) versus pq(2 - q). Since 2 - p > 2 - q, because q > p, then pq(2 - p) > pq(2 - q). Therefore, the probability is higher for F-M-F order.Wait, hold on, that seems counterintuitive. If the father is stronger, so playing the father twice would be harder, but here in F-M-F, he plays father first and third, whereas in M-F-M, he plays mother first and third, father in the middle. Wait, but according to the calculation, F-M-F gives higher probability.But let's check the calculations again.For F-M-F:Total probability = p*q + (1 - p)*q*p= pq + pq(1 - p)= pq(1 + 1 - p)= pq(2 - p)For M-F-M:Total probability = q*p + (1 - q)*p*q= pq + pq(1 - q)= pq(1 + 1 - q)= pq(2 - q)Since q > p, then (2 - q) < (2 - p). Therefore, pq(2 - p) > pq(2 - q). Therefore, F-M-F is better.Hmm, so according to this, playing father first and third, with mother in the middle, gives higher probability.But intuitively, since the father is stronger, playing him twice might seem worse. But maybe the positioning of the matches allows for more favorable sequences?Wait, let's think of specific numbers. Let me plug in some numbers for p and q where q > p. Let's say p = 0.4 (40% chance to beat father) and q = 0.6 (60% chance to beat mother).Then, for F-M-F:Total probability = 0.4*0.6*(2 - 0.4) = 0.24 * 1.6 = 0.384For M-F-M:Total probability = 0.6*0.4*(2 - 0.6) = 0.24 * 1.4 = 0.336So, 0.384 vs 0.336, so F-M-F is better. Hmm, interesting.Alternatively, take p = 0.3, q = 0.7.F-M-F: 0.3*0.7*(2 - 0.3) = 0.21 * 1.7 = 0.357M-F-M: 0.7*0.3*(2 - 0.7) = 0.21 * 1.3 = 0.273Again, F-M-F is better.Wait, so according to this, it's better to have the father first and third. But why?Looking at the two scenarios:In F-M-F, the two paths to victory are:1. Win first two matches (F then M). The first match is against the stronger opponent. If he can win that, then the next is against the weaker, which is more likely. So, winning F then M.2. Lose the first (against F), then win M and F. So, losing to father first, then beating mother and father. So, this path requires beating father in the third match.In M-F-M, the paths are:1. Win first two matches (M then F). Beating mother, then father.2. Lose the first (against M), then beat F and M. So, losing to mother, then beating father and mother.So, comparing the two:In F-M-F, the first path is beat F and M. The second path is lose F, then beat M and F.In M-F-M, the first path is beat M and F. The second path is lose M, then beat F and M.Since beating father is harder, maybe the second path in M-F-M (which requires beating father and then mother) is more probable than the second path in F-M-F (which requires beating mother then father)?Wait, let's look at the second terms:In F-M-F, the second term is (1 - p)*q*p. So, probability of losing to father, then beating mother, then beating father.In M-F-M, the second term is (1 - q)*p*q. So, probability of losing to mother, then beating father, then beating mother.Comparing (1 - p)*q*p vs (1 - q)*p*q. Since q > p, (1 - p) < (1 - q) if q > p. Because if q is bigger, then (1 - q) is smaller. Wait, no: if q = 0.6, (1 - q) = 0.4. If p = 0.4, (1 - p) = 0.6. So, (1 - p) > (1 - q) when p < q. So, (1 - p) is greater than (1 - q). Therefore, (1 - p)*q*p is greater than (1 - q)*p*q?Wait, let's compute:(1 - p)*q*p vs (1 - q)*p*q.Factor out p*q: we get [(1 - p) vs (1 - q)] * p*q.Since (1 - p) > (1 - q) because p < q, then indeed (1 - p)*q*p > (1 - q)*p*q. Therefore, the second term in F-M-F is larger than the second term in M-F-M.Therefore, even though the second term in F-M-F requires beating father in the third match, which is harder, the probability of getting to that scenario is higher because (1 - p) is higher than (1 - q). Because losing to father (first match) is more probable than losing to mother (first match in M-F-M).Therefore, even though the second path in F-M-F involves a harder match (father) at the end, the fact that losing the first match is more likely (since father is stronger) allows for a higher probability in the second term. So overall, the total probability is higher for F-M-F.So, according to the calculations, the boy should prefer the "father-mother-father" order. But this seems counterintuitive at first glance because he has to play the stronger opponent twice. But the key is that the sequence allows for more favorable paths where he can win two in a row, either by starting with a win against the father (which is hard, but if he does, then the next is easier) or by losing the first, then winning the next two (mother and father). The second term's higher probability due to higher chance of losing the first match (since it's against the stronger opponent) compensates for the difficulty in the third match.Alternatively, in M-F-M order, the first match is against the mother, which he is more likely to win. If he wins that, then he has to beat the father, which is harder. If he loses the first match against the mother (which is less probable), then he has to beat the father and mother in a row, which again is difficult because beating the father is tough.So, in M-F-M, even though he has a better chance of winning the first match, the subsequent matches are harder. Whereas in F-M-F, even though the first match is harder, the subsequent matches after a loss are easier, leading to a higher overall probability.Therefore, the answer is that the boy should prefer the "father-mother-father" order.Wait, but let me check with different values. Let's take p = 0.2, q = 0.8.F-M-F: 0.2*0.8*(2 - 0.2) = 0.16 * 1.8 = 0.288M-F-M: 0.8*0.2*(2 - 0.8) = 0.16 * 1.2 = 0.192So, F-M-F is still better. The difference increases as the gap between p and q widens.Another test: p = 0.5, q = 0.6.F-M-F: 0.5*0.6*(2 - 0.5) = 0.3 * 1.5 = 0.45M-F-M: 0.6*0.5*(2 - 0.6) = 0.3 * 1.4 = 0.42Still F-M-F is better.Wait, even when the mother is only slightly better than the father? Wait, no, in the problem statement, the father is better than the mother, so p < q. Wait, hold on. Wait, wait. Wait, wait a second. Wait, hold on, hold on. Wait, the problem says "the father plays better than the mother." So, the father is a better player than the mother, so the boy's probability of winning against the father is lower than against the mother. Therefore, p is the probability of beating father, q is the probability of beating mother, and since father is better, p < q.But in my previous examples, I assumed that. For example, p = 0.4, q = 0.6, which is p < q.But in the last test case, I set p = 0.5 and q = 0.6. Wait, that would mean the father is better, so the boy has a lower chance against the father. If p is the probability of beating the father, and q is the probability of beating the mother, then if father is better, p should be less than q. So, in that test case, p = 0.5 and q = 0.6 is still okay, as long as p < q.But even in that case, F-M-F is still better.Therefore, according to all these test cases, the F-M-F order is better. So, maybe the answer is that the boy should choose the "father-mother-father" order.But wait, let me think differently. Maybe the problem is about the arrangement of the matches such that the boy can get two consecutive wins. If he alternates opponents, perhaps playing the weaker opponent in the middle gives him a better chance to have two in a row.Alternatively, maybe if he plays the weaker opponent twice, but in the problem, he can't, because it's three matches alternating between parents. Wait, the two possible orders are F-M-F and M-F-M. So, in F-M-F, he plays father, then mother, then father. So, two fathers and one mother. Wait, no: the total matches are three. So, in F-M-F, he plays father, mother, father: two matches against father and one against mother. In M-F-M, he plays mother, father, mother: two against mother and one against father. Wait, but since the father is better, maybe playing the weaker opponent more times is better? But in the problem, he has to play three matches, alternating between parents.Wait, the problem says: "he will play three tennis matches with his parents." The order is either F-M-F or M-F-M. So, in one case, he plays father first, then mother, then father again. In the other case, mother first, then father, then mother again.So, in F-M-F, he has two matches against father and one against mother. In M-F-M, he has two against mother and one against father. Therefore, if he plays two matches against the weaker opponent (mother), that might be better. But according to our previous calculations, even with two matches against the father, the probability is higher. Wait, but in our model, we considered that the third match's result only matters in certain scenarios. For example, in the F-M-F case, if he wins the first two matches, the third doesn't matter. If he loses the first, then the third match is necessary.But perhaps, by having two matches against the father, he's increasing the chance of having to play the father in critical matches. But according to the math, it's still better. Let me think with p = 0.1 and q = 0.9.F-M-F: 0.1*0.9*(2 - 0.1) = 0.09 * 1.9 = 0.171M-F-M: 0.9*0.1*(2 - 0.9) = 0.09 * 1.1 = 0.099So, even with a very low chance against father, F-M-F is still better. So, in this case, playing father twice is better. Why?Because the two paths to victory are:1. Win first two matches: father then mother. Even though beating father is hard (0.1), then beating mother is easy (0.9), so 0.1*0.9 = 0.09.2. Lose first, then win mother and father: lose father (0.9), then beat mother (0.9) and father (0.1). So, 0.9*0.9*0.1 = 0.081. Total probability is 0.09 + 0.081 = 0.171.For M-F-M:1. Win first two matches: mother (0.9) then father (0.1). So, 0.9*0.1 = 0.09.2. Lose first (0.1), then win father (0.1) and mother (0.9). So, 0.1*0.1*0.9 = 0.009. Total probability is 0.09 + 0.009 = 0.099.So, even here, F-M-F is better. The reason is that even though the second path in F-M-F requires beating father at the end, the probability of getting to that scenario (losing first match against father) is high (0.9), which compensates for the low probability of the subsequent wins.Whereas in M-F-M, losing the first match (against mother) is low (0.1), so the second path has a low probability of occurring, despite the subsequent matches being relatively easier (father then mother).Therefore, the total probability ends up higher in F-M-F.Therefore, regardless of the specific probabilities (as long as p < q), the order F-M-F gives a higher chance of winning two consecutive matches. Therefore, the boy should prefer the "father-mother-father" order.But wait, the problem says "he must play three matches", does that mean he has to play all three regardless of earlier outcomes? For example, even if he wins the first two matches, he still has to play the third. But the problem says "he will be considered the winner if he wins two matches in a row." So, does the three matches are all played regardless, and he just needs two consecutive wins at any point? So, even if he wins the first two, he still plays the third, but he's already a winner. Similarly, if he loses the first, then wins the next two, he is a winner. But if he wins the first, loses the second, wins the third, he doesn't have two in a row. So, in that case, our previous calculations are correct, considering all three matches are played.Therefore, with that in mind, the order F-M-F is preferable.But let me just confirm once again with another perspective.Suppose the boy wants two consecutive wins. To maximize this, he should arrange the matches so that whenever he has a win, the next match is as easy as possible (against the mother) to maximize the chance of getting another win. Conversely, if he loses a match, the next match should be easier to get back on track.In F-M-F order:- First match against father: if he wins (hard), then next is mother (easy), high chance to get two in a row.- If he loses first (probable), then next is mother (easy). If he wins that, then next is father again (hard). So, even if he loses first, he has a path: lose (father), win (mother), win (father). Although beating father is hard, the chance is non-zero.In M-F-M order:- First match against mother: if he wins (easy), next is father (hard). So, lower chance to get the second win.- If he loses first (unlikely), next is father (hard). If he beats father (unlikely), then next is mother (easy). So, lose (mother), win (father), win (mother). But losing the first is less likely, and beating father is hard.Therefore, in F-M-F, the critical matches after a win are against easier opponents, which helps in getting the two in a row. Whereas in M-F-M, after a win against mother, he has to face father, which is harder. Similarly, after a loss in M-F-M, he has to face father, which is not helpful.Therefore, structuring the matches so that after a win, the next match is against the weaker player (mother) gives a better chance. That happens in F-M-F order. If he can win against father first, then mother is next, which is easier. But in M-F-M order, after beating mother, he has to face father, which is harder. So, the structure of F-M-F allows for a more favorable sequence after a win.Therefore, this further supports the conclusion that F-M-F is better.Hence, the answer should be that the boy should prefer the "father-mother-father" order.Final AnswerThe preferable order for the boy is boxed{father-mother-father}.</think>The boy needs to win two consecutive tennis matches out of three against his parents, with the father being a stronger player than the mother. We need to determine which order of matches, "father-mother-father" (F-M-F) or "mother-father-mother" (M-F-M), is preferable for the boy.To analyze this, we denote the probability of the boy winning against the father as ( p ) and against the mother as ( q ), where ( q > p ).For the F-M-F order:1. Winning the first two matches (F then M): Probability is ( p cdot q ).2. Losing the first match (F), then winning the next two (M and F): Probability is ( (1 - p) cdot q cdot p ).Total probability for F-M-F:[p cdot q + (1 - p) cdot q cdot p = pq + pq(1 - p) = pq(2 - p)]For the M-F-M order:1. Winning the first two matches (M then F): Probability is ( q cdot p ).2. Losing the first match (M), then winning the next two (F and M): Probability is ( (1 - q) cdot p cdot q ).Total probability for M-F-M:[q cdot p + (1 - q) cdot p cdot q = pq + pq(1 - q) = pq(2 - q)]Comparing the two probabilities, since ( q > p ), we have ( 2 - p > 2 - q ). Therefore, ( pq(2 - p) > pq(2 - q) ), indicating that the F-M-F order provides a higher probability of winning two consecutive matches.Thus, the preferable order for the boy is boxed{father-mother-father}.,<think>Okay, let's see. The problem is about a boy who needs to play three tennis matches against his parents. His goal is to win two matches in a row to be considered the winner. The father is a better player than the mother. The question is asking which order of matches is better for the boy: playing father first, then mother, then father again (F-M-F), or mother first, then father, then mother again (M-F-M). Hmm, so the key here is that the boy needs to win two consecutive matches. Since the father is a better player, the boy's chances of winning against the father are lower than against the mother. So, maybe the order affects the probability of getting two wins in a row. Let me think.First, let's consider both orders separately. Let's denote the probability of the boy winning against his father as p and against his mother as q. Since the father plays better than the mother, we know that p < q. So, p is less than q. Now, for each possible order, we need to calculate the probability that the boy wins two matches in a row. Let's start with the first order: Father-Mother-Father (F-M-F).In this case, the matches are in the order F, M, F. The boy needs to win two in a row. Since there are three matches, the possible sequences where he wins two in a row are:1. Win the first two matches (F and M). Then the third match doesn't matter because he already has two consecutive wins.2. Lose the first match, then win the next two (M and F). 3. Win the second and third matches. However, if he loses the first match but wins the second and third, that's two in a row (M and F). Wait, actually, in three matches, the possible two consecutive wins can be matches 1-2 or 2-3. So, if he wins matches 1 and 2, or 2 and 3. Wait, the problem says he must win two matches in a row. So, even if he wins all three, that's still two in a row. But does he need exactly two, or at least two? The problem states "he will be considered the winner if he wins two matches in a row." So, as soon as he has two consecutive wins, he is the winner. So, the matches might stop early if he wins the first two. But since there are three matches scheduled, maybe he has to play all three regardless? Wait, the problem says "he must play three tennis matches with his parents." So he has to play all three, but he is considered the winner if he wins two in a row at any point. So even if he loses the third match, as long as he has two consecutive wins in the three, he is the winner. Or does he need to have two consecutive wins within the three matches? So, if he wins matches 1 and 2, he is a winner. If he loses match 1, wins 2 and 3, he is a winner. If he wins match 1, loses match 2, wins match 3, then he doesn't have two in a row, so he loses. Similarly, if he loses matches 1 and 2, then wins match 3, he still doesn't have two in a row. So the total ways he can win are: either win the first two, or win the last two. Therefore, for the F-M-F order, the probability of winning is the probability of winning the first two matches plus the probability of losing the first match but winning the next two. But wait, if he wins the first two matches, he already has two in a row, so the third match is irrelevant. But he still has to play it? The problem states he must play three matches, so even if he wins the first two, he still plays the third. However, the condition is just that he has two wins in a row at any point. So, even if he wins the first two, he's already a winner, regardless of the third match. But the problem says "he will be considered the winner if he wins two matches in a row." So once he has two in a row, he is the winner. But does that stop the matches, or does he still have to play all three? The problem says "he must play three tennis matches," so I think all three are played regardless. Therefore, even if he wins the first two, he still has to play the third one. But since he already has two in a row, he is considered a winner regardless of the third match's outcome. Similarly, if he loses the first, then wins the next two, he is a winner. If he wins the first, loses the second, then wins the third, he only has one win, so he doesn't get two in a row. If he loses the first two, then wins the third, same thing. So the total probability is P(win first two) + P(lose first, win next two). Similarly for the other order.So let's formalize that. For the F-M-F order:Probability of winning = P(win F) * P(win M) + P(lose F) * P(win M) * P(win F)Similarly, for the M-F-M order:Probability of winning = P(win M) * P(win F) + P(lose M) * P(win F) * P(win M)Wait, but let's check that. For M-F-M:If he wins the first two matches (M and F), he's a winner. If he loses the first match (M), then he needs to win the next two (F and M) to have two in a row. So the probability is P(win M)*P(win F) + P(lose M)*P(win F)*P(win M)But both orders have similar expressions, but with different probabilities. Let me denote p as probability of beating father, q as probability of beating mother, with p < q.So for F-M-F:Probability = p*q + (1 - p)*q*pSimilarly, for M-F-M:Probability = q*p + (1 - q)*p*qSo let's compute both.First, F-M-F:= pq + (1 - p)qp= pq + pq(1 - p)= pq[1 + (1 - p)]= pq[2 - p]Wait, no, wait. Let me compute step by step.First term: p*qSecond term: (1 - p)*q*pSo total probability is pq + (1 - p)pq= pq[1 + (1 - p)] = pq(2 - p)Similarly, for M-F-M:First term: q*pSecond term: (1 - q)*p*qTotal probability: qp + (1 - q)pq= pq[1 + (1 - q)] = pq(2 - q)Therefore, comparing F-M-F probability pq(2 - p) versus M-F-M probability pq(2 - q)Since p < q, then (2 - p) > (2 - q), because since p is smaller, subtracting a smaller number gives a larger result. So pq(2 - p) > pq(2 - q). Therefore, the F-M-F order gives a higher probability of winning.Wait, but hold on. Wait, pq is common in both, so if we factor that out, the difference is in (2 - p) vs (2 - q). Since p < q, 2 - p > 2 - q. Therefore, F-M-F is better. Hmm, so according to this, the father-mother-father order is better? But that seems counter-intuitive because the father is harder to beat. Maybe my analysis is wrong.Wait, let's verify with an example. Let's suppose p = 0.4 and q = 0.6. Then:F-M-F probability = 0.4*0.6*(2 - 0.4) = 0.24*(1.6) = 0.384M-F-M probability = 0.6*0.4*(2 - 0.6) = 0.24*(1.4) = 0.336So 0.384 vs 0.336, so F-M-F is better. Hmm, interesting. Wait, but why? Even though the father is harder, arranging the matches as F-M-F gives a higher chance. Let me think why.Because in the F-M-F order, the middle match is against the mother, which is easier. So if he can win the first two (F and M), he's done. If he loses the first against the father, he can still win the next two (M and F). Since the middle match is easier, even if he loses the first hard match, he can still get two in a row by beating the mother and then the father. Whereas in the M-F-M order, the middle match is against the father, which is harder. So if he wins the first against the mother, he still has to beat the father in the middle, which is harder. If he loses the first against the mother, then he needs to beat the father and mother, which might be difficult.Wait, but in the M-F-M order, the two possible ways to get two in a row are:1. Win M and F2. Lose M, then win F and MSo the probability is q*p + (1 - q)*p*qIn the F-M-F order, it's p*q + (1 - p)*q*pSo, substituting numbers when p=0.4, q=0.6:F-M-F: 0.4*0.6 + 0.6*0.6*0.4 = 0.24 + 0.144 = 0.384M-F-M: 0.6*0.4 + 0.4*0.4*0.6 = 0.24 + 0.096 = 0.336So F-M-F is better. So in this case, even though the father is harder, playing him first and third with the mother in the middle is better. The key is that in F-M-F, the middle match is easier, so if you can win that, even if you lost the first, you have a chance to beat the father in the third. Whereas in M-F-M, the middle match is harder, so if you lose the first, you have to beat the father and then the mother again.Alternatively, maybe the difference comes from the fact that in F-M-F, the two possible winning sequences are F-M and M-F (if he loses the first). But since the middle match is easier, the M is easier to win, so the second term (lose F, win M-F) is more probable. Whereas in M-F-M, the two possible sequences are M-F and F-M. However, the middle match is F, which is harder, so the second term (lose M, win F-M) is less probable because winning F is harder. Therefore, even though starting with father is harder, having the mother in the middle provides a better chance to recover if you lose the first match. Whereas in the other order, starting with mother, if you lose the first, you have to win against father and mother, which is harder.Alternatively, maybe the formula is correct. Since in the F-M-F order, the probability is pq(2 - p), and in the other order pq(2 - q). Since p < q, (2 - p) > (2 - q), so F-M-F is better.But wait, let's think in another angle. The two scenarios where the boy can win are:1. Winning the first two matches.2. Losing the first match and winning the next two.In F-M-F:1. Win F and M: probability p*q2. Lose F, win M and F: (1 - p)*q*pTotal: p*q + (1 - p)*q*pIn M-F-M:1. Win M and F: q*p2. Lose M, win F and M: (1 - q)*p*qTotal: q*p + (1 - q)*p*qSo, the difference between the two total probabilities is:[ p*q + (1 - p)*q*p ] - [ q*p + (1 - q)*p*q ] = (1 - p)*q*p - (1 - q)*p*q= pq[ (1 - p) - (1 - q) ]= pq[ (1 - p - 1 + q) ]= pq[ q - p ]Since q > p, this difference is positive, so F-M-F is better.Therefore, regardless of the actual values of p and q (as long as p < q), the F-M-F order gives a higher probability.Wait, but this seems counterintuitive. If the father is harder, why would playing him twice (F-M-F) be better than playing the mother twice (M-F-M)? Because in F-M-F, the boy plays father twice, and mother once. In M-F-M, he plays mother twice, and father once. Since the father is harder, playing him more times would seem worse. But according to the calculation, it's better. Hmm.But in the calculation, even though he plays the father twice, the key is the position of the mother match. In F-M-F, the mother is in the middle, which allows a higher chance of getting two wins in a row, either by starting with F and M or by losing F then winning M and F. Since the middle match is easier, the chance of winning two in a row is higher. Whereas in M-F-M, the middle match is against the father, which is harder, so the chance of winning two in a row is lower.So even though he plays the father more times, the structure of the matches allows for a better recovery path.Alternatively, maybe the number of times he plays each parent is less important than the sequence. For example, in F-M-F, the critical middle match is easier, which can help bridge between the two harder matches. If he can win the middle match, he can connect a win from either the first or the third match. Since the middle is easier, that's better.In contrast, in M-F-M, the middle match is harder, so even if he starts with a win against the mother, he still has to face the father next. If he loses the middle match, then he needs to win the third, but the third is mother again, which is easier, but he needs two in a row, so he would have to win the third after losing the second. But two in a row would require winning the second and third, but if he lost the second, he can't get two in a row in that case.Wait, no. If he plays M-F-M:- If he wins the first two (M and F), he's a winner.- If he loses the first (M), then he needs to win F and M. But since he lost the first, even if he wins F and M, he has two wins in a row (F and M), but since the first was a loss, the only two in a row would be the second and third. Wait, no. If he loses the first, then the second and third are F and M. If he wins both, then he has two in a row (F and M). But in that case, the two in a row are matches 2 and 3, which counts. So, in M-F-M order, the two possible ways are matches 1-2 or 2-3. Similarly for F-M-F.So in both orders, the two possible two-in-a-rows are the first two or the last two. The difference is in the opponents. In F-M-F, the first two opponents are F and M, so winning two in a row requires beating F and then M. The last two are M and F, which requires beating M and then F. In M-F-M, the first two are M and F, and the last two are F and M. So, the probability of winning two in a row in F-M-F is p*q + (1 - p)*q*pIn M-F-M, it's q*p + (1 - q)*p*qSo, the difference between these two is:pq + (1 - p)pq - [qp + (1 - q)pq] = pq + pq - p²q - pq - pq + q²pWait, maybe expanding both expressions:F-M-F: pq + (1 - p)pq = pq + pq - p²q = 2pq - p²qM-F-M: qp + (1 - q)qp = qp + qp - q²p = 2qp - q²pTherefore, the difference is (2pq - p²q) - (2qp - q²p) = - p²q + q²p = pq(q - p)Since q > p, this is positive, so F-M-F is better. Therefore, the F-M-F order gives a higher probability of winning.So according to this, the answer is that the boy should prefer the father-mother-father order.But wait, let me check with different values. Suppose p = 0.3, q = 0.7.F-M-F: 0.3*0.7 + 0.7*0.7*0.3 = 0.21 + 0.147 = 0.357M-F-M: 0.7*0.3 + 0.3*0.3*0.7 = 0.21 + 0.063 = 0.273So again, F-M-F is better. So even if the probabilities are more different, the result still holds.Alternatively, let's take extreme values. Let p = 0.1 (almost no chance to beat father), q = 0.9 (almost certain to beat mother).F-M-F: 0.1*0.9 + 0.9*0.9*0.1 = 0.09 + 0.081 = 0.171M-F-M: 0.9*0.1 + 0.1*0.1*0.9 = 0.09 + 0.009 = 0.099Still, F-M-F is better, although both probabilities are low. The difference is 0.171 vs 0.099. So even in this case, F-M-F is preferable.Wait, so why is this happening? Even though the boy is very unlikely to beat the father, the F-M-F order gives a better chance. Because in F-M-F, he has two chances to get the father: first and third. But the key is the middle match against the mother. If he can win the middle match, which is very likely, then even if he loses the first, he can try to beat the father in the third. Since q is high, the (1 - p)*q*p term can be significant. Wait, with p=0.1, q=0.9:(1 - p)*q*p = 0.9*0.9*0.1 = 0.081Whereas in M-F-M:(1 - q)*p*q = 0.1*0.1*0.9 = 0.009So the second term in F-M-F is much larger, even though p is low. Because even though he's unlikely to beat the father, if he loses the first match (which is against the father, high probability), then he has a high chance of winning the middle match (mother) and then a small chance of beating the father again. But the small chance is multiplied by the high q. Whereas in M-F-M, if he loses the first match (mother, which is low probability), then he has to beat the father and mother again, which is very low probability. So in F-M-F, the second term (recovery path) is more probable because losing the first match (father) is likely, and then winning the next two (mother and father) is possible, even if low. Whereas in M-F-M, losing the first match (mother) is unlikely, but even if he does, the chance to recover is very low.Therefore, the F-M-F order allows for a more probable recovery path, even though it involves playing the father twice. Because the key is that the middle match is against the weaker parent, which allows a higher chance to link two wins across the harder parents.This seems to consistently give a higher probability. So according to the calculations, the answer should be father-mother-father order is preferable.But wait, the problem says "he must play three tennis matches with his parents. He will be considered the winner if he wins two matches in a row." So does the order matter in terms of when you play the easier opponent?Alternatively, maybe there's another approach. Let's model all possible sequences.For F-M-F:Possible sequences where he wins two in a row:1. WWW2. WWL3. LWW4. WLW (doesn't have two in a row)5. WLL (no)6. LWL (no)7. LLW (no)8. LLL (no)So winning sequences are WWW, WWL, LWW. So three outcomes. But actually, in terms of probability:- WWW: p*q*p- WWL: p*q*(1 - p)- LWW: (1 - p)*q*pSo total probability: p*q*p + p*q*(1 - p) + (1 - p)*q*p = p*q [p + (1 - p) + (1 - p)] ?Wait, no:Wait, WWW: p*q*pWWL: p*q*(1 - p)LWW: (1 - p)*q*pSo total probability is p*q*p + p*q*(1 - p) + (1 - p)*q*p = pq [p + (1 - p) + (1 - p)]Wait, let's compute:First term: p*q*pSecond term: p*q*(1 - p)Third term: (1 - p)*q*pSo combining:= p^2 q + pq(1 - p) + pq(1 - p)= p^2 q + 2 pq (1 - p)Similarly, for M-F-M:Winning sequences are WWW, WWL, LWWCalculating similarly:WWW: q*p*qWWL: q*p*(1 - q)LWW: (1 - q)*p*qTotal probability:= q*p*q + q*p*(1 - q) + (1 - q)*p*q= q^2 p + qp(1 - q) + pq(1 - q)= q^2 p + 2 pq(1 - q)So comparing F-M-F's p^2 q + 2 pq(1 - p) vs M-F-M's q^2 p + 2 pq(1 - q)Let me compute the difference:F-M-F - M-F-M = p^2 q + 2 pq(1 - p) - q^2 p - 2 pq(1 - q)= p^2 q - q^2 p + 2 pq [ (1 - p) - (1 - q) ]= pq(p - q) + 2 pq (q - p)= pq(p - q) - 2 pq(p - q)= - pq(p - q) [since (q - p) = -(p - q)]= pq(q - p)Since q > p, this is positive. Hence, F-M-F is better.Therefore, regardless of the method, F-M-F gives a higher probability.Therefore, the answer should be "father-mother-father" order is preferable.But wait, in the earlier calculation, we considered the probability as pq(2 - p) vs pq(2 - q). But according to this last calculation, the probabilities are p²q + 2pq(1 - p) vs q²p + 2pq(1 - q). Which also simplifies to:F-M-F: pq [ p + 2(1 - p) ] = pq [ 2 - p ]M-F-M: pq [ q + 2(1 - q) ] = pq [ 2 - q ]Which is the same as earlier. So indeed, F-M-F is better because (2 - p) > (2 - q)Therefore, the conclusion is that the boy should choose the father-mother-father sequence.But this feels counterintuitive because playing against the stronger opponent more times seems worse, but the mathematics shows that due to the positioning of the easier match in the middle, which allows a bridge between the harder matches, the overall probability is higher.Therefore, after careful consideration and calculation, the answer is that the father-mother-father order is preferable.Final AnswerThe preferable order is boxed{father-mother-father}.</think>The problem involves a boy who needs to win two consecutive tennis matches out of three against his parents, where the father is a better player than the mother. The goal is to determine which order of matches, "father-mother-father" (F-M-F) or "mother-father-mother" (M-F-M), is preferable for the boy.1. Define Probabilities: - Let ( p ) be the probability of winning against the father. - Let ( q ) be the probability of winning against the mother, with ( p < q ).2. Calculate Probabilities for Each Order: - F-M-F Order: - The boy can win by either winning the first two matches or losing the first and winning the next two. - Probability = ( p cdot q + (1 - p) cdot q cdot p ) - Simplifies to ( pq + pq(1 - p) = pq(2 - p) ) - M-F-M Order: - The boy can win by either winning the first two matches or losing the first and winning the next two. - Probability = ( q cdot p + (1 - q) cdot p cdot q ) - Simplifies to ( pq + pq(1 - q) = pq(2 - q) )3. Compare Probabilities: - Since ( p < q ), it follows that ( 2 - p > 2 - q ). - Therefore, ( pq(2 - p) > pq(2 - q) ).4. Conclusion: - The order "father-mother-father" (F-M-F) provides a higher probability of winning two consecutive matches due to the easier match (mother) being positioned in the middle, which allows for a better recovery path if the first match is lost.[boxed{father-mother-father}]]

🔑:14.49. Answer: "father-mother-father". (This answer seems strange at first glance because it involves playing against the stronger player twice. However, with a different order of matches, one must necessarily win the single match against the stronger player, rather than one of two matches.)Let the probability of winning a match against the father be p, and against the mother -q. According to the condition p2 p q-q p q.

🧠:[<think>Okay, let's see. I need to find the equation of the ellipse given some conditions. The problem states that the line y = x + √2 intersects the ellipse (x²/a²) + (y²/b²) = 1 at points M and N. Also, the vectors OM and ON are perpendicular, and the distance between M and N is √6. The ellipse has parameters a > b > 0, so it's a horizontal ellipse. First, I need to recall the standard equation of an ellipse and understand what the given conditions mean. The line intersects the ellipse at two points M and N. The origin O to these points forms vectors that are perpendicular. Additionally, the distance between M and N is √6. So, I need to use these conditions to find a and b.Let me start by finding the points of intersection between the line y = x + √2 and the ellipse. To do that, substitute y into the ellipse equation. So, substitute y = x + √2 into (x²/a²) + (y²/b²) = 1. That gives:x²/a² + (x + √2)²/b² = 1Expanding the (x + √2)² term:(x² + 2√2 x + 2)/b²So the equation becomes:x²/a² + x²/b² + (2√2 x)/b² + 2/b² = 1Combine like terms:x²(1/a² + 1/b²) + (2√2)/b² x + (2/b² - 1) = 0This is a quadratic in x. Let's denote coefficients as A, B, C:A = 1/a² + 1/b²B = 2√2 / b²C = 2/b² - 1So the quadratic equation is A x² + B x + C = 0. The solutions to this quadratic will give the x-coordinates of points M and N. Let me denote the roots as x₁ and x₂. Then, by Vieta's formulas:x₁ + x₂ = -B/Ax₁ x₂ = C/ASimilarly, the y-coordinates of M and N can be found using y = x + √2, so y₁ = x₁ + √2 and y₂ = x₂ + √2.Now, the vectors OM and ON are perpendicular. The vectors OM and ON are (x₁, y₁) and (x₂, y₂), respectively. The condition for perpendicularity is that their dot product is zero:x₁ x₂ + y₁ y₂ = 0Substituting y₁ and y₂:x₁ x₂ + (x₁ + √2)(x₂ + √2) = 0Let me expand this:x₁ x₂ + x₁ x₂ + √2 x₁ + √2 x₂ + (√2)(√2) = 0Simplify term by term:First term: x₁ x₂Second term: x₁ x₂ from expanding (x₁ + √2)(x₂ + √2)Third term: √2 x₁Fourth term: √2 x₂Fifth term: (√2)^2 = 2So combining:x₁ x₂ + x₁ x₂ + √2 x₁ + √2 x₂ + 2 = 0That is:2 x₁ x₂ + √2 (x₁ + x₂) + 2 = 0Now, from the quadratic equation, we can express x₁ + x₂ and x₁ x₂ in terms of a and b. So:x₁ + x₂ = -B/A = -(2√2 / b²) / (1/a² + 1/b²) = -(2√2 / b²) / ( (b² + a²) / (a² b²) )) = -(2√2 / b²) * (a² b² / (a² + b²)) ) = -2√2 a² / (a² + b²)Similarly, x₁ x₂ = C/A = (2/b² - 1) / (1/a² + 1/b²) = (2 - b²)/b² / ( (a² + b²) / (a² b²) )) = ( (2 - b²)/b² ) * (a² b² / (a² + b²)) ) = a² (2 - b²) / (a² + b²)So substitute these into the perpendicularity condition:2 * [a² (2 - b²)/(a² + b²)] + √2 * [ -2√2 a² / (a² + b²) ] + 2 = 0Let's compute each term:First term: 2 * [a² (2 - b²)/(a² + b²)] = 2a² (2 - b²)/(a² + b²)Second term: √2 * [ -2√2 a² / (a² + b²) ] = √2*(-2√2) a² / (a² + b²) = -2*2 a² / (a² + b²) = -4a²/(a² + b²)Third term: +2So combining all terms:[2a²(2 - b²) - 4a² + 2(a² + b²)] / (a² + b²) = 0Wait, maybe better to write each term over the same denominator:First term: 2a²(2 - b²)/(a² + b²)Second term: -4a²/(a² + b²)Third term: 2 = 2*(a² + b²)/(a² + b²)So:[2a²(2 - b²) - 4a² + 2(a² + b²)] / (a² + b²) = 0Multiply numerator:Let's compute numerator:2a²(2 - b²) - 4a² + 2a² + 2b²Expand the first term:4a² - 2a² b² - 4a² + 2a² + 2b²Combine like terms:(4a² - 4a² + 2a²) + (-2a² b²) + 2b²Which is:2a² - 2a² b² + 2b²Factor terms:2a²(1 - b²) + 2b²Wait, perhaps factor 2:2[ a²(1 - b²) + b² ]But let's check:Wait, 2a² - 2a² b² + 2b² = 2(a² - a² b² + b²) = 2[ a²(1 - b²) + b² ]Hmm, maybe not straightforward. Let me write it as:2a² + (-2a² b²) + 2b² = 2a² + 2b² - 2a² b²Factor 2:2(a² + b² - a² b²)So the numerator is 2(a² + b² - a² b²), denominator is (a² + b²). Therefore:2(a² + b² - a² b²)/(a² + b²) = 0Which implies that:a² + b² - a² b² = 0So:a² + b² = a² b²Hmm, interesting equation. Let's keep this as equation (1).Now, another condition is that the distance between M and N is √6. The distance MN is given by √[(x₁ - x₂)² + (y₁ - y₂)²] = √6.But y₁ - y₂ = (x₁ + √2) - (x₂ + √2) = x₁ - x₂. Therefore, MN distance becomes √[(x₁ - x₂)^2 + (x₁ - x₂)^2] = √[2(x₁ - x₂)^2] = √2 |x₁ - x₂|Therefore, √2 |x₁ - x₂| = √6 => |x₁ - x₂| = √6 / √2 = √3So |x₁ - x₂| = √3. But |x₁ - x₂| can be found from the quadratic equation. For a quadratic equation Ax² + Bx + C = 0, the difference of roots is √(D)/A, where D is the discriminant. Wait, actually, the difference is √[(x₁ + x₂)^2 - 4x₁x₂]. Because (x₁ - x₂)^2 = (x₁ + x₂)^2 - 4x₁x₂. Therefore, |x₁ - x₂| = √[(x₁ + x₂)^2 - 4x₁x₂]We already have expressions for x₁ + x₂ and x₁x₂ in terms of a and b.So:|x₁ - x₂| = √[ (x₁ + x₂)^2 - 4x₁x₂ ]From earlier:x₁ + x₂ = -2√2 a² / (a² + b²)x₁x₂ = a²(2 - b²)/(a² + b²)Compute (x₁ + x₂)^2:[ -2√2 a² / (a² + b²) ]² = 8 a^4 / (a² + b²)^2Compute 4x₁x₂:4 * [a²(2 - b²)/(a² + b²)] = 4a²(2 - b²)/(a² + b²)Therefore,(x₁ + x₂)^2 - 4x₁x₂ = 8a^4/(a² + b²)^2 - 4a²(2 - b²)/(a² + b²)Factor out 4a²/(a² + b²)^2:= [8a^4 - 4a²(2 - b²)(a² + b²)] / (a² + b²)^2Expand the numerator:8a^4 - 4a²(2 - b²)(a² + b²)First compute (2 - b²)(a² + b²):= 2a² + 2b² - a² b² - b^4Multiply by 4a²:= 4a²(2a² + 2b² - a² b² - b^4) = 8a^4 + 8a² b² - 4a^4 b² - 4a² b^4Subtract this from 8a^4:Numerator = 8a^4 - [8a^4 + 8a² b² - 4a^4 b² - 4a² b^4] =8a^4 -8a^4 -8a² b² +4a^4 b² +4a² b^4 =-8a² b² +4a^4 b² +4a² b^4Factor terms:= 4a² b²(-2 + a² + b²)Therefore, (x₁ + x₂)^2 - 4x₁x₂ = [4a² b²(-2 + a² + b²)] / (a² + b²)^2Therefore, |x₁ - x₂| = √[4a² b²(-2 + a² + b²)/(a² + b²)^2] = [2ab √(a² + b² - 2)] / (a² + b²)But we know |x₁ - x₂| = √3. So,[2ab √(a² + b² - 2)] / (a² + b²) = √3Square both sides:[4a² b² (a² + b² - 2)] / (a² + b²)^2 = 3Multiply both sides by (a² + b²)^2:4a² b² (a² + b² - 2) = 3(a² + b²)^2So now we have two equations:1) a² + b² = a² b²2) 4a² b² (a² + b² - 2) = 3(a² + b²)^2Let me try to substitute equation 1 into equation 2. Since from equation 1, a² + b² = a² b². Let me denote S = a² + b² and P = a² b². Then equation 1 is S = P. So equation 2 becomes:4P(S - 2) = 3 S²But since S = P, substitute S for P:4S(S - 2) = 3 S²Expand left side:4S² - 8S = 3S²Bring all terms to left:4S² -8S -3S² =0 => S² -8S =0 => S(S -8)=0Thus S=0 or S=8. Since S = a² + b² and a,b >0, S cannot be zero. Therefore, S=8. So a² + b²=8. But from equation 1, S = P = a² b². Therefore, a² b² =8.So we have:a² + b²=8anda² b²=8So we need to find a² and b² such that they add to 8 and multiply to 8. Let me set u = a² and v = b². Then u + v =8 and u v=8. So we can write this as:u + v =8uv=8These are the sums and products of roots of the quadratic equation t² -8t +8=0Solving for t:t = [8 ±√(64 -32)] /2 = [8 ±√32]/2 = [8 ±4√2]/2 =4 ±2√2Therefore, u and v are 4 +2√2 and 4 -2√2. Since a >b >0, then a² >b². Therefore, u=4 +2√2 and v=4 -2√2.Therefore, a²=4 +2√2 and b²=4 -2√2Hence, the equation of the ellipse is:x²/(4 +2√2) + y²/(4 -2√2) =1But we can rationalize the denominators if needed.For 4 +2√2, factor out 2: 2(2 +√2). Similarly, 4 -2√2 = 2(2 -√2). So,x²/(2(2 +√2)) + y²/(2(2 -√2)) =1Multiply numerator and denominator to rationalize:For x² denominator: 2(2 +√2). Multiply numerator and denominator by (2 -√2):[ (2 -√2) / ( (2 +√2)(2 -√2) ) ] = (2 -√2)/(4 -2) = (2 -√2)/2Similarly for y² denominator: 2(2 -√2). Multiply numerator and denominator by (2 +√2):(2 +√2)/ ( (2 -√2)(2 +√2) ) = (2 +√2)/2Therefore, the equation becomes:x² * (2 -√2)/(2*2) + y²*(2 +√2)/(2*2) =1Simplify denominators:x²*(2 -√2)/4 + y²*(2 +√2)/4 =1Multiply both sides by 4:x²(2 -√2) + y²(2 +√2) =4Alternatively, leave it as x²/(4 +2√2) + y²/(4 -2√2)=1. But perhaps the problem expects the answer in a simplified rationalized form. Let me check both forms.Alternatively, maybe even keep a² and b² as 4 +2√2 and 4 -2√2. But let's check if the denominators can be written as integers multiplied by radicals. For example, 4 +2√2 = 2*(2 +√2), so if we factor out 2, then the equation is x²/(2*(2 +√2)) + y²/(2*(2 -√2)) =1. However, rationalizing might be better.Alternatively, multiply numerator and denominator by the conjugate:For x² term:1/(4 +2√2) = (4 -2√2)/( (4 +2√2)(4 -2√2) ) = (4 -2√2)/(16 -8) = (4 -2√2)/8 = (2 -√2)/4Similarly, 1/(4 -2√2)= (4 +2√2)/8= (2 +√2)/4Therefore, the equation can be written as:x²*(2 -√2)/4 + y²*(2 +√2)/4 =1Which is equivalent to:[(2 -√2)x² + (2 +√2)y²]/4 =1Multiply both sides by 4:(2 -√2)x² + (2 +√2)y² =4This might be a more elegant form. Alternatively, perhaps even factor differently. Let me check if this is the simplest form. Alternatively, maybe they want the answer in the original form with denominators. The problem says "find the equation of the ellipse", so both forms are acceptable, but perhaps the first one with denominators 4 +2√2 and 4 -2√2 is acceptable. But maybe the answer expects rationalized denominators. Let me see.Wait, if the question says "find the equation", then both forms are correct, but perhaps the standard form with denominators. Let me check the problem statement again: "find the equation of the ellipse". The standard equation is (x²/a²) + (y²/b²)=1. So the answer is x²/(4 +2√2) + y²/(4 -2√2) =1, which can be written as x²/(2(2 +√2)) + y²/(2(2 -√2))=1, but unless they specify rationalized denominators, either is okay. However, in most cases, rationalizing the denominator is preferred. So perhaps we should present it as:x²/(2(2 +√2)) + y²/(2(2 -√2)) =1But even better to rationalize each term:x²/(4 +2√2) = x²*(4 -2√2)/[(4 +2√2)(4 -2√2)] = x²*(4 -2√2)/ (16 -8) = x²*(4 -2√2)/8 = x²*(2 -√2)/4Similarly for y² term: y²/(4 -2√2) = y²*(4 +2√2)/8 = y²*(2 +√2)/4Therefore, the equation becomes:(x²*(2 -√2) + y²*(2 +√2))/4 =1Which is the same as:(2 -√2)x² + (2 +√2)y² =4This seems like a cleaner form. Let me confirm that with substitution.Suppose a²=4 +2√2, then (2 -√2)/4 = 1/a²:1/a² =1/(4 +2√2)= (4 -2√2)/ ( (4 +2√2)(4 -2√2) )=(4 -2√2)/8=(2 -√2)/4. Yes, that's correct.Similarly, 1/b²=(2 +√2)/4. So the standard form is equivalent to (2 -√2)x²/4 + (2 +√2)y²/4 =1, which is equivalent to (2 -√2)x² + (2 +√2)y² =4. So either form is acceptable, but perhaps the problem expects the answer in the standard form with denominators. Let me check again.The original ellipse equation is given as x²/a² + y²/b²=1. So in the answer, we need to specify a² and b². However, in the problem statement, the answer should be the equation, so writing x²/(4 +2√2) + y²/(4 -2√2)=1 is correct, but maybe they want denominators rationalized.Alternatively, another approach: since we found that a²=4 +2√2 and b²=4 -2√2, just substitute these into the standard equation. So the answer is x²/(4 +2√2) + y²/(4 -2√2)=1. But let me rationalize the denominators if possible.Wait, 4 +2√2 can be written as 2*(2 +√2). Similarly, 4 -2√2=2*(2 -√2). Therefore, the equation becomes x²/[2*(2 +√2)] + y²/[2*(2 -√2)] =1. If we factor out the 2, then:(1/2)[x²/(2 +√2) + y²/(2 -√2)] =1Multiply both sides by 2:x²/(2 +√2) + y²/(2 -√2) =2But maybe this is not helpful. Alternatively, perhaps leave it as it is. Alternatively, if we rationalize each denominator:For x² term:1/(2 +√2) = (2 -√2)/[(2 +√2)(2 -√2)] = (2 -√2)/(4 -2) = (2 -√2)/2So x²/[2*(2 +√2)] =x²*(2 -√2)/[2*(2 +√2)(2 -√2)] =x²*(2 -√2)/[2*2] =x²*(2 -√2)/4Similarly for y² term:1/(2 -√2) = (2 +√2)/[(2 -√2)(2 +√2)] = (2 +√2)/2Therefore, y²/[2*(2 -√2)] = y²*(2 +√2)/4Therefore, combining:x²*(2 -√2)/4 + y²*(2 +√2)/4 =1Multiply numerator and denominator:[(2 -√2)x² + (2 +√2)y²]/4 =1Multiply both sides by 4:(2 -√2)x² + (2 +√2)y² =4This is a valid equation of the ellipse. So perhaps this is the most simplified form. However, the problem might expect the answer in the standard form with denominators expressed as simplified radicals. Alternatively, they might accept the initial form. Let me check with the problem statement. The problem says "find the equation of the ellipse". So both forms are correct, but perhaps the standard form with denominators 4 +2√2 and 4 -2√2 is acceptable. Alternatively, the problem might want integers in the denominators. Wait, but in this case, even after rationalization, the coefficients are fractions. Hmm.Wait, maybe the problem expects the answer in terms of a² and b² as integers, but given the conditions, they are not integers. So perhaps the answer is x²/(4 +2√2) + y²/(4 -2√2)=1.Alternatively, write a²=4 +2√2 and b²=4 -2√2, but since the problem asks for the equation, just plug these into the standard form. So yes, x² over (4 +2√2) plus y² over (4 -2√2) equals 1. Therefore, I think this is the answer. Let me check the calculations once more to make sure I didn't make any mistakes.First, solving the quadratic equation for x when substituting y = x +√2 into the ellipse. Then using Vieta's formulas. Then setting up the perpendicularity condition which led to equation a² + b² =a² b². Then using the distance MN=√6 to get another equation. Solving those gave a² +b²=8 and a² b²=8, leading to a² and b² being roots of t² -8t +8=0, which are 4 ±2√2. Since a >b, a²=4 +2√2 and b²=4 -2√2. Therefore, the equation is x²/(4 +2√2) + y²/(4 -2√2) =1. So this seems correct. Therefore, the final answer is this equation.Final Answerboxed{dfrac{x^2}{4 + 2sqrt{2}} + dfrac{y^2}{4 - 2sqrt{2}} = 1}</think>Given the line ( y = x + sqrt{2} ) intersecting the ellipse ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) at points ( M ) and ( N ), with ( OM perp ON ) and ( MN = sqrt{6} ), we need to find the equation of the ellipse.1. Substitute ( y = x + sqrt{2} ) into the ellipse equation: [ frac{x^2}{a^2} + frac{(x + sqrt{2})^2}{b^2} = 1 ] Expanding and simplifying, we get a quadratic equation in ( x ): [ x^2 left( frac{1}{a^2} + frac{1}{b^2} right) + frac{2sqrt{2}}{b^2} x + left( frac{2}{b^2} - 1 right) = 0 ]2. Using Vieta's formulas for roots ( x_1 ) and ( x_2 ): [ x_1 + x_2 = -frac{2sqrt{2} a^2}{a^2 + b^2}, quad x_1 x_2 = frac{a^2 (2 - b^2)}{a^2 + b^2} ]3. Perpendicularity condition ( x_1 x_2 + y_1 y_2 = 0 ): [ 2 x_1 x_2 + sqrt{2} (x_1 + x_2) + 2 = 0 ] Substituting Vieta's results, we derive: [ a^2 + b^2 = a^2 b^2 ]4. Distance condition ( MN = sqrt{6} ): [ sqrt{2} |x_1 - x_2| = sqrt{6} implies |x_1 - x_2| = sqrt{3} ] Using the discriminant of the quadratic equation, we find: [ 4a^2 b^2 (a^2 + b^2 - 2) = 3(a^2 + b^2)^2 ]5. Solving the system of equations: - From ( a^2 + b^2 = a^2 b^2 ) and substituting into the distance condition, we solve: [ a^2 + b^2 = 8 quad text{and} quad a^2 b^2 = 8 ] - Solving the quadratic equation ( t^2 - 8t + 8 = 0 ), we get ( a^2 = 4 + 2sqrt{2} ) and ( b^2 = 4 - 2sqrt{2} ).6. Equation of the ellipse: [ boxed{dfrac{x^2}{4 + 2sqrt{2}} + dfrac{y^2}{4 - 2sqrt{2}} = 1} ],<think>Okay, so I need to find the equation of the ellipse given some conditions. Let me start by understanding the problem step by step.First, the problem states that the line y = x + √2 intersects the ellipse x²/a² + y²/b² = 1 (with a > b > 0) at points M and N. Then, it's given that the vectors OM and ON are perpendicular to each other, where O is the origin. Additionally, the distance between points M and N is √6. I need to find the equation of the ellipse, which means I need to determine the values of a² and b².Alright, let's break this down.1. Find the points of intersection between the line and the ellipse. The line is y = x + √2. To find where this intersects the ellipse, substitute y into the ellipse equation. So substituting y = x + √2 into x²/a² + y²/b² = 1 gives: x²/a² + (x + √2)² / b² = 1. Let me expand that: x²/a² + (x² + 2√2 x + 2) / b² = 1. Combine like terms: (1/a² + 1/b²) x² + (2√2 / b²) x + (2 / b² - 1) = 0. So this is a quadratic equation in terms of x. Let me denote coefficients for clarity: Let A = 1/a² + 1/b², B = 2√2 / b², C = 2 / b² - 1. So the quadratic equation is A x² + B x + C = 0. The solutions to this equation will give the x-coordinates of points M and N. Let's call them x₁ and x₂.2. Use the condition that OM is perpendicular to ON. Vectors OM and ON are perpendicular, so their dot product is zero. If M is (x₁, y₁) and N is (x₂, y₂), then the vectors OM and ON are (x₁, y₁) and (x₂, y₂), respectively. Their dot product is x₁ x₂ + y₁ y₂ = 0. Since y = x + √2 for both points, y₁ = x₁ + √2 and y₂ = x₂ + √2. Therefore, the dot product becomes: x₁ x₂ + (x₁ + √2)(x₂ + √2) = 0. Let's expand this: x₁ x₂ + x₁ x₂ + √2 x₁ + √2 x₂ + 2 = 0. Combine like terms: 2 x₁ x₂ + √2 (x₁ + x₂) + 2 = 0. So, 2 x₁ x₂ + √2 (x₁ + x₂) + 2 = 0. --- (1) Now, from the quadratic equation A x² + B x + C = 0, we can recall that: x₁ + x₂ = -B/A, x₁ x₂ = C/A. So let me compute x₁ + x₂ and x₁ x₂ in terms of a and b. From above: A = 1/a² + 1/b², B = 2√2 / b², C = 2 / b² - 1. So, x₁ + x₂ = -B/A = -(2√2 / b²) / (1/a² + 1/b²), x₁ x₂ = C/A = (2 / b² - 1) / (1/a² + 1/b²). Let me write this more neatly: Let’s denote S = x₁ + x₂ = - (2√2 / b²) / (1/a² + 1/b²), and P = x₁ x₂ = (2 / b² - 1) / (1/a² + 1/b²). Then equation (1) becomes: 2P + √2 S + 2 = 0. So substituting S and P into this equation will give a relation between a and b. Let me compute that. First, substitute S and P: 2 * [(2 / b² - 1) / (1/a² + 1/b²)] + √2 * [ - (2√2 / b²) / (1/a² + 1/b²) ] + 2 = 0. Since all terms have a denominator of (1/a² + 1/b²), except the last term 2. Let's factor out 1/(1/a² + 1/b²): [2(2 / b² - 1) - √2 * 2√2 / b²] / (1/a² + 1/b²) + 2 = 0. Simplify the numerator: First, expand the terms inside the numerator: 2*(2 / b² - 1) = 4 / b² - 2, √2 * 2√2 / b² = 2*(√2 * √2)/b² = 2*2 / b² = 4 / b². But since there is a negative sign in front, it's -4 / b². So the numerator becomes: (4 / b² - 2) - 4 / b² = (4 / b² - 4 / b²) - 2 = 0 - 2 = -2. Therefore, the equation becomes: (-2) / (1/a² + 1/b²) + 2 = 0. Let me write that as: -2 / (1/a² + 1/b²) + 2 = 0. Move the first term to the other side: -2 / (1/a² + 1/b²) = -2. Multiply both sides by -1: 2 / (1/a² + 1/b²) = 2. Divide both sides by 2: 1 / (1/a² + 1/b²) = 1. Which implies: 1/a² + 1/b² = 1. --- (2) So that's one equation relating a and b. Wait, that's interesting. So from the perpendicularity condition, we get that 1/a² + 1/b² = 1. Let me note that down as equation (2).3. Now, use the other condition: MN = √6. The distance between points M and N is √6. Since M and N are points on the line y = x + √2, their coordinates are (x₁, x₁ + √2) and (x₂, x₂ + √2). The distance between M and N can be found using the distance formula: MN = sqrt[(x₂ - x₁)^2 + (y₂ - y₁)^2]. Since y = x + √2, then y₂ - y₁ = (x₂ + √2) - (x₁ + √2) = x₂ - x₁. Therefore, MN = sqrt[(x₂ - x₁)^2 + (x₂ - x₁)^2] = sqrt[2*(x₂ - x₁)^2] = sqrt(2)*|x₂ - x₁|. Given that MN = √6, then: sqrt(2)*|x₂ - x₁| = √6. Divide both sides by sqrt(2): |x₂ - x₁| = √6 / sqrt(2) = sqrt(3). So |x₂ - x₁| = sqrt(3). But |x₂ - x₁| can be expressed in terms of the quadratic equation's coefficients. Remember that for a quadratic equation Ax² + Bx + C = 0, the difference of roots is sqrt(D)/A, where D is the discriminant? Wait, actually, the difference of roots is sqrt((x₁ + x₂)^2 - 4x₁x₂). Wait, the discriminant is B² - 4AC, which equals (x₁ - x₂)^2 * A². Wait, no, let me recall. The roots of Ax² + Bx + C = 0 are x = [-B ± sqrt(B² - 4AC)]/(2A). Therefore, the difference between roots is [sqrt(B² - 4AC)/A] (since the denominator is 2A, the difference would be 2*sqrt(B² - 4AC)/(2A) = sqrt(B² - 4AC)/A. Wait, let me do that step by step. The roots are x₁ and x₂ = [-B ± sqrt(B² - 4AC)]/(2A). Therefore, x₂ - x₁ = [ -B + sqrt(B² - 4AC) - (-B - sqrt(B² - 4AC)) ] / (2A) = [2 sqrt(B² - 4AC)] / (2A) = sqrt(B² - 4AC)/A. Therefore, |x₂ - x₁| = sqrt(B² - 4AC)/|A|. Since A is positive (because 1/a² + 1/b² is positive), we can drop the absolute value. Therefore, |x₂ - x₁| = sqrt(B² - 4AC)/A. So, according to the problem, sqrt(B² - 4AC)/A = sqrt(3). Therefore, sqrt(B² - 4AC) = A*sqrt(3). Square both sides: B² - 4AC = 3A². So, B² - 4AC - 3A² = 0. --- (3) Now, recall the coefficients: A = 1/a² + 1/b², B = 2√2 / b², C = 2 / b² - 1. Let me substitute A, B, C into equation (3): (2√2 / b²)² - 4*(1/a² + 1/b²)*(2 / b² - 1) - 3*(1/a² + 1/b²)^2 = 0. Let me compute each term step by step. First term: (2√2 / b²)^2 = 8 / b⁴. Second term: -4*(1/a² + 1/b²)*(2 / b² - 1). Let me expand this: Let’s denote D = (1/a² + 1/b²), E = (2 / b² - 1). Then, -4 D E. Multiply D and E: D*E = (1/a² + 1/b²)(2 / b² - 1) = (1/a²)(2 / b²) - (1/a²) + (1/b²)(2 / b²) - (1/b²) = 2/(a² b²) - 1/a² + 2/b⁴ - 1/b². So multiplying by -4: -4*[2/(a² b²) - 1/a² + 2/b⁴ - 1/b²] = -8/(a² b²) + 4/a² - 8/b⁴ + 4/b². Third term: -3*(1/a² + 1/b²)^2. Let me expand (1/a² + 1/b²)^2 = 1/a⁴ + 2/(a² b²) + 1/b⁴. Multiply by -3: -3/a⁴ - 6/(a² b²) - 3/b⁴. Now, adding all three terms together: First term: 8 / b⁴, Second term: -8/(a² b²) + 4/a² - 8/b⁴ + 4/b², Third term: -3/a⁴ - 6/(a² b²) - 3/b⁴. Let me combine them: 8/b⁴ + (-8/(a² b²) + 4/a² - 8/b⁴ + 4/b²) + (-3/a⁴ - 6/(a² b²) - 3/b⁴). Combine like terms: -3/a⁴, (4/a²), (-8/(a² b²) - 6/(a² b²)) = -14/(a² b²), (4/b²), (8/b⁴ -8/b⁴ -3/b⁴) = (-3/b⁴). So altogether: -3/a⁴ + 4/a² -14/(a² b²) + 4/b² -3/b⁴ = 0. Hmm, this seems complicated. Maybe there's a better way to approach this. Wait, perhaps we can use equation (2): 1/a² + 1/b² = 1. Let me denote that as D = 1/a² + 1/b² = 1. Let me use this to simplify the expression. Let me see: Let’s let u = 1/a², v = 1/b². Then equation (2) is u + v = 1. Then, the coefficients: A = u + v = 1, B = 2√2 v, C = 2v - 1. So equation (3) becomes B² - 4AC - 3A² = 0. Since A = 1, B = 2√2 v, C = 2v - 1. Substituting into equation (3): (2√2 v)^2 - 4*1*(2v -1) - 3*(1)^2 = 0. Compute each term: (2√2 v)^2 = 8 v², -4*1*(2v -1) = -8v +4, -3*1 = -3. So overall equation: 8v² -8v +4 -3 =0 => 8v² -8v +1=0. So, 8v² -8v +1=0. Solve for v: Quadratic equation: 8v² -8v +1=0. Solutions: v = [8 ± sqrt(64 -32)] / 16 = [8 ± sqrt(32)] /16 = [8 ± 4√2]/16 = [2 ± √2]/4. Therefore, v = (2 + √2)/4 or (2 - √2)/4. Since v =1/b² and b² must be positive, both solutions are acceptable as they are positive. Let me compute them: (2 + √2)/4 ≈ (2 +1.414)/4 ≈ 3.414/4 ≈0.8535, (2 - √2)/4 ≈ (2 -1.414)/4 ≈0.586/4≈0.1465. Now, since a > b >0, and from equation (2): u + v =1, where u=1/a², v=1/b². Since a > b, then 1/a² <1/b², so u <v. Therefore, since u + v =1, if u <v, then v >0.5. Let's check the two values of v: First solution: (2 + √2)/4 ≈0.8535>0.5, Second solution: (2 - √2)/4≈0.1465 <0.5. Therefore, since v must be greater than u, and u =1 -v, so if v>0.5, then u =1 -v <0.5 <v, which satisfies u <v. If v <0.5, then u =1 -v >0.5 >v, which would mean u >v, contradicting a >b (since u =1/a² <1/b² =v). Therefore, the only valid solution is v=(2 +√2)/4. Therefore, v=(2 +√2)/4. Then u =1 -v =1 - (2 +√2)/4=(4 -2 -√2)/4=(2 -√2)/4. Therefore, u=(2 -√2)/4=1/a², v=(2 +√2)/4=1/b². Therefore, a²=4/(2 -√2), b²=4/(2 +√2). To rationalize denominators: For a²: 4/(2 -√2)=4*(2 +√2)/[(2 -√2)(2 +√2)]=4*(2 +√2)/(4 -2)=4*(2 +√2)/2=2*(2 +√2)=4 +2√2. Similarly, for b²: 4/(2 +√2)=4*(2 -√2)/[(2 +√2)(2 -√2)]=4*(2 -√2)/(4 -2)=4*(2 -√2)/2=2*(2 -√2)=4 -2√2. Therefore, a²=4 +2√2, and b²=4 -2√2. Let me check these values to see if they satisfy the original conditions. First, check equation (2): 1/a² +1/b²= (2 -√2)/4 + (2 +√2)/4= (2 -√2 +2 +√2)/4=4/4=1. Correct. Now, check if MN=√6. Let's recall that |x₂ -x₁|=sqrt(3). Since we derived that when substituting the conditions, and we found the values of a and b through that, so it should hold. But let's verify. From the quadratic equation, the difference |x₂ -x₁|=sqrt(3). Then MN= sqrt(2)*sqrt(3)=sqrt(6). So it matches. Therefore, the equation of the ellipse is x²/(4 +2√2) + y²/(4 -2√2)=1. However, it's better to rationalize the denominators if possible, or write in simpler forms. Wait, 4 +2√2 can be written as 2*(2 +√2), but maybe it's acceptable as is. Alternatively, since a²=4 +2√2 and b²=4 -2√2, we can write the equation as: x²/(4 +2√2) + y²/(4 -2√2) =1. But sometimes, problems prefer denominators without radicals. Let me check if these denominators can be simplified. Let me see: 4 +2√2=2*(2 +√2), 4 -2√2=2*(2 -√2). So maybe factor out 2: x²/[2*(2 +√2)] + y²/[2*(2 -√2)] =1, which can be written as: x²/(2*(2 +√2)) + y²/(2*(2 -√2)) =1. Alternatively, if we rationalize the denominators: For the term x²/(4 +2√2): Multiply numerator and denominator by (4 -2√2): (x²*(4 -2√2))/[(4 +2√2)(4 -2√2)] = (x²*(4 -2√2))/(16 -8) = (x²*(4 -2√2))/8. Similarly, for y²/(4 -2√2): Multiply numerator and denominator by (4 +2√2): (y²*(4 +2√2))/[(4 -2√2)(4 +2√2)] = (y²*(4 +2√2))/8. Therefore, the equation becomes: [ (4 -2√2)/8 ] x² + [ (4 +2√2)/8 ] y² =1, which simplifies to: (4 -2√2)x² + (4 +2√2)y² =8. Maybe that's a cleaner form. But both forms are correct. However, since the problem asks for the equation of the ellipse, which is typically presented in the standard form with denominators. So, likely, the answer is x²/(4 +2√2) + y²/(4 -2√2) =1. But perhaps to rationalize the denominators, as I did above. Wait, let me check if 4 +2√2 and 4 -2√2 can be expressed as squares or something else. For example, (√2 +1)^2=2 +2√2 +1=3 +2√2, which is not 4 +2√2. Similarly, (√3 +1)^2=3 +2√3 +1=4 +2√3, which is different. So probably not. Therefore, the denominators are okay as they are. Alternatively, if they want integer coefficients, the equation can be written as: [ (4 -2√2)x² + (4 +2√2)y² ] /8 =1, which would be multiplying both sides by 8: (4 -2√2)x² + (4 +2√2)y² =8. But I think the standard form is better. So, I think the answer is x²/(4 +2√2) + y²/(4 -2√2) =1. However, the problem might expect simplifying the denominators. Let me see: 4 +2√2 = 2*(2 +√2), 4 -2√2 = 2*(2 -√2). So, another way is: x²/[2(2 +√2)] + y²/[2(2 -√2)] =1. Or, even further simplifying by rationalizing: For example, 2 +√2 multiplied by (2 -√2) gives 4 -2=2. So 2(2 +√2)= (2 +√2)*2, but I don't know if that helps. Alternatively, maybe we can write it as: x²/(4 +2√2) + y²/(4 -2√2) =1. To check which form is more standard. Since 4 +2√2 and 4 -2√2 are just constants, and they are simplified, so I think that's acceptable. Let me compute 4 +2√2 and 4 -2√2 numerically to verify if they make sense. 4 +2√2 ≈4 +2*1.414≈4 +2.828≈6.828, 4 -2√2≈4 -2.828≈1.172. So, a²≈6.828, b²≈1.172, so a≈2.614, b≈1.083. Indeed, a >b, which matches the condition a >b >0. Therefore, these values are valid. Therefore, the equation of the ellipse is x²/(4 +2√2) + y²/(4 -2√2)=1. However, to make sure, maybe we can rationalize the denominators for better form. Let me rationalize each denominator: For x²/(4 +2√2): Multiply numerator and denominator by (4 -2√2): x²(4 -2√2)/[(4 +2√2)(4 -2√2)] =x²(4 -2√2)/(16 -8)=x²(4 -2√2)/8=(4 -2√2)/8 x²=(2 -√2)/4 x². Similarly for y²/(4 -2√2): Multiply numerator and denominator by (4 +2√2): y²(4 +2√2)/[(4 -2√2)(4 +2√2)]= y²(4 +2√2)/8=(4 +2√2)/8 y²=(2 +√2)/4 y². Therefore, the equation becomes: (2 -√2)/4 x² + (2 +√2)/4 y² =1, which can be written as: [(2 -√2)x² + (2 +√2)y²]/4 =1, so multiplying both sides by 4: (2 -√2)x² + (2 +√2)y² =4. This might be a nicer form because it has integer coefficients. Let me check if this is acceptable. Let me compute: (2 -√2)x² + (2 +√2)y² =4. Alternatively, we can write it as: (2 -√2)x² + (2 +√2)y² =4. Either way, both forms are correct. But in terms of standard form, the first version with denominators 4 +2√2 and 4 -2√2 is standard because it's expressed as x²/a² + y²/b²=1. However, the problem might expect the answer in the simplest radical form, which could be either. Let me check the problem statement again. It says "find the equation of the ellipse". Since the standard form is x²/a² + y²/b²=1 with a and b given, so unless there's a need to rationalize, both forms are acceptable. However, in most cases, rationalizing the denominator is preferred. Therefore, perhaps writing it as [(2 -√2)x² + (2 +√2)y²]/4 =1, but that's not the standard form. Alternatively, keeping the denominators as 4 +2√2 and 4 -2√2 is better since they directly represent a² and b². Let me check the two forms: Original form: x²/(4 +2√2) + y²/(4 -2√2)=1. Simplified with rational denominators: [(2 -√2)x² + (2 +√2)y²]/4 =1. Which one is preferred? Since the problem doesn't specify, but in Olympiad problems, often they prefer the form without radicals in the denominators, so the second form. However, in standard ellipse equation, the first form is standard. Hmm. Let me check if the problem gives any previous examples. The problem is written in Chinese, but the instruction here is in English. Since in the problem statement, the ellipse is given as x²/a² + y²/b²=1, so they probably expect the answer in that form, with a² and b² substituted. So since we found a²=4 +2√2 and b²=4 -2√2, then the equation is x²/(4 +2√2) + y²/(4 -2√2)=1. Therefore, this is the answer. Therefore, the equation of the ellipse is x²/(4 +2√2) + y²/(4 -2√2) =1. Alternatively, if we want to write this as x² over 2*(2 +√2) plus y² over 2*(2 -√2) equals 1, but I think leaving it as x²/(4 +2√2) + y²/(4 -2√2) =1 is fine. Let me verify once again with the initial conditions to make sure. Let’s take a²=4 +2√2≈6.828, b²=4 -2√2≈1.172. The line y =x +√2 intersects the ellipse. Let's pick a point on the line and see if it lies on the ellipse. Let me compute the points M and N. From the quadratic equation: A x² + B x + C =0, with A=1/a² +1/b²=1/(4 +2√2)+1/(4 -2√2). Compute 1/(4 +2√2)= (4 -2√2)/[(4 +2√2)(4 -2√2)]=(4 -2√2)/(16 -8)=(4 -2√2)/8=(2 -√2)/4. Similarly, 1/(4 -2√2)= (4 +2√2)/8=(2 +√2)/4. Therefore, A=(2 -√2)/4 + (2 +√2)/4=(4)/4=1. Which matches our previous result. B=2√2 / b²=2√2/(4 -2√2). Let me compute that: 2√2/(4 -2√2)=2√2*(4 +2√2)/[(4 -2√2)(4 +2√2)]=2√2*(4 +2√2)/8= [8√2 +4*2]/8= (8√2 +8)/8= √2 +1. Therefore, B=√2 +1. C=2/b² -1=2/(4 -2√2) -1=2*(4 +2√2)/8 -1=(4 +2√2)/4 -1=1 + (√2)/2 -1=√2/2. Therefore, the quadratic equation is: x² + (√2 +1)x + √2/2 =0. Wait, since A=1, B=√2 +1, C=√2 /2. Let me compute discriminant D=B² -4AC=(√2 +1)^2 -4*1*(√2 /2). Expand (√2 +1)^2=2 +2√2 +1=3 +2√2. Then, 4AC=4*1*(√2 /2)=2√2. Therefore, D=3 +2√2 -2√2=3. Therefore, the roots are x= [-B ±√D]/(2A)= [ - (√2 +1) ±√3 ] /2. Therefore, the x-coordinates of M and N are [ - (√2 +1) +√3 ] /2 and [ - (√2 +1) -√3 ] /2. Then, the difference between the roots is [ - (√2 +1) +√3 - ( - (√2 +1) -√3 ) ] /2= [2√3]/2=√3. Which matches the previous result that |x₂ -x₁|=√3, so MN=√2*√3=√6. Correct. Also, check if OM is perpendicular to ON. Let's compute the dot product of vectors OM and ON. Let me denote M=(x₁, x₁ +√2) and N=(x₂, x₂ +√2). Then, the dot product is x₁x₂ + (x₁ +√2)(x₂ +√2). We have already done this earlier, and derived the condition 1/a² +1/b²=1, which holds. Therefore, the calculation is consistent. Therefore, the answer is correct. Hence, the equation of the ellipse is x²/(4 +2√2) + y²/(4 -2√2)=1.Final AnswerThe equation of the ellipse is boxed{dfrac{x^2}{4 + 2sqrt{2}} + dfrac{y^2}{4 - 2sqrt{2}} = 1}.</think>Given the line ( y = x + sqrt{2} ) intersecting the ellipse ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) at points ( M ) and ( N ), with ( OM perp ON ) and ( MN = sqrt{6} ), we need to find the equation of the ellipse.1. Substitute ( y = x + sqrt{2} ) into the ellipse equation: [ frac{x^2}{a^2} + frac{(x + sqrt{2})^2}{b^2} = 1 ] Expanding and simplifying, we get a quadratic equation in ( x ): [ left( frac{1}{a^2} + frac{1}{b^2} right)x^2 + frac{2sqrt{2}}{b^2}x + left( frac{2}{b^2} - 1 right) = 0 ]2. Condition ( OM perp ON ): The dot product of vectors ( OM ) and ( ON ) is zero: [ x_1 x_2 + (x_1 + sqrt{2})(x_2 + sqrt{2}) = 0 ] Expanding and using Vieta's formulas, we derive: [ 2x_1 x_2 + sqrt{2}(x_1 + x_2) + 2 = 0 ] Substituting ( x_1 + x_2 = -frac{B}{A} ) and ( x_1 x_2 = frac{C}{A} ), we find: [ frac{1}{a^2} + frac{1}{b^2} = 1 ]3. Condition ( MN = sqrt{6} ): The distance between ( M ) and ( N ) is ( sqrt{6} ), leading to: [ |x_2 - x_1| = sqrt{3} ] Using the discriminant of the quadratic equation, we solve for ( a ) and ( b ): [ left( frac{2sqrt{2}}{b^2} right)^2 - 4 left( frac{1}{a^2} + frac{1}{b^2} right) left( frac{2}{b^2} - 1 right) - 3 left( frac{1}{a^2} + frac{1}{b^2} right)^2 = 0 ] Solving this with ( frac{1}{a^2} + frac{1}{b^2} = 1 ), we find: [ a^2 = 4 + 2sqrt{2}, quad b^2 = 4 - 2sqrt{2} ]The equation of the ellipse is:[boxed{dfrac{x^2}{4 + 2sqrt{2}} + dfrac{y^2}{4 - 2sqrt{2}} = 1}]]

🔑:Three, let points Mleft(x_{1}, y_{1}right), Nleft(x_{2}, y_{2}right).Substituting y=x+sqrt{2} into the ellipse equation and eliminating y givesbegin{array}{l}frac{x^{2}}{a^{2}}+frac{(x+sqrt{2})^{2}}{b^{2}}=1 Rightarrowleft(a^{2}+b^{2}right) x^{2}+2 sqrt{2} a^{2} x+a^{2}left(2-b^{2}right)=0 .end{array}Since points M, N are on the ellipse, i.e., x_{1}, x_{2} are the roots of the equation, we havex_{1}+x_{2}=frac{-2 sqrt{2} a^{2}}{a^{2}+b^{2}}, x_{1} x_{2}=frac{a^{2}left(2-b^{2}right)}{a^{2}+b^{2}} text {. }Given O M perp O Nbegin{aligned}Rightarrow & x_{1} x_{2}+y_{1} y_{2}=0 Rightarrow & 0=x_{1} x_{2}+left(x_{1}+sqrt{2}right)left(x_{2}+sqrt{2}right) & =2 x_{1} x_{2}+sqrt{2}left(x_{1}+x_{2}right)+2 .end{aligned}Substituting conclusion (1) into the above equation givesbegin{array}{l}frac{2 a^{2}left(2-b^{2}right)}{a^{2}+b^{2}}-frac{4 a^{2}}{a^{2}+b^{2}}+2=0 Rightarrow a^{2}+b^{2}=a^{2} b^{2} .end{array}Thus, conclusion (1) becomesx_{1}+x_{2}=-frac{2 sqrt{2}}{b^{2}}, x_{1} x_{2}=frac{2}{b^{2}}-1 text {. }Given M N=sqrt{6}, we haveleft(x_{1}-x_{2}right)^{2}+left(y_{1}-y_{2}right)^{2}=6 text {. }Noting that points M, N lie on the line y=x+sqrt{2}.begin{array}{l}text { then } x_{1}-x_{2}=y_{1}-y_{2} Rightarrowleft(x_{1}-x_{2}right)^{2}=3 Rightarrowleft(x_{1}+x_{2}right)^{2}-4 x_{1} x_{2}=3 .end{array}From conclusion (3), we getbegin{array}{l}frac{8}{b^{4}}-4left(frac{2}{b^{2}}-1right)=3 Rightarrow b^{4}-8 b^{2}+8=0 .end{array}From equations (2), (4), and a^{2}>b^{2}, we geta^{2}=4+2 sqrt{2}, b^{2}=4-2 sqrt{2} text {. }Thus, the equation of the ellipse is frac{x^{2}}{4+2 sqrt{2}}+frac{y^{2}}{4-2 sqrt{2}}=1.

🧠:[<think>Okay, let's tackle this problem step by step. The question is about hexadecimal numbers and figuring out how many of the first 1000 positive integers, when represented in hexadecimal, only use digit characters (0-9). Then, we need to find the sum of the digits of that count (n) and choose the correct option from A to E.First, let's make sure I understand hexadecimal numbering correctly. Hexadecimal is base 16, right? So each digit represents a value from 0 to 15. The digits 0-9 are the same as in decimal, and then A-F represent 10-15. So, a hexadecimal number like "1A" in decimal would be 1*16 + 10 = 26.The problem is asking for the first 1000 positive integers. Wait, does that mean numbers from 1 to 1000 in decimal? Or is it the first 1000 hexadecimal numbers? Hmm. The wording says "the first 1000 positive integers," which are typically in decimal unless specified otherwise. So, we're considering numbers from 1 to 1000 (decimal) and converting each to hexadecimal, then checking if their hexadecimal representations only consist of digits 0-9 (i.e., no letters A-F). Then, we need to count how many such numbers there are (that's n), find the sum of the digits of n, and select the answer from the options given.So the steps are:1. For each number from 1 to 1000 (decimal), convert it to hexadecimal.2. Check if the hexadecimal representation contains only digits 0-9 (no A-F).3. Count how many numbers satisfy this condition. Let's call this count n.4. Compute the sum of the digits of n.5. Choose the correct option based on the sum.Alright. Let's start by figuring out how hexadecimal numbers are structured. Since hexadecimal is base 16, each digit represents a power of 16. For example, the hexadecimal number "ABC" is 10*16² + 11*16 + 12 in decimal.But we need to convert numbers from decimal to hexadecimal. Let's recall the conversion process: divide the decimal number by 16, take the remainder as the least significant digit, then repeat with the quotient until the quotient is zero. The hexadecimal digits are the remainders, read in reverse order.However, since we need to check if all digits in the hexadecimal representation are 0-9, we need to ensure that none of the digits in the hexadecimal form are A-F (i.e., 10-15). Therefore, for a number to satisfy the condition, all digits in its hexadecimal form must be between 0 and 9.So, another way to think about this is: how many numbers from 1 to 1000 (decimal) have hexadecimal representations where each digit is less than 10. That is, each hexadecimal digit is 0-9. So, we can model this as numbers in base 16 where each digit is 0-9, but since we're converting from decimal, we need to ensure that their hexadecimal form doesn't have A-F.Alternatively, perhaps it's easier to model this as numbers in base 10 that, when converted to base 16, only have digits 0-9. So, the question becomes: how many numbers up to 1000 in decimal are such that when written in base 16, all their digits are 0-9. Then, count those numbers, find n, then sum the digits of n.Hmm. Let's first consider the range of numbers. 1000 in decimal is... let's see what that is in hexadecimal. To find the hexadecimal representation of 1000.Divide 1000 by 16:1000 ÷ 16 = 62.5, so quotient is 62, remainder is 8. So the least significant digit is 8.62 ÷ 16 = 3.875, quotient 3, remainder 14, which is E in hex.3 ÷ 16 = 0.1875, quotient 0, remainder 3.So reading the remainders from last to first: 3, E, 8. So 1000 decimal is 3E8 in hexadecimal.Therefore, numbers from 1 to 3E8 in hexadecimal. But the numbers we are considering are 1 to 1000 in decimal, which is 1 to 3E8 in hex.But we need to find how many numbers in 1 to 3E8 (hex) have hexadecimal digits only 0-9. Wait, but since we are converting decimal numbers to hexadecimal, maybe it's better to approach this problem by considering the hexadecimal digits of each number from 1 to 1000 (decimal) and check if any digit is A-F.Alternatively, perhaps we can model this as numbers in base 16 with digits 0-9 and find how many such numbers are less than or equal to 1000 (decimal), which is 3E8 in hex. So, we need to count all hexadecimal numbers composed only of digits 0-9 (i.e., digits 0-9 in each place) that are less than or equal to 3E8 (hex).But since 3E8 is the hexadecimal equivalent of 1000, we need to count all numbers in base 16, with digits 0-9 only, from 1 up to 3E8 (hex). Hmm. This might be a way to model it.But perhaps breaking it down by the number of digits in the hexadecimal representation.First, note that hexadecimal numbers can have 1, 2, 3, or more digits. Since 1000 decimal is 3E8 hex, which is a 3-digit hexadecimal number (digits: 3, E, 8). So hexadecimal numbers up to 3 digits.But since we need to go up to 3E8, we need to consider numbers with 1, 2, or 3 hexadecimal digits, but the 3-digit ones can't exceed 3E8.But since hexadecimal digits are 0-F, but our numbers can only use digits 0-9. So, for each number of digits (1, 2, 3), count the numbers where each digit is 0-9, and the entire number is less than or equal to 3E8 (hex).But since hexadecimal is base 16, we need to compare these numbers in hexadecimal. So, perhaps it's better to handle each digit place by place.First, let's find how many 1-digit hexadecimal numbers (1 to F) are composed only of digits 0-9. Since we are considering positive integers, 1-digit numbers are from 1 to F (1 to 15 in decimal). But since we need only digits, it's 1 to 9. So there are 9 numbers.Wait, but in hexadecimal, 1-digit numbers are from 0 to F (0 to 15), but since we are considering positive integers, it's 1 to 15. However, numbers with hexadecimal digits only 0-9 would be 1 to 9. So 9 numbers. Okay.Then, 2-digit hexadecimal numbers: from 10 (hex) to FF (hex). But in decimal, that's 16 to 255. But in terms of hexadecimal digits, the first digit (16's place) can be 1 to F, and the second digit (1's place) can be 0 to F. However, since we need both digits to be 0-9, the first digit can be 1 to 9 (since leading zeros aren't allowed for numbers), and the second digit can be 0 to 9. So the count is 9 (for first digit) * 10 (for second digit) = 90 numbers.Similarly, 3-digit hexadecimal numbers: from 100 (hex) to 3E8 (hex). So first, 100 (hex) is 256 in decimal, and 3E8 (hex) is 1000 decimal. So, for 3-digit hexadecimal numbers that use only digits 0-9, we need to consider numbers from 100 (hex) to 3E8 (hex), but all digits must be 0-9. However, since 3E8 in hex has digits 3, E, 8, and E is 14, which is a letter, so the highest number we can have with all digits 0-9 in hex is 999 (hex). But 999 (hex) is 9*16² + 9*16 + 9 = 9*256 + 9*16 + 9 = 2304 + 144 + 9 = 2457 decimal. But 1000 decimal is 3E8 hex, which is way less than 999 hex.Therefore, when considering 3-digit hexadecimal numbers with all digits 0-9, we can't go up to 999 (hex) but only up to 3E8 (hex). Therefore, we need to count all 3-digit hexadecimal numbers with digits 0-9 where the number is less than or equal to 3E8 (hex).But since the first digit of the hexadecimal number is the hundreds place (16²). Let's analyze the 3-digit hexadecimal numbers:Each 3-digit hexadecimal number is of the form ABC (hex), where A, B, C are digits. Here, A can be 1 to F (since leading zeros are not allowed), but in our case, A can be 1 to 9 (since we can't use A-F). However, since our upper limit is 3E8 (hex), which is a 3-digit number starting with 3, followed by E, followed by 8.So, in hexadecimal, to be less than or equal to 3E8, the first digit (A) can be 1, 2, or 3. But since we are restricted to digits 0-9, the first digit can be 1, 2, or 3 (as 3 is allowed). If the first digit is 1 or 2, then the remaining two digits can be anything from 0-9. If the first digit is 3, then the second digit must be less than or equal to E (14 in decimal), but since we are restricted to digits 0-9, the second digit can be 0-9 (but E is 14, so digits 0-9 are all less than E). Wait, no. Wait, in hexadecimal, 3E8 is 3*16² + 14*16 + 8. So, if we have a 3-digit number starting with 3, then the second digit is in the 16's place. If the second digit is 0-9, which in decimal is 0-9, but in hexadecimal, they are digits 0-9. So when comparing hexadecimal numbers, 3 followed by A (10) would be 3A0, which is higher than 3E8? Wait, no, 3A0 is 3*16² + 10*16 + 0 = 3*256 + 160 = 768 + 160 = 928. 3E8 is 1000. So 3A0 is 928, which is less than 1000.Wait, but if we have a 3-digit hexadecimal number starting with 3, then the maximum second digit is E (14), but since our numbers can only have digits 0-9, then even if the second digit is 9, which is less than E (14), so all such numbers starting with 3 and followed by 0-9 in the second digit would be less than 3E8 (hex). But wait, let's check.For example, 399 (hex) is 3*16² + 9*16 + 9 = 3*256 + 144 + 9 = 768 + 144 + 9 = 921. Which is less than 1000 (decimal). But 3E8 (hex) is 1000 (decimal). So even if the third digit is 9, the entire number is 399 hex = 921 decimal. Wait, so even with all digits maxed out at 9 in the first three digits (but first digit is 3), it's still 399 hex = 921 decimal. So all 3-digit hexadecimal numbers starting with 3 followed by two digits 0-9 are from 300 (hex) to 399 (hex), which is 768 to 921 decimal. But 3E8 (hex) is 1000 decimal, so actually, numbers starting with 3 in the first digit, and any digits 0-9 in the next two digits, are all less than 3E8 (hex). Wait, 3E8 is hexadecimal, which is 1000 decimal, so 399 hex is 921 decimal, which is less than 1000. Therefore, all 3-digit hexadecimal numbers starting with 3 and followed by two digits 0-9 are all less than 3E8 hex. Therefore, in that case, the count of 3-digit numbers would be:First digit: 3 (only 1 possibility, since it can't be more than 3 as 4 would be 400 hex which is 1024 decimal, over 1000)Second digit: 0-9 (10 possibilities)Third digit: 0-9 (10 possibilities)Therefore, 1*10*10 = 100 numbers.But wait, if the first digit is 3, second and third digits can be 0-9, so that's 100 numbers. However, 3E8 hex is 1000 decimal. So, in hexadecimal, numbers starting with 3 can go up to 3E8. However, in our case, since the second digit can only be 0-9 (since we can't use A-F), the maximum second digit is 9. So, 39 (hex) is 57 decimal. Wait, but in the 3-digit number, the second digit is multiplied by 16. So, 3 (first digit) * 256 + 9*16 + 9 = 921 decimal. So, even if we take the maximum possible with digits 0-9 in the first three digits (399 hex), it's 921 decimal, which is less than 1000. Therefore, all numbers starting with 3 followed by two digits 0-9 are within 1 to 1000 decimal.But wait, 3*16² + 9*16 + 9 = 768 + 144 + 9 = 921, as above. So 921 is less than 1000, so all 3-digit numbers starting with 3 and digits 0-9 are okay.Therefore, the total number of 3-digit hexadecimal numbers with digits only 0-9 is:First digit: 1-3. Wait, no. Wait, in the 3-digit numbers, first digit can be 1-9 (since leading zeros are not allowed), but since we need to be less than or equal to 3E8 (hex), which is 1000 decimal, but if first digit is 1, 2, or 3.Wait, actually, in hexadecimal, numbers from 100 to 3E8 (hex). So, 100 hex is 256 decimal. So, numbers from 256 to 1000 decimal. But in hexadecimal, the 3-digit numbers start at 100 (hex) = 256 (decimal). So, how do the 3-digit numbers break down:If the first digit is 1, the number is from 100 (hex) to 1FF (hex) (which is 256 to 511 decimal). If the first digit is 2, it's from 200 (hex) to 2FF (hex) (512 to 767 decimal). If the first digit is 3, it's from 300 (hex) to 3FF (hex) (768 to 1023 decimal). But our upper limit is 3E8 (hex) = 1000 decimal. So, in the first digit 3, the numbers go up to 3E8 (hex), which is 1000 decimal.But since we need numbers only up to 1000 decimal, the 3-digit numbers starting with 3 can go up to 3E8 (hex). But since our digits can only be 0-9, the first digit when it's 3 can be followed by digits 0-9. However, in reality, even if the first digit is 3 and the next two digits are 9, as we saw earlier, that's 399 hex = 921 decimal. But 3E8 hex is 1000 decimal, so 921 is still less than 1000. Wait, so actually, even if we use all digits 9, the maximum number is 399 hex, which is 921 decimal. Therefore, all 3-digit numbers starting with 3 and followed by digits 0-9 are within 256 to 921 decimal, which is within 1 to 1000. Therefore, actually, all 3-digit numbers with first digit 1, 2, or 3, followed by two digits 0-9, would be within the range. But wait, first digit 1, 2, or 3 in hexadecimal 3-digit numbers.But wait, hexadecimal digits can go up to F, but we're restricted to 0-9. So, the first digit can be 1-9 (since leading zeros are not allowed), but since the maximum first digit in our case (to not exceed 3E8 hex) is 3. Because if the first digit is 4, that's 400 hex which is 1024 decimal, which is beyond 1000.Therefore, first digit can only be 1, 2, or 3. So, for 3-digit numbers:- First digit: 1, 2, 3 (3 options)- Second digit: 0-9 (10 options)- Third digit: 0-9 (10 options)So, total 3*10*10 = 300 numbers.But wait, hold on. Let's verify:If first digit is 1: numbers from 100 (hex) to 199 (hex) which is 256 to 415 decimal. All these numbers would have hexadecimal digits 1 followed by two digits 0-9, so all within the allowed digits. Similarly, first digit 2: 200 (hex) to 299 (hex) which is 512 to 767 decimal. Again, all digits 0-9. First digit 3: 300 (hex) to 399 (hex) which is 768 to 921 decimal. Again, all digits 0-9. So, 3*10*10 = 300 numbers. However, our upper limit is 1000 decimal which is 3E8 hex, which is higher than 399 hex (921 decimal). So, there's a gap between 399 hex (921 decimal) and 3E8 hex (1000 decimal). So, numbers from 922 to 1000 decimal. But these numbers in hexadecimal would start with 3A0 (hex) which is 928 decimal. Wait, 3A0 hex is 3*256 + 10*16 = 768 + 160 = 928. Then 3A1, 3A2, ..., up to 3E8 (1000). However, since these numbers (from 3A0 to 3E8 hex) contain the digit A to E in their hexadecimal representation, which are not allowed (we can only use digits 0-9). Therefore, even though these numbers are within 1 to 1000 decimal, their hexadecimal representations include letters, so they are excluded. Therefore, the maximum allowed 3-digit hexadecimal number with digits 0-9 is 399 hex (921 decimal). Therefore, all 3-digit numbers with first digit 1, 2, or 3 and digits 0-9 are allowed, totaling 300 numbers. The numbers from 3A0 (928 decimal) to 3E8 (1000 decimal) are excluded because they contain letters.But wait, 3-digit hexadecimal numbers starting with 3 followed by digits 0-9 (in the second and third positions) go up to 399 hex, which is 921 decimal. So, numbers from 300 hex (768 decimal) to 399 hex (921 decimal). Then numbers from 922 to 1000 decimal (3E8 hex) are not included because their hexadecimal representations would have A-F digits. Therefore, in the 3-digit numbers, only 300 numbers (1-3 in first digit, 0-9 in second and third) are allowed. So total for 3-digit numbers is 300.So, combining all:- 1-digit numbers: 1-9 (9 numbers)- 2-digit numbers: 10-99 (hex), which is 16-159 decimal. Wait, hold on. Wait, 2-digit hexadecimal numbers range from 10 (hex) to FF (hex), which is 16 to 255 decimal. But since we need only digits 0-9, the first digit can be 1-9 and the second digit 0-9, as leading zeros are not allowed for numbers. Therefore, 9*10 = 90 numbers. Each of these is from 10 (hex) = 16 decimal to 99 (hex) = 153 decimal. Wait, 99 hex is 9*16 + 9 = 144 + 9 = 153 decimal. Wait, so 2-digit hexadecimal numbers with digits 0-9 are 10 (hex) to 99 (hex), which is 16 to 153 decimal. But the 2-digit numbers in hexadecimal go up to FF (255 decimal), but our upper limit is 1000 decimal. However, we need to count all 2-digit hexadecimal numbers with digits 0-9, which are 90 numbers. However, their decimal equivalents go up to 153. But since our total range is up to 1000, we need to include all of them. So, even though 153 is less than 1000, all 90 numbers are within 1-1000. So, 90 numbers.Therefore, total numbers:1-digit: 92-digit: 903-digit: 300Total n = 9 + 90 + 300 = 399Wait, that would be 9 + 90 = 99, plus 300 is 399. So n is 399. Then, the sum of the digits of n is 3 + 9 + 9 = 21. But 21 is option E. However, the options given are up to 21, so E is 21. But the answer choices don't list 21 as correct? Wait, wait the options are (A)17, (B)18, (C)19, (D)20, (E)21. So 21 is an option. But is 399 the correct count?Wait, hold on. Let's check this again. Maybe there's a mistake here.Wait, when we considered 3-digit numbers, we assumed first digit 1, 2, 3. But hexadecimal numbers with 3 digits start at 100 hex (256 decimal) and go up. But when we converted 399 hex to decimal, we got 921. However, 1000 decimal is 3E8 hex, which is beyond that. But all numbers from 1 to 1000 decimal that in hexadecimal have only digits 0-9 are:- 1 to 9 (1-digit)- 10 to 99 (hex) which is 16 to 153 decimal (2-digit)- 100 to 399 (hex) which is 256 to 921 decimal (3-digit)But wait, 399 hex is 921 decimal. Then, from 922 to 1000 decimal, the numbers in hex would start with 3A0 to 3E8, which include letters A-E. Therefore, those are excluded. So, the total count would be numbers from:- 1-9: 9 numbers- 16-153: 153 - 16 + 1 = 138 numbers? Wait, no. Wait, in hexadecimal, 2-digit numbers with digits 0-9 are from 10 (hex) to 99 (hex). Let's convert 10 hex to decimal: 16. 99 hex is 153 decimal. So the count is 153 - 16 + 1 = 138 numbers. Wait, but previously we thought of 9*10=90 numbers. So, which is correct?Wait, this seems like a conflict. Wait, perhaps here's the mistake. If we model the 2-digit hexadecimal numbers as 1-9 for the first digit and 0-9 for the second digit, that gives 9*10 = 90 numbers. But in decimal, those numbers start at 10 hex (16 decimal) and go up to 99 hex (153 decimal). The total count is 153 - 16 + 1 = 138 numbers. Wait, so there's a discrepancy here. Why the difference?Because when you model it as 9*10=90 numbers, each first digit (1-9) and second digit (0-9), each combination corresponds to a unique number. So, 10 hex is 16, 11 hex is 17, ..., 19 hex is 25, 1A hex is 26... Wait, no. Wait, in our case, the second digit is only 0-9, so 10 hex is 16, 11 hex is 17, ..., 19 hex is 25, 1A hex would be 26, but we are not including numbers with A-F. Therefore, the 2-digit numbers with first digit 1-9 and second digit 0-9 would be 10 hex (16) to 99 hex (153). Let's calculate how many numbers that is.From 16 to 153 inclusive. The number of numbers is 153 - 16 + 1 = 138. But according to the multiplication principle, it's 9*10=90. But 90 ≠ 138. Wait, that can't be. There must be an error here.Wait, hold on. Wait, hexadecimal numbers from 10 to FF in hex are from 16 to 255 in decimal. But we are only considering those with digits 0-9, which would be from 10 (hex) to 99 (hex) (16 to 153 decimal). So, the numbers are 16,17,...,153. How many numbers is that? Let's compute 153 - 16 + 1 = 138. But if we model it as first digit 1-9 (9 options) and second digit 0-9 (10 options), that's 9*10=90. So, why the discrepancy?Wait, here's the problem. The first digit in hexadecimal can be 1-9 (9 options) and the second digit 0-9 (10 options). Each combination gives a unique number. For example:First digit 1: 10,11,...,19 (hex) which is 16,17,...,25 decimal (10 numbers)First digit 2: 20,21,...,29 (hex) which is 32,33,...,41 decimal (10 numbers)...First digit 9: 90,91,...,99 (hex) which is 144,145,...,153 decimal (10 numbers)So each first digit gives 10 numbers. There are 9 first digits, so 9*10=90 numbers. So, the decimal equivalents are from 16 to 153, but not all numbers in that range are included. Because between 25 and 32 decimal, there's a gap. For example, 26 decimal is 1A hex, which is excluded. Similarly, between 41 and 48 decimal (hex 29 is 41, hex 30 is 48), but hex 30 is 48, which is allowed. Wait, but 30 hex is a 2-digit number with digits 3 and 0, so allowed. Wait, no. Wait, hex 30 is 48 decimal. Wait, 3*16 + 0 = 48. So, 30 hex is allowed. Wait, but if the first digit is 3, second digit 0, that's allowed. So, the numbers would go:First digit 1: 10-19 hex (16-25 decimal)First digit 2: 20-29 hex (32-41 decimal)First digit 3: 30-39 hex (48-57 decimal)...First digit 9: 90-99 hex (144-153 decimal)So, each block is 10 numbers, but there are gaps between the blocks. For example, between 25 and 32 decimal, there's 26-31 decimal which are represented as 1A-1F hex, which are excluded. Similarly, between 41 and 48 decimal, there's 42-47 decimal which are 2A-2F hex, excluded.Therefore, in decimal, the numbers included are 16-25, 32-41, 48-57, ..., 144-153. Each block of 10 numbers, but spaced apart by 6 numbers in between (the ones with A-F in hex). Therefore, the total count is 9 blocks * 10 numbers = 90 numbers, but in decimal, they are non-consecutive ranges.Therefore, the correct count is 90 numbers, not 138. The 138 would be if we were counting all numbers from 16 to 153 inclusive, but since many of those numbers (those that have A-F in their hex digits) are excluded, the actual count is 90.Therefore, going back to the original calculation:1-digit numbers: 1-9 (9 numbers)2-digit numbers: 90 numbers (as above)3-digit numbers: first digit 1-3, second and third digits 0-9. So:First digit 1: 100-199 hex (256-415 decimal). Each first digit 1, second and third digits 0-9. So 10*10=100 numbers.First digit 2: 200-299 hex (512-767 decimal). Similarly, 100 numbers.First digit 3: 300-399 hex (768-921 decimal). Similarly, 100 numbers.Total 3-digit numbers: 100 + 100 + 100 = 300 numbers.But wait, 100-199 hex is 256-415 decimal. However, 199 hex is 1*16² + 9*16 + 9 = 256 + 144 + 9 = 409 decimal. So, the range is 256-409 decimal. Similarly, 200-299 hex is 512-767 decimal, and 300-399 hex is 768-921 decimal. But each of these ranges has 100 numbers (10*10). However, in decimal, they cover:First digit 1: 256-409 (154 numbers) but 100 numbers in hex. Wait, this seems conflicting. Wait, 100-199 in hex is 256-409 in decimal. The difference is 409 - 256 + 1 = 154 numbers. But since we are only allowing digits 0-9 in hex, each of the 3-digit numbers with first digit 1, second and third digits 0-9 would be 100-199 in hex (digits 1,0-9,0-9). So in decimal, each is unique and counts to 100 numbers. But the decimal range 256-409 includes numbers outside of the 100-199 hex range. Wait, no. Actually, 100 hex is 256, 199 hex is 409. So, numbers from 256 to 409 decimal correspond to 100-199 hex. But since we are only allowing digits 0-9 in hex, the numbers from 100-199 hex with digits 0-9 are exactly 1 (first digit) * 10 (second digit) * 10 (third digit) = 100 numbers. Similarly for 200-299 hex and 300-399 hex.Therefore, in decimal, each 3-digit hexadecimal number with digits 0-9 is unique and translates to a unique decimal number. Therefore, the count is indeed 100 per first digit (1,2,3), totaling 300 numbers. However, these correspond to decimal ranges:- 100-199 hex: 256-409 decimal (100 numbers)- 200-299 hex: 512-767 decimal (100 numbers)- 300-399 hex: 768-921 decimal (100 numbers)Thus, total 300 numbers.Therefore, total n = 9 (1-digit) + 90 (2-digit) + 300 (3-digit) = 399.But 399 in decimal. Then, the sum of the digits of n (399) is 3 + 9 + 9 = 21. So, the answer would be E) 21.But the options given include 21 as E. However, let's check again whether our count is correct.Wait, but hold on. When we count the 3-digit numbers, we considered first digit 1, 2, 3. But in hexadecimal, 3-digit numbers can start with 1-9 (since leading zeros are not allowed), but since our upper limit is 3E8 hex (1000 decimal), first digits can only go up to 3. So that's correct.But another way to calculate it is to consider that the total numbers with hexadecimal digits only 0-9 up to 3E8 hex can be calculated using the positional digits.In hexadecimal, each digit can be 0-9, and we need numbers less than or equal to 3E8 (hex). So, this is similar to counting numbers in base 16 with digits 0-9, up to 3E8.But since 3E8 has a digit 'E' which is beyond 9, we can model this as follows:For numbers with 1 digit: 1-9 (9 numbers)For numbers with 2 digits: first digit 1-9, second digit 0-9 (9*10=90 numbers)For numbers with 3 digits:- If the first digit is 1 or 2: the remaining two digits can be 0-9 (2*10*10=200 numbers)- If the first digit is 3: the second digit can be 0-9 (since E in 3E8 is 14, which is higher than 9, so any second digit 0-9 is allowed). However, the third digit must be 0-8 because the original number is 3E8, and the third digit in 3E8 is 8. Wait, but hold on. In hexadecimal, 3E8 is 3*16² + 14*16 + 8. If we fix the first digit as 3, then the second digit can be 0-9 (since 9 < E (14)), but the third digit must be 0-8, because if the second digit is 9, the third digit can only go up to 8 to not exceed 3E8. Wait, but this is a mistake.Wait, let's think carefully. If the first digit is 3, then in hexadecimal, the maximum number is 3E8. So, if we fix the first digit as 3, the second digit can be 0-14 (0-E), but since we are restricted to digits 0-9, the second digit can only be 0-9. Similarly, the third digit: if the second digit is less than E (14), which it is (since second digit is 0-9), then the third digit can be 0-15 (0-F), but again restricted to 0-9. However, since the original number is 3E8, even if the second digit is less than E, the third digit can still be up to 15, but we're limited to 0-9.Wait, no. Actually, in order for the hexadecimal number to be less than or equal to 3E8 (hex), when the first digit is 3:- If the second digit < E (14), then the third digit can be anything (0-9 in our case).- If the second digit = E (14), then the third digit must be <= 8. But since we can't use E in the second digit (only 0-9 allowed), the second digit can't be E. Therefore, all numbers with first digit 3, second digit 0-9, and third digit 0-9 are allowed, because even with second digit 9 (which is < E), the third digit can be 0-9, and the total value would be 3 9 9 (hex) = 3*256 + 9*16 + 9 = 921 decimal, which is less than 1000. Therefore, all numbers starting with 3 followed by two digits 0-9 are allowed. Therefore, 10*10=100 numbers for first digit 3.Therefore, total 3-digit numbers:First digit 1: 100-199 (hex) → 100 numbersFirst digit 2: 200-299 (hex) → 100 numbersFirst digit 3: 300-399 (hex) → 100 numbersTotal 300 numbers.Thus, total n = 9 + 90 + 300 = 399.Sum of digits: 3 + 9 + 9 = 21.But the answer options go up to 21 (option E). However, looking back at the problem statement: "the first 1000 positive integers". So numbers from 1 to 1000 inclusive. Wait, but 399 hex is 921 decimal. So numbers up to 921 decimal would be included. However, numbers from 922 to 1000 decimal (which are 399 hex +1 to 3E8 hex) are excluded because their hexadecimal representations include A-F. Therefore, our count of 399 is correct.But wait, let's verify with another approach. Let's model this as numbers in base 10 that when converted to base 16 do not contain any letters. So, for each number from 1 to 1000, check if all digits in hex are 0-9.An alternative way is to note that numbers that in hexadecimal have only digits 0-9 are exactly the numbers that can be written in decimal using digits 0-9 in base 16. So, they are numbers that in base 16, each digit is 0-9. So, these numbers are a subset of decimal numbers.Another approach is to model it as numbers in base 10 which are also valid in base 16 with digits 0-9. But this might complicate.Alternatively, let's think of it as numbers from 1 to 1000 where when divided by 16 successively, all quotients and remainders are less than 10.For a number N, divide by 16:- If the remainder (digit) is >=10, reject.- Then take the quotient and divide by 16 again.- Repeat until quotient is 0.If at any step, the remainder is >=10, reject the number. Otherwise, accept it.So, we need to count all numbers from 1 to 1000 where all digits in their hexadecimal representation are 0-9.But how to compute this?Alternatively, think of each number as a 3-digit hexadecimal number (since 1000 is 3E8 hex). So, numbers can have 1, 2, or 3 hexadecimal digits. For each number of digits:- 1-digit: 1-9 (9 numbers)- 2-digit: first digit 1-9, second digit 0-9 (9*10=90)- 3-digit: first digit 1-3 (since 3E8 is the upper limit). If first digit is 1 or 2, the remaining two digits can be 0-9. If first digit is 3, the remaining two digits can be 0-9 (since 399 hex is 921 decimal, which is <=1000). Therefore, first digit 1: 100-199 hex (256-409 decimal) → 100 numbersFirst digit 2: 200-299 hex (512-767 decimal) → 100 numbersFirst digit 3: 300-399 hex (768-921 decimal) → 100 numbersSo 300 numbers.Total n=9+90+300=399. Sum of digits: 3+9+9=21.So, option E.But wait, looking back at the problem statement, the answer choices are (A)17, (B)18, (C)19, (D)20, (E)21. So 21 is an option. Therefore, the answer should be E)21.But let's verify with a small example.For instance, numbers from 1 to 16:1-9: valid (1-9 in hex)10-15: in hex, A-F → invalid16: 10 hex → valid (digits 1 and 0)So numbers from 1-16 decimal:Valid numbers: 1-9 (9 numbers) and 16 (1 number). Total 10.But according to our formula:1-digit: 92-digit: 1 (only 10 hex, which is 16 decimal)But in our earlier calculation for 2-digit numbers, we have 9*10=90. But in this case, up to 16 decimal, which is 10 hex, the count for 2-digit numbers is 1. But according to 9*10, it's 90. This discrepancy is because in the case of the full range up to 1000 decimal, the 2-digit numbers are fully covered, but in the smaller range up to 16 decimal, only one 2-digit number exists.Therefore, our formula works for the full range but not for partial ranges. However, since the problem is about numbers up to 1000 decimal (3E8 hex), which is a full range where 3-digit numbers start at 256 decimal, the formula holds.Alternatively, another way to check is to code a small script to count the numbers from 1 to 1000 that have hexadecimal digits only 0-9. But since we can't do that here, we need to trust the mathematical approach.But let's test another small example. Let's take numbers up to 30 decimal.Numbers from 1-30 decimal.Valid hexadecimal numbers (digits 0-9):1-9 (9 numbers)10-15 decimal: in hex, A-F → invalid16 decimal: 10 hex → valid17-25 decimal: 11-19 hex → valid (digits 1-1 to 1-9)26 decimal: 1A hex → invalid27-31 decimal: 1B-1F hex → invalid32 decimal: 20 hex → validSo, up to 30 decimal:Valid numbers:1-9 (9)16 (10 hex)17-25 (11-19 hex) → 9 numbers (17 to 25 is 9 numbers)Total: 9 + 1 + 9 = 19.But according to our formula:1-digit: 92-digit: 9*10=90 (but up to 30 decimal, we can't have all 90). So, clearly, the formula works only when the entire range of 2-digit hex numbers (16-153 decimal) is considered. However, in partial ranges, it doesn't. But in our problem, the upper limit is 1000 decimal, which is beyond the 2-digit hex numbers (up to 153 decimal), so all 2-digit hex numbers with digits 0-9 are included. Therefore, the formula applies.Therefore, with that verification, I think our original calculation is correct: n=399, sum of digits=21.However, let's check one more thing. If n=399, sum is 3+9+9=21. But perhaps the problem is in Chinese, and there might be a translation error. Wait, the original problem says: "the first 1000 positive integers". So numbers from 1 to 1000 inclusive. Then, the hexadecimal numbers using only digits 0-9. So, when converted to hexadecimal, they don't have A-F.But if we count n as 399, then sum is 21. However, the options include 21. But let me check once more.Alternative approach:We can model the allowed numbers as numbers in base 10 that, when written in base 16, contain only digits 0-9. So, each digit in base 16 must be 0-9. Therefore, the numbers are those where all digits in base 16 are <=9.This is equivalent to numbers that can be expressed as sums of powers of 16 with coefficients 0-9.So, the count of such numbers up to 1000 (decimal) is equivalent to the count of numbers in base 16 with digits 0-9 that are <=3E8 (hex).So, another way to compute this is:Let's represent 1000 decimal in hex, which is 3E8.So, we need numbers in base 16 with digits 0-9, less than or equal to 3E8.This is a classic digit dynamic programming problem, where we count numbers with certain digit constraints.But since time is limited, let's try to compute it.First, consider numbers with 1 digit: 1-9 (9 numbers)Numbers with 2 digits: from 10 (16) to 99 (153) hex. Each digit 0-9, leading digit 1-9. Total 9*10=90.Numbers with 3 digits: from 100 (256) to 3E8 (1000). Now, since the upper limit is 3E8, we need to consider numbers with 3 digits:The first digit can be 1,2, or 3 (since 3E8 starts with 3). If the first digit is 1 or 2, then the remaining two digits can be 00-99 (hex), i.e., 0-9 in each. If the first digit is 3, then the next digits must be <= E8 (hex). However, since we can only use digits 0-9, the second digit can be 0-9 (which is <= E=14), but then the third digit must be <=8 (since in the upper limit it's 8). Wait, no:Wait, the upper limit when first digit is 3 is 3E8 (hex). So, if the first digit is 3, the second digit can be from 0-E (but we can only use 0-9), and the third digit from 0-8 (if the second digit is E). But since we can't use E in the second digit, the second digit is 0-9. So even if the second digit is 9, which is less than E, the third digit can be 0-9. However, 3 9 9 hex is 3*256 +9*16 +9=768+144+9=921 decimal, which is less than 1000. Therefore, even with second digit 9 and third digit 9, it's still 921. Therefore, all numbers starting with 3 followed by two digits 0-9 are allowed. Therefore, first digit 3, second digit 0-9, third digit 0-9: 10*10=100 numbers.Therefore, total 3-digit numbers: first digit 1 (100-199 hex): 100 numbers, first digit 2 (200-299 hex): 100 numbers, first digit 3 (300-399 hex): 100 numbers. Total 300 numbers.Thus, n=9+90+300=399, sum of digits 3+9+9=21, option E.However, looking back at the answer choices, the options given are up to 21, which is E. Therefore, the answer should be E)21.But wait, the options in the problem are:(A) 17(B) 18(C) 19(D) 20(E) 21Therefore, the correct answer is E)21.But I recall that in some similar problems, the count is different, but I think in this case, our step-by-step approach is correct. Therefore, the answer is E)21.Final Answerboxed{E}</think>To determine the number of hexadecimal numbers among the first 1000 positive integers that use only digits 0-9 (i.e., no letters A-F), we need to consider the hexadecimal representations of these numbers. 1. 1-digit hexadecimal numbers: These are 1 to 9 in decimal, giving 9 numbers.2. 2-digit hexadecimal numbers: These range from 10 (hex) to 99 (hex), which converts to 16 to 153 in decimal. Each digit must be 0-9, leading to 9 (first digit) * 10 (second digit) = 90 numbers.3. 3-digit hexadecimal numbers: These range from 100 (hex) to 3E8 (hex) (which is 1000 in decimal). - If the first digit is 1 or 2, the remaining two digits can be 0-9, giving 2 * 10 * 10 = 200 numbers. - If the first digit is 3, the second and third digits must be 0-9 (since 3E8 in hex is 1000 in decimal, and all digits 0-9 in this range are valid), giving 1 * 10 * 10 = 100 numbers. - Total 3-digit numbers: 200 + 100 = 300.Summing these, we get:- 1-digit: 9- 2-digit: 90- 3-digit: 300Total count ( n = 9 + 90 + 300 = 399 ).The sum of the digits of 399 is ( 3 + 9 + 9 = 21 ).Thus, the answer is (boxed{E}).,<think>Okay, so I have this problem here about hexadecimal numbers. Let me try to understand what it's asking. Hexadecimal numbers use digits 0-9 and letters A-F, which represent 10-15. The question says: among the first 1000 positive integers, find all hexadecimal numbers that are represented using only digits. Then, determine the number of these numbers, n, and the sum of the digits of n is which option (A to E). The answer choices are numbers 17 to 21. Hmm.First, I need to clarify what exactly the problem is asking. It says "hexadecimal numbers that are represented using only digits." So, in hexadecimal, the digits can be 0-9 and A-F, but they want numbers that, when represented in hexadecimal, only use the digit characters (0-9), not the letters A-F. Then, among the first 1000 positive integers, how many such numbers are there? Then, once we find that number n, compute the sum of the digits of n and select the correct option.So, step by step:1. Convert the numbers from 1 to 1000 into hexadecimal.2. Check if each hexadecimal representation contains only digit characters (0-9), no letters A-F.3. Count how many such numbers there are; that's n.4. Compute the sum of the digits of n.5. Choose the correct option based on that sum.Alright, let's break this down.First, the first 1000 positive integers. That is, numbers from 1 to 1000 in decimal. But we need to convert them to hexadecimal and check if their hexadecimal representations consist solely of digits 0-9.So, for each decimal number from 1 to 1000, convert it to hexadecimal. If in that hexadecimal number, there are no letters A-F, then count it. The total count is n. Then sum the digits of n.So, the key here is figuring out how many numbers from 1 to 1000 (decimal) have hexadecimal representations without any letters A-F. Then, once we have that count, sum its digits.First, let's recall how hexadecimal works. Each digit represents 4 bits, so hexadecimal digits go from 0 to 15, with 0-9 being digits and 10-15 being A-F. So, if a hexadecimal number has digits only from 0-9, that means each digit in its hexadecimal representation is between 0 and 9.Therefore, such a number in decimal must be representable in hexadecimal with digits 0-9. So, for example, decimal 15 is hexadecimal F, which would be excluded. Decimal 10 is hexadecimal A, excluded. But decimal 9 is hexadecimal 9, included. Decimal 16 is hexadecimal 10, which uses digits 1 and 0, so that's included.Therefore, we need to find all numbers from 1 to 1000 (decimal) such that when converted to hexadecimal, all their hexadecimal digits are 0-9. So, how can we count these numbers?One approach is to model this as numbers in base 10 that, when converted to base 16, only have digits 0-9 in each position. Alternatively, it's equivalent to numbers in decimal that, when divided by 16 repeatedly, give remainders only between 0 and 9 at each step.Alternatively, perhaps it's easier to model this as numbers in base 16 with digits 0-9. So, numbers in base 16 with digits only 0-9 are numbers that in base 10 are equal to the sum of digits multiplied by powers of 16. For example, a 1-digit hexadecimal number is 0-9, so decimal 0-9. A 2-digit hexadecimal number is from 0*16 + 0 to 9*16 + 9, which is decimal 0 to 153. A 3-digit hexadecimal number is 0*256 + 0*16 + 0 to 9*256 + 9*16 + 9, which is 0 to 2457. But since we are only going up to decimal 1000, we need to consider hexadecimal numbers up to 3 digits (since 16^3 = 4096, which is more than 1000). Wait, but 16^3 is 4096, so the first 1000 decimal numbers would be up to hexadecimal 3E8 (since 3E8 in hexadecimal is 3*16^2 + 14*16 + 8 = 3*256 + 14*16 + 8 = 768 + 224 + 8 = 1000). So, hexadecimal numbers up to 3E8.But we need numbers in hexadecimal that use only digits 0-9. So, how many such numbers are there from 1 to 3E8 in hexadecimal?But since we need to count decimal numbers from 1 to 1000 (which is hexadecimal 1 to 3E8), and check if their hexadecimal representations only contain digits 0-9.So, another way to think about this: the numbers we want are numbers from 1 to 1000 in decimal, whose hexadecimal digits are all 0-9. So, each digit in the hexadecimal representation is between 0 and 9.Therefore, to count these numbers, we can model this as numbers in base 10 that, when expressed in base 16, have all digits <=9. So, we need to count numbers in the range [1, 1000] that don't have any digits in their hexadecimal representation equal to A-F (10-15).So, how can we compute this?Perhaps we can model this as a digit-based problem. Let's consider how many digits the hexadecimal number has. Since 16^3 = 4096, and 1000 is less than 4096, hexadecimal numbers up to 3 digits (but the 3-digit hexadecimal numbers go up to 4095). However, since we are only going up to 1000 decimal, which is 3E8 in hexadecimal, we need to consider 1 to 3E8 in hexadecimal.So, hexadecimal numbers from 1 to 3E8 (hex). So, how many hexadecimal numbers in that range use only digits 0-9.This is similar to counting numbers in base 16 with certain digit restrictions. Let's consider how many 1-digit, 2-digit, and 3-digit hexadecimal numbers there are in this range, with all digits 0-9.First, let's note that 3E8 in hexadecimal is 3 digits: '3', 'E', '8'. So, the first digit (from left) can be 1, 2, or 3 (since hexadecimal 3E8 is 1000 decimal). The second digit can be up to E (14) if the first digit is 3, but since we want digits only 0-9, the second digit can only go up to 9. Wait, but 3E8 is hexadecimal. So, in the range 1 to 3E8 (hex), how many numbers have all digits 0-9.Wait, perhaps it's better to split into cases based on the number of digits in hexadecimal.First, 1-digit hexadecimal numbers: 1 to F (1 to 15 decimal). But we need only those with digits 0-9. So, 1 to 9. That's 9 numbers.Then, 2-digit hexadecimal numbers: 10 (A) to FF (255 decimal). But we need only those 2-digit hexadecimal numbers where both digits are 0-9. So, from 10 (which is '10' in hex, decimal 16) to 99 (hex), which is 153 decimal. So, how many numbers is that? In hexadecimal, the first digit can be 1-9 (since it's 2-digit), and the second digit can be 0-9. So, 9*10=90 numbers. But wait, in decimal terms, that's 16 to 153.But wait, the range in hexadecimal is 10 to 99, which is 16 to 153 in decimal. So, 153 - 16 +1 = 138 numbers? Wait, no. Wait, 10 in hex is 16 decimal, 11 is 17, ..., 99 in hex is 153. So, the count is 153 - 16 +1 = 138. But we calculated 9*10=90. Wait, that's a discrepancy. Wait, actually, the numbers from 10 to 99 in hexadecimal are 9*16 + 0 to 9*16 +15? Wait, no. Wait, hexadecimal numbers:Each 2-digit hexadecimal number is from 10 to FF in hex. But if we restrict both digits to 0-9, then the first digit is 1-9 (hex), and the second digit is 0-9 (hex). So, in decimal, that's 1*16 +0=16 up to 9*16 +9= 153. So, how many numbers? For the first digit (hex), 1-9 (9 choices), second digit 0-9 (10 choices), so 9*10=90 numbers. So, 90 numbers from 16 to 153 decimal. Therefore, 90 numbers. Then why does 153-16+1=138? That can't be. Wait, 153-16 is 137, so 137+1=138. So, that's conflicting with the 90. Wait, maybe I'm confusing decimal and hexadecimal counts.Wait, no. Wait, in hexadecimal, the numbers 10 to 99 (hex) correspond to decimal 16 to 153. However, if we restrict the hexadecimal digits to 0-9, then the numbers are 10 (hex) to 99 (hex), but with each digit 0-9. So, that's exactly 1-9 in the first digit (hex) and 0-9 in the second digit (hex). Therefore, 9*10=90 numbers. But 10 (hex) is 16 decimal, and 99 (hex) is 153 decimal. So, decimal numbers from 16 to 153, but only 90 numbers. Because in hexadecimal, there are numbers between 10 (16) and 99 (153) that have letters in their digits, which are excluded. For example, 1A (26 decimal) is excluded because of the A. So, the count is 90 numbers.So, okay, for 2-digit hexadecimal numbers with digits only 0-9, it's 90 numbers. Then, moving on.3-digit hexadecimal numbers: 100 (hex) to 3E8 (hex). But again, only those with digits 0-9. So, 3-digit hexadecimal numbers where each digit is 0-9, but also not exceeding 3E8 (hex). Wait, 3E8 is 1000 decimal. So, hexadecimal numbers from 100 (hex) to 3E8 (hex) (decimal 256 to 1000), but only those with digits 0-9 in each position.But in hexadecimal, the first digit can be 1, 2, or 3. However, 3E8 has the second digit as E (14), which is a letter. So, if we restrict to digits 0-9, then the maximum first digit is 3, but the second digit can be up to 9 if the first digit is 3. Wait, let's break it down.For 3-digit hexadecimal numbers with all digits 0-9:First digit: 1-9 (hex), but since our upper limit is 3E8 (hex), which is 3 in the first digit. However, since we can't exceed 3E8, we have to consider that when the first digit is 3, the second digit can only be up to E (14), but since we need digits 0-9, the second digit can only be up to 9. Therefore, for first digit 3, the second digit can be 0-9, and third digit can be 0-9, but the maximum number would be 399 (hex) which is 3*16^2 +9*16 +9= 768 + 144 +9= 921 decimal. But 3E8 (hex) is 1000 decimal. So, 399 (hex) is 921, which is less than 1000. Therefore, even if we consider numbers up to 3 digits in hexadecimal with all digits 0-9, the maximum is 921, but we need to go up to 1000. However, since 1000 is 3E8 in hex, which has a digit 'E' in the second place, which is not allowed. Therefore, the maximum allowed number in hexadecimal with all digits 0-9 would be 399 (hex) = 921 decimal. Wait, but 921 is less than 1000, so actually, numbers from 256 (100 hex) to 921 (399 hex) in decimal, but only those with hexadecimal digits 0-9. So, 3-digit hexadecimal numbers with digits 0-9, but even the maximum such number is 399 hex, which is 921 decimal. Therefore, the numbers from 256 to 921 decimal, but only those that in hexadecimal have all digits 0-9.But wait, the problem says "the first 1000 positive integers", so decimal numbers from 1 to 1000. So, we need to include all numbers from 1 to 1000 that in hexadecimal have only digits 0-9. So, 1-digit, 2-digit, and 3-digit hexadecimal numbers as above, but up to decimal 1000. However, as we saw, 3-digit hexadecimal numbers with digits 0-9 only go up to 399 hex = 921 decimal. So, numbers from 256 to 921 decimal that are 3-digit in hexadecimal with all digits 0-9, plus numbers from 1-9 (1-digit), 16-153 (2-digit). But wait, 1-9 decimal are 1-digit hexadecimal numbers 1-9. Then 10-15 decimal are hexadecimal A-F, which are excluded. Then 16 decimal is hexadecimal 10, which is allowed (digits 1 and 0). So, 16 is included. Then up to 153 decimal is hexadecimal 99. Then 154 decimal is hexadecimal 9A, which is excluded.Then 3-digit numbers from 256 (100 hex) to 921 (399 hex). So, how many numbers is that?So, breaking down:1-digit hexadecimal numbers (digits 1-9): 9 numbers (decimal 1-9).2-digit hexadecimal numbers (digits 1-9 followed by 0-9): 9*10=90 numbers (decimal 16-153).3-digit hexadecimal numbers (digits 1-9 followed by 0-9 followed by 0-9): So, first digit can be 1-9, second digit 0-9, third digit 0-9. But wait, in hexadecimal, 3-digit numbers start at 100 hex (256 decimal). However, if we allow first digit to be 1-9, that would be hexadecimal numbers 100 to 999 (hex). But 999 hex is 2457 decimal, which is beyond our limit of 1000. Therefore, we have to restrict the 3-digit hexadecimal numbers to those that, when converted to decimal, are ≤1000.Wait, so perhaps instead of 9*10*10=900 numbers (for 3-digit hexadecimal numbers with digits 0-9), we have to limit them to numbers ≤1000 decimal.But 3-digit hexadecimal numbers with digits 0-9 start at 100 hex (256 decimal) and go up to 399 hex (921 decimal). Wait, but 399 hex is 3*16^2 + 9*16 +9 = 768 + 144 +9=921. So, decimal 256 to 921. So, how many numbers is that?Well, in hexadecimal, from 100 to 399 (hex). Each digit:First digit: 1, 2, 3 (since 3 is the first digit of the upper bound 399). But since in hexadecimal, 399 is the upper limit. Wait, but hexadecimal 399 is 921, which is less than 1000. So, for 3-digit numbers:First digit: 1, 2, 3 (hex digits). Wait, but in hexadecimal, the digits go up to F, but since we are only allowing digits 0-9, the first digit can be 1-9 in hexadecimal. However, since our upper limit is 399 hex (921 decimal), but hexadecimal 399 is part of the allowed numbers (digits 3,9,9), which are all 0-9. Therefore, all 3-digit hexadecimal numbers with digits 0-9, first digit from 1-3, but with the constraint that when first digit is 3, the second and third digits can only go up to 9. Wait, but in hexadecimal, 3 as the first digit is allowed, but we need to ensure that the entire number doesn't exceed 399 hex (921 decimal). Wait, but 3-digit hexadecimal numbers with first digit 3 can have second and third digits 0-9, resulting in numbers from 300 hex (768 decimal) to 399 hex (921 decimal). So, those are all allowed because 921 is less than 1000. Therefore, total 3-digit numbers would be:First digit: 1, 2, 3 (since 1-9 in hex, but our upper limit is 3 as the first digit). Wait, no. Wait, in hexadecimal, first digit can be 1-9, but since we are only going up to 3E8 (hex) which is 1000 decimal, and 3 is the first digit. However, if the first digit is 4-9 in hexadecimal, that would correspond to decimal 1024-... which is beyond 1000. So, actually, the first digit can only be 1, 2, or 3 (hex digits). Because 4 in the first digit is 4*256=1024, which is over 1000. Therefore, first digit is 1, 2, or 3.So, first digit: 1, 2, 3 (3 choices).Second digit: 0-9 (10 choices).Third digit: 0-9 (10 choices).So, total 3-digit numbers: 3*10*10=300 numbers.But wait, 3-digit hexadecimal numbers from 100 (256 decimal) to 399 (921 decimal). The total count is 921 -256 +1=666 numbers. But according to the calculation above, it's 300 numbers. Again, conflicting results. So, where is the mistake?Ah, here's the problem. The 3-digit hexadecimal numbers with digits 0-9 only can have first digit 1-9, but we are limited to numbers up to 3E8 (1000 decimal). However, since 3-digit hexadecimal numbers with digits 0-9 can go up to 399 hex (921 decimal), which is less than 1000. Therefore, all 3-digit hexadecimal numbers with digits 0-9 (i.e., first digit 1-9, second and third digits 0-9) up to 399 hex are included. However, in our problem, the decimal range is up to 1000, but the maximum hexadecimal number with all digits 0-9 is 399 hex (921 decimal). Therefore, the numbers from 256 decimal (100 hex) to 921 decimal (399 hex) that have all digits 0-9 in hexadecimal.But how many are these? Let's compute:First, 1-digit: 1-9 (9 numbers)2-digit: 10-99 hex (16-153 decimal), 90 numbers3-digit: 100-399 hex (256-921 decimal), but only with digits 0-9 in each position.Wait, but 399 hex is 921 decimal, which is less than 1000. So, 3-digit hexadecimal numbers with all digits 0-9 are from 100 hex (256 decimal) to 399 hex (921 decimal). So, how many numbers is that?In hexadecimal, 100 to 399 is 399 -100 +1 = 3*16^2 + 9*16 +9 -1*16^2 +1 = but perhaps easier in decimal.100 hex is 256 decimal.399 hex is 3*16^2 +9*16 +9 = 768 + 144 +9=921.So, numbers from 256 to 921 decimal. The total numbers are 921 -256 +1 = 666. But how many of these have hexadecimal digits only 0-9?But since the hexadecimal numbers in this range with digits only 0-9 are exactly the numbers where each digit is 0-9. So, first digit can be 1,2,3 (since 4 would be 1024 decimal which is over 1000). Wait, in hexadecimal, 3-digit numbers with first digit 1, 2, or 3 (hex), second and third digits 0-9. So, first digit: 1,2,3 (3 choices), second digit: 0-9 (10 choices), third digit: 0-9 (10 choices). So, 3*10*10=300 numbers. Therefore, in decimal, these numbers range from 100 hex (256) to 399 hex (921). So, 300 numbers. However, 399 hex is 921 decimal, which is less than 1000. So, the total numbers we have are:1-digit: 9 (1-9)2-digit: 90 (16-153)3-digit: 300 (256-921)Total: 9 +90 +300= 399 numbers.Wait, but 399? But 9+90=99, 99+300=399. Hmm.But decimal numbers from 1-9, 16-153, and 256-921. But 9 numbers, 90 numbers, 300 numbers. So total 399 numbers. But is that correct?Wait, but wait, for example, decimal 10 is hexadecimal A, which is excluded. Decimal 15 is hexadecimal F, excluded. Decimal 16 is hexadecimal 10, included. So, 16 is included. So, 1-9: included. 10-15: excluded. 16-153: included (as hexadecimal 10-99). 154-255: excluded (hexadecimal 9A-FF). 256-921: included (hexadecimal 100-399). 922-1000: excluded (hexadecimal 39A-3E8). So, yes, the total included numbers are 9 +90 +300=399.Therefore, n=399. Then, the sum of the digits of n. So, 399. The digits are 3, 9, 9. Sum: 3+9+9=21. But the options are (A)17, (B)18, (C)19, (D)20, (E)21. So, the answer would be E)21. But wait, the options are given in Chinese, probably, but here it's written as (A)17, etc. So, E is 21, which would be the answer.But wait, wait. Wait, in the problem statement, it says "the first 1000 positive integers". So, decimal 1 to 1000. But when we calculated n=399, we considered numbers up to 921 decimal (399 hex). But 921 is less than 1000. So, the numbers from 922 to 1000 decimal would be hexadecimal 39A to 3E8. However, in their hexadecimal representation, they have letters (since 39A has 'A', etc.), so they are excluded. Therefore, all numbers from 922 to 1000 are excluded. Therefore, n is indeed 399. Therefore, sum of digits is 3+9+9=21. Therefore, the answer is E)21.But wait, hold on. Wait, the problem says "the first 1000 positive integers", which includes decimal 1 to 1000. But when converted to hexadecimal, some of these numbers (decimal 1-1000) have hexadecimal representations with only digits 0-9. We calculated 399 such numbers. Then, the sum of the digits of n=399 is 3+9+9=21, which is option E.But the answer choices are (A)17, (B)18, (C)19, (D)20, (E)21. So, 21 is option E. Therefore, the answer is E.But wait, let me double-check my calculations, because sometimes when I do these digit counts, I might miscount.So, first, 1-digit hexadecimal numbers: 1-9 (decimal 1-9): 9 numbers.2-digit hexadecimal numbers: first digit 1-9 (hex), second digit 0-9 (hex). So, 9*10=90 numbers. Their decimal equivalents are 16 (10 hex) to 153 (99 hex).3-digit hexadecimal numbers: first digit 1-3 (hex) because first digit 4 would be 400 hex=1024 decimal, which is over 1000. So, first digit 1-3 (hex), second digit 0-9 (hex), third digit 0-9 (hex). So, 3*10*10=300 numbers. Their decimal equivalents are 256 (100 hex) to 921 (399 hex). So, 300 numbers.Total numbers: 9+90+300=399. So, n=399.Sum of digits: 3+9+9=21. So, answer is E)21.But wait, the options don't have 21 as the answer. Wait, no, wait, looking back:(A)17(B)18(C)19(D)20(E)21So, 21 is option E. So, the answer is E.But wait, according to the problem statement, it's a Chinese competition problem, maybe from the original poster. But perhaps I made a mistake in considering the 3-digit numbers. Let me verify.Wait, another way to think: each digit in the hexadecimal number can be 0-9. So, the numbers we are counting are numbers in decimal that can be written in hexadecimal with digits 0-9. So, essentially, numbers in base 16 where each digit is 0-9, which is equivalent to numbers in base 10 that are combinations of base 16 digits 0-9. So, how many such numbers are there up to 1000.Alternatively, we can model this as numbers in base 10 that do not contain any hexadecimal letters. So, numbers that in base 16 have digits only 0-9.But how to count them.Alternatively, using inclusion-exclusion. Wait, but maybe another approach. Let's consider hexadecimal digits.For numbers from 1 to 1000 in decimal, convert each to hexadecimal, and check if any digit is A-F. But since 1000 is 3E8 in hex, numbers from 1 to 3E8 in hex. So, numbers with 1, 2, or 3 hexadecimal digits (since 3E8 is 3 digits). But 3E8 has an 'E', so numbers up to that need to be considered.But let's think of it as numbers in base 16 with up to 3 digits, where each digit is 0-9, and the number is less than or equal to 3E8 (hex).But perhaps breaking it down digit by digit.First, 1-digit numbers: 1-9 (hex) = 1-9 decimal: 9 numbers.Second, 2-digit numbers: first digit 1-9 (hex), second digit 0-9 (hex). So, 9*10=90 numbers. In decimal, 16-153.Third, 3-digit numbers: here it gets trickier because of the upper limit 3E8 (hex). So, 3-digit numbers in hexadecimal where each digit is 0-9 and the number is ≤3E8 (hex).So, let's split this into 3-digit numbers:First digit: 1, 2, or 3 (hex). Because 4 and above would exceed 3E8 (since 400 hex is 1024 decimal, which is over 1000). So, first digit is 1, 2, or 3.If first digit is 1 or 2 (hex), then the remaining two digits can be 0-9 (hex), since 2FF hex is 767 decimal, which is less than 1000. So, for first digit 1 or 2:First digit: 1 or 2 (2 choices)Second digit: 0-9 (10 choices)Third digit: 0-9 (10 choices)So, 2*10*10=200 numbers.If the first digit is 3 (hex), then we need to ensure that the number is ≤3E8 (hex). Since 3E8 is 3*16^2 +14*16 +8=1000 decimal.But since the first digit is 3, and the remaining digits can only be 0-9 (hex), the maximum number we can get is 399 hex=921 decimal. So, even if we set second and third digits to 9 (hex), we get 399 hex=921 decimal, which is less than 1000. Therefore, when first digit is 3 (hex), second and third digits can be 0-9 (hex) without exceeding 3E8 (hex). Therefore, for first digit 3:Second digit: 0-9 (10 choices)Third digit: 0-9 (10 choices)So, 1*10*10=100 numbers.Therefore, total 3-digit numbers: 200 +100=300 numbers.Therefore, total numbers: 9 (1-digit) +90 (2-digit) +300 (3-digit)= 399 numbers.Therefore, n=399.Sum of digits: 3 +9 +9=21. So, answer is E)21.Therefore, the answer is E.Final Answerboxed{E}</think>To solve the problem, we need to determine how many of the first 1000 positive integers (in decimal) can be represented in hexadecimal using only digits 0-9 (i.e., without any letters A-F). We then find the sum of the digits of this count.1. 1-digit hexadecimal numbers: These range from 1 to 9 (decimal 1 to 9). There are 9 such numbers.2. 2-digit hexadecimal numbers: These range from 10 (hex) to 99 (hex). In decimal, this is 16 to 153. Each digit can be 0-9, giving 9 choices for the first digit (1-9) and 10 choices for the second digit (0-9). This results in (9 times 10 = 90) numbers.3. 3-digit hexadecimal numbers: These range from 100 (hex) to 399 (hex). In decimal, this is 256 to 921. The first digit can be 1, 2, or 3 (3 choices), and each of the next two digits can be 0-9 (10 choices each). This results in (3 times 10 times 10 = 300) numbers.Summing these counts:- 1-digit: 9- 2-digit: 90- 3-digit: 300Total count (n = 9 + 90 + 300 = 399).The sum of the digits of 399 is (3 + 9 + 9 = 21).Thus, the final answer is (boxed{E}).]

🔑:18. E.Notice that the hexadecimal representation of 1000 is (3E8)16. The number we are looking for is less than 1000 and is a hexadecimal number (a b c)_{16}, where a can be 0, 1, 2, 3; b can be 0 sim 9; c can also be 0 sim 9, resulting in a total of 400 numbers. However, 0=(000)_{16} is not a positive integer, so it is excluded.Thus, n=399.Therefore, the answer is 3+9+9=21.

🧠:[<think>Okay, so I need to simplify this expression: |a - b - c| + |b - c + a| + |c - a - b|. Hmm, the problem says that a, b, and c are the sides of a triangle. Let me recall some properties of triangles. In any triangle, the sum of any two sides must be greater than the third side. That is, a + b > c, a + c > b, and b + c > a. This is called the triangle inequality theorem. I think this will be useful here because the absolute value expressions might involve these kinds of sums and differences.Let me look at each term in the expression one by one. The first term is |a - b - c|. Let me rearrange that as |a - (b + c)|. Similarly, the second term is |b - c + a|, which can be written as |(a + b) - c|. The third term is |c - a - b|, which is the same as |c - (a + b)|. Wait, actually, that's the same as |(a + b) - c| with the absolute value, so maybe terms 2 and 3 are related? Let me check again.Wait, the second term is |b - c + a|. If I rearrange the terms inside the absolute value, that's |a + b - c|. The third term is |c - a - b|, which is the same as |c - (a + b)|. But absolute value makes |x| = |-x|, so |c - (a + b)| = |(a + b) - c|. Therefore, the second and third terms are actually the same? Wait, no. Let me see. The second term is |a + b - c| and the third term is |a + b - c| as well, because |c - a - b| = |-(a + b - c)| = |a + b - c|. So actually, terms 2 and 3 are the same. So that's interesting, so the original expression simplifies to |a - (b + c)| + 2|a + b - c|. Wait, is that right? Let me verify.Original expression: |a - b - c| + |b - c + a| + |c - a - b|.First term: |a - (b + c)|.Second term: |(a + b) - c|.Third term: |c - (a + b)|, which equals |(a + b) - c|. So yes, the second and third terms are the same. Therefore, the expression becomes |a - b - c| + 2|a + b - c|.Now, since a, b, c are sides of a triangle, by the triangle inequality, a + b > c, so a + b - c is positive. Therefore, |a + b - c| = a + b - c. Similarly, the first term is |a - b - c|. Let's analyze that. Since a, b, c are sides of a triangle, we have b + c > a. So, a - b - c would be a - (b + c) which is negative because b + c > a. Therefore, |a - (b + c)| = |-(b + c - a)| = b + c - a. Similarly, the first term is b + c - a.So putting that all together, the first term is (b + c - a), the second term is 2*(a + b - c). So the entire expression is (b + c - a) + 2*(a + b - c). Let me compute that.First term: b + c - aSecond term: 2*(a + b - c) = 2a + 2b - 2cAdding them together:(b + c - a) + (2a + 2b - 2c) = ( -a + 2a ) + (b + 2b) + (c - 2c )Which simplifies to:a + 3b - cWait, that can't be right. Let me check my arithmetic again.Wait, (b + c - a) + (2a + 2b - 2c) = combine like terms:For a: -a + 2a = aFor b: b + 2b = 3bFor c: c - 2c = -cSo total is a + 3b - c. Hmm. But this seems strange. Let me think. If the original expression is |a - b - c| + |a + b - c| + |a + b - c|, but that can't be. Wait, no, the original expression was |a - b - c| + |a + b - c| + |c - a - b|. But we saw that the second and third terms are both |a + b - c|. So the total expression is |a - b - c| + 2|a + b - c|.But when we substitute, since a < b + c, then |a - b - c| = b + c - a, and since a + b > c, then |a + b - c| = a + b - c. Therefore, the expression becomes (b + c - a) + 2*(a + b - c). Then, as before, adding these terms:(b + c - a) + 2a + 2b - 2c = ( -a + 2a ) + (b + 2b) + (c - 2c ) = a + 3b - c.But this seems counterintuitive. Let me test with actual triangle sides. Let me pick a triangle where a = 3, b = 4, c = 5. That's a right triangle. Then compute the original expression and see if it equals a + 3b - c.First, compute |a - b - c| + |b - c + a| + |c - a - b|.Substitute a=3, b=4, c=5:|3 - 4 -5| + |4 -5 +3| + |5 -3 -4|Compute each term:First term: |3 -4 -5| = |-6| = 6Second term: |4 -5 +3| = |2| = 2Third term: |5 -3 -4| = |-2| = 2Total expression: 6 + 2 + 2 = 10.Now compute a + 3b - c: 3 + 12 -5 = 10. So it matches here.But let's try another triangle. Let's take an equilateral triangle, a = b = c = 1.Original expression:|1 -1 -1| + |1 -1 +1| + |1 -1 -1| = |-1| + |1| + |-1| = 1 +1 +1 =3.a + 3b -c =1 +3*1 -1=1+3-1=3. It matches again.Wait, another example. Let's take a=5, b=5, c=5.Original expression: |5-5-5| + |5-5+5| + |5-5-5| = |-5| + |5| + |-5| =5+5+5=15.a + 3b -c=5 +15 -5=15. Correct.Hmm, so the expression simplifies to a + 3b -c? But how does this hold in general?Wait, but let's check another triangle where the sides are different. Let's take a=2, b=3, c=4.Compute original expression:|2 -3 -4| + |3 -4 +2| + |4 -2 -3| = |-5| + |1| + |-1| =5+1+1=7.a + 3b -c=2 +9 -4=7. Correct.Wait, so this seems to hold. But why? How does the expression |a - b - c| + 2|a + b - c| equal a + 3b - c? Let me see.Wait, if we start with the expression (b + c - a) + 2*(a + b - c), then:(b + c - a) + 2a + 2b - 2c = (-a + 2a) + (b + 2b) + (c - 2c) = a + 3b - c.So algebraically, that's correct. But why does this happen? Maybe it's due to the triangle inequalities.But is there a more straightforward way to simplify the original expression without substituting?Alternatively, maybe there's a symmetry here. Let me consider the original expression again:|a - b - c| + |b - c + a| + |c - a - b|But we noticed that the second and third terms are the same. So the expression is |a - b - c| + 2|a + b - c|.But given the triangle inequalities:a + b > c, so a + b - c >0, so |a + b - c| = a + b - c.Similarly, since a < b + c, then a - b - c = a - (b + c) <0, so |a - b - c| = b + c - a.Therefore, substituting:|a - b - c| + 2|a + b - c| = (b + c - a) + 2*(a + b - c) = b + c - a + 2a + 2b - 2c = ( -a + 2a ) + (b + 2b ) + (c - 2c ) = a + 3b - c.So algebraically, that's correct.But the problem says "simplify the expression". Is there a way to write this in terms of the triangle's sides?Wait, but in all the examples I tried, the expression simplifies to a + 3b -c, but that seems arbitrary. Wait, but in the examples, if I switch the roles of a, b, c, does the formula still hold?Wait, but in the original problem, the expression is |a - b - c| + |b - c + a| + |c - a - b|. If the variables are permuted, the expression would change. However, since a, b, c are sides of a triangle, they are interchangeable. But in our simplification, we ended up with a + 3b - c, which is not symmetric. That seems odd. How come?Wait, maybe I made a mistake in the simplification. Let me check again.Original expression: |a - b - c| + |a + b - c| + |a + b - c| = |a - (b + c)| + 2|a + b - c|. But is this correct?Wait, the original expression is |a - b - c| + |b - c + a| + |c - a - b|. Let's parse each term again.First term: |a - b - c| = |a - (b + c)|.Second term: |b - c + a| = |a + b - c|.Third term: |c - a - b| = |c - (a + b)| = |(a + b) - c|, which is the same as |a + b - c|. Therefore, the third term is |a + b - c|.Therefore, the expression is |a - (b + c)| + 2|a + b - c|. So yes, that is correct.But then, given triangle inequalities, a + b > c, so |a + b - c| = a + b - c. Also, since a < b + c, then |a - (b + c)| = b + c - a.Therefore, substituting those in:(b + c - a) + 2*(a + b - c) = b + c - a + 2a + 2b - 2c = ( -a + 2a ) + (b + 2b ) + (c - 2c ) = a + 3b - c.Therefore, the expression simplifies to a + 3b - c.But this is not symmetric in a, b, c. However, in the problem statement, a, b, c are sides of a triangle but not specified in any particular order. So if the problem had permuted the variables, would we get a different expression?Wait, for example, if we take the original expression and swap a and b, we would get |b - a - c| + |a - c + b| + |c - b - a|. Following the same steps, this would simplify to b + 3a - c. But in the original problem, since the expression isn't symmetric, the result isn't symmetric either. Therefore, the simplification depends on how the variables are ordered in the original expression.Therefore, in the given problem, the expression |a - b - c| + |b - c + a| + |c - a - b| simplifies to a + 3b - c. But wait, in our first example, when a=3, b=4, c=5, the expression equals 10, which is 3 + 3*4 -5=3+12-5=10. When a=5, b=5, c=5, it's 5+15-5=15. When a=2, b=3, c=4, it's 2+9-4=7. All these match. So algebraically, this seems to hold.But wait, perhaps I misapplied the triangle inequalities? Let me confirm. For triangle sides a, b, c:a + b > ca + c > bb + c > aTherefore, in each absolute value term:First term: |a - (b + c)| = b + c - a, because a < b + c.Second term: |a + b - c| = a + b - c, because a + b > c.Third term: |c - (a + b)| = a + b - c, same as above.Thus, substituting these in, the expression becomes (b + c - a) + 2*(a + b - c) = a + 3b - c. So this is correct.But why does this happen? It's not a symmetric expression. The answer is a linear combination of the sides. But maybe there's another way to think about it. For example, maybe all terms can be expressed in terms of the triangle's perimeter or something else. However, perimeter is a + b + c, but here we have a + 3b - c, which is different.Alternatively, is there a way to express this in terms of the triangle's semiperimeter or other properties? Maybe not directly.Alternatively, maybe there's a geometric interpretation. The absolute values might represent distances or something else. However, since a, b, c are sides, perhaps not. Alternatively, think of the expression as related to the triangle inequality.But regardless of the interpretation, algebraically we have shown that the expression simplifies to a + 3b - c. However, in the problem statement, the answer is requested as a number, but since it's in terms of a, b, c, perhaps the answer is a + 3b - c. But when I tested with actual triangles, it worked. However, this seems to depend on the order of a, b, c. Let me check with a different permutation.Suppose I have a triangle with sides a=4, b=3, c=5. Then compute the expression:Original expression: |4 - 3 -5| + |3 -5 +4| + |5 -4 -3| = | -4 | + | 2 | + | -2 | =4+2+2=8.Compute a + 3b -c=4 +9 -5=8. It matches. So even if we permute a and b, but keep the same order in the expression, it still works. Wait, but in this case, in the original problem, the variables are fixed as a, b, c. So if we swap a and b in the problem, the expression would be different. Therefore, the answer is dependent on the order of terms in the original expression. Therefore, the simplified expression is indeed a + 3b - c.But the problem statement doesn't specify any particular order for the sides, just that a, b, c are the sides. However, in the given expression, the order is fixed. Therefore, the answer must be a + 3b - c. But to verify once more, let's take another triangle where a=5, b=4, c=3. Then compute the original expression:|5 -4 -3| + |4 -3 +5| + |3 -5 -4| = | -2 | + |6| + | -6 | =2+6+6=14.Compute a + 3b -c=5 +12 -3=14. Correct. So even when c is the smallest side, the expression still holds.Wait, but here, a=5, b=4, c=3. Then a +3b -c=5+12-3=14. So the expression is equal to that. Therefore, regardless of the order of the sides, as long as the variables in the expression are fixed as |a - b - c| + |b - c + a| + |c - a - b|, then it simplifies to a + 3b - c. So even if the sides are such that b is not the middle length, the formula still holds.Therefore, despite the asymmetry in the variables, the expression indeed simplifies to a + 3b - c. Therefore, the answer is a + 3b - c, which can be written as boxed{a + 3b - c}.Wait, but wait a second. Let me check another triangle where b is the smallest side. Let me take a=5, b=2, c=4. Check if this is a valid triangle: 5 + 2 >4 (7>4), 5 +4 >2 (9>2), 2 +4 >5 (6>5). Yes, it's valid.Original expression: |5 -2 -4| + |2 -4 +5| + |4 -5 -2|.Compute each term:First term: |5 -2 -4| = |-1| =1.Second term: |2 -4 +5| = |3| =3.Third term: |4 -5 -2| = |-3| =3.Total: 1 +3 +3=7.Compute a +3b -c=5 +6 -4=7. Correct.Therefore, even when b is the smallest side, the formula holds. Therefore, regardless of the relative sizes of a, b, c, as long as they are sides of a triangle, the expression simplifies to a +3b -c.Alternatively, let's test a case where b is the largest side. Let's take a=3, b=5, c=4. Then compute:Original expression: |3 -5 -4| + |5 -4 +3| + |4 -3 -5| = | -6 | + |4| + |-4| =6+4+4=14.Compute a +3b -c=3 +15 -4=14. Correct.So in all these cases, it's working. Therefore, despite the asymmetry, the expression simplifies to a +3b -c. Therefore, the answer is boxed{a + 3b - c}.But wait, the problem says "simplify the expression", and usually, in such problems, there might be a more symmetric answer or a numerical constant. However, based on the algebra and the examples, the expression simplifies to a +3b -c, which is a linear combination of the sides.Is there any alternative approach that might lead to a different simplification? Let's think.Alternatively, maybe combine the absolute values by considering different cases based on the values of a, b, c. But since a, b, c are sides of a triangle, we already know the necessary inequalities (triangle inequalities), so we don't need to consider all cases. We can directly substitute based on the known inequalities.Therefore, the conclusion is that the expression simplifies to a +3b -c. Hence, the answer is boxed{a + 3b - c}.Final Answerboxed{a + 3b - c}</think>Given that (a), (b), and (c) are the three side lengths of a triangle, we need to simplify the expression: (|a - b - c| + |b - c + a| + |c - a - b|).First, we use the triangle inequality theorem, which states that the sum of any two sides of a triangle must be greater than the third side. This gives us the inequalities:- (a + b > c)- (a + c > b)- (b + c > a)We analyze each term in the expression:1. (|a - b - c|) can be rewritten as (|a - (b + c)|). Since (b + c > a), this term simplifies to (b + c - a).2. (|b - c + a|) can be rewritten as (|a + b - c|). Since (a + b > c), this term simplifies to (a + b - c).3. (|c - a - b|) can be rewritten as (|c - (a + b)|). Since (a + b > c), this term also simplifies to (a + b - c).Thus, the expression becomes:[|a - (b + c)| + 2|a + b - c|]Substituting the simplified values based on the triangle inequalities:[(b + c - a) + 2(a + b - c)]Combining like terms:[(b + c - a) + 2a + 2b - 2c = a + 3b - c]Therefore, the simplified expression is:[boxed{a + 3b - c}],<think>Okay, so I need to simplify this expression: |a - b - c| + |b - c + a| + |c - a - b|. Hmm, the problem states that a, b, and c are the sides of a triangle. That probably means they satisfy the triangle inequalities. Let me recall, the triangle inequality says that the sum of any two sides must be greater than the third side. So, for a triangle with sides a, b, c, we must have:1. a + b > c2. a + c > b3. b + c > aRight, so each side is less than the sum of the other two. That's going to be important here because the expressions inside the absolute values might be positive or negative, and the triangle inequalities can help determine their signs.Let's look at each absolute value term one by one.First term: |a - b - c|Let me rearrange that: |a - (b + c)|. According to the triangle inequality, the sum of any two sides must be greater than the third. So, b + c > a. Therefore, a - (b + c) would be negative. So, the absolute value of a negative number is its opposite. So, |a - b - c| = |a - (b + c)| = (b + c) - a.Second term: |b - c + a|Let me rearrange that: |(a + b) - c|. Again, using triangle inequality, a + b > c, so (a + b) - c is positive. Therefore, the absolute value of a positive number is itself. So, |b - c + a| = |(a + b) - c| = (a + b) - c.Wait a second, hold on. Let me check that again. The expression is |b - c + a|. Let's group terms differently. Is it |(b + a) - c|? Yes. Since a + b > c, as per triangle inequality. So, same as before, that's positive. So, the absolute value doesn't change anything here. So, that term simplifies to (a + b) - c.Third term: |c - a - b|Similarly, rearrange: |c - (a + b)|. Again, by triangle inequality, a + b > c, so c - (a + b) is negative. Therefore, the absolute value would be (a + b) - c.Wait, so let's write all three terms:First term: |a - b - c| = (b + c) - aSecond term: |b - c + a| = (a + b) - cThird term: |c - a - b| = (a + b) - cWait, wait, the third term is |c - a - b|. Let me check again. Yes, that's |c - a - b| = |c - (a + b)|, which, since a + b > c, is negative, so absolute value becomes (a + b) - c.But hold on, that's the same as the second term. Wait, is that correct?Wait, first term is |a - b - c| = (b + c) - aSecond term: |b - c + a| = |(a + b) - c| = (a + b) - cThird term: |c - a - b| = |c - (a + b)| = (a + b) - cWait, so actually, the first term is (b + c - a), and the second and third terms are both (a + b - c). Wait, no, that's not symmetric. Let me check again.Wait, actually, maybe I made a mistake in grouping terms for the second expression. Let's take the second term: |b - c + a|Addition is commutative, so b - c + a is the same as a + b - c. So, that's |a + b - c|. Since a + b > c, as per triangle inequality, then a + b - c is positive, so absolute value is a + b - c.Similarly, the third term is |c - a - b| = |c - (a + b)| = same as | - (a + b - c)| = |a + b - c|, which is again a + b - c. So, the third term is also a + b - c.But wait, that seems like the second and third terms are the same. But let's check again.Wait, the third term is |c - a - b|. Let's rearrange that:c - a - b = c - (a + b). Since a + b > c, then c - (a + b) is negative. So, absolute value is (a + b) - c. So yes, same as the second term.So, substituting back into the original expression:|a - b - c| + |b - c + a| + |c - a - b| = (b + c - a) + (a + b - c) + (a + b - c)Wait, hold on, first term is (b + c - a), second term is (a + b - c), third term is (a + b - c). So, adding them together:First term: (b + c - a)Second term: (a + b - c)Third term: (a + b - c)So, let's compute term by term:First term: b + c - aSecond term: + a + b - cThird term: + a + b - cAdding them up:Let's collect like terms:For a:- a (from first term) + a (second term) + a (third term) = (-a + a + a) = aFor b:b (first term) + b (second term) + b (third term) = 3bFor c:c (first term) - c (second term) - c (third term) = c - c - c = -cSo total expression:a + 3b - cWait, that seems odd. Let me check again step by step.First term: (b + c - a)Second term: (a + b - c)Third term: (a + b - c)Adding these together:(b + c - a) + (a + b - c) + (a + b - c)Combine the first two terms first:(b + c - a) + (a + b - c) = b + c - a + a + b - cThe -a and +a cancel, c and -c cancel, so left with b + b = 2b.Then add the third term: 2b + (a + b - c) = 2b + a + b - c = a + 3b - c.Hmm, same result. But this seems problematic because the answer would depend on a, b, c. But given that a, b, c are sides of a triangle, maybe this can be simplified further?Wait, but maybe I made a mistake in the sign of the first term.Wait, first term is |a - b - c|. Since a < b + c, as per triangle inequality, so a - b - c is negative, so absolute value is (b + c - a). That's correct.Second term is |b - c + a|. Which is |a + b - c|. Since a + b > c, so it's (a + b - c). Correct.Third term is |c - a - b|, which is same as |a + b - c|, again (a + b - c). Correct.So adding up (b + c - a) + (a + b - c) + (a + b - c).Yes, as above. So combining:First term: b + c - aSecond term: a + b - cThird term: a + b - cAdding together:b + c - a + a + b - c + a + b - cNow, let's combine like terms:a terms: -a + a + a = ab terms: b + b + b = 3bc terms: c - c - c = -cSo total: a + 3b - c.Wait, but this seems to depend on the variables, but the problem says to simplify the expression. Maybe it can be expressed in terms of the triangle's perimeter or something else?Wait, the perimeter is a + b + c, but here we have a + 3b - c. Hmm, not sure. Maybe I made a mistake here.Wait, perhaps I messed up the signs when expanding the terms. Let me try again step by step.First term: (b + c - a)Second term: (a + b - c)Third term: (a + b - c)So adding all three:(b + c - a) + (a + b - c) + (a + b - c)Let's distribute the addition:= b + c - a + a + b - c + a + b - cNow, group the a terms:- a + a + a = ( -a + a ) + a = 0 + a = aGroup the b terms:b + b + b = 3bGroup the c terms:c - c - c = (c - c) - c = 0 - c = -cTherefore, total is a + 3b - c. Hmm. So that's the result. But wait, this seems strange because if the problem is symmetric in a, b, c, then the answer should be symmetric as well, but here we have a + 3b - c, which isn't symmetric. That suggests I made a mistake.Wait, the original expression is |a - b - c| + |b - c + a| + |c - a - b|. Is this expression symmetric in a, b, c? Let's check.First term: |a - b - c|, which is the same as |a - (b + c)|Second term: |b - c + a|, which is |a + b - c|Third term: |c - a - b|, which is |c - (a + b)|So, the first and third terms are similar, but the second term is different. So, actually, the expression isn't symmetric. So, the result might not be symmetric either. But according to our calculation, it's a + 3b - c, which is not symmetric, but maybe that's correct?Wait, but let's test with specific values. Let's take a triangle with sides a = 3, b = 4, c = 5. Since 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3, it's a valid triangle.Compute the original expression:First term: |3 - 4 - 5| = |3 - 9| = | -6 | = 6Second term: |4 - 5 + 3| = |2| = 2Third term: |5 - 3 - 4| = | -2 | = 2Sum: 6 + 2 + 2 = 10Now, according to our formula a + 3b - c: 3 + 3*4 -5 = 3 +12 -5 = 10. So it matches.But let's check another triangle to see if the formula holds. Let's take an equilateral triangle with sides a = b = c = 2.Original expression:First term: |2 - 2 - 2| = | -2 | = 2Second term: |2 - 2 + 2| = |2| = 2Third term: |2 - 2 - 2| = | -2 | = 2Sum: 2 + 2 + 2 = 6Using the formula a + 3b - c: 2 + 3*2 -2 = 2 +6 -2=6. Correct again.Wait, but in this case, since all sides are equal, the formula gives 6. If we swap a and c, for example, let a=5, b=4, c=3 in the first example. Then the original expression:First term: |5 -4 -3| = | -2 | =2Second term: |4 -3 +5|= |6|=6Third term: |3 -5 -4|=| -6 | =6Sum:2 +6 +6=14Now, the formula a +3b -c:5 +12 -3=14. Correct. So even when a and c are swapped, the formula still works. Wait, but in this case, a was 5, b=4, c=3. So in this case, the formula gives 5 + 12 -3=14, which is correct, but the formula is dependent on which variable is which.So, the expression is not symmetric, so the answer is dependent on the order of the variables. But in the problem statement, it's just given that a, b, c are sides of a triangle, but the expression is written as |a -b -c| + |b -c +a| + |c -a -b|. Therefore, the answer would be a + 3b - c, but that seems to depend on the order of the variables. However, in the problem statement, the variables are given as a, b, c, so the answer is specific to that order.Wait, but maybe there's a different way to simplify this. Let me think again. Is there a different approach?Alternatively, maybe we can consider all the absolute values and use the triangle inequalities to determine the sign of each expression inside.Let me re-express each term:1. |a - b - c| = |a - (b + c)|. Since in a triangle, b + c > a, so this is (b + c - a).2. |b - c + a| = |a + b - c|. Since a + b > c, this is (a + b - c).3. |c - a - b| = |c - (a + b)|. Since a + b > c, this is (a + b - c).So, substituting back into the original expression:(b + c - a) + (a + b - c) + (a + b - c)Wait, this is the same as before, leading to a + 3b - c.But if we take another triangle where, say, a = 5, b = 5, c = 5 (equilateral). Then:Original expression:First term: |5 -5 -5| = 5Second term: |5 -5 +5| =5Third term: |5 -5 -5|=5Sum:15Formula: a + 3b -c =5 +15 -5=15. Correct.Another example: a=2, b=3, c=4.Original expression:First term: |2 -3 -4|=5Second term: |3 -4 +2|=1Third term: |4 -2 -3|=1Sum:5 +1 +1=7Formula: a +3b -c=2 +9 -4=7. Correct.So, seems like the formula a + 3b - c holds for these examples. Therefore, maybe that is the simplified expression. However, the problem asks to simplify the expression. But in the original problem statement, the answer is supposed to be a numerical value or a more simplified expression. But a + 3b - c is already simplified. However, in all test cases, the answer equals a + 3b - c, which depends on the sides. Wait, but the problem states that a, b, c are sides of a triangle, so maybe there's a way to express this in terms of other variables or find a relation?Wait, but the problem is to simplify the expression, not to find a numerical answer. So perhaps the answer is a + 3b - c. But before concluding, let me check once more.Alternatively, maybe there is a different approach.Wait, another way to think about absolute values is that they represent distances. But in this case, maybe we can use the triangle inequalities to rewrite each term.Alternatively, note that in any triangle, the sum of two sides minus the third side is positive. So, expressions like (a + b - c), (a + c - b), (b + c - a) are all positive. So, these are often used in triangle geometry, like in Heron's formula for the area.But in our problem, the first term is |a - b - c|, which is (b + c - a). The second term is |a + b - c|, which is (a + b - c). The third term is |c - a - b|, which is (a + b - c). So, adding all three terms:(b + c - a) + (a + b - c) + (a + b - c)Which simplifies to:First term cancels with parts of the other terms:(b + c - a) + (a + b - c) + (a + b - c) = [ (b + c - a) + (a + b - c) ] + (a + b - c)= [ (b + c - a + a + b - c) ] + (a + b - c)= [ 2b ] + (a + b - c)= 2b + a + b - c= a + 3b - cSame result. So this seems to be the correct simplification.But the problem says "simplify the expression". So unless there's a different interpretation, the answer is a + 3b - c.Wait, but in the problem statement, the sides are given as a, b, c of a triangle, but there's no more information. So, perhaps the answer is supposed to be in terms of something else, but I don't see how.Alternatively, maybe the problem is expecting a different approach. Let me check again.Wait, the expression is |a - b - c| + |b - c + a| + |c - a - b|.Let me note that the terms inside the absolute values can be represented as follows:First term: a - b - c = a - (b + c)Second term: b - c + a = (a + b) - cThird term: c - a - b = c - (a + b)But given that in a triangle, each side is less than the sum of the other two, so:1. a < b + c => a - b - c < 0 => |a - b - c| = b + c - a2. (a + b) - c > 0 => |b - c + a| = a + b - c3. c - (a + b) < 0 => |c - a - b| = a + b - cThus, sum is (b + c - a) + (a + b - c) + (a + b - c) = a + 3b - c. So, same as before.But maybe there's a typo in the problem? Or maybe the user expects a different answer. Alternatively, perhaps the problem is symmetric in some way that I didn't notice.Wait, perhaps if I group terms differently. Let's see:Original expression:|a - b - c| + |b - c + a| + |c - a - b|Let me rewrite the second term as |a + b - c| and the third term as |a + b - c|. So, two terms are the same.So, expression becomes |a - (b + c)| + 2|a + b - c|.But since a - (b + c) is negative, |a - (b + c)| = (b + c) - a, and |a + b - c| = a + b - c.Therefore, expression is (b + c - a) + 2(a + b - c) = b + c - a + 2a + 2b - 2c = ( -a + 2a ) + ( b + 2b ) + ( c - 2c ) = a + 3b - c. Same result.Alternatively, is there a way to write this as a combination of other expressions? For example, in a triangle, the perimeter is a + b + c. But our expression is a + 3b - c. Not sure.Alternatively, if we factor something out. Let's see:a + 3b - c = (a + b + c) + 2b - 2c = (a + b + c) + 2(b - c). But I don't see how this helps. Maybe not useful.Alternatively, the problem might have a typo, but given the test cases worked, maybe the answer is indeed a + 3b - c. However, I need to check if there is a different interpretation.Wait, perhaps the problem is written in Chinese, and the translation might have different ordering? Wait, no, the problem is stated in English. Let me check the original problem again."Given that a, b, and c are the three side lengths of a triangle, simplify the expression: |a - b - c| + |b - c + a| + |c - a - b| = ______."No, the problem is as written. So based on the calculation, the answer is a + 3b - c.Wait, but maybe the problem is expecting a numerical answer. For example, maybe all the terms cancel out to a constant? But in our examples, the answer varied: 10, 6, 14, 7, 15, so it's not a constant. Therefore, the answer must be expressed in terms of a, b, c.But the question says "simplify the expression". If the original expression is |a - b - c| + |b - c + a| + |c - a - b|, and we simplified it to a + 3b - c, then that is the simplified form. So maybe that's the answer.Alternatively, perhaps there's a miscalculation here. Let me check once again.Original expression:1. |a - b - c| = b + c - a2. |b - c + a| = a + b - c3. |c - a - b| = a + b - cSo sum is (b + c - a) + (a + b - c) + (a + b - c)Simplify term by term:First term: b + c - aSecond term: a + b - cThird term: a + b - cAdding:For a: -a + a + a = aFor b: b + b + b = 3bFor c: c - c - c = -cTotal: a + 3b - cYes, that's correct. Therefore, the simplified expression is a + 3b - c.But let me check another triangle where a = 4, b = 5, c = 6.Original expression:First term: |4 -5 -6| = 7Second term: |5 -6 +4| = 3Third term: |6 -4 -5| = 3Sum:7 +3 +3=13Formula: a +3b -c=4 +15 -6=13. Correct.Alternatively, a=1, b=1, c=1. But wait, 1,1,1 is a valid triangle.Original expression:First term: |1 -1 -1|=1Second term: |1 -1 +1|=1Third term: |1 -1 -1|=1Sum=3Formula:1 +3*1 -1=3. Correct.So, all test cases confirm that the simplified expression is a + 3b - c.Therefore, the answer is a + 3b - c, which can be written as boxed{a + 3b - c}.But wait, the problem says "simplify the expression", but the original expression has three absolute values, and the simplified expression is a linear combination of the sides. So, unless there's another step I'm missing, this should be the answer.Alternatively, maybe the problem expects further simplification using perimeter or something else. But since perimeter is a + b + c, and semiperimeter is s = (a + b + c)/2, but I don't see a connection here. So, unless there's a different approach, this is the simplified form.Therefore, after thorough calculation and checking with multiple examples, I conclude that the simplified expression is a + 3b - c.Final Answerboxed{a + 3b - c}</think>Given that (a), (b), and (c) are the three side lengths of a triangle, we need to simplify the expression (|a - b - c| + |b - c + a| + |c - a - b|).First, we use the triangle inequalities, which state that the sum of any two sides must be greater than the third side. This implies:1. (a + b > c)2. (a + c > b)3. (b + c > a)We analyze each term in the expression:1. (|a - b - c|): This can be rewritten as (|a - (b + c)|). Since (b + c > a), the expression inside the absolute value is negative, so (|a - (b + c)| = (b + c - a)).2. (|b - c + a|): This can be rewritten as (|(a + b) - c|). Since (a + b > c), the expression inside the absolute value is positive, so (|(a + b) - c| = (a + b - c)).3. (|c - a - b|): This can be rewritten as (|c - (a + b)|). Since (a + b > c), the expression inside the absolute value is negative, so (|c - (a + b)| = (a + b - c)).Substituting these results back into the original expression, we get:[|a - b - c| + |b - c + a| + |c - a - b| = (b + c - a) + (a + b - c) + (a + b - c)]Combining like terms:- For (a): (-a + a + a = a)- For (b): (b + b + b = 3b)- For (c): (c - c - c = -c)Thus, the simplified expression is:[a + 3b - c]Therefore, the final answer is (boxed{a + 3b - c}).]

🔑:Given that a, b, and c are the three side lengths of a triangle, we need to simplify the expression: |a-b-c|+|b-c+a|+|c-a-b|.Step 1: Utilize the triangle inequality theorem, which states that for any triangle with sides a, b, and c, the sum of the lengths of any two sides must be greater than the length of the remaining side. Therefore, we have:- a + b > c- b + c > a- c + a > bStep 2: From the triangle inequality, we can derive the following inequalities:- Since a + b > c, it follows that a - b - c a, it implies b - c + a > 0.- Since c + a > b, we deduce c - a - b 0.- |c - a - b| = -(c - a - b) because c - a - b < 0.Step 4: Simplify the expression by substituting the values from step 3:begin{align*}|a-b-c|+|b-c+a|+|c-a-b| &= -(a - b - c) + (b - c + a) - (c - a - b) &= -a + b + c + b - c + a - c + a + b &= a + 3b - c.end{align*}Therefore, the simplified expression is boxed{a+3b-c}.

🧠:[<think>Okay, so I have this problem here: Among the natural numbers from 1 to 2019, if any 809 odd numbers are taken and multiplied together, what is the unit digit of the product? Hmm, interesting. Let me try to figure this out step by step.First, let me make sure I understand the problem correctly. We have the numbers from 1 to 2019. All of these are natural numbers, so that's straightforward. The task is to take any 809 odd numbers from this range, multiply them all together, and find the unit digit (i.e., the last digit) of the resulting product. The question is asking for the unit digit regardless of which 809 odd numbers are chosen, right? So no matter which 809 odd numbers you pick from 1 to 2019, when you multiply them, the unit digit will always be the same. So I need to find that consistent unit digit.Alright, let's break this down. First, I need to know how many odd numbers there are between 1 and 2019. Then, since we are selecting 809 of them, perhaps the number of available odd numbers is relevant. Let me check that first.Numbers from 1 to 2019: total numbers is 2019. Since 2019 is odd, the number of odd numbers should be (2019 + 1)/2 = 2020/2 = 1010. Let me verify this. If we have numbers from 1 to an odd number N, the count of odd numbers is (N + 1)/2. So here, (2019 + 1)/2 is indeed 1010. Therefore, there are 1010 odd numbers between 1 and 2019. So we are selecting 809 out of these 1010. Got it.Now, the problem is to multiply any 809 of these odd numbers and find the unit digit. Since the unit digit of a product depends only on the unit digits of the numbers being multiplied, maybe I can analyze the possible unit digits of the odd numbers and how they contribute to the overall product's unit digit.First, let's recall that the unit digit of a product of numbers is determined by the product of their unit digits modulo 10. So if I have numbers ending with digits d1, d2, ..., dn, then the unit digit of their product is (d1 * d2 * ... * dn) mod 10. So even though the numbers themselves can be large, only their last digits matter for the unit digit of the product.Since we are dealing with odd numbers, the possible unit digits can only be 1, 3, 5, 7, or 9. So each odd number ends with one of these digits. Therefore, when multiplying 809 such numbers, we can consider the product modulo 10. The challenge is to figure out what this product is, regardless of which 809 numbers are chosen. Since the question states "any 809 odd numbers," the unit digit must be the same no matter the selection. Therefore, perhaps there's a pattern or some invariance here.Wait, but how can that be? If we can choose different combinations of numbers with different unit digits, wouldn't the product modulo 10 vary? Unless there's some constraint that forces the product to always be a certain value modulo 10. Let me think.Wait, maybe the key here is the number of factors of 2 and 5 in the product, but since all numbers are odd, there are no factors of 2. However, there might be factors of 5, which end with 5. If there's at least one multiple of 5 in the selected numbers, then the product would be a multiple of 5, and since there are no factors of 2, the product would end with 5. If there are no multiples of 5, then the product would be an odd number not divisible by 5, so the unit digit would have to be 1, 3, 7, or 9. But the problem states "any 809 odd numbers," so we have to consider whether in any selection of 809 odd numbers, there is necessarily at least one multiple of 5. If that's the case, then the unit digit would always be 5. Otherwise, it might depend.Therefore, the critical question is: when selecting 809 odd numbers from 1 to 2019, is it possible to avoid selecting any multiples of 5? If it's possible, then the unit digit could be 1, 3, 7, or 9 depending on the selection. But if it's impossible to avoid all multiples of 5 when selecting 809 numbers, then every product would include at least one multiple of 5, hence the unit digit would be 5.So let's check how many multiples of 5 are there among the odd numbers from 1 to 2019. Since 5 divides 5, 15, 25, ..., up to 2015. Let me calculate that. The number of multiples of 5 in 1 to 2019 is floor(2019/5) = 403. But among these, the odd multiples of 5 are those where the multiple is odd. Since 5 is odd, multiplying by an odd number gives an odd multiple. So the odd multiples of 5 are 5*1, 5*3, 5*5, ..., up to the largest odd multiple less than or equal to 2019.So, the nth odd multiple is 5*(2n-1). So to find the number of odd multiples, we can set 5*(2n -1) <= 2019. Let's solve for n.2n -1 <= 2019/5 = 403.8. So 2n -1 <= 403.8, so 2n <= 404.8, so n <= 202.4. Therefore, n=202. So there are 202 odd multiples of 5 between 1 and 2019. Let me verify this. The largest odd multiple of 5 less than or equal to 2019 is 5*(2*202 -1) = 5*(403) = 2015. Indeed, 2015 is less than 2019, and the next one would be 2015 + 10 = 2025, which is over. So there are 202 odd multiples of 5.Therefore, the total number of odd numbers not divisible by 5 is total odd numbers minus odd multiples of 5: 1010 - 202 = 808. Wait, 1010 - 202 is 808. So that means there are 808 odd numbers from 1 to 2019 that are not divisible by 5.But the problem says we are selecting 809 odd numbers. Since there are only 808 odd numbers not divisible by 5, if we try to select 809 odd numbers, we must include at least one multiple of 5. Because the maximum number of non-multiples of 5 we can select is 808, so any selection of 809 must include at least one multiple of 5. Therefore, the product will have at least one factor of 5. Since all numbers are odd, there are no factors of 2. Therefore, the product will be divisible by 5 but not by 2. Therefore, the unit digit must be 5. Because numbers divisible by 5 but not by 2 end with 5 or 0, but since the product is odd, it cannot end with 0. Therefore, the unit digit must be 5.So that seems to be the answer. Let me recap to make sure. There are 1010 odd numbers from 1 to 2019. Among them, 202 are multiples of 5, so 1010 - 202 = 808 are not multiples of 5. Therefore, if you select 809 odd numbers, you have to include at least one multiple of 5. Therefore, the product will have 5 as a factor. Since all numbers are odd, there are no factors of 2, so the product is an odd multiple of 5. Therefore, the unit digit is 5. Hence, regardless of which 809 numbers you choose, the unit digit will always be 5.Therefore, the answer should be 5, boxed as boxed{5}.Final Answerboxed{5}</think>To determine the unit digit of the product of any 809 odd numbers taken from 1 to 2019, we first note that there are 1010 odd numbers in this range. Among these, the number of odd multiples of 5 is calculated as follows:1. The largest odd multiple of 5 less than or equal to 2019 is 2015.2. The number of odd multiples of 5 is given by solving (5 times (2n - 1) leq 2019), which results in (n = 202).Thus, there are 202 odd multiples of 5. The remaining odd numbers are (1010 - 202 = 808). Since we need to select 809 odd numbers, we must include at least one multiple of 5. A product containing a multiple of 5 (and no factors of 2, as all numbers are odd) will have a unit digit of 5. Therefore, the unit digit of the product of any 809 odd numbers from 1 to 2019 is (boxed{5}).,<think>Okay, let me try to figure out this problem. So, the question is: Among the natural numbers from 1 to 2019, if we take any 809 odd numbers and multiply them together, what is the unit digit of the product? Hmm, interesting. Let's break this down step by step.First, let me understand what's being asked. We have numbers from 1 to 2019. All these numbers are natural numbers. We need to pick any 809 odd numbers from this range, multiply them all together, and find the unit digit of the resulting product. The key here is that we're dealing with odd numbers and their multiplication, and specifically, we want the unit digit. Since the unit digit of a product depends only on the unit digits of the numbers being multiplied, maybe there's a pattern or something we can exploit here.Let me start by recalling that when you multiply numbers, the unit digit of the product is determined only by the unit digits of the factors. For example, 13 * 27 has the same unit digit as 3 * 7, which is 1. So, maybe instead of dealing with all the numbers from 1 to 2019, I can focus on their unit digits. Since we are only interested in odd numbers, all the numbers we choose will have unit digits 1, 3, 5, 7, or 9. Because even numbers end with 0, 2, 4, 6, or 8, but we are only selecting odd numbers.So, the possible unit digits for our selected numbers are 1, 3, 5, 7, 9. Now, when multiplying these together, each multiplication step will affect the unit digit. Let me think about how these digits multiply.First, let's note that multiplying by 1 doesn't change the unit digit. For example, if the current product has a unit digit of 3, multiplying by 1 will keep it 3. Multiplying by 5 will always result in a unit digit of 5 or 0, but since all numbers are odd, multiplying by 5 will result in 5. Wait, no. If you multiply 5 by any odd number, the unit digit will be either 5 or 5 multiplied by some odd number. Wait, 5*1=5, 5*3=15 (unit digit 5), 5*5=25 (unit digit 5), 5*7=35 (unit digit 5), 5*9=45 (unit digit 5). Oh, interesting! So, multiplying by 5, regardless of the other odd digits, the unit digit will always be 5. So, if there's at least one 5 in the numbers being multiplied, the unit digit of the entire product will be 5. Is that right?Wait, but 5 times 5 is 25, which ends with 5. 25 times 5 is 125, which also ends with 5. So, once you have a 5 in the factors, multiplying by more 5s keeps the unit digit as 5. But even multiplying by other odd numbers, like 3, 7, 9, after a 5 is present, the unit digit remains 5. Let's check: 5*3=15 (unit digit 5), 15*7=105 (unit digit 5), 105*9=945 (unit digit 5). So, yeah, once there's a single 5 in the factors, the unit digit is 5 and stays 5 regardless of subsequent multiplications. That seems important.Therefore, if among the 809 odd numbers we pick, if there is at least one number ending with 5, then the unit digit of the product will be 5. So, the critical question is: How many numbers ending with 5 are there in the range from 1 to 2019? Because if there are at least one, and since we are picking 809 numbers, unless all numbers ending with 5 are excluded, the product will have a unit digit of 5. But if all the numbers we pick don't include any 5, then the unit digit will be something else. So, maybe we need to check whether it's possible to pick 809 odd numbers without any number ending with 5. If that's possible, then the unit digit could be different. Otherwise, if it's impossible to pick 809 odd numbers without including at least one number ending with 5, then the unit digit must be 5.So, first, let's figure out how many odd numbers there are from 1 to 2019. Since 2019 is odd, the total number of odd numbers is (2019 + 1)/2 = 2020/2 = 1010. So, there are 1010 odd numbers between 1 and 2019.Out of these 1010 odd numbers, how many end with 5? Let's see. The numbers ending with 5 between 1 and 2019 are 5, 15, 25, ..., up to the largest one less than or equal to 2019.To find how many such numbers there are, we can note that these numbers form an arithmetic sequence where the first term is 5, the common difference is 10 (since we add 10 each time to get the next number ending with 5). The nth term of this sequence is given by a_n = 5 + (n - 1)*10. We need to find the largest n such that a_n ≤ 2019.Let's solve for n:5 + (n - 1)*10 ≤ 2019(n - 1)*10 ≤ 2014n - 1 ≤ 201.4n ≤ 202.4Since n must be an integer, the maximum n is 202. Therefore, there are 202 numbers ending with 5 between 1 and 2019.Therefore, in total, there are 1010 odd numbers, 202 of which end with 5, and 1010 - 202 = 808 that do not end with 5. So, if we want to pick 809 odd numbers, and there are only 808 odd numbers that don't end with 5, then by the pigeonhole principle, we must pick at least one number that ends with 5. Therefore, any selection of 809 odd numbers from 1 to 2019 must include at least one number ending with 5. Therefore, the product will have a unit digit of 5.Wait, let me verify that again. If there are 808 non-5-ending odd numbers, then if we try to pick 809 odd numbers, we can pick all 808 non-5-ending ones and then have to pick one from the 5-ending numbers. Therefore, yes, exactly. Therefore, in any such selection, there must be at least one number ending with 5, which when multiplied, gives a unit digit of 5. Therefore, regardless of which 809 numbers we pick, the unit digit must be 5.Therefore, the answer should be 5.But wait, let me check again. Let's recap:Total odd numbers from 1 to 2019: 1010.Numbers ending with 5: 202.Numbers not ending with 5: 1010 - 202 = 808.Therefore, if we need to pick 809 numbers, even if we pick all 808 non-5-ending numbers, we have to pick 1 more, which must be from the 5-ending numbers. Therefore, in any such selection, there is at least one 5. Therefore, the product must have a unit digit of 5.Therefore, the answer is 5. Therefore, the unit digit is 5.But wait, let me check if there's another perspective. Suppose that even if you have a 5 in the product, but if there's an even number of 5s, would that affect anything? Wait, no. Because 5 multiplied by any odd number still ends with 5, as we saw earlier. So, once you have one 5, the unit digit becomes 5. Subsequent multiplications by other numbers (even if they include more 5s or other odd digits) will keep the unit digit as 5. For example, 5*5=25, unit digit 5. 25*3=75, unit digit 5. 75*7=525, unit digit 5. So, yeah, regardless of how many 5s or other digits you multiply, as long as there's at least one 5, the unit digit remains 5.Therefore, since any selection of 809 numbers must include at least one 5, the unit digit is necessarily 5.Therefore, the answer is 5.Wait, but just to be thorough, let me check the counts again.Total odd numbers from 1 to 2019: Since 2019 is odd, the number of odd numbers is (2019 + 1)/2 = 1010. Correct.Numbers ending with 5: From 5 to 2015. Let's check 2015 is the last number ending with 5 before 2019. Then, the numbers are 5, 15, 25,...,2015. The formula for the nth term is 5 + (n-1)*10 = 10n -5. So, set 10n -5 ≤2019:10n ≤ 2024n ≤ 202.4So, n=202. Therefore, there are 202 numbers. So 202 numbers ending with 5. Therefore, 1010 -202=808 numbers that are odd and not ending with 5. So, if you pick 809 numbers, you can take all 808 non-5s and then one 5. Therefore, in any case, you must have at least one 5. Therefore, the unit digit is 5. So, the answer is 5.Therefore, the unit digit is 5. So, the final answer should be boxed{5}.Final Answerboxed{5}</think>To determine the unit digit of the product of any 809 odd numbers taken from 1 to 2019, we need to consider the unit digits of the selected numbers. The possible unit digits for odd numbers are 1, 3, 5, 7, and 9. Key observations:1. Multiplying any number by 5 results in a unit digit of 5 if the other factor is odd.2. If there is at least one number ending in 5 among the selected numbers, the unit digit of the product will be 5.First, we calculate the total number of odd numbers between 1 and 2019:[text{Total odd numbers} = frac{2019 + 1}{2} = 1010]Next, we determine how many of these odd numbers end in 5. These numbers form an arithmetic sequence: 5, 15, 25, ..., 2015. The nth term of this sequence is given by:[a_n = 5 + (n - 1) times 10]Solving for (n) when (a_n leq 2019):[5 + (n - 1) times 10 leq 2019 implies (n - 1) times 10 leq 2014 implies n - 1 leq 201.4 implies n leq 202.4]Thus, there are 202 numbers ending in 5.The remaining odd numbers not ending in 5 are:[1010 - 202 = 808]Using the pigeonhole principle, if we select 809 odd numbers, we must include at least one number ending in 5 (since there are only 808 non-5-ending odd numbers). Therefore, the unit digit of the product will be 5.The final answer is:[boxed{5}]]

🔑:Reference answer: 5Key points: Mathematical principle -- Pigeonhole principle, Last digit problem