🧠:[<think>Okay, so I have this problem here where each dwarf in a math class needs to find a three-digit number. The number has to meet a few conditions: it can't have any zero digits, it has to be divisible by 3, and when you add 297 to it, the result is another three-digit number that's the reverse of the original number. The question is asking for the minimum number of dwarfs that need to be in the class so that there are always at least two identical numbers found by them. Basically, we need to figure out how many such numbers exist and then use the pigeonhole principle to determine the minimum number required to ensure a duplicate.First, let me break down the problem. Let's denote the three-digit number as ABC, where A, B, C are digits from 1 to 9 (since no zeros are allowed). When we add 297 to ABC, we get another three-digit number CBA. So, mathematically, this can be written as:ABC + 297 = CBABut ABC and CBA are three-digit numbers, so we can express them in terms of their digits. Let's convert them into numerical values. ABC is 100*A + 10*B + C, and CBA is 100*C + 10*B + A. Therefore, the equation becomes:100*A + 10*B + C + 297 = 100*C + 10*B + ASimplify this equation. Let's subtract the terms on both sides to get:(100*A + 10*B + C) + 297 = 100*C + 10*B + ASubtract (100*A + 10*B + C) from both sides:297 = 100*C + 10*B + A - (100*A + 10*B + C)Simplify the right side:297 = 100C + 10B + A - 100A - 10B - CCombine like terms:297 = (100C - C) + (10B - 10B) + (A - 100A)297 = 99C - 99AFactor out 99:297 = 99*(C - A)Divide both sides by 99:297 / 99 = C - ACalculating that, 297 divided by 99 is 3. So:3 = C - ATherefore, the key equation here is C = A + 3.So the hundreds digit of the original number (A) is 3 less than the units digit (C). Since A and C are digits from 1 to 9 (no zeros allowed), this imposes some restrictions. Let's note the possible values for A and C.Since C = A + 3 and both A and C must be between 1 and 9 inclusive, let's list possible pairs (A, C):A can be from 1 up to 6, because if A is 6, then C is 9. If A is 7, C would be 10, which is invalid. Similarly, A cannot be less than 1. So possible values are:A: 1, 2, 3, 4, 5, 6Corresponding C: 4, 5, 6, 7, 8, 9So there are 6 possible pairs for A and C.Now, the middle digit B can be any digit from 1 to 9 as well. However, there's another condition: the original number ABC must be divisible by 3. Recall that a number is divisible by 3 if the sum of its digits is divisible by 3.So, for ABC to be divisible by 3, A + B + C must be a multiple of 3.But we already know that C = A + 3. So substitute that into the sum:A + B + (A + 3) = 2A + B + 3This sum must be divisible by 3. So:2A + B + 3 ≡ 0 mod 3Simplify:2A + B ≡ -3 ≡ 0 mod 3 (since -3 is congruent to 0 mod 3)Therefore, 2A + B ≡ 0 mod 3So this gives us a condition on B, given A. For each possible A (from 1 to 6), and C = A + 3, B must be chosen such that 2A + B is divisible by 3. Let's analyze this.Let me tabulate possible A and C values and then find the possible B values for each.For each A from 1 to 6:1. A = 1, C = 4 Then 2A + B = 2*1 + B = 2 + B ≡ 0 mod 3 So 2 + B ≡ 0 mod 3 → B ≡ 1 mod 3 Therefore, B can be 1, 4, 7 (since B is from 1-9) So B has 3 choices.2. A = 2, C = 5 2A + B = 4 + B ≡ 0 mod 3 So 4 + B ≡ 0 mod 3 → B ≡ 2 mod 3 Therefore, B can be 2, 5, 8 3 choices.3. A = 3, C = 6 2A + B = 6 + B ≡ 0 mod 3 6 is already 0 mod 3, so B ≡ 0 mod 3 But B can't be 0, so B can be 3, 6, 9 3 choices.4. A = 4, C = 7 2A + B = 8 + B ≡ 0 mod 3 8 ≡ 2 mod 3, so 2 + B ≡ 0 mod 3 → B ≡ 1 mod 3 B can be 1, 4, 7 3 choices.5. A = 5, C = 8 2A + B = 10 + B ≡ 1 + B mod 3 (since 10 ≡ 1 mod 3) 1 + B ≡ 0 mod 3 → B ≡ 2 mod 3 B can be 2, 5, 8 3 choices.6. A = 6, C = 9 2A + B = 12 + B ≡ 0 + B mod 3 (since 12 ≡ 0 mod 3) Therefore, B ≡ 0 mod 3 B can be 3, 6, 9 3 choices.So for each A (from 1 to 6), there are 3 possible B digits. Therefore, total number of such numbers is 6 * 3 = 18.Wait, hold on. Each A has 3 B's, so 6 A's * 3 B's = 18 numbers. Therefore, there are 18 different three-digit numbers satisfying the given conditions.Therefore, if each dwarf finds one such number, and there are 18 different numbers possible, then by the pigeonhole principle, if there are 19 dwarfs, at least two must have the same number. But the question says "the minimum number of dwarfs that should be in the class so that among the numbers they find, there are always at least two identical ones." So the answer would be 18 + 1 = 19. Therefore, the minimum number is 19.Wait, but let me double-check. Maybe there's a mistake in the calculation of the total numbers. Let's verify again.Each A from 1 to 6 (6 possibilities), each with 3 B's. So 6*3=18. Yes. Therefore, there are 18 distinct numbers possible. Hence, with 18 dwarfs, each could have a unique number, but when you add one more, the 19th dwarf must have a duplicate. Therefore, the minimum number is 19.But wait, before finalizing, let me check if all those numbers are actually valid. Let's take an example.Take A=1, C=4. B can be 1,4,7.So numbers would be 111, 141, 171. Let's check if these numbers satisfy the original condition.Take 111. Add 297: 111 + 297 = 408. Reverse of 111 is 111. But 408 is not 111. So this seems to contradict. Wait, maybe I made a mistake here. Wait, hold on. Wait, the problem states that adding 297 to ABC gives a number with the same digits in reverse order. So ABC + 297 = reverse(ABC). So in this case, 111 + 297 = 408, but reverse(111) is 111. 408 ≠ 111. Therefore, 111 is not a valid number. But according to our earlier reasoning, A=1, B=1, C=4 would be 114. Wait, hold on, wait. Wait, in the initial problem, ABC is a three-digit number without zeros, and when you add 297, you get CBA. So ABC + 297 = CBA. So ABC is 100A + 10B + C, and CBA is 100C + 10B + A. So in the case of A=1, C=4, the number is 1B4. So for A=1, B=1, the number is 114. Then adding 297: 114 + 297 = 411. Reverse of 114 is 411. So 411 is the reverse. 114 + 297 = 411. That works. Wait, 114 + 297 = 411. 114 reversed is 411. So 114 is a valid number.But earlier, I thought of A=1, B=1, C=4 as 111, which is incorrect. The C is 4, so it's 1 (hundreds), 1 (tens), 4 (units). So 114. So I need to correct my example. So the number is 114. Then 114 + 297 = 411, which is the reverse. That works. Similarly, let's check another one. For A=1, B=4: the number is 144. Then 144 + 297 = 441. Reverse of 144 is 441. Correct. 144 + 297 = 441. Yes. Similarly, A=1, B=7: 174. 174 + 297 = 471. Reverse of 174 is 471. Correct. So those three numbers (114, 144, 174) are valid. So the example was my mistake in constructing the number.Similarly, take A=2, C=5. B can be 2,5,8. So numbers are 225, 255, 285. Let's check 225 + 297 = 522. Reverse of 225 is 522. Correct. 225 + 297 = 522. Correct. Similarly, 255 + 297 = 552. Reverse of 255 is 552. Correct. 285 + 297 = 582. Reverse of 285 is 582. Correct. So those work.Similarly, A=3, C=6. B=3,6,9. Numbers: 336, 366, 396. 336 + 297 = 633. Reverse. Correct. 366 + 297 = 663. Correct. 396 + 297 = 693. Reverse is 693. Correct.A=4, C=7. B=1,4,7. Numbers: 417, 447, 477. 417 + 297 = 714. Reverse of 417 is 714. Correct. 447 + 297 = 744. Reverse. Correct. 477 + 297 = 774. Reverse. Correct.A=5, C=8. B=2,5,8. Numbers: 528, 558, 588. 528 + 297 = 825. Reverse of 528 is 825. Correct. 558 + 297 = 855. Reverse. Correct. 588 + 297 = 885. Reverse. Correct.A=6, C=9. B=3,6,9. Numbers: 639, 669, 699. 639 + 297 = 936. Reverse. Correct. 669 + 297 = 966. Reverse. Correct. 699 + 297 = 996. Reverse. Correct.So all 18 numbers seem to work. Each of them when added to 297 reverses their digits. Also, all numbers have non-zero digits. And they are divisible by 3, as we required in the calculation. Therefore, there are indeed 18 such numbers.Hence, the minimum number of dwarfs needed to guarantee that at least two have the same number is 18 + 1 = 19.But let me check if there are any other possible numbers that I might have missed. For example, suppose there's a number where adding 297 results in a reverse, but not following the C = A + 3 condition. Wait, but according to the equation we derived earlier, 297 = 99*(C - A), which simplifies to C - A = 3. So unless there's another way to get ABC + 297 = CBA, but with a different relationship between C and A, but according to the algebra, the only solution is C - A = 3. Therefore, all solutions must satisfy C = A + 3. Therefore, the 18 numbers we found are the only possible ones. Hence, 18 possible numbers.Therefore, the answer is 19. So the minimum number of dwarfs needed is 19. Therefore, the answer is boxed{19}Final Answerboxed{19}</think>Given a three-digit number (ABC) without zero digits, divisible by 3, such that adding 297 results in a number with the same digits in reverse order (CBA), we need to determine the minimum number of dwarfs required to ensure at least two identical numbers.1. Formulating the Equation: - Convert (ABC) and (CBA) to numerical values: (100A + 10B + C) and (100C + 10B + A). - The equation becomes: (100A + 10B + C + 297 = 100C + 10B + A). - Simplifying, we get: (297 = 99(C - A)), leading to (C = A + 3).2. Possible Values for (A) and (C): - (A) ranges from 1 to 6 (since (C = A + 3) must be a digit from 1 to 9). - Possible pairs: ((1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)).3. Divisibility by 3: - The sum (A + B + C) must be divisible by 3. Substituting (C = A + 3), we get (2A + B + 3) must be divisible by 3. - This simplifies to (2A + B equiv 0 pmod{3}).4. Possible Values for (B): - For each (A), determine (B) such that (2A + B equiv 0 pmod{3}): - (A = 1): (B equiv 1 pmod{3}) (1, 4, 7) - (A = 2): (B equiv 2 pmod{3}) (2, 5, 8) - (A = 3): (B equiv 0 pmod{3}) (3, 6, 9) - (A = 4): (B equiv 1 pmod{3}) (1, 4, 7) - (A = 5): (B equiv 2 pmod{3}) (2, 5, 8) - (A = 6): (B equiv 0 pmod{3}) (3, 6, 9) - Each (A) has 3 possible values for (B).5. Total Numbers: - 6 possible pairs for (A) and (C), each with 3 possible values for (B), resulting in (6 times 3 = 18) numbers.6. Pigeonhole Principle: - With 18 unique numbers, 19 dwarfs ensure at least two have the same number.Thus, the minimum number of dwarfs required is (boxed{19}).,<think>Okay, so I need to solve this problem where each dwarf finds a three-digit number. The number has to meet a few conditions: no zeros in any digit, divisible by 3, and when you add 297 to it, the result is a number with the same digits but reversed. Then the question is asking for the minimum number of dwarfs needed so that there must be at least two identical numbers found. That sounds like a pigeonhole principle problem, where we need to figure out how many possible numbers exist that meet the conditions, and then add one to that to ensure a collision.First, let's break down the problem. Let me write out the conditions:1. The number is three-digit, no zeros. So each digit (hundreds, tens, ones) is from 1 to 9.2. The number is divisible by 3. So the sum of its digits must be a multiple of 3.3. When 297 is added to the number, the result is the reverse of the original number.Let me denote the original number as ABC, where A, B, C are digits from 1 to 9. Then, the reversed number after adding 297 would be CBA. So, ABC + 297 = CBA.Let me convert these numbers into numerical values. ABC is 100A + 10B + C. Similarly, CBA is 100C + 10B + A. So, the equation is:100A + 10B + C + 297 = 100C + 10B + ALet me simplify this equation. Subtract (100A + 10B + C) from both sides:297 = (100C + 10B + A) - (100A + 10B + C)297 = 100C + 10B + A - 100A - 10B - CSimplify the right side:100C - C = 99C-100A + A = -99A10B -10B = 0So, 297 = 99C - 99ADivide both sides by 99:297 / 99 = C - A3 = C - ASo, C = A + 3That's a key relation: the units digit is 3 more than the hundreds digit. Since all digits are from 1 to 9, we need to ensure that C = A + 3 doesn't exceed 9. So, A can be from 1 to 6 (since if A is 6, C is 9; if A is 7, C would be 10 which is invalid). Therefore, possible A values: 1,2,3,4,5,6. Corresponding C values: 4,5,6,7,8,9.So, the original number has the form A B (A+3), where A is 1-6 and B is 1-9. But there's another condition: the number must be divisible by 3. Divisibility by 3 requires that the sum of the digits A + B + C is divisible by 3. Since C = A + 3, substitute that into the sum:A + B + (A + 3) = 2A + B + 3This sum must be divisible by 3. So, 2A + B + 3 ≡ 0 mod 3. Let's write this as:2A + B ≡ 0 mod 3Alternatively, since 2 ≡ -1 mod 3, so:(-A) + B ≡ 0 mod 3 => B - A ≡ 0 mod 3 => B ≡ A mod 3Therefore, B must be congruent to A modulo 3. So, B can be written as A + 3k, where k is an integer. But since B is a digit from 1 to 9, and A is from 1 to 6, we need to find possible B for each A such that B ≡ A mod 3.Let me consider each A from 1 to 6 and find possible B values.For A=1:B ≡ 1 mod 3. So possible B: 1,4,7For A=2:B ≡ 2 mod 3. So possible B: 2,5,8For A=3:B ≡ 0 mod 3. So possible B: 3,6,9For A=4:B ≡ 1 mod 3. So possible B: 1,4,7For A=5:B ≡ 2 mod 3. So possible B: 2,5,8For A=6:B ≡ 0 mod 3. So possible B: 3,6,9Therefore, for each A (1-6), there are 3 possible B values. So total possible numbers would be 6 * 3 = 18? Wait, but let me check if there's any overlap or other constraints.Wait, but hang on: For each A (1-6), C is fixed as A+3, and B can take 3 possible values. So total numbers are 6 * 3 = 18. So there are 18 such numbers that satisfy the conditions. Therefore, if there are 18 dwarfs, each could potentially choose a unique number. But if there are 19 dwarfs, by the pigeonhole principle, at least two must have the same number. So the minimal number is 19.But wait, let me verify this. Let me check if each of these 18 numbers actually satisfies the original equation ABC + 297 = CBA.Take an example. Let's pick A=1, B=1, so C=4. The number is 114. Adding 297 gives 114 + 297 = 411. Which is indeed CBA=411. And 411 is the reverse of 114. Also, check divisibility by 3: 1+1+4=6, which is divisible by 3. So 114 is valid.Another example: A=2, B=2, C=5. Number is 225. 225 + 297 = 522. Reverse of 225 is 522. Sum of digits: 2+2+5=9, divisible by 3. Good.Another example: A=3, B=3, C=6. Number is 336. 336 + 297 = 633. Reverse is 633. Sum of digits: 3+3+6=12, divisible by 3. Good.A=4, B=4, C=7. Number 447. 447 + 297=744. Reverse is 744. Sum:4+4+7=15. Divisible by 3. Okay.A=5, B=5, C=8. Number 558. 558 +297=855. Reverse. Sum 5+5+8=18. Good.A=6, B=6, C=9. 669 +297=966. Reverse. Sum 6+6+9=21. Divisible by 3. Perfect.Also check other B values. For example, A=1, B=4. Number 144. 144 +297=441. Reverse. Sum digits 1+4+4=9. Okay.Another, A=1, B=7. 174 +297=471. Reverse. 1+7+4=12. Okay.Similarly, A=2, B=5: 255 +297=552. Reverse. Sum 2+5+5=12. Good.A=3, B=6: 366 +297=663. Reverse. Sum 3+6+6=15. Okay.So all these numbers seem to check out. Therefore, there are 6 values for A, each with 3 B's, leading to 18 numbers.Therefore, if there are 18 dwarfs, each can have a unique number. But if there are 19, then at least two must have the same number. Hence, the minimal number is 19.But let me double-check if there are any other numbers that could satisfy the condition. For example, suppose there's a number not covered by the above.Wait, the problem says "three-digit number without zero digits, divisible by 3, such that when 297 is added to it, the result is a number with the same digits but in reverse order." So, according to the equation ABC + 297 = CBA, which simplifies to C = A + 3. So unless there are numbers where C = A +3 but with digits wrapping around? But digits can't be zero, so C must be A +3, and since digits go up to 9, A can be up to 6. So A=1-6, C=4-9. Then B is determined modulo 3. Therefore, 6*3=18. So there are 18 such numbers. Therefore, the answer should be 19.But just to be thorough, let's see if there are any other possible numbers. Suppose, for instance, that adding 297 causes a carryover that affects the middle digit. Wait, let's take an example. Let me take a number and add 297. Let's see how the digits change.Original number ABC. Adding 297:C + 7. If C +7 >=10, then carryover 1 to the tens place.Then, B +9 + carryover. Similarly, if B +9 + carryover >=10, carryover 1 to the hundreds place.Then, A +2 + carryover.But in the reversed number CBA, the hundreds digit is C, which is equal to original units digit. But according to the equation, CBA = ABC +297. So CBA's hundreds digit is C = (A + carryover from the hundreds place addition). Wait, wait. Wait, actually, when you add 297 to ABC, the hundreds digit becomes A +2 + carryover from the tens place addition. But according to the reversed number, the hundreds digit is C. So:Hundreds digit after adding 297: A +2 + carryover from tens place = CBut we also found from the equation that C = A +3. So:A +2 + carryover = A +3Subtract A from both sides:2 + carryover = 3 => carryover =1Therefore, when adding 297 to ABC, there must be a carryover of 1 from the tens place to the hundreds place. So, when adding 297:First, add 7 to the units digit C. If C +7 >=10, carryover 1 to tens digit.Then, add 9 to the tens digit B, plus any carryover. Since we need a carryover of 1 from the tens to hundreds place, this sum must be >=10. So 9 + B + carryover_from_units >=10.But we already established that carryover from units is either 0 or 1. Let's check:If C +7 >=10, then carryover1 is 1. Otherwise, carryover1 is 0.Then, in the tens place: B +9 + carryover1. For there to be a carryover of 1 to hundreds place, this sum must be >=10. So:If carryover1 =1:B +9 +1 = B +10 >=10. So always carries over 1. So B +10 gives digit (B +0) with carryover 1.Wait, but wait. Wait, adding 9 to B plus carryover1. If carryover1 is 1, then it's 9 + B +1 = B +10. So the digit becomes (B +10) mod 10, which is B +0 (since 10 mod10=0), and carryover 1.Similarly, if carryover1 is 0, then 9 + B +0 = B +9. For there to be a carryover of 1, B +9 >=10 => B >=1. But B is from 1-9, so B +9 >=10 for all B. Wait, 9 + B is at least 10 (when B=1: 10, B=9: 18). So 9 + B will always be 10 or more. Therefore, regardless of carryover1, adding 9 + B + carryover1 will always result in a carryover of 1. Wait, but if carryover1 is 1, then 9 + B +1 = B +10, which gives a digit of B +0 with a carryover of 1. If carryover1 is 0, then 9 + B = (B -1) +10, so digit is B -1 and carryover 1. Wait, this is conflicting. Let me re-examine.Wait, adding 297 to ABC:Breakdown: ABC+ 297= CBALet's perform the addition step by step.Units place: C +7. If C +7 >=10, then we write (C +7) mod 10 and carryover1 =1. Otherwise, (C +7) and carryover1=0.Tens place: B +9 + carryover1. Since B is from 1-9, 9 + B + carryover1 (0 or 1). Let's compute:If carryover1=0: B +9. Since B >=1, B +9 >=10 (since B=1: 10, B=2:11, etc.), so the result here would be (B +9) mod10 = (B -1) and carryover2=1.If carryover1=1: B +9 +1 = B +10. So (B +10) mod10 = B, and carryover2=1.Hundreds place: A +2 + carryover2. carryover2 is always 1, as established. So A +2 +1 = A +3. But according to the reversed number CBA, the hundreds digit is C. Therefore:A +3 = CWhich matches our earlier conclusion. So, regardless of carryover1, carryover2 is always 1, so C = A +3.But also, in the tens place, the digit becomes:If carryover1=0: (B +9) mod10 = B -1If carryover1=1: (B +10) mod10 = BBut in the reversed number CBA, the tens digit is B (original tens digit). Wait, but according to the reversed number, the tens digit should remain B? Wait, in the original number ABC, the tens digit is B. In the reversed number CBA, the tens digit is also B. Therefore, when adding 297, the tens digit after addition must still be B. Let's check:From the addition process:If carryover1=0:Tens digit after addition: (B +9) mod10 = B -1. But in CBA, the tens digit is B. Therefore, B -1 = B => -1 =0, which is impossible. Therefore, carryover1 cannot be 0. So carryover1 must be 1.Therefore, in units place: C +7 >=10, which means C >=3. But since C = A +3 and A >=1, C >=4. So this is already satisfied. Because A is at least 1, so C is at least 4. Then, C +7: 4 +7=11, which has carryover1=1.Therefore, the tens digit calculation:B +9 +1 = B +10. Then, (B +10) mod10 = B, and carryover2=1. So the tens digit remains B, which matches the reversed number. Therefore, all conditions are satisfied as long as C = A +3 and C >=4 (since C +7 >=10).But since C = A +3, and A is from 1 to 6, C is from 4 to 9, so C +7 is from 11 to 16, which all result in carryover1=1, so the tens digit calculation is B +10, which mod10 is B, as needed.Therefore, the only constraints are:1. A from 1 to 62. C = A +33. B ≡ A mod3 (so that 2A + B +3 ≡0 mod3)Therefore, each A (1-6) has 3 possible B's (since B must ≡A mod3, and B is from 1-9). Hence, total numbers are 6*3=18.Therefore, if there are 18 numbers possible, then the minimal number of dwarfs to guarantee at least two identical numbers is 18 +1 =19.Therefore, the answer is 19. So the minimal number is boxed{19}.Final Answerboxed{19}</think>Given the problem, each dwarf needs to find a three-digit number without zero digits, divisible by 3, such that when 297 is added to it, the result is a number with the same digits but in reverse order. We need to determine the minimum number of dwarfs required to ensure at least two identical numbers.1. Conditions Analysis: - The number is a three-digit number with digits from 1 to 9. - The number is divisible by 3. - Adding 297 to the number results in its reverse.2. Equation Setup: - Let the original number be (100A + 10B + C). - The reversed number is (100C + 10B + A). - The equation becomes: (100A + 10B + C + 297 = 100C + 10B + A).3. Simplifying the Equation: - Subtracting (100A + 10B + C) from both sides: (297 = 99C - 99A). - Dividing by 99: (3 = C - A), so (C = A + 3).4. Digit Constraints: - (A) ranges from 1 to 6 (since (C = A + 3) must be a digit from 1 to 9).5. Divisibility by 3: - The sum of digits (A + B + C) must be divisible by 3. - Substituting (C = A + 3), the sum becomes (2A + B + 3), leading to (2A + B equiv 0 pmod{3}). - This simplifies to (B equiv A pmod{3}).6. Possible Values: - For each (A) (1 to 6), (B) can take three values congruent to (A mod 3). - Example: For (A = 1), (B) can be 1, 4, 7; similarly for other values of (A).7. Counting Valid Numbers: - Each (A) (6 values) has 3 possible (B) values, resulting in (6 times 3 = 18) valid numbers.8. Pigeonhole Principle: - To ensure at least two identical numbers, the minimum number of dwarfs is (18 + 1 = 19).Thus, the minimum number of dwarfs required is (boxed{19}).]

🔑:Answer: 19.Solution: Let the number of numbers that satisfy the condition of the problem be N. Then, by the Pigeonhole Principle, the minimum number of gnomes must be N+1.Let's write a three-digit number as overline{x y z}, where x is the number of hundreds, y is the number of tens, and z is the number of units. Since along with the number overline{x y z}, the number overline{z y x} is also considered, then x>0 and z>0. According to the condition, 100 x+10 y+z+297=100 z+10 y+x. Let's transform this:99 z-99 x=297 Leftrightarrow x=z-3Since 0<x=z-3 leq 9 and z leq 9, then 0<x leq 6. Thus, we have 6 suitable values for x. Each x corresponds to one value of z, and y is any number from 1 to 9 for which overline{x y z} is divisible by 3. Note that0=(x+y+z) quad bmod 3=(2 x+y) quad bmod 3which means any pair (x, z) uniquely determines y bmod 3. Among the numbers from 1 to 9, there are exactly 3 numbers with a given remainder when divided by 3. Therefore, the condition of the problem is satisfied by 6 cdot 3=18 numbers, and the minimum number of gnomes is 19.

🧠:[<think>Okay, let's try to tackle this equation: (x^{2}(x^{2} + 1) = 21^{y} - 1). So, we need to find all integer solutions ((x, y)) for this. Hmm, let's see. First, I should note that both sides of the equation need to be non-negative because the left side is (x^2) times something, and the right side is (21^y - 1). Since (21^y) is always positive, (21^y - 1) must be non-negative. Therefore, (21^y geq 1), which means (y) must be a non-negative integer. Also, since (21^0 = 1), if (y = 0), then the right side is (1 - 1 = 0). Let me check if the left side can be zero. The left side is (x^2(x^2 + 1)). For this to be zero, (x^2) must be zero, which implies (x = 0). So, ((0, 0)) is a solution. But let's verify that: (0^2(0^2 + 1) = 0 * 1 = 0), and (21^0 - 1 = 1 - 1 = 0). Yes, that works. So, one solution is ((0, 0)). Now, let's check for (y = 1). Then, (21^1 - 1 = 20). So, the left side must equal 20. Let's solve (x^2(x^2 + 1) = 20). Let me let (a = x^2), so the equation becomes (a(a + 1) = 20). That's a quadratic in (a): (a^2 + a - 20 = 0). Solving this quadratic equation, discriminant is (1 + 80 = 81), so (a = [-1 pm 9]/2). Since (a) must be non-negative (because it's (x^2)), we take (a = (-1 + 9)/2 = 4). Therefore, (x^2 = 4), so (x = pm2). Let's check: (x^2(x^2 +1) = 4*5 = 20), which matches (21^1 -1 = 20). So, ((pm2, 1)) are solutions. Next, check (y = 2). Then, (21^2 -1 = 441 -1 = 440). So, the left side must be 440. Let's solve (x^2(x^2 +1) = 440). Again, let (a = x^2), so (a(a +1) = 440). Then, (a^2 + a - 440 = 0). Discriminant is (1 + 1760 = 1761). Let me check what the square root of 1761 is. 42^2 = 1764, which is 3 more, so sqrt(1761) is approximately 41.96. Not an integer. Therefore, the roots are not integers. Thus, (a = [-1 pm sqrt{1761}]/2), which are not integers. Therefore, no integer solutions for (x) when (y=2). What about (y=3)? (21^3 = 9261), so (21^3 -1 = 9260). Then, solving (x^2(x^2 +1) = 9260). Let (a = x^2), so (a^2 + a - 9260 = 0). Discriminant is (1 + 4*9260 = 1 + 37040 = 37041). Let's see if this is a perfect square. What's 192^2? 192^2 = 36864, 193^2 = 37249. Hmm, between 192 and 193. So sqrt(37041) is not an integer. Hence, no integer solutions here. Similarly, for higher values of (y), the right-hand side grows exponentially, while the left-hand side is a quartic in (x). So, likely, there are no solutions for (y geq 2). But maybe we need to check if there's some (y) where (21^y -1) can be written as the product of two consecutive integers (since (a(a+1)) is two consecutive integers). Alternatively, let's think modulo some numbers to see if there are contradictions. First, note that the left-hand side is (x^2(x^2 + 1)), which is two consecutive integers multiplied together, so one of them must be even. Therefore, the product is even. Therefore, the right-hand side (21^y -1) must be even. Let's check: (21) is odd, so (21^y) is odd, so (21^y -1) is even. So, that's consistent. Another approach: let's analyze the equation modulo 3 or 4. Let me try modulo 3. Left side: (x^2(x^2 +1)). Let's see possible values of (x^2) modulo 3. The squares modulo 3 are 0 and 1. If (x^2 equiv 0 mod 3), then left side is (0*(0 +1) = 0 mod 3). If (x^2 equiv 1 mod 3), then left side is (1*(1 +1) = 2 mod 3). Right side: (21^y -1 = (3*7)^y -1 equiv 0^y -1 equiv (-1) mod 3). Wait, because 21 ≡ 0 mod 3, so 21^y ≡ 0 mod 3. Therefore, 21^y -1 ≡ -1 mod 3. Therefore, left side must be ≡ -1 mod 3. But left side is either 0 or 2 mod 3. Therefore, 0 ≡ -1 mod 3? No. 2 ≡ -1 mod 3? Yes, because 2 +1 = 3 ≡0. So 2 ≡ -1 mod 3. So when the left side is 2 mod 3, which occurs when x^2 ≡1 mod 3, which is when x ≡ ±1 mod 3. Therefore, the right side is -1 mod 3, which is 2 mod 3, so that's consistent. Therefore, the equation is possible only when the left side is 2 mod 3, which requires that x is not divisible by 3, i.e., x ≡ ±1 mod 3. Therefore, modulo 3 analysis tells us that x cannot be a multiple of 3. Another modulus: maybe 4. Let's check modulo 4. Left side: (x^2(x^2 +1)). Squares modulo 4 are 0 and 1. If x is even, x^2 ≡0 mod4, then left side is 0*(0+1)=0 mod4. If x is odd, x^2 ≡1 mod4, then left side is 1*(1 +1)=2 mod4. Right side: 21^y -1. 21 ≡1 mod4, so 21^y ≡1^y=1 mod4. Therefore, 21^y -1 ≡0 mod4. Therefore, left side must be ≡0 mod4. Therefore, for left side to be 0 mod4, x must be even. Because if x is odd, left side is 2 mod4, which contradicts right side being 0 mod4. Therefore, x must be even. So, we can write x=2k for some integer k. Therefore, substituting back into the equation:( (2k)^2[(2k)^2 +1] = 21^y -1 )Simplify:(4k^2(4k^2 +1) = 21^y -1)So, left side is 4k^2*(4k^2 +1). Let's note that 4k^2 and 4k^2 +1 are consecutive integers, so they are coprime (since consecutive integers are coprime). Therefore, the product is 4k^2*(4k^2 +1). Since they are coprime, each factor must be a power of 21 minus 1? Wait, but 4k^2 and 4k^2 +1 are coprime, and their product is 21^y -1. Since they are coprime, each must be a power of 21 or 1, but 21 is composite. Wait, but 21^y -1 is one less than a power of 21, so it's not necessarily a multiple of 21. Wait, but since 4k^2 and 4k^2 +1 are coprime, their product is 21^y -1, so each of them must be a power of 21's factors. But 21 = 3*7. So, since 4k^2 and 4k^2 +1 are coprime, one must be a power of 3 and the other a power of 7. Because 21^y -1 = (3*7)^y -1, but maybe factorization here is tricky. Alternatively, since 4k^2 and 4k^2 +1 are coprime, and their product is 21^y -1. Therefore, 4k^2 = a and 4k^2 +1 = b where a*b =21^y -1 and gcd(a,b)=1. So, 21^y -1 must be a product of two coprime numbers differing by 1. The only two coprime numbers differing by 1 are 1 and 2. Wait, but 21^y -1 for y ≥1 is 20, 440, etc. For y=1, 21^1 -1=20. So 20 can be written as 4*5, which are consecutive integers? 4 and 5 are consecutive, yes. So 4*5=20. Which matches with the left side when k^2=1, so 4*1*(4*1 +1)=4*5=20. So that works. For y=2, 21^2 -1=440. Let's see if 440 can be written as a product of two coprime numbers differing by 1. Let's factor 440. 440=8*55=8*5*11. Hmm, 440= 20*22, but 20 and 22 are not coprime. 440= 11*40, 11 and 40 are coprime, but 40 -11=29, not 1. 440= 5*88, difference 83. 8*55, difference 47. 10*44, difference 34. 4*110, difference 106. 5*88, etc. So no two factors differing by 1. Therefore, no solution for y=2. Similarly, for higher y, 21^y -1 will be a larger number, but the factors 4k^2 and 4k^2 +1 need to differ by 1, which would require that 21^y -1 is a product of two consecutive integers. But 21^y -1 is equal to (4k^2)(4k^2 +1). For y ≥1, 21^y -1 increases rapidly. Since 4k^2 and 4k^2 +1 are consecutive integers, their product is between (4k^2)^2 and (4k^2 +1)^2. So, for each y, we can estimate k. But perhaps there's a better approach. Let me try to see if for y ≥2, 21^y -1 can be written as the product of two consecutive integers. For that, suppose that there exists integers m and m+1 such that m(m+1) =21^y -1. Then, m^2 + m - (21^y -1)=0. Solving for m, discriminant is 1 +4*(21^y -1)=4*21^y -3. So, m = [-1 ± sqrt(4*21^y -3)]/2. For m to be integer, sqrt(4*21^y -3) must be integer. Let me check for y=1: 4*21 -3 =81, sqrt(81)=9, which gives m=( -1 +9)/2=4, which works. For y=2:4*441 -3=1764-3=1761, which is not a perfect square (as checked earlier). For y=3:4*9261 -3= 37044 -3=37041, which is not a square. So, it's only a square when y=1, which gives m=4. Therefore, the only solutions are when y=0 and y=1. Wait, but we have y=0 as a solution (0,0). However, when y=0, the equation becomes x^2(x^2 +1)=0, which requires x=0. So, that's another case. So, combining these, the solutions are (0,0), (2,1), (-2,1). Wait, but how does y=0 fit into the previous analysis? When y=0, 21^0 -1=0, so x^2(x^2 +1)=0, which gives x=0. So, even though the previous reasoning for y≥1 gives solutions for y=1, and none beyond, y=0 is a separate case. Therefore, the integer solutions are (0,0), (2,1), (-2,1). Let me check if there are other possibilities. Wait, what if x is zero? Then, the left side is zero, so we need 21^y -1=0, which implies y=0. So, (0,0) is the only solution with x=0. For x non-zero, we must have y ≥1. But earlier analysis shows that for y ≥1, the only solutions are y=1, x=±2. So, in conclusion, the integer solutions are (0,0), (2,1), (-2,1). But let me verify for y=1 again. x=2: 2^2(2^2 +1)=4*5=20. 21^1 -1=20. Correct. x=-2: same as x=2. Correct. For y=0, x=0: 0=21^0 -1=0. Correct. Are there any other solutions? Let's check for x=1. Then left side is 1*(1+1)=2. So, 21^y -1=2 implies 21^y=3. No solution. x= -1 same as x=1. x=3: 9*10=90. 21^y=91. 91 is not a power of 21. Similarly, x=4:16*17=272. 21^y=273. 273 divided by 21 is 13. So, 21*13=273. Not a power. x=5:25*26=650. 21^y=651. 651/21=31. So, 21*31=651. Not a power. Similarly, x=6:36*37=1332. 21^y=1333. 1333/21≈63.47. Not an integer. So, no solution. Therefore, the only solutions are the ones found earlier. Another angle: perhaps use inequalities. For x ≥3, x^2(x^2 +1) ≥9*10=90. Let's see 21^y -1. For y=2, 21^2 -1=440. Which is much bigger than 90. But when x increases, the left side grows as x^4, and the right side as 21^y. So, perhaps for x ≥3 and y ≥2, the left side will outgrow the right side? Wait, x^4 vs 21^y. It's not straightforward. Let's check for x=3: 81*82=6642. 21^y=6643. What's 21^5=4084101... Wait, 21^4=194481, 21^5=4084101. So, 6643 is between 21^3=9261 and 21^4=194481. So, no. Similarly, x=4: 16*17=272, which is less than 21^3 -1=9260. Hmm, but when x increases, the left side x^4 +x^2 grows polynomially, while 21^y grows exponentially. So, for sufficiently large y, 21^y will surpass x^4, but for a given y, there might be a corresponding x. But since the left side must equal exactly 21^y -1, maybe there are no solutions beyond y=1. Alternatively, we can bound y. For example, let's check for y=1: solution exists. For y=2, no solution. y=3, no. For y=4, 21^4 -1=194481 -1=194480. Let's see if this can be written as x^2(x^2 +1). Let a =x^2, so a^2 +a -194480=0. Discriminant sqrt(1 +4*194480)=sqrt(777921). sqrt(777921)=approx 882. Because 882^2=777924, which is 3 more. So sqrt(777921)=882 - tiny bit. Not integer. Thus, no solution. Similarly, higher y would result in even larger discriminants that are unlikely to be perfect squares. Therefore, the only integer solutions are (0,0), (2,1), (-2,1). Final AnswerThe integer solutions are (boxed{(0, 0)}), (boxed{(2, 1)}), and (boxed{(-2, 1)}).</think>To solve the equation (x^{2}(x^{2} + 1) = 21^{y} - 1) in the set of integers, we analyze both sides modulo different numbers and check possible values for (x) and (y).1. Case (y = 0): - The equation becomes (x^2(x^2 + 1) = 0). - This implies (x = 0). - Solution: ((0, 0)).2. Case (y = 1): - The equation becomes (x^2(x^2 + 1) = 20). - Let (a = x^2), then (a(a + 1) = 20). - Solving the quadratic equation (a^2 + a - 20 = 0), we get (a = 4) (since (a) must be non-negative). - Thus, (x^2 = 4) gives (x = pm 2). - Solutions: ((pm 2, 1)).3. Case (y geq 2): - For (y = 2), (21^2 - 1 = 440). Solving (x^2(x^2 + 1) = 440) yields no integer solutions. - For higher (y), the right-hand side grows exponentially, while the left-hand side grows polynomially. Checking modulo 3 and 4, we find contradictions or non-integer solutions for higher (y).4. Modulo Analysis: - Modulo 3: The right-hand side must be 2 mod 3, which requires (x notequiv 0 mod 3). - Modulo 4: The right-hand side must be 0 mod 4, requiring (x) to be even.5. Conclusion: - The only integer solutions are found for (y = 0) and (y = 1).The integer solutions are (boxed{(0, 0)}), (boxed{(2, 1)}), and (boxed{(-2, 1)}).,<think>Okay, so I need to solve the equation ( x^{2}(x^{2} + 1) = 21^{y} - 1 ) where ( x ) and ( y ) are integers. Hmm, let's see. First, maybe I should start by trying small integer values for ( y ) and see if the right-hand side becomes something that can be expressed as ( x^{2}(x^{2} + 1) ). That product on the left is two consecutive integers multiplied together, right? Because ( x^2 ) and ( x^2 + 1 ) are consecutive. So the left-hand side is essentially ( n(n + 1) ) where ( n = x^2 ). So maybe that could help.Let me first compute ( 21^y - 1 ) for small ( y ):- For ( y = 0 ): ( 21^0 - 1 = 1 - 1 = 0 ). Then left-hand side would be ( x^2(x^2 + 1) = 0 ). Since ( x^2 ) is non-negative, the only solution is ( x = 0 ). So (0, 0) could be a solution. Wait, but ( y = 0 ). Is that allowed? The problem says integers, so ( y ) can be 0. So maybe (0,0) is a solution.- For ( y = 1 ): ( 21^1 - 1 = 21 - 1 = 20 ). So we need ( x^2(x^2 + 1) = 20 ). Let's check possible ( x ). Let's try ( x^2 = 4 ): then 4*5 = 20. Yes, that works. So ( x^2 = 4 ) implies ( x = pm 2 ). So (2,1) and (-2,1) are solutions.- For ( y = 2 ): ( 21^2 - 1 = 441 - 1 = 440 ). Now check if 440 can be written as ( n(n + 1) ). Let's see. Let's solve ( n^2 + n - 440 = 0 ). Discriminant is 1 + 1760 = 1761. The square root of 1761 is approximately 41.96, which isn't an integer. So no integer solution here. So no solutions for ( y = 2 ).- For ( y = 3 ): ( 21^3 = 9261 ), so 9261 - 1 = 9260. Check if 9260 is a product of two consecutive integers. Let's approximate the square root: ( sqrt{9260} approx 96.23 ). So testing n = 96: 96*97 = 9312, which is higher than 9260. 95*96 = 9120, which is lower. So between 95 and 96, but neither works. So no solution here.- For ( y = 4 ): ( 21^4 = 194481 ), so 194481 - 1 = 194480. Let's approximate square root: around 440. So 440*441 = 194040, which is less than 194480. Then 441*442 = 194, 440 + 441 = 194, 440*442 = let's compute 440*442: 440*(440 + 2) = 440^2 + 880 = 193600 + 880 = 194480. Wait, exactly! So 440*441 is 194040, but 440*442 is 194480? Wait, no, that can't be. Wait, 440*442: yes, 440*(441 + 1) = 440*441 + 440 = 194040 + 440 = 194480. So the right-hand side is 194480, which is 440*442. But these are two even numbers, not consecutive. But we need two consecutive integers. So 440*442 is two numbers with a difference of 2, so not consecutive. Therefore, that's not possible. So 194480 cannot be expressed as n(n + 1). Therefore, no solution here.Wait, maybe I made a mistake here. Let me check. If ( n(n + 1) = 194480 ), then discriminant would be ( 1 + 4 * 194480 = 1 + 777920 = 777921 ). What's the square root of 777921? Let me check. 882 squared is 777,924. Oh, wait, 882^2 = (880 + 2)^2 = 880² + 4*880 + 4 = 774,400 + 3,520 + 4 = 777,924. So discriminant is 777,921, which is 882^2 - 3. So not a perfect square. Therefore, no solution here. So y=4 doesn't work.Similarly, trying higher y would lead to even larger numbers, so maybe there's no solution beyond y=1. But we need to check for negative y as well? Wait, the problem says integers. So y can be negative? Let me check. If y is negative, 21^y would be a fraction, so 21^y - 1 would be negative. But the left-hand side is x²(x² + 1), which is non-negative since x² is non-negative. So negative y would lead to negative right-hand side, which can't equal the left-hand side. So y must be non-negative integers. Therefore, only y >= 0.Wait, but y=0 gives 0, which we have x=0. So (0,0) is a solution. Wait, let me check that again. If y=0, then 21^0 = 1, so 1 - 1 = 0. Then left-hand side is x²(x² + 1) = 0. Since x² is non-negative, x² = 0. So x=0. Therefore, (0,0) is a solution. So that's valid. So that's another solution. So so far, we have solutions at (0,0), (2,1), (-2,1).Wait, but let me check if there are other possible y. For example, maybe y=1 is the only other possible one. Let's check y=5: 21^5 is 4084101, 4084101 -1 = 4084100. Check if this is n(n+1). Let's approximate sqrt(4084100). The square root of 4,084,100 is 2021, because 2021^2 = (2000 + 21)^2 = 2000² + 2*2000*21 + 21² = 4,000,000 + 84,000 + 441 = 4,084,441. So 2021² = 4,084,441, which is more than 4,084,100. So n(n+1) = 2020*2021 = 2020² + 2020 = 4,080,400 + 2020 = 4,082,420, which is less than 4,084,100. So no solution here.Alternatively, maybe there's a pattern here where for higher y, the RHS is not a product of two consecutive integers. So maybe only y=0 and y=1 give solutions. But we need to make sure there's no other solutions.Alternatively, maybe we can approach this equation algebraically. Let's let ( n = x^2 ). Then the equation becomes ( n(n + 1) = 21^{y} - 1 ). So ( n^2 + n + 1 = 21^{y} ).Wait, that's an interesting way to rewrite it. So ( n^2 + n + 1 = 21^y ). Hmm. So we have a quadratic in n: ( n^2 + n + 1 = 21^y ). Maybe this form can be more helpful. Let's see.We need to find integers n and y such that ( n^2 + n + 1 = 21^y ). Since n is non-negative (as n = x²), we can restrict n to non-negative integers.So let's try to analyze this equation modulo some numbers to see if there are contradictions or possible restrictions.First, consider modulo 3. Since 21 is divisible by 3, 21^y is congruent to 0 modulo 3 for y >=1. So for y >=1, the RHS is 0 mod 3. Let's check the LHS: ( n^2 + n + 1 ). Compute this mod 3:- If n ≡ 0 mod 3: 0 + 0 + 1 ≡ 1 mod 3- If n ≡ 1 mod 3: 1 + 1 + 1 ≡ 3 ≡ 0 mod 3- If n ≡ 2 mod 3: 4 + 2 + 1 ≡ 7 ≡ 1 mod 3So LHS is congruent to 0 mod 3 only when n ≡ 1 mod 3. But RHS for y >=1 is 0 mod 3. Therefore, for the equation to hold when y >=1, n must ≡1 mod 3.So that gives a condition on n. So n ≡1 mod3.Similarly, maybe check modulo 4. Let's see.21 is congruent to 1 mod 4, so 21^y ≡1^y ≡1 mod4. So RHS ≡1 -1=0 mod4 when y >=1. Wait, no: wait, 21^y mod4: 21 ≡1 mod4, so 21^y ≡1 mod4. Therefore, 21^y -1 ≡0 mod4. So the RHS is 0 mod4 for y >=1. Let's check the LHS: n(n + 1). Since n and n +1 are consecutive integers, one of them is even. Therefore, their product is even. But 0 mod4 means divisible by 4. So the product n(n +1) must be divisible by4. Therefore, either n or n +1 must be divisible by4.Since n and n +1 are consecutive, one of them is even. For their product to be divisible by4, the even number must be divisible by4. So either n ≡0 mod4 or n ≡3 mod4 (since n +1 ≡0 mod4 in that case). So that's another condition. So for the equation to hold when y >=1, n must be ≡0 or3 mod4. But also, from earlier, n ≡1 mod3.So combining these, n ≡1 mod3 and n ≡0 or3 mod4.We can use the Chinese remainder theorem to find possible residues for n modulo 12.Case1: n ≡0 mod4 and n≡1 mod3.Find n ≡0 mod4, n≡1 mod3.Let’s write n=4k. Then 4k≡1 mod3 => k≡1 mod3 (since 4≡1 mod3, so k≡1 mod3). Therefore, n=4*(3m +1)=12m +4. So n≡4 mod12.Case2: n≡3 mod4 and n≡1 mod3.n=4k +3. Then 4k +3 ≡1 mod3 =>4k≡-2≡1 mod3 =>4k≡1 mod3 =>k≡1 mod3 (since 4≡1 mod3). Therefore, k=3m +1, so n=4*(3m +1) +3=12m +7. So n≡7 mod12.Therefore, possible residues are 4 or7 mod12.So n must be ≡4 or7 mod12.Alternatively, maybe other moduli can give more restrictions.Alternatively, modulo 7. Let's see. 21 is divisible by7, so 21^y ≡0 mod7 for y>=1. Therefore, RHS is -1 mod7. Wait, wait: 21^y -1. 21≡0 mod7, so 21^y ≡0 mod7. Therefore, 21^y -1≡-1 mod7. So LHS is n^2 +n +1 ≡-1 mod7. So n^2 +n +2≡0 mod7.So solve n^2 +n +2 ≡0 mod7.Let's check all n mod7:n=0: 0 +0 +2=2≡2≠0n=1:1 +1 +2=4≡4≠0n=2:4 +2 +2=8≡1≠0n=3:9 +3 +2=14≡0 mod7. So n≡3 mod7.n=4:16 +4 +2=22≡1≠0n=5:25 +5 +2=32≡4≠0n=6:36 +6 +2=44≡2≠0So only n≡3 mod7 satisfies the congruence. Therefore, combining with previous results, n must ≡3 mod7, and ≡4 or7 mod12.So now, let's solve these congruence systems.First, for n≡4 mod12 and n≡3 mod7.Let n=12k +4. Then 12k +4 ≡3 mod7 =>12k ≡-1 mod7. 12≡5 mod7, so 5k ≡-1 mod7 =>5k≡6 mod7. Multiply both sides by inverse of5 mod7. Since5*3=15≡1 mod7, inverse is3. So k≡6*3=18≡4 mod7. Therefore, k=7m +4. Therefore, n=12*(7m +4)+4=84m +52. So n≡52 mod84.Second case: n≡7 mod12 and n≡3 mod7.Let n=12k +7. Then 12k +7≡3 mod7 =>12k≡-4 mod7. 12≡5 mod7, so 5k≡-4≡3 mod7. Multiply by inverse of5 mod7, which is3. So k≡3*3=9≡2 mod7. Therefore, k=7m +2. Thus, n=12*(7m +2)+7=84m +24 +7=84m +31. So n≡31 mod84.Therefore, n≡52 mod84 or n≡31 mod84.Therefore, any solution with y>=1 must have n≡31 or52 mod84.So possible solutions would be n=31,52,115,136,... etc.But since n =x², n must be a perfect square. Wait, no, wait: n =x², so x² must be ≡31 or52 mod84. Wait, but x² is a square. Let's check if squares can be ≡31 or52 mod84.First, check possible squares modulo84. Since 84=4*3*7. Let's see:First, modulo4: squares can be 0 or1 mod4.Modulo3: squares can be0 or1 mod3.Modulo7: squares can be0,1,2,4 mod7.So let's check n≡31 mod84: 31 mod4=3, which is not a square (since squares mod4 are 0 or1). Therefore, n≡31 mod84 cannot be a square. Similarly, n≡52 mod84: 52 mod4=0, which is okay. 52 mod3: 52/3=17*3 +1 =>52≡1 mod3. Which is a square (since squares mod3 are 0 or1). 52 mod7: 52=7*7 + 3 =>52≡3 mod7. Now, squares mod7 are 0,1,2,4. So 3 is not a square mod7. Therefore, 52 mod7≡3 is not a square. Therefore, n≡52 mod84 cannot be a square. Hence, there's no solution with y>=1 except possibly small y which we already checked (y=1). Because for higher y, n must be ≡31 or52 mod84, but these are not squares. Therefore, the only possible solutions are y=0 and y=1. Which we already found: (0,0), (2,1), (-2,1).Wait, but let me check y=1 case. When y=1, n(n +1)=20. We found n=4, which is a square. So x=±2. So yes, that works. For n=4, 4*5=20. So that's correct. And n=4≡4 mod12, which is one of the possible residues. But 4 mod84 is 4, which is different from 52 mod84, but maybe because 4 is less than84. Wait, but the general solution is n≡52 mod84, but 4≡52 mod84? 52-4=48, which is not a multiple of84. So 4 is not ≡52 mod84. Hmm. Maybe the congruence applies only for larger n. Wait, perhaps the congruence n≡52 mod84 or31 mod84 comes from the conditions that y>=1, but when y=1, n=4, which doesn't satisfy these congruences. So maybe our analysis is correct in that for y>=1, solutions must satisfy n≡52 or31 mod84, but since n=4 doesn't, it's a special case. But in reality, n=4 is a solution. So perhaps our modular approach is missing something. Maybe because when y=1, 21^y -1=20, which is a small number, and our congruence conditions start applying for higher y.Alternatively, maybe the modular constraints only apply when y >=2. Let's check. For y=1, RHS=20, which is 0 mod4. The LHS n(n+1)=4*5=20, which is indeed 0 mod4. So n=4≡0 mod4, which is allowed. Also, n=4≡1 mod3 (since 4 mod3=1). So it's compatible with the previous conditions. But in our Chinese remainder theorem solution, we found that n≡4 mod12 or7 mod12. So n=4 is≡4 mod12, which fits. But when we considered modulo7, we found that n must≡3 mod7, but 4 mod7=4, which is not≡3 mod7. But wait, when y=1, RHS=20, which is 21^1 -1=20. Then 20 mod7=20-2*7=6. So RHS=20≡6 mod7. The LHS is n(n+1)=4*5=20≡6 mod7. So actually, in this case, the congruence modulo7 is satisfied. Wait, but earlier, when we considered modulo7, we found that for y>=1, n must≡3 mod7. But here, n=4≡4 mod7. Hmm. So there's a contradiction here. Wait, let's re-examine that step.Earlier, we considered the equation n^2 +n +1=21^y. Then, modulo7, for y>=1, 21^y≡0 mod7, so n^2 +n +1≡0 mod7. Therefore, n^2 +n +1≡0 mod7. But in reality, when y=1, the equation is n(n+1)=21^y -1=20. So n(n+1)=20. Therefore, n^2 +n -20=0. So in that case, the equation is different. Wait, I think I made a mistake here. Wait, in the original equation, when we let n=x^2, then x^2(x^2 +1)=21^y -1. So that is n(n +1)=21^y -1. Then, rearranged, we get n^2 +n -21^y +1=0. But earlier, I thought of rearranging to n^2 +n +1=21^y, but that would be incorrect. Wait, no. Let me check:Original equation: n(n +1) =21^y -1. Therefore, n² +n =21^y -1. Then, n² +n +1=21^y. Oh, yes, that's correct. So n² +n +1=21^y. So that part is correct. Then, when considering modulo7, 21^y ≡0 mod7 for y>=1, so n² +n +1≡0 mod7. Then, solving n² +n +1≡0 mod7 gave n≡3 mod7. However, when y=1, n=4. Then, n=4 mod7 is4. Then, 4² +4 +1=16 +4 +1=21≡0 mod7. Ah, right! So 21≡0 mod7. So actually, 4² +4 +1=21≡0 mod7, which matches. Therefore, even though n=4≡4 mod7, when plugged into n² +n +1, it gives 21≡0 mod7. So actually, n=4 mod7 also satisfies the congruence n² +n +1≡0 mod7. Wait, but earlier when we solved n² +n +1≡0 mod7, we found only n≡3 mod7. But here, n=4 gives 0 mod7. So there's a mistake in my earlier solution of the congruence.Let me re-examine solving n² +n +1≡0 mod7.Compute n from0 to6:n=0:0 +0 +1=1≡1≠0n=1:1 +1 +1=3≡3≠0n=2:4 +2 +1=7≡0 mod7. So n=2 mod7.n=3:9 +3 +1=13≡6≠0n=4:16 +4 +1=21≡0 mod7. So n=4 mod7.n=5:25 +5 +1=31≡3≠0n=6:36 +6 +1=43≡1≠0Therefore, solutions are n≡2 mod7 and n≡4 mod7. So I had a mistake earlier. The solutions modulo7 are n≡2 or4 mod7, not3. So that changes things.Therefore, earlier analysis was wrong. So now, correct solutions modulo7 are n≡2 or4 mod7. Therefore, when combining with the previous congruences.Earlier, we had:From modulo3: n≡1 mod3.From modulo4: n≡0 or3 mod4.From modulo7: n≡2 or4 mod7.Therefore, we need to combine these congruences.So there are multiple cases:Case1: n≡1 mod3, n≡0 mod4, n≡2 mod7Case2: n≡1 mod3, n≡0 mod4, n≡4 mod7Case3: n≡1 mod3, n≡3 mod4, n≡2 mod7Case4: n≡1 mod3, n≡3 mod4, n≡4 mod7Each case needs to be solved using Chinese Remainder Theorem.Let's solve each case:Case1: n≡1 mod3, n≡0 mod4, n≡2 mod7Find n ≡0 mod4, ≡1 mod3, ≡2 mod7.Let’s write n=4k. Then, 4k ≡1 mod3 => k≡1 mod3 (since4≡1 mod3). So k=3m +1. Then n=4*(3m +1)=12m +4.Now, n≡12m +4 ≡2 mod7. So 12m +4 ≡2 mod7 =>12m ≡-2 ≡5 mod7. 12≡5 mod7, so5m≡5 mod7 =>m≡1 mod7. Therefore, m=7t +1. Then, n=12*(7t +1)+4=84t +12 +4=84t +16. So n≡16 mod84.Case2: n≡1 mod3, n≡0 mod4, n≡4 mod7Similarly, n=4k, 4k≡1 mod3 =>k≡1 mod3. So k=3m +1, n=12m +4.Now, n≡12m +4 ≡4 mod7 =>12m ≡0 mod7 =>5m≡0 mod7 (since12≡5 mod7). So m≡0 mod7. Therefore, m=7t, so n=12*(7t)+4=84t +4. Thus, n≡4 mod84.Case3: n≡1 mod3, n≡3 mod4, n≡2 mod7.n=4k +3. Then, 4k +3≡1 mod3 =>4k ≡-2 mod3 =>k≡1 mod3 (since4≡1 mod3). So k=3m +1. Then n=4*(3m +1)+3=12m +7.Now, n=12m +7 ≡2 mod7 =>12m +7 ≡2 mod7 =>12m ≡-5 ≡2 mod7. 12≡5 mod7, so5m≡2 mod7. Multiply by inverse of5 mod7, which is3. So m≡2*3=6 mod7. So m=7t +6. Then, n=12*(7t +6)+7=84t +72 +7=84t +79. Thus, n≡79 mod84.Case4: n≡1 mod3, n≡3 mod4, n≡4 mod7.n=4k +3. 4k +3≡1 mod3 =>4k≡-2 mod3 =>k≡1 mod3. So k=3m +1. Then n=12m +7.n=12m +7 ≡4 mod7 =>12m +7 ≡4 mod7 =>12m ≡-3≡4 mod7. 12≡5 mod7, so5m≡4 mod7. Multiply by inverse of5 mod7, which is3. So m≡4*3=12≡5 mod7. Therefore, m=7t +5. Then, n=12*(7t +5)+7=84t +60 +7=84t +67. Thus, n≡67 mod84.Therefore, the solutions are n≡16,4,79,67 mod84.Therefore, n can be 4,16,67,79,148,160,… etc.But n must be a perfect square, because n=x². Let's check these residues modulo84:Check if these residues are quadratic residues modulo84. However, since 84 factors into 4*3*7, we can check each modulus.For n≡4 mod84: 4 is a square (2²).For n≡16 mod84: 16=4².For n≡67 mod84: 67 mod4=3, which is not a square modulo4 (squares modulo4 are0 or1). So 67 can't be a square.Similarly, n≡79 mod84:79 mod4=3, also not a square modulo4.Therefore, only n≡4 or16 mod84 can be squares. So possible solutions are n=4,16,84+4=88,84+16=100, etc. Let's check these.First, n=4: yes, x=±2, which we already found for y=1.n=16: Check if 16*17=272=21^y -1. So 21^y=273. But 273=21*13. 21^1=21, 21^2=441. 273 isn't a power of21. So no solution here.n=88: Check if 88*89=7832=21^y -1. Then 21^y=7833. 7833 divided by21: 7833 ÷21=373. So 21*373=7833, which is not a power of21. So no.n=100:100*101=10100=21^y -1. So21^y=10101. Check 10101 ÷21=481. So21*481=10101, not a power. So no.Similarly, n=84*2 +4=172:172*173=172*(170 +3)=172*170 +172*3=29,240 +516=29,756. 21^y=29,757. Divide by21:29,757 ÷21≈1,417. So21*1,417=29,757. Not a power. So no.Therefore, the only possible n that is a square and satisfies the congruence conditions is n=4, leading to y=1.Therefore, the only integer solutions are x=0,y=0 and x=±2,y=1.But wait, let me check y=0 case. When y=0, RHS=0, so x²(x² +1)=0. Which gives x=0. So (0,0) is a solution. Are there any other possible solutions?Wait, n=0:0*1=0=21^0 -1=1-1=0. So yes. So x=0,y=0.Any other possible n?If y=0, then 21^y -1=0. So n(n +1)=0. Therefore, n=0 orn=-1. But since n=x² >=0, only n=0 is valid.Therefore, the only solutions are (0,0), (2,1), (-2,1).But let's check if there's a possibility of x=1. Let's test x=1:1²*(1 +1)=1*2=2. So RHS=21^y -1=2. So21^y=3. But 3 is not a power of21. So no.x=3:9*10=90. So21^y=91. 91=13*7. Not a power of21.x=4:16*17=272. 21^y=273=21*13. Not a power.x=5:25*26=650. 21^y=651=21*31. Not a power.Similarly for negative x, same as positive.Thus, indeed, the only solutions are (0,0), (2,1), (-2,1).But let me check y=0 case again. y=0 is allowed, since the problem says integers. So y=0 is a non-negative integer. So (0,0) is valid.Therefore, final solutions are:x=0, y=0;x=±2, y=1.Final AnswerThe integer solutions are (boxed{(0, 0)}) and (boxed{(pm 2, 1)}).</think>To solve the equation ( x^{2}(x^{2} + 1) = 21^{y} - 1 ) in the set of integers, we start by analyzing small values of ( y ):1. For ( y = 0 ): - The equation becomes ( x^{2}(x^{2} + 1) = 0 ). - The only solution is ( x = 0 ). - This gives the solution ((0, 0)).2. For ( y = 1 ): - The equation becomes ( x^{2}(x^{2} + 1) = 20 ). - Solving ( x^{2} = 4 ), we get ( x = pm 2 ). - This gives the solutions ((pm 2, 1)).3. For ( y geq 2 ): - We analyze the equation modulo 3, 4, and 7 to find constraints on ( x ) and ( y ). - Modulo 3 analysis shows ( n equiv 1 mod 3 ) where ( n = x^2 ). - Modulo 4 analysis shows ( n equiv 0 mod 4 ) or ( n equiv 3 mod 4 ). - Modulo 7 analysis shows ( n equiv 2 mod 7 ) or ( n equiv 4 mod 7 ). - Combining these congruences using the Chinese Remainder Theorem, we find that ( n ) must satisfy specific residues modulo 84. However, these residues do not yield perfect squares for ( n ) except for the cases already handled by ( y = 0 ) and ( y = 1 ).4. Checking higher values of ( y ): - For ( y geq 2 ), the right-hand side ( 21^y - 1 ) becomes very large and does not factor into the product of two consecutive integers, confirming no further solutions.Thus, the integer solutions are:[boxed{(0, 0)} quad text{and} quad boxed{(pm 2, 1)}]]

🔑:Solution: (Lucian Petrescu) We can observe that y in mathbb{N}, and if left(x_{0}, y_{0}right) is a solution, then left(-x_{0}, y_{0}right) is also a solution of the equation. Moreover, (0,0) is a solution. We rewrite the equation as left(x^{2}-x+1right)left(x^{2}+x+1right)=21^{y}. We notice that x^{2}-x+1 and x^{2}+x+1 are coprime. It follows that x^{2}-x+1=1 and x^{2}+x+1=21^{y} with the solution (0,0) or x^{2}-x+1=3^{y} and x^{2}+x+1=7^{y}. From the first relation, we have (2 x-1)^{2}=3left(4 cdot 3^{y-1}-1right), which means 3 mid (2 x-1)^{2}, and more specifically, 9 mid (2 x-1)^{2}, so 9 mid left(4 cdot 3^{y-1}-1right), which gives the solution y=1, then x=2. Therefore, the set of solutions is (-2,1),(0,0),(2,1).

🧠:[<think>Okay, so I need to find the natural number k where the expression k squared divided by 1.001 to the power of k reaches its maximum. Hmm, let's think about how to approach this. First, I remember that when dealing with functions that have both polynomial and exponential components, the exponential usually grows faster eventually, but for smaller values, the polynomial might dominate. Since 1.001 is just slightly larger than 1, the denominator grows slowly, so maybe the maximum isn't too large a number? Not sure yet.Let me consider the function f(k) = k² / 1.001^k. I need to find the integer k where this is maximized. For continuous functions, we might take the derivative and set it to zero, but since k is discrete (natural numbers), maybe I should look at the ratio of consecutive terms and see when it changes from increasing to decreasing. That is, find where f(k+1)/f(k) transitions from being greater than 1 to less than 1. That point should be around the maximum.Let's compute the ratio f(k+1)/f(k):f(k+1)/f(k) = [( (k+1)^2 / 1.001^{k+1} ) / ( k² / 1.001^k )] = [(k+1)^2 / k²] / 1.001Simplify that: [(1 + 1/k)^2] / 1.001So the ratio is (1 + 2/k + 1/k²) / 1.001. We want this ratio to be equal to 1 when the terms switch from increasing to decreasing. So set (1 + 2/k + 1/k²)/1.001 = 1. Then solving for k:1 + 2/k + 1/k² = 1.001Subtract 1 from both sides:2/k + 1/k² = 0.001Multiply both sides by k² to eliminate denominators:2k + 1 = 0.001k²Bring all terms to one side:0.001k² - 2k - 1 = 0Multiply both sides by 1000 to eliminate the decimal:k² - 2000k - 1000 = 0Now, solving this quadratic equation for k. The quadratic formula is k = [2000 ± sqrt(2000² + 4000)] / 2Wait, the discriminant is b² - 4ac. Here, a = 1, b = -2000, c = -1000. So discriminant D = (-2000)^2 - 4*1*(-1000) = 4,000,000 + 4,000 = 4,004,000So sqrt(D) = sqrt(4,004,000). Let me approximate that. sqrt(4,004,000) ≈ sqrt(4,000,000 + 4,000) = 2000 + 4,000/(2*2000) by linear approximation? Wait, that might not be precise enough. Alternatively, factor out sqrt(4,004,000) = sqrt(1000 * 4004) = sqrt(1000) * sqrt(4004) ≈ 31.622 * 63.277 ≈ 31.622 * 63.277. Let's calculate that:31.622 * 60 = 1897.3231.622 * 3.277 ≈ 31.622*3 + 31.622*0.277 ≈ 94.866 + 8.76 ≈ 103.626Total ≈ 1897.32 + 103.626 ≈ 2000.946So sqrt(4,004,000) ≈ 2000.946Therefore, k = [2000 ± 2000.946]/2Since we want positive k, take the positive root:k = (2000 + 2000.946)/2 ≈ (4000.946)/2 ≈ 2000.473So the critical point is around 2000.473. Since k must be a natural number, the maximum is either at k=2000 or k=2001.But wait, let's check the ratio at k=2000. Plugging back into the ratio f(k+1)/f(k):(1 + 2/2000 + 1/(2000)^2)/1.001 ≈ (1 + 0.001 + 0.00000025)/1.001 ≈ (1.00100025)/1.001 ≈ 1.00000025. Which is slightly greater than 1. So f(k+1)/f(k) >1 at k=2000, meaning that f(k) is still increasing at k=2000. Then at k=2001, compute the ratio:(1 + 2/2001 + 1/(2001)^2)/1.001 ≈ (1 + 0.0009995 + 0.000000249)/1.001 ≈ (1.000999749)/1.001 ≈ approx 0.99999925. Which is less than 1. So at k=2001, the ratio is less than 1, meaning f(k) starts decreasing. Therefore, the maximum occurs at k=2001.Wait, but the critical point was approximately 2000.473. So between 2000 and 2001. Since the function is increasing before the critical point and decreasing after, then the maximum should be at k=2000 if the critical point is at 2000.473, because 2000.473 is closer to 2000 than 2001? Wait, but since the ratio at k=2000 is still greater than 1, so f(2001)/f(2000) ≈1.00000025, which is just over 1, so f(2001) is slightly larger than f(2000). Then at k=2001, the ratio becomes less than 1, so f(2002) < f(2001). Therefore, the maximum is at k=2001.Therefore, the natural number k where the expression attains its maximum is 2001.But wait, let me verify by calculating f(2000) and f(2001):f(2000) = 2000² / 1.001^2000f(2001) = 2001² / 1.001^2001So f(2001)/f(2000) = (2001² / 2000²) / 1.001 = ( (1 + 1/2000)^2 ) / 1.001 ≈ (1 + 2/2000 + 1/2000²) / 1.001 ≈ (1.001 + 0.00000025)/1.001 ≈ 1.00000025, which is just over 1, as before. Therefore, f(2001) is indeed slightly larger than f(2000). Then f(2002)/f(2001) would be (2002² / 2001²)/1.001 ≈ (1 + 2/2001 + 1/2001²)/1.001 ≈ (1.0009995 + 0.00000025)/1.001 ≈ 1.00099975/1.001 ≈ approximately 0.99999925, which is less than 1. So f(2002) is less than f(2001). Therefore, the maximum is at k=2001.But let me check with another approach. Maybe using logarithms to simplify the maximization.Taking natural logarithm of f(k): ln(f(k)) = 2 ln k - k ln(1.001)To maximize ln(f(k)), take the derivative with respect to k (treating k as continuous) and set to zero.The derivative is 2/k - ln(1.001). Setting equal to zero:2/k - ln(1.001) = 0 => 2/k = ln(1.001) => k = 2 / ln(1.001)Compute ln(1.001). Since ln(1+x) ≈ x - x²/2 + x³/3 - ..., for small x. Here, x = 0.001, so ln(1.001) ≈ 0.001 - 0.0000005 + ... ≈ approximately 0.0009995.Therefore, k ≈ 2 / 0.0009995 ≈ 2001.001. So this gives k ≈ 2001.001, which is very close to 2001. Therefore, the maximum occurs around k=2001. Since k must be an integer, check k=2001. This aligns with the previous method. Therefore, the maximum is at k=2001.But wait, when we solved the quadratic equation earlier, we had k≈2000.473. But here, the calculus approach gives approximately 2001.001. Hmm, discrepancy here. Why?Wait, the quadratic equation was derived from setting the ratio equal to 1, which gives the approximate critical point, but perhaps the quadratic approximation wasn't precise? Let me check.Wait, the original ratio was (1 + 2/k + 1/k²)/1.001 = 1. When we set that equal to 1, we approximated (1 + 2/k) /1.001 ≈1, but actually, even with the 1/k² term. But the quadratic equation approach gave k≈2000.473, while the derivative approach gave k≈2001.001. The two methods give slightly different approximate solutions.This is because the ratio method is a discrete approach, while the derivative method is a continuous approximation. So when dealing with discrete terms, the exact maximum might be at a different integer. But given that the ratio at k=2000 is just over 1, leading to an increase, and at k=2001 the ratio is just under 1, leading to a decrease, the maximum should be at k=2001. The continuous critical point is around 2001.001, which would round to 2001. So both methods confirm that k=2001 is the maximum.Therefore, the answer is 2001.Final AnswerThe natural number ( k ) for which the expression attains its maximum value is boxed{2001}.</think>To find the natural number ( k ) where the expression ( frac{k^2}{1.001^k} ) attains its maximum value, we analyze the ratio of consecutive terms ( frac{f(k+1)}{f(k)} ). This ratio is given by:[frac{f(k+1)}{f(k)} = frac{(k+1)^2 / 1.001^{k+1}}{k^2 / 1.001^k} = frac{(k+1)^2}{k^2} cdot frac{1}{1.001} = left(1 + frac{2}{k} + frac{1}{k^2}right) / 1.001]We set this ratio equal to 1 to find the critical point:[left(1 + frac{2}{k} + frac{1}{k^2}right) / 1.001 = 1]Simplifying, we get:[1 + frac{2}{k} + frac{1}{k^2} = 1.001][frac{2}{k} + frac{1}{k^2} = 0.001]Multiplying through by ( k^2 ):[2k + 1 = 0.001k^2]Rearranging into a quadratic equation:[0.001k^2 - 2k - 1 = 0]Multiplying by 1000 to clear the decimal:[k^2 - 2000k - 1000 = 0]Solving this quadratic equation using the quadratic formula ( k = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[k = frac{2000 pm sqrt{2000^2 + 4 cdot 1000}}{2} approx frac{2000 pm 2000.946}{2}]The positive root is approximately ( k approx 2000.473 ). Since ( k ) must be a natural number, we check ( k = 2000 ) and ( k = 2001 ).For ( k = 2000 ):[frac{f(2001)}{f(2000)} approx left(1 + frac{2}{2000} + frac{1}{2000^2}right) / 1.001 approx 1.00000025 > 1]For ( k = 2001 ):[frac{f(2002)}{f(2001)} approx left(1 + frac{2}{2001} + frac{1}{2001^2}right) / 1.001 approx 0.99999925 < 1]Since ( f(2001) ) is greater than ( f(2000) ) and ( f(2002) ) is less than ( f(2001) ), the maximum value occurs at ( k = 2001 ).Additionally, using the continuous approach by taking the natural logarithm of ( f(k) ) and setting its derivative to zero confirms the critical point is near ( k approx 2001.001 ).Thus, the natural number ( k ) for which the expression attains its maximum value is (boxed{2001}).,<think>Okay, so I need to find the natural number k where the expression k² divided by 1.001 to the power of k is the largest. Hmm, let's start by understanding the problem. First, the expression is k² / (1.001)^k. I need to find the value of k in natural numbers (so positive integers) where this fraction is maximized. That means I want to find the k where increasing or decreasing k from that point makes the fraction smaller. Let me think about how the numerator and denominator behave as k increases. The numerator is k squared, which grows quadratically. The denominator is 1.001 to the power of k, which is an exponential function. Exponential functions eventually outpace polynomial functions, so even though 1.001 is a small base, it's still exponential. So, initially, the numerator might grow faster, but after a certain point, the denominator will dominate, making the whole expression decrease. So there must be a maximum somewhere in between.To find the maximum, I can treat k as a continuous variable and use calculus to find the critical points, then check the integers around that point. Since k has to be a natural number, the maximum could be at the floor or ceiling of the critical point found through calculus.Let me set up the function f(k) = k² / (1.001)^k. To find its maximum, I can take the derivative with respect to k and set it equal to zero. But since it's a function of a discrete variable, calculus gives an approximation. The actual maximum will be near where the derivative is zero.First, let's take the natural logarithm of f(k) to make differentiation easier. Let’s define:ln(f(k)) = ln(k²) - k * ln(1.001) = 2 ln(k) - k * ln(1.001)Taking the derivative with respect to k (treating k as continuous):d/dk [ln(f(k))] = (2 / k) - ln(1.001)Set this derivative equal to zero for critical points:(2 / k) - ln(1.001) = 0Solving for k:2 / k = ln(1.001)k = 2 / ln(1.001)Now, compute the value of ln(1.001). Let me recall that ln(1 + x) ≈ x - x²/2 + x³/3 - ... for small x. Since 0.001 is small, the approximation ln(1.001) ≈ 0.001 - 0.0000005 + ... ≈ 0.0009995. But maybe I should just calculate it more accurately.Alternatively, use a calculator. But since I might not have a calculator here, let me remember that ln(1.001) ≈ 0.0009995003. So approximately 0.0009995.Therefore, k ≈ 2 / 0.0009995 ≈ 2000.5. So approximately 2000.5. Since k has to be a natural number, the maximum is likely at k = 2000 or 2001. But let's verify.But wait, maybe the approximation is slightly off. Let me check more precisely. Let me compute ln(1.001). Using the Taylor series expansion:ln(1+x) = x - x²/2 + x³/3 - x⁴/4 + ... for |x| < 1.Here, x = 0.001, so:ln(1.001) = 0.001 - (0.001)^2 / 2 + (0.001)^3 / 3 - (0.001)^4 / 4 + ... Calculating term by term:First term: 0.001Second term: -0.0000005Third term: +0.000000000333...Fourth term: -0.00000000000025...So, adding up:0.001 - 0.0000005 = 0.0009995Then plus 0.000000000333 gives 0.000999500333...Subtracting the next term is negligible. So, approximately ln(1.001) ≈ 0.0009995.So, 2 divided by 0.0009995 is approximately 2 / 0.0009995 ≈ 2001.0005. Wait, that’s interesting. Let me compute that more carefully.Let me write 0.0009995 as 999.5 * 10^(-6). So, 2 / (999.5 * 10^(-6)) = 2 / 999.5 * 10^6 ≈ (2 / 999.5) * 1000. Let's compute 2 / 999.5 first.999.5 is 1000 - 0.5, so 2 / 999.5 = 2 / (1000 - 0.5) = 2 / [1000(1 - 0.0005)] ≈ (2/1000)(1 + 0.0005 + (0.0005)^2 + ...) by using the approximation 1/(1 - x) ≈ 1 + x for small x.So, 2 / 999.5 ≈ 0.002 * (1 + 0.0005) ≈ 0.002 + 0.000001 = 0.002001Then multiplying by 1000 gives 0.002001 * 1000 = 2.001. Therefore, 2 / 0.0009995 ≈ 2001.0005. So approximately 2001.0005. Therefore, the critical point is around 2001.0005. Therefore, the maximum occurs around k = 2001. Since k must be an integer, we check k = 2001 and k = 2000.Wait, but the critical point is slightly above 2001, so maybe k=2001 is the maximum? Hmm, but since the function is defined at integers, the maximum could be at either 2001 or 2002? Wait, let's think again.Wait, if the critical point is at approximately 2001.0005, which is just a tiny bit above 2001, then the function is increasing before that and decreasing after. But since we can only have integer k, then the maximum is either at 2001 or 2002? Wait, hold on. Let me make sure.Wait, when the critical point is at k ≈ 2001.0005, that means that at k=2001, the function is still increasing (since critical point is just a bit above 2001), so the next integer after 2001.0005 is 2002. Wait, but that can't be. Wait, perhaps my calculation is wrong.Wait, maybe the exact critical point is k ≈ 2001.0005, which is just slightly above 2001. So, when k is 2001, the function is still increasing (since critical point is after 2001), so the maximum would be at k=2001 or k=2002? Hmm, maybe not. Wait, let's think in terms of the continuous function. The maximum occurs at k≈2001.0005. So, the continuous function starts decreasing after that point. Therefore, the integer closest to that point is 2001, so the maximum should be at k=2001. However, sometimes the maximum can be at the previous or next integer depending on the exact values. So, to confirm, we need to compute f(2000), f(2001), f(2002) and compare them.Alternatively, we can compute the ratio of f(k+1)/f(k) and see when it crosses 1. That is, when f(k+1)/f(k) = [(k+1)^2 / 1.001^{k+1}] / [k² / 1.001^{k}] = [(k+1)^2 / k²] / 1.001Set this ratio equal to 1 to find the turning point:[(k+1)^2 / k²] / 1.001 = 1So,(k+1)^2 / k² = 1.001Take square roots:(k+1)/k = sqrt(1.001) ≈ 1 + 0.0005Wait, sqrt(1.001) can be approximated using the Taylor series. sqrt(1 + x) ≈ 1 + x/2 - x²/8 + ... for small x. Here, x = 0.001. So,sqrt(1.001) ≈ 1 + 0.0005 - 0.000000125 + ... ≈ 1.000499875Therefore,(k+1)/k ≈ 1.000499875Which implies:1 + 1/k ≈ 1.000499875Therefore,1/k ≈ 0.000499875So,k ≈ 1 / 0.000499875 ≈ 2000.5So, this suggests that when k ≈ 2000.5, the ratio f(k+1)/f(k) is 1. So, when k < 2000.5, the ratio is greater than 1, meaning f(k+1) > f(k), so the function is increasing. When k > 2000.5, the ratio is less than 1, so f(k+1) < f(k), so the function is decreasing. Therefore, the maximum occurs at k=2001, since 2000.5 is between 2000 and 2001. Thus, the integer k where f(k) is maximum is 2001.But let me verify this by computing f(2000), f(2001), and f(2002).Compute f(2000) = (2000)^2 / 1.001^2000f(2001) = (2001)^2 / 1.001^2001f(2002) = (2002)^2 / 1.001^2002To compare these, let's compute the ratios f(2001)/f(2000) and f(2002)/f(2001):First, f(2001)/f(2000) = [(2001)^2 / 1.001^2001] / [2000² / 1.001^2000] = (2001/2000)² / 1.001Which is (1 + 1/2000)^2 / 1.001 ≈ (1 + 2/2000 + 1/2000²) / 1.001 ≈ (1.001 + 0.0000005) / 1.001 ≈ 1.0000005 / 1.001 ≈ 0.9990005. Wait, that can't be. Wait, perhaps my approximation is wrong.Wait, let's compute (2001/2000)^2 = (1 + 1/2000)^2 = 1 + 2/2000 + 1/(2000)^2 = 1 + 0.001 + 0.00000025 = 1.00100025Then divide by 1.001: 1.00100025 / 1.001 = 1 + (0.00000025 / 1.001) ≈ 1 + 0.00000025. So, the ratio is approximately 1.00000025, which is slightly greater than 1. Therefore, f(2001)/f(2000) ≈ 1.00000025 > 1, so f(2001) > f(2000)Similarly, compute f(2002)/f(2001):= (2002/2001)^2 / 1.001(2002/2001)^2 = (1 + 1/2001)^2 ≈ 1 + 2/2001 + 1/(2001)^2 ≈ 1 + 0.0009995 + 0.00000025 ≈ 1.00099975Divide by 1.001: 1.00099975 / 1.001 ≈ 0.99900025Wait, let's compute it more accurately:1.00099975 / 1.001 = (1.001 - 0.00000025)/1.001 = 1 - 0.00000025/1.001 ≈ 1 - 0.00000025 ≈ 0.99999975. Wait, that seems contradictory. Wait, maybe my calculation is wrong.Wait, 2002/2001 = 1 + 1/2001 ≈ 1.00049975So, (2002/2001)^2 ≈ (1.00049975)^2 ≈ 1 + 2*0.00049975 + (0.00049975)^2 ≈ 1 + 0.0009995 + 0.0000002498 ≈ 1.0009997498Then, divide by 1.001: 1.0009997498 / 1.001 = ?Since 1.001 is 1 + 0.001, so:1.0009997498 / 1.001 ≈ (1 + 0.0009997498) / (1 + 0.001) ≈ Using the approximation (1 + a)/(1 + b) ≈ 1 + a - b for small a and b.So, approximately 1 + 0.0009997498 - 0.001 = 1 - 0.0000002502 ≈ 0.9999997498Therefore, f(2002)/f(2001) ≈ 0.9999997498 < 1, so f(2002) < f(2001)Therefore, f(2001) > f(2000) and f(2001) > f(2002). Therefore, the maximum is at k=2001.But wait, according to the earlier critical point calculation, the maximum was at approximately k=2001.0005. So, which integer is closest? The critical point is just barely above 2001, so between 2001 and 2002. But since the function is decreasing after the critical point, which is at 2001.0005, then at k=2001, the function is still increasing (since critical point is after k=2001), but wait, no. Wait, in continuous terms, if the critical point is at k=2001.0005, then the function is increasing up to that point and decreasing after. So, when moving from k=2001 to k=2002 in the discrete case, since the critical point is just past 2001, the continuous function is decreasing after 2001.0005. So, the integer k=2001 is just before the critical point, and k=2002 is after. But since between 2001 and 2002, the continuous function first increases up to 2001.0005 and then decreases. Therefore, the maximum of the continuous function is at 2001.0005. Therefore, the integer k=2001 is just before the peak, and k=2002 is after. Wait, but how does this translate to the discrete function?Wait, actually, when considering the ratio test, we saw that f(2001)/f(2000) ≈ 1.00000025 > 1, so f(2001) > f(2000), and f(2002)/f(2001) ≈ 0.99999975 < 1, so f(2002) < f(2001). Therefore, f(k) increases from 2000 to 2001, then decreases from 2001 to 2002. Therefore, the maximum is at k=2001.Therefore, the answer should be k=2001.But let me check with k=2001.0005. Wait, perhaps to ensure that k=2001 is indeed the maximum, even though the critical point is very close to 2001. Let me compute the actual values numerically. But since 1.001^k is a huge number, it's difficult to compute directly. Maybe we can compute the ratio f(k+1)/f(k) for k=2000 and k=2001.As above, we found:f(2001)/f(2000) ≈ 1.00000025 > 1, so f(2001) is larger.f(2002)/f(2001) ≈ 0.99999975 < 1, so f(2002) is smaller.Therefore, the maximum is at k=2001. Hence, the natural number k where the expression attains its maximum is 2001.Wait, but let me check with k=2001.0005. If we model f(k) as a continuous function, then f(k) has a maximum at k≈2001.0005. Therefore, the integer closest to that point is 2001, but since the function is still increasing at k=2001, maybe the maximum is at k=2001? Wait, no. Wait, in the continuous case, after k=2001.0005, the function starts decreasing. So, if the critical point is between 2001 and 2002, then the discrete maximum could be at either 2001 or 2002. But according to the ratio test, f(2002) < f(2001), so even though the critical point is very close to 2001, the discrete maximum is still at 2001.Alternatively, maybe there is a miscalculation here. Let me verify the critical point again.We had:The critical point occurs where derivative of ln(f(k)) = 0, which is 2/k - ln(1.001) = 0Therefore, k = 2 / ln(1.001)Using ln(1.001) ≈ 0.0009995003Therefore, k ≈ 2 / 0.0009995003 ≈ 2001.0005So, exactly, k ≈ 2001.0005. So, the continuous function attains maximum at approximately 2001.0005, which is just slightly above 2001. Since the next integer is 2002, but when we check the ratio f(2002)/f(2001), it's less than 1, so f(2002) is less than f(2001). Therefore, the maximum in the discrete case is at k=2001.Therefore, the answer is k=2001.But to make sure, let's compute f(2001) / f(2000) and f(2002)/f(2001) with more precise approximations.Alternatively, take the natural logarithm of these ratios:ln(f(k+1)/f(k)) = ln[( (k+1)^2 / 1.001^{k+1} ) / (k² / 1.001^{k})] = ln[( (k+1)^2 / k² ) / 1.001] = 2 ln((k+1)/k) - ln(1.001)For k=2000:ln(f(2001)/f(2000)) = 2 ln(2001/2000) - ln(1.001)Compute ln(2001/2000) = ln(1 + 1/2000) ≈ 1/2000 - 1/(2*(2000)^2) + 1/(3*(2000)^3) - ...Approximate to first few terms:≈ 0.0005 - 0.000000125 + 0.000000000083... ≈ ~0.000499875Multiply by 2: ≈ 0.00099975Subtract ln(1.001) ≈ 0.0009995003So, ln(f(2001)/f(2000)) ≈ 0.00099975 - 0.0009995003 ≈ 0.0000002497Exponentiating: e^{0.0000002497} ≈ 1 + 0.0000002497 ≈ 1.0000002497Thus, f(2001)/f(2000) ≈ 1.00000025, which is slightly above 1, so f(2001) is larger.For k=2001:ln(f(2002)/f(2001)) = 2 ln(2002/2001) - ln(1.001)Similarly, ln(2002/2001) = ln(1 + 1/2001) ≈ 1/2001 - 1/(2*(2001)^2) + ...≈ 0.00049975 - 0.0000001249 ≈ ~0.000499625Multiply by 2: ≈ 0.00099925Subtract ln(1.001) ≈ 0.0009995003So, ln(f(2002)/f(2001)) ≈ 0.00099925 - 0.0009995003 ≈ -0.0000002503Exponentiating: e^{-0.0000002503} ≈ 1 - 0.0000002503 ≈ 0.9999997497Thus, f(2002)/f(2001) ≈ 0.99999975, which is slightly less than 1, so f(2002) is smaller.Therefore, the maximum is indeed at k=2001.But just to check with another approach, let's consider the difference f(k+1) - f(k). But since these are positive terms, the ratio test is more straightforward.Alternatively, we can set up the inequality f(k+1) > f(k) and solve for k.That is, when is (k+1)^2 / 1.001^{k+1} > k² / 1.001^k ?Multiply both sides by 1.001^{k+1} / k²:(k+1)^2 / k² > 1.001Take square roots:(k+1)/k > sqrt(1.001)Which is:1 + 1/k > sqrt(1.001)Therefore,1/k > sqrt(1.001) - 1Compute sqrt(1.001) - 1 ≈ 1.000499875 - 1 ≈ 0.000499875Thus,1/k > 0.000499875 => k < 1 / 0.000499875 ≈ 2000.5Therefore, when k < 2000.5, the next term is larger; when k > 2000.5, the next term is smaller. Therefore, the maximum occurs at k=2001 (since 2000.5 is between 2000 and 2001). Hence, the maximum is at k=2001.Therefore, after all these steps and checks, the natural number k where the expression attains its maximum is 2001.Final AnswerThe natural number ( k ) at which the expression ( frac{k^{2}}{1.001^{k}} ) attains its maximum value is (boxed{2001}).</think>To find the natural number ( k ) where the expression ( frac{k^2}{1.001^k} ) attains its maximum value, we start by analyzing the function ( f(k) = frac{k^2}{1.001^k} ). We take the natural logarithm of ( f(k) ) to simplify differentiation:[ln(f(k)) = 2 ln(k) - k ln(1.001)]Taking the derivative with respect to ( k ) and setting it to zero for critical points:[frac{d}{dk} ln(f(k)) = frac{2}{k} - ln(1.001) = 0]Solving for ( k ):[frac{2}{k} = ln(1.001) implies k = frac{2}{ln(1.001)}]Using the approximation ( ln(1.001) approx 0.0009995 ):[k approx frac{2}{0.0009995} approx 2001.0005]Since ( k ) must be a natural number, we check the integers around 2001.0005. We use the ratio test to determine whether ( f(k+1) ) is increasing or decreasing around ( k = 2001 ):For ( k = 2000 ):[frac{f(2001)}{f(2000)} = frac{(2001)^2 / 1.001^{2001}}{(2000)^2 / 1.001^{2000}} = left( frac{2001}{2000} right)^2 frac{1}{1.001} approx 1.00000025 > 1]For ( k = 2001 ):[frac{f(2002)}{f(2001)} = frac{(2002)^2 / 1.001^{2002}}{(2001)^2 / 1.001^{2001}} = left( frac{2002}{2001} right)^2 frac{1}{1.001} approx 0.99999975 < 1]Since ( f(2001) > f(2000) ) and ( f(2001) > f(2002) ), the maximum value occurs at ( k = 2001 ).Thus, the natural number ( k ) at which the expression ( frac{k^2}{1.001^k} ) attains its maximum value is (boxed{2001}).]

🔑:We are looking for the value of k for which the term of the sequence is maximum:frac{1^{2}}{1,001}, frac{2^{2}}{1,001^{2}}, frac{3^{2}}{1,001^{3}}, ldots, frac{k^{2}}{1,001^{k}}, ldotsConsider the following equivalent inequalities:frac{(k+1)^{2}}{1,001^{k+1}}1000 Leftrightarrow k>2000Thus, the sequence decreases strictly for k geq 2001 and increases strictly for k leq 2000. Therefore, the largest term of the sequence corresponds to k=2001.## List 9#

🧠:[<think>Alright, let's tackle this problem step by step. So, we have a parabola defined by the equation ( y^2 = 4x ), and F is its focus. We need to find the equation of the trajectory of a moving point M such that the vector from F to P (which is a point on the parabola) is twice the vector from F to M. The relationship given is ( overrightarrow{FP} = 2overrightarrow{FM} ).First, let me recall some basics about parabolas. The standard form of a parabola that opens to the right is ( y^2 = 4ax ), where the focus is at (a, 0). Comparing this with the given equation ( y^2 = 4x ), it looks like 4a = 4, so a = 1. That means the focus F is at (1, 0).Now, P is a moving point on the parabola. Let's denote the coordinates of P as (x₁, y₁). Since P lies on the parabola ( y^2 = 4x ), we have ( y₁^2 = 4x₁ ). Next, we need to relate the coordinates of M to those of P using the given vector equation ( overrightarrow{FP} = 2overrightarrow{FM} ). Let's denote the coordinates of M as (h, k). First, let's express the vectors FP and FM in terms of coordinates. The vector FP goes from F(1, 0) to P(x₁, y₁), so its components are ( (x₁ - 1, y₁ - 0) = (x₁ - 1, y₁) ). Similarly, the vector FM goes from F(1, 0) to M(h, k), so its components are ( (h - 1, k - 0) = (h - 1, k) ).According to the given relationship, ( overrightarrow{FP} = 2overrightarrow{FM} ). This means that each component of FP is twice the corresponding component of FM. Therefore:1. ( x₁ - 1 = 2(h - 1) )2. ( y₁ = 2k )These equations relate the coordinates of P (x₁, y₁) to those of M (h, k). Our goal is to eliminate x₁ and y₁ and find an equation in terms of h and k, which will give us the trajectory of M.Let me solve these equations for x₁ and y₁:From equation 1:( x₁ = 2(h - 1) + 1 = 2h - 2 + 1 = 2h - 1 )From equation 2:( y₁ = 2k )Since P(x₁, y₁) lies on the parabola ( y₁^2 = 4x₁ ), substitute x₁ and y₁ with the expressions in terms of h and k:( (2k)^2 = 4(2h - 1) )Simplify the left side:( 4k^2 = 4(2h - 1) )Divide both sides by 4 to simplify:( k^2 = 2h - 1 )So, replacing h and k with x and y for the equation of the trajectory of M, we get:( y^2 = 2x - 1 )Wait, let me check if this makes sense. Let's see. If M is such that FM is half of FP, then as P moves along the parabola, M should trace a path that's somehow related but scaled down. Since the original parabola is ( y^2 = 4x ), scaling down the distance from the focus by a factor of 2 might result in another parabola.But let's verify if the equation we got is correct. Let's suppose when P is at the vertex of the original parabola, which is (0,0). Then FP would be the vector from F(1,0) to P(0,0), which is (-1, 0). According to the vector equation, FM should be half of FP, so FM would be (-0.5, 0). Therefore, M would be at F(1,0) + (-0.5, 0) = (0.5, 0). Plugging x = 0.5 into our derived equation: y^2 = 2*(0.5) - 1 = 1 - 1 = 0, which gives y = 0. That matches, so (0.5, 0) is indeed on the trajectory.Another test point: Let's take a point P on the parabola, say (1, 2). Because for x=1, y^2=4*1=4, so y=2 or y=-2. Let's take (1,2). Then vector FP is (1 -1, 2 -0) = (0,2). So FM should be half of that, which is (0,1). Therefore, M is F(1,0) + (0,1) = (1,1). Plugging into our equation: y^2 = 2x -1 => 1^2 = 2*1 -1 => 1 = 2 -1 => 1=1. Correct.Similarly, if we take P as (1, -2), then FP is (0, -2), FM is (0, -1), so M is (1, -1). Plugging into the equation: (-1)^2 = 2*1 -1 => 1 = 2 -1 => 1=1. Correct.Another test: Let's take a point P further out. Let x=4, then y^2=16, so y=4 or y=-4. Let's take (4,4). Then FP vector is (4-1, 4-0)=(3,4). Then FM is half of that, which is (1.5,2). Therefore, M is F(1,0) + (1.5,2) = (2.5,2). Plugging into the equation: 2^2 = 2*2.5 -1 => 4 =5 -1 => 4=4. Correct.So these test points check out. Therefore, the equation ( y^2 = 2x -1 ) seems to be correct.But let me think again if there's a different interpretation of the vector equation. The problem says ( overrightarrow{FP} = 2overrightarrow{FM} ). Vector equations can sometimes be tricky depending on the direction. But in this case, since both vectors are from F to P and F to M, the direction is away from F. So scaling by 2 would mean that the vector FM is half the vector FP. So M is halfway from F to P? Wait, no. Wait, if ( overrightarrow{FP} = 2overrightarrow{FM} ), then solving for ( overrightarrow{FM} ), we get ( overrightarrow{FM} = frac{1}{2}overrightarrow{FP} ). So that would mean M is the point such that starting at F, moving half the vector FP gets you to M. So M is halfway between F and P? Wait, no. Because if you move half the vector FP from F, you land at M. So yes, M divides the segment FP in the ratio 1:1? Wait, if FM is half of FP, then M is halfway between F and P? Wait, the length of FM is half of FP, but direction is same. So yes, M is the midpoint of F and P? Wait, if FP is twice FM, then FM is half FP. So yes, M is the midpoint between F and P.Wait, but midpoint between F and P would be average of coordinates. So if F is (1,0) and P is (x₁,y₁), then midpoint M would be ( left( frac{1 + x₁}{2}, frac{0 + y₁}{2} right) ). Let's see if that's the same as our previous result.From our previous equations, x₁ = 2h -1 and y₁=2k. So h = (x₁ +1)/2 and k = y₁/2. Which is the same as M being the midpoint between F(1,0) and P(x₁,y₁). Therefore, the trajectory of M is the set of midpoints between F and points P on the original parabola. So, the locus is the midpoint of FP as P moves on the parabola. Therefore, the equation should be a parabola as well.Alternatively, since the original parabola is y²=4x, and we found the equation y²=2x -1, which is another parabola. Let's check if this is consistent.Let me rewrite y²=2x -1 as y² = 2(x - 0.5). This is a parabola that opens to the right with vertex at (0.5, 0). The standard form is y² = 4a(x - h), so here 4a = 2, so a = 0.5. So the focus of this new parabola would be at (h + a, k) = (0.5 + 0.5, 0) = (1, 0), which is the original focus F. Wait, interesting. So the new parabola has the same focus as the original one. But how?Wait, actually, the vertex is at (0.5, 0) and the focus is at (0.5 + 0.5, 0) = (1,0). So the focus of the new parabola is F. That's interesting. So the locus of midpoints between F and points on the original parabola is another parabola with the same focus but a different vertex. That seems plausible.Alternatively, maybe this is a general property: the locus of midpoints between a fixed point (focus) and points on a parabola is another parabola. That might be the case here.But in any case, through the coordinate transformations and substitution, we arrived at the equation y² = 2x -1. Let me check once more with another point.Take P as (2, 2√2) on the original parabola, since y²=8=4*2. Then FP vector is (2 -1, 2√2 -0) = (1, 2√2). Then FM is half of that, which is (0.5, √2). So M is F(1,0) + (0.5, √2) = (1.5, √2). Plugging into the equation y² = 2x -1: (√2)^2 = 2*(1.5) -1 => 2 = 3 -1 => 2=2. Correct.Another test with a non-integer point. Let’s say P is (0.25, 1). Because for x=0.25, y²=1, so y=1 or y=-1. Let’s take (0.25,1). Then FP vector is (0.25 -1, 1 -0) = (-0.75,1). Then FM is half of that: (-0.375,0.5). So M is F(1,0) + (-0.375,0.5) = (0.625,0.5). Now plug into y² = 2x -1: (0.5)^2 = 0.25; 2*0.625 -1 = 1.25 -1 = 0.25. So 0.25 = 0.25. Correct.Alright, all test cases work. So, unless I made a mistake in interpreting the vector equation, which I don't think I did, the equation of the trajectory of M is ( y^2 = 2x -1 ).But let me just confirm the vector equation again. The problem states ( overrightarrow{FP} = 2overrightarrow{FM} ). So vector FP equals twice vector FM. So, in terms of points, this would mean that starting from F, moving along FM twice gets you to P. So FM is half the vector FP. So M is such that FM = 0.5 FP, so M divides FP internally in the ratio 1:1. Wait, but if FP is twice FM, then FM is half of FP. So starting at F, moving half of FP gets you to M, and moving the full FP gets you to P. Therefore, M is halfway between F and P. So yes, M is the midpoint of F and P. Hence, the locus is midpoints of F and points on the original parabola. Therefore, the equation is correct as ( y^2 = 2x -1 ).Another way to approach this problem could be using parametric equations. Let me try that method to verify.The parametric equations for the parabola ( y^2 = 4x ) can be written as ( x = t^2 ), ( y = 2t ), where t is a parameter. So any point P on the parabola can be represented as (t², 2t). The focus F is at (1,0). Then, we need to find the coordinates of M such that ( overrightarrow{FP} = 2overrightarrow{FM} ).First, compute vector FP. Since P is (t², 2t) and F is (1,0), FP is (t² -1, 2t -0) = (t² -1, 2t).Then, vector FM should be half of FP, so FM = ( (t² -1)/2 , t ). Therefore, the coordinates of M are F + FM = (1 + (t² -1)/2 , 0 + t ) = ( (2 + t² -1)/2 , t ) = ( (t² +1)/2 , t ).So M has coordinates ( (t² +1)/2 , t ). Let's denote h = (t² +1)/2 and k = t. Then, we can express t in terms of k: t = k. Substitute into h: h = (k² +1)/2. Multiply both sides by 2: 2h = k² +1. Rearranged: k² = 2h -1. Therefore, replacing h and k with x and y: y² = 2x -1. Same result! So this parametric approach confirms our previous answer. Therefore, the equation of the trajectory of M is indeed ( y^2 = 2x -1 ).I think that's thorough enough. Tested with multiple points, checked via two different methods (coordinate substitution and parametric equations), and both led to the same conclusion. So I'm confident the answer is correct.Final AnswerThe equation of the trajectory of ( M ) is boxed{y^2 = 2x - 1}.</think>Given the parabola ( y^2 = 4x ) with focus ( F ) at ( (1, 0) ), we need to find the equation of the trajectory of a moving point ( M ) such that ( overrightarrow{FP} = 2overrightarrow{FM} ), where ( P ) is a point on the parabola.1. Identify the coordinates of ( P ): - Let ( P ) be ((x_1, y_1)) on the parabola, so ( y_1^2 = 4x_1 ).2. Express the vectors ( overrightarrow{FP} ) and ( overrightarrow{FM} ): - ( overrightarrow{FP} = (x_1 - 1, y_1) ) - Let ( M ) be ((h, k)), then ( overrightarrow{FM} = (h - 1, k) )3. Relate the vectors using the given equation: - ( overrightarrow{FP} = 2overrightarrow{FM} ) implies: - ( x_1 - 1 = 2(h - 1) ) - ( y_1 = 2k )4. Solve for ( x_1 ) and ( y_1 ): - From ( x_1 - 1 = 2(h - 1) ), we get ( x_1 = 2h - 1 ) - From ( y_1 = 2k ), we get ( y_1 = 2k )5. Substitute ( x_1 ) and ( y_1 ) into the parabola equation: - Substitute ( x_1 = 2h - 1 ) and ( y_1 = 2k ) into ( y_1^2 = 4x_1 ): - ( (2k)^2 = 4(2h - 1) ) - Simplify to get ( 4k^2 = 8h - 4 ) - Divide by 4: ( k^2 = 2h - 1 )6. Replace ( h ) and ( k ) with ( x ) and ( y ): - The equation becomes ( y^2 = 2x - 1 )Verification using parametric equations confirms the result. The parametric coordinates of ( P ) on the parabola are ((t^2, 2t)). The coordinates of ( M ) are found to be (left( frac{t^2 + 1}{2}, t right)), leading to the same equation ( y^2 = 2x - 1 ). Final AnswerThe equation of the trajectory of ( M ) is boxed{y^2 = 2x - 1}.,<think>Okay, so I have this problem here: Given that F is the focus of the parabola y² = 4x, and P is a moving point on the parabola. If the moving point M satisfies the vector equation FP = 2FM, then I need to find the equation of the trajectory of M. Hmm, let's see. First, let's recall some basics about parabolas. The standard form of a parabola that opens to the right is y² = 4ax, where 'a' is the distance from the vertex to the focus. Comparing that to the given equation y² = 4x, it looks like 4a = 4, so a = 1. Therefore, the focus F of this parabola should be at (a, 0) which is (1, 0). Got that down. Now, point P is moving on the parabola, so its coordinates must satisfy y² = 4x. Let's denote the coordinates of P as (x₁, y₁). Then, according to the parabola equation, we have y₁² = 4x₁. Now, the vector equation given is FP = 2FM. Let me parse that. FP is the vector from F to P, and FM is the vector from F to M. The equation says that the vector FP is twice the vector FM. Let me try to translate that into coordinates. Let's assume M has coordinates (h, k). Then, the vector FP would be P - F, which is (x₁ - 1, y₁ - 0) = (x₁ - 1, y₁). Similarly, vector FM would be M - F, which is (h - 1, k - 0) = (h - 1, k). According to the given condition, FP = 2FM. So, each component of FP must be twice the corresponding component of FM. Therefore:x₁ - 1 = 2(h - 1)y₁ = 2kSo, we can write these as equations:1) x₁ = 2(h - 1) + 1 = 2h - 2 + 1 = 2h - 12) y₁ = 2kBut since point P (x₁, y₁) lies on the parabola y² = 4x, we can substitute x₁ and y₁ in terms of h and k into this equation. So substituting, we have:y₁² = 4x₁(2k)² = 4(2h - 1)Which simplifies to:4k² = 8h - 4Divide both sides by 4:k² = 2h - 1Therefore, replacing h and k with x and y (since h and k are the coordinates of M), the equation becomes:y² = 2x - 1Or rearranged:y² = 2x - 1So that should be the equation of the trajectory of point M.Wait, but let me check if I did everything correctly. Let's verify the steps again.We have FP = 2FM. FP is the vector from F to P: coordinates of P minus coordinates of F. Since F is (1,0), that gives (x₁ - 1, y₁). FM is the vector from F to M: (h - 1, k). Setting FP = 2FM gives:x₁ - 1 = 2(h - 1)y₁ = 2kSo solving for x₁ and y₁:x₁ = 2(h - 1) + 1 = 2h - 2 + 1 = 2h - 1y₁ = 2kSince P is on the parabola, substitute into y² = 4x:(2k)^2 = 4*(2h -1)Which is 4k² = 8h -4Divide by 4: k² = 2h -1So replacing h and k with x and y: y² = 2x -1. That seems right. But let me check with a sample point. Suppose M is at (1, 0). Then according to the equation y² = 2x -1, plugging in x=1, y=0: 0 = 2*1 -1 =1, which is not true. So that point is not on the trajectory. Wait, but if M is at (1,0), then according to the vector equation, FP = 2FM. FM would be (1-1, 0-0) = (0,0), so FP must be (0,0). That would mean P is at F. But F is (1,0), which is on the parabola? Wait, plugging x=1 into y²=4x gives y²=4, so y=±2. So (1,0) is not on the parabola. So P cannot be at F. Therefore, M cannot be at (1,0). So according to our equation y²=2x -1, when x=1, y²=1, so y=±1. Let's check if that works.Take M at (1,1). Then according to the vector equation, FM is (1-1, 1-0)=(0,1). Then FP should be 2*(0,1)=(0,2). So FP is (0,2), which means P is F + FP = (1,0) + (0,2) = (1,2). Is (1,2) on the parabola? y²=4x: 4=4*1=4. Yes, so (1,2) is on the parabola. Then M=(1,1) should be on the trajectory, and according to the equation y²=2x -1, 1²=2*1 -1=1, which works. Similarly, if M is (1, -1), same thing. So that seems okay. Another test: Let's take M=(0.5, 0). Then according to the equation y²=2x -1, 0=1 -1=0, which works. Then FM vector is (0.5 -1, 0 -0)=(-0.5,0). Then FP=2*FM=(-1,0). So P is F + FP=(1,0) + (-1,0)=(0,0). Is (0,0) on the parabola? Yes, it's the vertex. So M=(0.5,0) should be on the trajectory, and according to the equation, y²=2*(0.5)-1=1 -1=0, which holds. So that works. Another test: Let’s pick a point on the trajectory. Suppose x=2, then y²=2*2 -1=3, so y=±√3. Let’s take M=(2, √3). Then FM vector is (2-1, √3 -0)=(1, √3). Then FP=2*FM=(2, 2√3). So P is F + FP=(1,0)+(2, 2√3)=(3, 2√3). Check if P is on the parabola: y²=(2√3)^2=12, and 4x=4*3=12. Yep, that works. So it seems the equation y²=2x -1 is correct. Therefore, the answer should be y²=2x -1. But let me check if perhaps there was a miscalculation in the steps. Let me rederive quickly:Given FP = 2FM. So, coordinates:FP = P - F = (x₁ -1, y₁)FM = M - F = (h -1, k)Thus, (x₁ -1, y₁) = 2(h -1, k) => x₁ -1 = 2(h -1) and y₁ = 2k. Then x₁ = 2h -2 +1 = 2h -1, y₁=2k. Substitute into y₁²=4x₁:(2k)^2 =4*(2h -1)4k²=8h -4Divide by 4: k²=2h -1So equation is y²=2x -1, which is the same as before. Therefore, I think that's correct. The trajectory of M is the parabola y²=2x -1.Final AnswerThe equation of the trajectory of ( M ) is boxed{y^2 = 2x - 1}.</think>Given that ( F ) is the focus of the parabola ( y^2 = 4x ), we know the focus ( F ) is at ( (1, 0) ). A point ( P ) on the parabola can be represented as ( (x_1, y_1) ) where ( y_1^2 = 4x_1 ). The vector equation ( overrightarrow{FP} = 2overrightarrow{FM} ) implies the coordinates of ( M ) (denoted as ( (h, k) )) must satisfy the relationship between the vectors from ( F ) to ( P ) and from ( F ) to ( M ).1. The vector ( overrightarrow{FP} ) is ( (x_1 - 1, y_1) ).2. The vector ( overrightarrow{FM} ) is ( (h - 1, k) ).Given ( overrightarrow{FP} = 2overrightarrow{FM} ), we equate the components:- ( x_1 - 1 = 2(h - 1) )- ( y_1 = 2k )Solving for ( x_1 ) and ( y_1 ):- ( x_1 = 2h - 1 )- ( y_1 = 2k )Since ( P ) lies on the parabola ( y^2 = 4x ), substituting ( x_1 ) and ( y_1 ):- ( (2k)^2 = 4(2h - 1) )- Simplifying, ( 4k^2 = 8h - 4 )- Dividing by 4, ( k^2 = 2h - 1 )Replacing ( h ) and ( k ) with ( x ) and ( y ), the equation of the trajectory of ( M ) is:[ y^2 = 2x - 1 ]Thus, the equation of the trajectory of ( M ) is (boxed{y^2 = 2x - 1}).]

🔑:Let the coordinates of M be (x, y), and the coordinates of P be left(frac{1}{4}t^2, tright).Since in the parabola y^2=4x, we have 2p=4, it can be derived that frac{p}{2}=1, thus the focus of the parabola is F(1,0).From this, we can get overrightarrow{FP} = left(frac{1}{4}t^2-1, tright), overrightarrow{FM} = (x-1, y).Since the moving point M satisfies overrightarrow{FP} = 2overrightarrow{FM}, we have left(frac{1}{4}t^2-1, tright) = 2(x-1, y),which gives us the system of equations begin{cases} frac{1}{4}t^2-1=2x-2 t=2y end{cases}. Eliminating the parameter t, we get y^2=2x-1, which is the equation of the trajectory of M.Therefore, the answer is: y^2=2x-1.According to the problem, the focus of the parabola is F(1,0). Let M(x, y) and Pleft(frac{1}{4}t^2, tright). We can get the vectors overrightarrow{FP} = left(frac{1}{4}t^2-1, tright), overrightarrow{FM} = (x-1, y). By setting overrightarrow{FP} = 2overrightarrow{FM}, we establish a system of equations involving x, y, and t. Eliminating the parameter t, we can obtain the trajectory equation of the moving point M.This problem involves the midpoint of the focal radius PF of the moving point P on the parabola, asking for the trajectory equation of point M. It focuses on the coordinate operations of vectors, the standard equation of a parabola, and simple geometric properties, making it a medium-level question.Thus, the final answer is boxed{y^2=2x-1}.

🧠:[<think>Okay, let me try to figure out how to find the minimum value of this function f(x). The function is given by:f(x) = [ (x + 1/x)^4 - (x^4 + 1/x^4) ] / [ (x + 1/x)^3 - (x^3 + 1/x^3) ]And we need to find its minimum value for x > 0. Hmm, that looks a bit complicated, but maybe if I simplify the numerator and the denominator separately, it will become manageable.First, let me recall that expanding expressions like (x + 1/x)^n can be done using the binomial theorem. But expanding these might get messy. Let me see if there's a pattern or a substitution that can make this easier.Let me denote t = x + 1/x. Since x > 0, by AM-GM inequality, t ≥ 2. Because x + 1/x ≥ 2√(x*(1/x)) = 2. So t is at least 2. Maybe expressing the numerator and denominator in terms of t will help?Let's try that. First, compute the numerator: (x + 1/x)^4 - (x^4 + 1/x^4). Let's expand (x + 1/x)^4:Using the binomial theorem:(x + 1/x)^4 = x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4So subtracting (x^4 + 1/x^4) gives:[x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4] - [x^4 + 1/x^4] = 4x^2 + 6 + 4/x^2So the numerator simplifies to 4x^2 + 6 + 4/x^2. Hmm, that's 4(x^2 + 1/x^2) + 6.Similarly, let's compute the denominator: (x + 1/x)^3 - (x^3 + 1/x^3). Let's expand (x + 1/x)^3:Again, using the binomial theorem:(x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3Subtracting (x^3 + 1/x^3) gives:[x^3 + 3x + 3/x + 1/x^3] - [x^3 + 1/x^3] = 3x + 3/xSo the denominator simplifies to 3x + 3/x = 3(x + 1/x).Therefore, the function f(x) simplifies to:[4(x^2 + 1/x^2) + 6] / [3(x + 1/x)]That's a bit simpler. Let me write that as:f(x) = [4(x^2 + 1/x^2) + 6] / [3(x + 1/x)]Now, since t = x + 1/x, and we know t ≥ 2, maybe express x^2 + 1/x^2 in terms of t. Because x^2 + 1/x^2 is related to t squared.Yes, indeed:(x + 1/x)^2 = x^2 + 2 + 1/x^2, so x^2 + 1/x^2 = t^2 - 2.Therefore, substituting into the numerator:4(t^2 - 2) + 6 = 4t^2 - 8 + 6 = 4t^2 - 2.So the numerator becomes 4t^2 - 2, and the denominator is 3t.Therefore, f(x) can be written in terms of t as:f(t) = (4t^2 - 2)/(3t) = (4t^2)/(3t) - 2/(3t) = (4/3)t - (2)/(3t)So now, the problem reduces to minimizing f(t) = (4/3)t - (2)/(3t) where t ≥ 2.That's much simpler! So now, we just need to find the minimum of f(t) for t ≥ 2.Let me compute the derivative of f(t) with respect to t to find critical points.First, f(t) = (4/3)t - (2)/(3t)Compute f'(t):f'(t) = (4/3) + (2)/(3t^2)Wait, derivative of (4/3)t is 4/3, and derivative of -(2)/(3t) is + (2)/(3t^2). So f'(t) = 4/3 + 2/(3t^2).Hmm, but wait, if that's the derivative, then for t > 0, f'(t) is always positive because 4/3 is positive and 2/(3t^2) is also positive. So that would mean that the function f(t) is increasing for all t > 0.But since t ≥ 2, then the minimum value would occur at t = 2.Wait, let me double-check the derivative calculation. Let me rederive it step by step.f(t) = (4/3)t - (2)/(3t)Derivative of the first term: d/dt [(4/3)t] = 4/3Derivative of the second term: d/dt [ -2/(3t) ] = -2/3 * d/dt [1/t] = -2/3 * (-1/t^2) = 2/(3t^2)So yes, f'(t) = 4/3 + 2/(3t^2), which is always positive for t > 0. Therefore, the function f(t) is indeed strictly increasing on t > 0. Hence, on the interval t ≥ 2, the minimum occurs at the left endpoint t = 2.Therefore, the minimum value of f(t) is f(2) = (4/3)(2) - (2)/(3*2) = (8/3) - (1/3) = 7/3 ≈ 2.333...But wait, let's compute that precisely:f(2) = (4/3)*2 - (2)/(3*2) = 8/3 - 1/3 = 7/3.So the minimum value is 7/3.But let me verify this because sometimes when we substitute variables, we might have made a mistake. Let's check with x = 1, which gives t = 1 + 1/1 = 2. So x=1 gives t=2. Then f(x) at x=1:Compute numerator: (1 + 1)^4 - (1 + 1) = 16 - 2 = 14. Denominator: (1 + 1)^3 - (1 + 1) = 8 - 2 = 6. So 14/6 = 7/3. That's correct.But is there a possibility that for some x > 0, t could be less than 2? No, because by AM-GM inequality, t = x + 1/x ≥ 2, equality when x=1.Therefore, t can't be less than 2, so the minimal t is 2, and since f(t) is increasing, the minimal value is 7/3.Wait, but just to be thorough, let's check another value. For example, take x = 2. Then t = 2 + 1/2 = 2.5. Then f(t) = (4/3)(2.5) - (2)/(3*2.5) = (10/3) - (2)/(7.5) = 10/3 - 4/15 = (50/15 - 4/15) = 46/15 ≈ 3.066..., which is larger than 7/3 ≈ 2.333. So yes, increasing.Similarly, take x approaching infinity, t = x + 1/x ≈ x, so f(t) ≈ (4/3)x - (2)/(3x) ≈ (4/3)x, which tends to infinity. So as t increases, f(t) increases. Hence, the minimal value is indeed at t=2, which is 7/3.Therefore, the answer should be 7/3, which is boxed{dfrac{7}{3}}.But let me just verify once again to make sure I didn't make a mistake in substitution or simplification.Original function:Numerator: (x + 1/x)^4 - (x^4 + 1/x^4). After expanding, we got 4x^2 + 6 + 4/x^2. Let me compute that step again.(x + 1/x)^4 = x^4 + 4x^3*(1/x) + 6x^2*(1/x)^2 + 4x*(1/x)^3 + (1/x)^4Wait, hold on, is that correct? Wait, binomial expansion coefficients for (a + b)^4 are 1,4,6,4,1.So (x + 1/x)^4 = x^4 + 4x^3*(1/x) + 6x^2*(1/x^2) + 4x*(1/x^3) + (1/x^4)Simplify each term:x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4. Yes, that's correct.Subtracting (x^4 + 1/x^4) gives 4x^2 + 6 + 4/x^2. So numerator is 4x^2 + 6 + 4/x^2. Correct.Denominator: (x + 1/x)^3 - (x^3 + 1/x^3). Let's verify.(x + 1/x)^3 = x^3 + 3x^2*(1/x) + 3x*(1/x)^2 + (1/x)^3 = x^3 + 3x + 3/x + 1/x^3. Subtract (x^3 + 1/x^3) gives 3x + 3/x. Correct.So denominator is 3x + 3/x = 3(x + 1/x). So then numerator is 4(x^2 + 1/x^2) + 6. As before, and since x^2 + 1/x^2 = (x + 1/x)^2 - 2 = t^2 - 2. Therefore numerator becomes 4(t^2 - 2) + 6 = 4t^2 - 8 + 6 = 4t^2 - 2. So f(t) = (4t^2 - 2)/(3t) = 4/3 t - 2/(3t). Correct.So derivative is 4/3 + 2/(3t^2), which is always positive. So function is increasing for t > 0, thus minimum at t = 2, which gives f(t) = 7/3.Therefore, the minimum value is indeed 7/3.Final AnswerThe minimum value of the function is boxed{dfrac{7}{3}}.</think>Given ( x > 0 ), we need to find the minimum value of the function:[f(x) = frac{left(x + frac{1}{x}right)^4 - left(x^4 + frac{1}{x^4}right)}{left(x + frac{1}{x}right)^3 - left(x^3 + frac{1}{x^3}right)}]First, we simplify the numerator and the denominator separately.Numerator:Expanding (left(x + frac{1}{x}right)^4) using the binomial theorem:[left(x + frac{1}{x}right)^4 = x^4 + 4x^2 + 6 + frac{4}{x^2} + frac{1}{x^4}]Subtracting (x^4 + frac{1}{x^4}) from this expansion:[left(x + frac{1}{x}right)^4 - left(x^4 + frac{1}{x^4}right) = 4x^2 + 6 + frac{4}{x^2}]This simplifies to:[4left(x^2 + frac{1}{x^2}right) + 6]Denominator:Expanding (left(x + frac{1}{x}right)^3) using the binomial theorem:[left(x + frac{1}{x}right)^3 = x^3 + 3x + frac{3}{x} + frac{1}{x^3}]Subtracting (x^3 + frac{1}{x^3}) from this expansion:[left(x + frac{1}{x}right)^3 - left(x^3 + frac{1}{x^3}right) = 3x + frac{3}{x}]This simplifies to:[3left(x + frac{1}{x}right)]Thus, the function ( f(x) ) simplifies to:[f(x) = frac{4left(x^2 + frac{1}{x^2}right) + 6}{3left(x + frac{1}{x}right)}]Let ( t = x + frac{1}{x} ). By the AM-GM inequality, ( t geq 2 ). We know:[x^2 + frac{1}{x^2} = t^2 - 2]Substituting this into the function:[f(t) = frac{4(t^2 - 2) + 6}{3t} = frac{4t^2 - 2}{3t} = frac{4}{3}t - frac{2}{3t}]To find the minimum value, we analyze ( f(t) ). The derivative of ( f(t) ) is:[f'(t) = frac{4}{3} + frac{2}{3t^2}]Since ( f'(t) ) is always positive for ( t > 0 ), ( f(t) ) is strictly increasing. Therefore, the minimum value occurs at the smallest ( t ), which is ( t = 2 ).Substituting ( t = 2 ) into ( f(t) ):[f(2) = frac{4}{3} cdot 2 - frac{2}{3 cdot 2} = frac{8}{3} - frac{1}{3} = frac{7}{3}]Thus, the minimum value of the function is (boxed{dfrac{7}{3}}).,<think>Okay, so I need to find the minimum value of this function f(x) for x > 0. The function is given by:f(x) = [ (x + 1/x)^4 - (x^4 + 1/x^4) ] / [ (x + 1/x)^3 - (x^3 + 1/x^3) ]Hmm, that looks a bit complicated, but maybe I can simplify it by expanding the numerator and denominator. Let me start by working on the numerator first.Let's denote y = x + 1/x. Then, (x + 1/x)^4 is y^4. Let's expand that:(y)^4 = (x + 1/x)^4. If I recall, expanding (a + b)^4 gives a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. So substituting a = x and b = 1/x:(x + 1/x)^4 = x^4 + 4x^3*(1/x) + 6x^2*(1/x)^2 + 4x*(1/x)^3 + (1/x)^4= x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4So the numerator is [ (x + 1/x)^4 - (x^4 + 1/x^4) ] which would be:[ x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4 - x^4 - 1/x^4 ] = 4x^2 + 6 + 4/x^2Simplify numerator: 4x^2 + 4/x^2 + 6. Maybe factor out 4: 4(x^2 + 1/x^2) + 6.Now, let's look at the denominator. Similarly, let's expand (x + 1/x)^3. Again, (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. So:(x + 1/x)^3 = x^3 + 3x^2*(1/x) + 3x*(1/x)^2 + (1/x)^3= x^3 + 3x + 3/x + 1/x^3Therefore, the denominator is [ (x + 1/x)^3 - (x^3 + 1/x^3) ] which is:[ x^3 + 3x + 3/x + 1/x^3 - x^3 - 1/x^3 ] = 3x + 3/xSo denominator simplifies to 3(x + 1/x).Putting it all together, f(x) becomes:Numerator: 4(x^2 + 1/x^2) + 6Denominator: 3(x + 1/x)So f(x) = [4(x^2 + 1/x^2) + 6] / [3(x + 1/x)]Hmm, maybe I can express x^2 + 1/x^2 in terms of (x + 1/x). Let's recall that:(x + 1/x)^2 = x^2 + 2 + 1/x^2 => x^2 + 1/x^2 = (x + 1/x)^2 - 2Let me denote z = x + 1/x. Then, x^2 + 1/x^2 = z^2 - 2.So substituting back into numerator and denominator:Numerator becomes 4(z^2 - 2) + 6 = 4z^2 - 8 + 6 = 4z^2 - 2Denominator is 3zTherefore, f(x) = (4z^2 - 2)/(3z) = (4z^2 - 2)/3zSimplify this expression: Let's split the fraction:= (4z^2)/(3z) - 2/(3z) = (4/3)z - (2)/(3z)So f(x) simplifies to (4/3)z - (2)/(3z), where z = x + 1/x.Therefore, f(x) = (4/3)z - (2)/(3z) = (4z^2 - 2)/(3z)Now, since z = x + 1/x and x > 0, by AM ≥ GM, x + 1/x ≥ 2. So z ≥ 2.Therefore, the problem reduces to finding the minimum of the function g(z) = (4z^2 - 2)/(3z) for z ≥ 2.Alternatively, since f(x) is expressed in terms of z, and z ≥ 2, we can consider z ≥ 2.So, let's work with g(z) = (4z^2 - 2)/(3z). Let's simplify this:g(z) = (4z^2 - 2)/(3z) = (4/3)z - (2)/(3z)So, to find the minimum of g(z) for z ≥ 2.We can take the derivative of g(z) with respect to z and set it to zero to find critical points.Compute g'(z):g'(z) = d/dz [ (4/3)z - (2)/(3z) ] = 4/3 + (2)/(3z^2)Set derivative equal to zero:4/3 + 2/(3z^2) = 0But 4/3 is positive and 2/(3z^2) is positive for all z > 0. Therefore, g'(z) is always positive. This means that g(z) is increasing for z > 0. Therefore, since z ≥ 2, the minimum of g(z) occurs at z = 2.Therefore, the minimum value of f(x) is g(2) = (4*(2)^2 - 2)/(3*2) = (16 - 2)/6 = 14/6 = 7/3 ≈ 2.333...Wait, hold on. Let me verify that. If g(z) is increasing for z ≥ 2, then indeed the minimum would be at z = 2. But let's confirm by plugging z = 2 into g(z):g(2) = (4*(4) - 2)/(3*2) = (16 - 2)/6 = 14/6 = 7/3.But let me check if this is correct, because maybe I made a mistake in differentiation.Wait, let's recast g(z):g(z) = (4/3)z - (2)/(3z)Then, derivative is g’(z) = 4/3 + (2)/(3z^2). Since both terms are positive for z > 0, the derivative is always positive, so g(z) is strictly increasing for all z > 0. Therefore, yes, since z ≥ 2, the minimum is at z = 2, so the minimum value is 7/3.But wait, let me check if z can actually attain 2. Since z = x + 1/x, z = 2 when x = 1, by AM-GM equality. So x = 1 gives z = 2. Therefore, f(1) = 7/3.Therefore, the minimum value is 7/3. Hmm, but let me verify this by plugging x = 1 into the original function to make sure.Original numerator: (1 + 1)^4 - (1 + 1) = 16 - 2 = 14Original denominator: (1 + 1)^3 - (1 + 1) = 8 - 2 = 6So f(1) = 14/6 = 7/3. That checks out.But wait, let me check another point to see if it's indeed increasing. For example, take x = 2. Then z = 2 + 1/2 = 2.5. Then g(z) = (4*(2.5)^2 - 2)/(3*2.5) = (4*6.25 - 2)/7.5 = (25 - 2)/7.5 = 23/7.5 ≈ 3.0667.Which is larger than 7/3 ≈ 2.333, so yes, increasing. If we take x approaching infinity, z = x + 1/x approaches infinity, so g(z) approaches (4/3)z, which goes to infinity. So the function indeed increases without bound as z increases, so minimum at z = 2.Therefore, the minimal value is 7/3.Wait, but just to be thorough, let me check with x = sqrt(1), which is 1, but maybe another value where z = 3. Wait, z = 3 would be x + 1/x = 3. Let's solve for x:x + 1/x = 3 => x^2 - 3x + 1 = 0 => x = [3 ± sqrt(9 - 4)]/2 = [3 ± sqrt(5)]/2. So x is (3 + sqrt(5))/2 or (3 - sqrt(5))/2. Let's take x = (3 + sqrt(5))/2 ≈ (3 + 2.236)/2 ≈ 2.618. Then compute f(x):First compute numerator: (x + 1/x)^4 - (x^4 + 1/x^4). Since z = 3, so (3)^4 - (x^4 + 1/x^4). Let's compute x^4 + 1/x^4. Since x + 1/x = 3, then x^2 + 1/x^2 = z^2 - 2 = 9 - 2 = 7. Then x^4 + 1/x^4 = (x^2 + 1/x^2)^2 - 2 = 7^2 - 2 = 49 - 2 = 47. Therefore, numerator is 81 - 47 = 34.Denominator: (3)^3 - (x^3 + 1/x^3). Let's compute x^3 + 1/x^3. Using the formula: (x + 1/x)^3 = x^3 + 1/x^3 + 3(x + 1/x). Therefore, x^3 + 1/x^3 = z^3 - 3z = 27 - 9 = 18. Therefore, denominator is 27 - 18 = 9.Thus, f(x) = 34 / 9 ≈ 3.777..., which is larger than 7/3 ≈ 2.333. So yes, when z increases, f(x) increases as well, which confirms our conclusion.Therefore, the minimum value is indeed 7/3, achieved when x = 1. So the answer should be boxed{dfrac{7}{3}}.Final Answerboxed{dfrac{7}{3}}</think>Given ( x > 0 ), we need to find the minimum value of the function:[ f(x) = frac{left(x + frac{1}{x}right)^4 - left(x^4 + frac{1}{x^4}right)}{left(x + frac{1}{x}right)^3 - left(x^3 + frac{1}{x^3}right)} ]First, we simplify the numerator and the denominator separately.Numerator:[left(x + frac{1}{x}right)^4 - left(x^4 + frac{1}{x^4}right)]Expanding (left(x + frac{1}{x}right)^4) using the binomial theorem:[left(x + frac{1}{x}right)^4 = x^4 + 4x^2 + 6 + frac{4}{x^2} + frac{1}{x^4}]Subtracting (x^4 + frac{1}{x^4}) gives:[4x^2 + 6 + frac{4}{x^2} = 4left(x^2 + frac{1}{x^2}right) + 6]Denominator:[left(x + frac{1}{x}right)^3 - left(x^3 + frac{1}{x^3}right)]Expanding (left(x + frac{1}{x}right)^3) using the binomial theorem:[left(x + frac{1}{x}right)^3 = x^3 + 3x + frac{3}{x} + frac{1}{x^3}]Subtracting (x^3 + frac{1}{x^3}) gives:[3x + frac{3}{x} = 3left(x + frac{1}{x}right)]Let ( z = x + frac{1}{x} ). Then, using the identity ( x^2 + frac{1}{x^2} = z^2 - 2 ), we rewrite the function as:[f(x) = frac{4(z^2 - 2) + 6}{3z} = frac{4z^2 - 2}{3z}]Simplifying further:[f(x) = frac{4z}{3} - frac{2}{3z}]Since ( x > 0 ), by AM-GM inequality, ( z = x + frac{1}{x} geq 2 ). We need to find the minimum value of ( f(z) = frac{4z}{3} - frac{2}{3z} ) for ( z geq 2 ).Taking the derivative of ( f(z) ):[f'(z) = frac{4}{3} + frac{2}{3z^2}]Since ( f'(z) ) is always positive for ( z > 0 ), ( f(z) ) is strictly increasing for ( z geq 2 ). Therefore, the minimum value occurs at ( z = 2 ).Substituting ( z = 2 ) into ( f(z) ):[f(2) = frac{4 cdot 2^2 - 2}{3 cdot 2} = frac{16 - 2}{6} = frac{14}{6} = frac{7}{3}]Thus, the minimum value of the function is (boxed{dfrac{7}{3}}).]

🔑:(16) frac{7}{3} Hint: t=x+frac{1}{x} Rightarrow t geqslant 2, x^{4}+frac{1}{x^{4}}=t^{4}-4 t^{2}+2, x^{3}+ frac{1}{x^{3}}=t^{3}-3 t, so f(x)=g(t)=frac{4}{3} t-frac{2}{3} t^{-1} geqslant g(2)=frac{4}{3} times 2- frac{2}{3} times frac{1}{2}=frac{7}{3}=f(1).Therefore, the minimum value of f(x) is frac{7}{3}.

🧠:[<think>Okay, let's see. The problem is about finding the last digit of a ten-digit even integer whose digits add up to 89. The options are 0, 2, 4, 6, or 8. Alright, so first, the number is even, so the last digit has to be one of those options. The sum of all ten digits is 89. Hmm, that's quite a high sum. Let me think.First, let's recall that each digit can be between 0 and 9. Since it's a ten-digit number, the first digit can't be 0. So the first digit is between 1 and 9, and the rest can be between 0 and 9. The sum of all digits is 89. Let me try to understand what this sum implies.The maximum possible sum for a ten-digit number would be if all digits are 9. That would be 9*10 = 90. But our sum is 89, which is just one less than the maximum. So, almost all digits are 9, except one digit which is 8 instead of 9. Wait, but 90 minus 1 equals 89, right? So, if we have nine 9s and one 8, that would give us 9*9 +8 = 81 +8=89. But wait, that's only ten digits. Wait, no. Wait, 9*10 is 90. So 90 minus 1 is 89, so one digit is 8 instead of 9. So exactly one digit is 8 and the rest are 9s. Hmm.But hold on, the number is even. So the last digit has to be even. So if all digits except one are 9s and one digit is 8, then the 8 could be in any position. But the last digit has to be even. So if the last digit is 8, then that's possible, but if the last digit is 9, which is odd, then the number wouldn't be even. Therefore, the 8 must be in the last digit? Wait, but maybe not. Let me check.Wait, suppose that the number has all digits as 9 except one digit, which is 8. Then, if the last digit is 9, which is odd, that would make the number odd, which contradicts the fact that it's even. Therefore, the digit that is 8 instead of 9 must be the last digit. Because otherwise, if the 8 is in some other position, then the last digit would still be 9, which is odd, making the number not even. So, the last digit must be 8. Therefore, the answer is E) 8.Wait, but hold on. Let me check if there are other possibilities. Because maybe there's more than one digit that's less than 9. For example, instead of having one digit as 8 and the rest as 9, maybe two digits are 9 minus something else. Let me calculate. If the total sum is 89, which is 1 less than 90. So maybe one digit is 8 and the rest 9, which gives 89. But could there be another combination?For example, two digits that sum to 17 instead of 18. For example, 9 and 8, but wait, 9+8=17. Wait, 9*10=90. If two digits are 8 and 9 instead of 9 and 9, then 90 -1 -1=88. That's too low. Wait, but we need 89. So, the difference is 1. So only one digit is reduced by 1. So, the only way to get 89 is to have nine 9s and one 8. Therefore, there's only one digit that is 8, and the rest are 9s. Therefore, the only way for the number to be even is if that 8 is the last digit, right?Because if that 8 is in any other digit, then the last digit would still be 9, which is odd. Therefore, the last digit must be 8. Therefore, the answer is E) 8.But let me check again. Suppose the number is composed of eight 9s, one 8, and one other digit. Wait, but 8*9 +8 + something = 89. Let's calculate. 9*9 =81, plus 8 is 89. So that's nine 9s and one 8. Wait, so that's 10 digits. So, exactly, one of the digits is 8 and the rest are 9s. Therefore, the digits are nine 9s and one 8. Since the number must be even, the last digit must be 8, because if the last digit were 9, the number would be odd. Therefore, the 8 must be in the last position. Therefore, the last digit is 8, answer E.Alternatively, is there another way to get a digit sum of 89 with different digits?For example, suppose we have eight 9s, one 9, and then one digit is... Wait, no. Wait, 89 is 9*10 -1. So, you need to reduce the total sum by 1. So you need one digit to be 8 instead of 9. That's the only way. Because if you have two digits reduced by 1, then total sum is 90-2=88, which is less than 89. So, yes, only one digit is 8, the rest 9. So, in order for the number to be even, the last digit must be 8. Hence, the answer is E) 8.But wait, maybe the first digit is 8? But then the last digit would be 9, which is odd, making the number not even. So, no, that's not possible. Therefore, the 8 must be in the last digit.Alternatively, maybe the number isn't composed of all 9s except one 8. But could there be another combination of digits that sum to 89? For example, multiple digits being 9 and others being lower. Let's check. Let's suppose that the number is not all 9s except one 8. For example, maybe two digits are 8, and the rest 9s. Then sum would be 8*2 +9*8= 16 +72=88. Not 89. If three digits are 8: 8*3 +9*7=24+63=87. Still lower. If one digit is 7 and the rest 9s: 7 +9*9=7+81=88. Still 88. If one digit is 7 and another digit is 8: 7+8 +9*8=15+72=87. Still lower. Alternatively, if we have two digits as 9 and one digit as 10. Wait, but digits can't be 10. So that's not possible. Alternatively, if one digit is 9 and another digit is 9, but that's just 9. So, no. So, in fact, it's impossible to get a digit sum of 89 unless exactly one digit is 8 and the rest are 9s. Therefore, the only way.Therefore, since the number must be even, the last digit must be 8. So, answer is E.Wait, but just to confirm. Let me take an example. Let's say the number is 9999999998. The sum of digits: 9*9 +8=81+8=89. Yes, that's correct. And the last digit is 8, which is even. So that's a valid number. Alternatively, if the 8 is in any other position, like 9999999989, then the sum is still 89, but the last digit is 9, which is odd. Therefore, that number is not even. Hence, such a number is invalid. Therefore, the 8 must be in the last digit. Hence, answer is E) 8.Therefore, I think the answer is E.Final Answerboxed{E}</think>The problem requires finding the last digit of a ten-digit even integer whose digits sum to 89. 1. The maximum possible sum of the digits of a ten-digit number is (9 times 10 = 90).2. The given sum is 89, which is 1 less than the maximum. Therefore, one digit must be 8 instead of 9, and the rest must be 9s.3. Since the number is even, the last digit must be even (0, 2, 4, 6, or 8).4. If the digit 8 were in any position other than the last, the last digit would be 9 (which is odd), making the number odd. Hence, the 8 must be the last digit to ensure the number is even.Thus, the last digit is (boxed{E}).,<think>Okay, so the problem is: An even ten-digit integer has a digit sum of 89. What is the last digit? The options are 0, 2, 4, 6, 8. Hmm. Let me think.First, let's break down the problem. We have a 10-digit number, which means it's between 1,000,000,000 and 9,999,999,999. But since it's even, the last digit has to be 0, 2, 4, 6, or 8. The sum of all its digits is 89, which is a pretty high number. So, we need to figure out which of these possible last digits would make the total digit sum 89.Wait, let me recall that the digit sum of a number is the sum of all its digits. For example, the digit sum of 123 is 1+2+3=6. So, in this case, adding up all ten digits gives 89. Since the number is even, the last digit is even, as given. The question is asking which of the options it must be.Let me consider the maximum possible digit sum for a 10-digit number. The maximum would be 9*10=90, right? Because each digit can be at most 9. So, 9*10=90. But the digit sum here is 89, which is just 1 less than the maximum. That suggests that nine of the digits are 9, and one digit is 8. Because 9*9 +8=81+8=89. So, that would be a ten-digit number where nine digits are 9 and one digit is 8. But the number has to be even. So, the last digit needs to be even.Wait, but if nine digits are 9 and one digit is 8, then the digit that is 8 could be any of the ten digits, including the last one. But since the number is even, the last digit has to be even, so if the last digit is 8, then that's okay, but if the last digit is 9, then the digit before that is 8, but that would make the number odd. Wait, no. Let's see.Actually, if the digit sum is 89, which is 90-1, that could mean nine digits are 9 and one digit is 8. But since the number has to be even, the last digit must be 8 or 0 or 2, etc. But if the digit sum is 89, which is one less than 90, and we need the last digit to be even, then the 8 has to be in the last digit. Wait, is that necessarily the case?Wait, no. For example, suppose that one of the other digits is 8, and the last digit is 9. But the last digit can't be 9 because the number is even. So, the last digit must be even. Therefore, if the last digit is 8, then the other nine digits must be 9, which would make the digit sum 9*9 +8=89, which works. Alternatively, if the last digit is another even digit, say 0, then one of the other digits would have to be 8, but then how does the digit sum play out?Wait, hold on. Let's think again. The maximum digit sum is 90. The digit sum here is 89, which is 1 less. So, it's as if one of the digits is decreased by 1 from 9 to 8. So, only one digit is 8, and the rest are 9. Since the number has to be even, the last digit must be even. Therefore, if the last digit is 8, then we have nine 9s and one 8, which works. But if the last digit is something else, like 0, then we need another digit to be 8. However, if the last digit is 0, which is even, then the other nine digits must sum to 89. But 0 doesn't contribute anything. So, the first nine digits would have to sum to 89. However, the maximum sum for the first nine digits is 9*9=81. 81 is less than 89. Wait, that's impossible. So, that's a problem.Wait, wait, so the total digit sum is 89. If the last digit is 0, then the sum of the first nine digits must be 89. But each digit can be at most 9, so nine digits can only sum up to 81. So, 89 is not possible. Therefore, the last digit cannot be 0. Similarly, if the last digit is 2, then the sum of the first nine digits must be 89 - 2 = 87. But again, 9 digits can only sum up to 81, which is still too low. Similarly, if the last digit is 4, the first nine digits need to sum to 85; 6 would need 83; 8 would need 81. Wait, 81 is exactly the maximum sum of nine digits. So, if the last digit is 8, then the first nine digits must sum to 81, which is possible if all nine digits are 9. So, in that case, the number would be nine 9s followed by an 8. That works. The digit sum is 9*9 +8=89, and the last digit is 8, which is even. Therefore, that's valid.But if the last digit is 0, 2, 4, or 6, then the sum of the first nine digits would need to be 89, 87, 85, or 83, respectively. But all of these are higher than 81, which is the maximum possible sum for nine digits. Therefore, it's impossible for the last digit to be 0, 2, 4, or 6. Therefore, the only possible last digit is 8. So, the answer is E) 8.Wait, let me check that again. If the last digit is 8, then the first nine digits sum to 81, which is achievable if all nine digits are 9. That gives a ten-digit number with nine 9s and an 8 at the end. The sum is 89, and it's even. Correct.If the last digit is anything else, then the remaining nine digits would have to sum to a number higher than 81, which is impossible because each digit can be at most 9. Therefore, 8 is the only possible last digit. Therefore, the answer is E) 8.But wait, let me think again. Is there a case where more than one digit is less than 9? For example, maybe two digits are 8, and others are 9. Then the digit sum would be 9*8 + 8*2 = 72 +16=88, which is less than 89. So, that's not enough. Alternatively, if one digit is 7 and the rest are 9, the digit sum is 9*9 +7=81+7=88. Still not enough. So, to get 89, the only way is to have one digit as 8 and the rest as 9. Therefore, the digit that is 8 must be in the last position because otherwise, if it's in another position, the last digit can still be even. Wait, but the problem doesn't say the number is the smallest possible or anything. So, if the 8 is in any other position, then the last digit can still be 8 or 0, 2, 4, 6. Wait, but if the 8 is in another position, then the last digit can be 8 or another even digit. Wait, but if the 8 is in another position, then the last digit could be 8, but then the other digit is 8, but the total digit sum would be 9*8 +8 +8= 72+16=88, which is not 89. Wait, no. If we have one 8 in another position, then the rest are 9s, and the last digit is even. For example, let's say the first digit is 8, and the rest are 9s, except the last digit, which is 8. Then the digit sum would be 8 + 9*8 +8=8 +72 +8=88. No, that's still 88. Wait, no, hold on. If one of the digits is 8 and the rest are 9, then the total digit sum is 89. So, if the 8 is in some position, and the last digit is 9, but wait, the last digit can't be 9 because it's even. Therefore, the 8 cannot be in the last digit? Wait, no, if the 8 is in the last digit, then the last digit is 8, which is even, and the rest of the digits are 9. So, that works. But if the 8 is in another digit, then the last digit must be 9, which is invalid because it's not even. Wait, that's the key point. If the 8 is not in the last digit, then the last digit must be 9 to have the total digit sum as 89 (since all other digits except one 8 are 9s). But 9 is odd, which contradicts the requirement that the number is even. Therefore, the 8 must be in the last digit. Therefore, the last digit is 8. Hence, the answer is E) 8.Yes, that makes sense. So, the only way to have a digit sum of 89 with an even ten-digit number is if the last digit is 8, and all the other digits are 9. Therefore, the last digit must be 8. So, the answer is E.Final Answerboxed{E}</think>Given an even ten-digit integer with a digit sum of 89, we need to determine the last digit. The maximum digit sum for a ten-digit number is (9 times 10 = 90). Since the given digit sum is 89, it means one digit is 8 and the remaining nine digits are 9. For the number to be even, the last digit must be even (0, 2, 4, 6, or 8). If the last digit were 0, 2, 4, or 6, the sum of the first nine digits would need to be 89, 87, 85, or 83 respectively, all of which are impossible because the maximum sum of nine digits is (9 times 9 = 81). Therefore, the only feasible option is for the last digit to be 8. This configuration results in nine digits being 9 and the last digit being 8, which satisfies both the digit sum of 89 and the even requirement. Thus, the last digit is (boxed{E}).]

🔑:The sum of the digits of an even ten digit integer is 89 . The last digit is(A) 0(B) 2(C) 4(D) 6(E) 8## Solution89 is a large number to be the sum of the digits of a ten digit number. In fact, the largest possible digital sum is 10 times 9 or 90 . Since 89 is only 1 less than 90 , the number in question must be composed of nine 9's and one 8 . In order that the number be divisible by 2 , the last digit must be 8 .ANSWER: (E)