🧠:[<think>Okay, let's see. I need to find the value of n where the sum of the first n terms of this sequence is maximized. The sequence starts at 19, and each subsequent term decreases by 3. So, the first term is 19, the second is 16, then 13, and so on. The options given are 6, 7, 8, or 9. Hmm.First, I should recall that the sum of an arithmetic sequence can be calculated using the formula: S_n = n/2 * (2a_1 + (n - 1)d), where a_1 is the first term, d is the common difference, and n is the number of terms. Alternatively, since this is a decreasing sequence, the terms will eventually become negative. The sum will be maximized just before the terms turn negative, right? Because once the terms become negative, adding them would decrease the total sum.Wait, but let's verify that. Let's first figure out when the terms become zero or negative. The nth term of the sequence can be found using the formula a_n = a_1 + (n - 1)d. But here, the common difference d is -3 because each term is 3 less than the previous. So, a_n = 19 + (n - 1)(-3) = 19 - 3(n - 1). Let's simplify that: a_n = 19 - 3n + 3 = 22 - 3n.So, when does a_n become zero or negative? Let's set 22 - 3n ≤ 0. Solving for n: 22 ≤ 3n => n ≥ 22/3 ≈ 7.333. Since n has to be a natural number, the first term that is negative is when n = 8. Therefore, the 8th term is 22 - 3*8 = 22 - 24 = -2. So, the 7th term is 22 - 21 = 1, which is still positive. The 8th term is the first negative term.Therefore, if we sum up to the 7th term, all terms are positive, but adding the 8th term (which is negative) will decrease the sum. So, the maximum sum should occur at n = 7. Wait, but let me confirm this by calculating the sum at n = 7 and n = 8.Alternatively, maybe there's a way to model the sum as a quadratic function and find its maximum. Since the sequence is arithmetic, the sum S_n is a quadratic function of n. Let's write that out.Using the sum formula: S_n = n/2 * [2a_1 + (n - 1)d]. Substituting the given values, a_1 = 19, d = -3. So,S_n = n/2 [2*19 + (n - 1)(-3)] = n/2 [38 - 3(n - 1)] = n/2 [38 - 3n + 3] = n/2 [41 - 3n].So, S_n = (n/2)(41 - 3n) = (41n - 3n²)/2 = (-3n² + 41n)/2. This is a quadratic function in terms of n, which opens downward (since the coefficient of n² is negative), so the maximum occurs at the vertex.The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). Here, a = -3/2 and b = 41/2. So, n = - (41/2) / (2*(-3/2)) = -41/2 / (-3) = (41/2)/3 = 41/6 ≈ 6.833.Since n must be an integer, the maximum sum occurs either at n = 6 or n = 7. To determine which one gives a higher sum, we need to compute S_6 and S_7.Calculating S_6:Using the formula S_n = (-3n² + 41n)/2.S_6 = (-3*(6)^2 + 41*6)/2 = (-3*36 + 246)/2 = (-108 + 246)/2 = 138/2 = 69.Calculating S_7:S_7 = (-3*(7)^2 + 41*7)/2 = (-3*49 + 287)/2 = (-147 + 287)/2 = 140/2 = 70.So, S_7 is 70, which is higher than S_6 (69). Then, if we calculate S_8:S_8 = (-3*(8)^2 + 41*8)/2 = (-3*64 + 328)/2 = (-192 + 328)/2 = 136/2 = 68.So, S_8 is 68, which is less than S_7. Therefore, the maximum sum occurs at n = 7. Hence, the answer should be B: 7.But let's check by another method. Since the terms start at 19 and decrease by 3 each time, the terms are 19, 16, 13, 10, 7, 4, 1, -2, etc. So, the 7th term is 1, which is still positive. Adding up to 7 terms:19 + 16 = 35; 35 +13=48; 48+10=58; 58+7=65; 65+4=69; 69+1=70. So, sum at 7 terms is 70. Then the 8th term is -2, so sum becomes 70 -2=68, which is indeed less. So, 7 is the correct answer.But just to be thorough, maybe the problem is designed to trick us. Let's make sure that the quadratic formula gives the vertex at 41/6 ≈6.833, which is approximately 6.83. Since the parabola is opening downward, the maximum is at n≈6.83, so the integer closest to it is 7. Hence, n=7. So, the answer is B:7.Alternatively, another way to think about it is that the sum increases as long as the terms are positive. Once the term becomes negative, adding it would decrease the sum. So, the last positive term is at n=7 (since 22 -3*7=1), and the next term is negative. Therefore, the maximum sum is achieved just before adding the negative term, so n=7. So, that's consistent with the previous results.Therefore, the correct answer should be B:7.Final Answerboxed{B}</think>Given the sequence ({a_n}) where (a_1 = 19) and (a_{n+1} = a_n - 3) for (n in mathbb{N}^*), we need to find the value of (n) that maximizes the sum of the first (n) terms.First, we determine the (n)-th term of the sequence:[a_n = 19 - 3(n - 1) = 22 - 3n]We find when the terms become non-positive by solving (22 - 3n leq 0):[22 leq 3n implies n geq frac{22}{3} approx 7.33]Thus, the 7th term is the last positive term ((a_7 = 1)) and the 8th term is negative ((a_8 = -2)).Next, we calculate the sum of the first (n) terms using the formula for the sum of an arithmetic sequence:[S_n = frac{n}{2} [2a_1 + (n - 1)d] = frac{n}{2} [38 - 3(n - 1)] = frac{n}{2} [41 - 3n]]This simplifies to:[S_n = frac{-3n^2 + 41n}{2}]The vertex of this quadratic function, which gives the maximum value, occurs at:[n = frac{-b}{2a} = frac{-41/2}{2(-3/2)} = frac{41}{6} approx 6.83]Since (n) must be an integer, we check (n = 6) and (n = 7):[S_6 = frac{-3(6)^2 + 41(6)}{2} = frac{-108 + 246}{2} = 69][S_7 = frac{-3(7)^2 + 41(7)}{2} = frac{-147 + 287}{2} = 70](S_7 = 70) is higher than (S_6 = 69). Checking (S_8) confirms the sum decreases:[S_8 = frac{-3(8)^2 + 41(8)}{2} = frac{-192 + 328}{2} = 68]Thus, the maximum sum occurs at (n = 7).[boxed{B}],<think>Okay, let's see. I have this problem about a sequence where the first term is 19, and each subsequent term is 3 less than the previous one. So it's an arithmetic sequence with a common difference of -3. The question is asking for the value of n where the sum of the first n terms is maximized. The options are 6, 7, 8, or 9. Alright, let me think through this step by step.First, I remember that the sum of an arithmetic sequence can be calculated using the formula: S_n = n/2 * (2a_1 + (n - 1)d), where a_1 is the first term, d is the common difference, and n is the number of terms. Alternatively, since the sequence is decreasing by 3 each time, maybe there's another way to approach it. But let's stick with the formula for now.Given that a_1 = 19 and d = -3, let's plug those into the formula. The sum S_n would be n/2 * [2*19 + (n - 1)*(-3)]. Simplifying that, we get S_n = n/2 * [38 - 3(n - 1)]. Let's expand that a bit more: 38 - 3n + 3, so 41 - 3n. Then, multiplying by n/2, we get S_n = (n/2)(41 - 3n). So S_n = (41n - 3n²)/2. Hmm, this is a quadratic function in terms of n. Since the coefficient of n² is negative (-3/2), the parabola opens downward, so the maximum occurs at the vertex. The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). Here, our quadratic is -3/2 n² + (41/2)n. So, a = -3/2 and b = 41/2. Plugging into the vertex formula, n = - (41/2) / (2*(-3/2)) = -41/2 / (-3) = 41/6 ≈ 6.8333.Since n has to be a positive integer, the maximum sum would occur either at n = 6 or n = 7. But the options include 6,7,8,9. So the possible answers are A:6, B:7. But wait, let me check again. The vertex is approximately 6.83, which is between 6 and 7. Since the parabola is opening downward, the maximum value is at the vertex. So since we can't have n as a fraction, we need to check which integer n gives a higher sum, n=6 or n=7. Therefore, the sum should be higher at n=7. But wait, let's verify this by actually computing S_6 and S_7.Alternatively, maybe there's a different approach. Since each term is decreasing by 3, the terms go 19, 16, 13, 10, 7, 4, 1, -2, -5, etc. The sum will increase as long as the terms are positive, and once the terms become negative, adding them would decrease the total sum. Therefore, the maximum sum occurs just before the terms become negative. Let's see when the terms become negative.The nth term of the sequence is a_n = a_1 + (n - 1)d. Here, a_1 = 19, d = -3. So a_n = 19 + (n - 1)(-3) = 19 - 3(n - 1) = 22 - 3n. Wait, let me check that. Let's compute a_n:a_1 = 19a_2 = 19 - 3 = 16a_3 = 16 - 3 = 13a_4 = 10a_5 = 7a_6 = 4a_7 = 1a_8 = -2a_9 = -5So the 7th term is 1, which is still positive, but the 8th term is -2, which is negative. Therefore, up to the 7th term, all terms are non-negative. So adding the 8th term would subtract from the total. Therefore, the maximum sum occurs at n=7. Therefore, the answer is B:7.But wait, let me check using the sum formula. Let's compute S_6 and S_7.Using the formula S_n = (n/2)(2a_1 + (n - 1)d)For n=6:S_6 = 6/2 [2*19 + 5*(-3)] = 3 [38 - 15] = 3 [23] = 69.For n=7:S_7 = 7/2 [2*19 + 6*(-3)] = 7/2 [38 - 18] = 7/2 [20] = 7*10 = 70.So S_7 is 70, which is higher than S_6=69.For n=8:S_8 = 8/2 [38 + 7*(-3)] = 4 [38 -21] = 4*17 = 68.Which is lower than 70.Therefore, indeed, the maximum is at n=7.Alternatively, using the quadratic formula, the vertex is at n ≈6.83, which is closer to 7, so n=7.Therefore, the answer should be B:7.Final Answerboxed{B}</think>Given the sequence ({a_n}) where (a_1 = 19) and (a_{n+1} = a_n - 3) for (n in mathbb{N}^*), we need to find the value of (n) that maximizes the sum of the first (n) terms.The sequence is an arithmetic sequence with the first term (a_1 = 19) and common difference (d = -3). The sum of the first (n) terms of an arithmetic sequence is given by:[S_n = frac{n}{2} left[2a_1 + (n - 1)dright]]Substituting the given values:[S_n = frac{n}{2} left[2 cdot 19 + (n - 1)(-3)right] = frac{n}{2} left[38 - 3(n - 1)right] = frac{n}{2} left[41 - 3nright]]This simplifies to:[S_n = frac{41n - 3n^2}{2}]The quadratic function (S_n = frac{-3n^2 + 41n}{2}) opens downwards (since the coefficient of (n^2) is negative), so the maximum occurs at the vertex. The vertex of a quadratic (ax^2 + bx + c) is at (n = -frac{b}{2a}). Here, (a = -3/2) and (b = 41/2), so:[n = -frac{41/2}{2 cdot -3/2} = frac{41}{6} approx 6.83]Since (n) must be an integer, we check (n = 6) and (n = 7). Calculating the sums:- For (n = 6): [ S_6 = frac{6}{2} left[38 - 3 cdot 5right] = 3 left[38 - 15right] = 3 cdot 23 = 69 ]- For (n = 7): [ S_7 = frac{7}{2} left[38 - 3 cdot 6right] = frac{7}{2} left[38 - 18right] = frac{7}{2} cdot 20 = 7 cdot 10 = 70 ]The sum (S_7 = 70) is higher. Checking subsequent terms, the 8th term is negative, confirming the sum decreases after (n = 7).Thus, the value of (n) that maximizes the sum is (boxed{B}).]

🔑:Since a_1=19, a_{n+1}=a_n-3 (ninmathbb{N}^*), the sequence {a_n} is an arithmetic sequence with the first term 19 and common difference -3, thus a_n=19+(n-1)times(-3)=22-3n, from a_n=22-3ngeqslant 0, we get nleqslant 7 frac{1}{3}, therefore, when the sum of the first n terms of the sequence {a_n} is maximized, the value of n is 7. Hence, the correct choice is boxed{text{B}}. First, we find a_n=19+(n-1)times(-3)=22-3n from the given conditions, then from a_n=22-3ngeqslant 0, we get nleqslant 7 frac{1}{3}, which leads us to the value of n when the sum of the first n terms of the sequence {a_n} is maximized. This problem tests the properties and applications of arithmetic sequences and is a basic question. It requires careful reading and meticulous solving.

🧠:[<think>Okay, so I have this problem here where I need to find the smallest natural number that starts with 4, and when you move that first 4 to the end of the number, the new number is 1/4 of the original. Hmm, let me try to wrap my head around this.First off, let's break down the problem. The original number has its first digit as 4. Let's denote this number as N. When we move the first digit (which is 4) to the end, we get a new number, let's call it M. The problem states that M is 1/4 of N. So, M = N / 4. Our task is to find the smallest such N.Let me think about how numbers change when moving digits. For example, if N is a 4-digit number starting with 4, say 4abc, then moving the 4 to the end would give abc4. But in this case, the transformed number abc4 is supposed to be 1/4 of 4abc. But how do I model this mathematically?Let's consider N as a k-digit number starting with 4. So, N can be written as 4 * 10^{k-1} + x, where x is the remaining number after the first digit. For example, if N is 4123, then x would be 123, and k is 4. Then, moving the first digit to the end would result in M = x * 10 + 4. According to the problem, M = N / 4. Therefore, we have the equation:x * 10 + 4 = (4 * 10^{k-1} + x) / 4Let me write this equation out again:10x + 4 = (4 * 10^{k-1} + x) / 4Multiply both sides by 4 to eliminate the denominator:40x + 16 = 4 * 10^{k-1} + xSubtract x from both sides:39x + 16 = 4 * 10^{k-1}So, rearranged:39x = 4 * 10^{k-1} - 16Then:x = (4 * 10^{k-1} - 16) / 39Since x must be a natural number (as it's the remaining part of N after the first digit), this fraction must result in an integer. Therefore, (4 * 10^{k-1} - 16) must be divisible by 39.So, the key is to find the smallest k such that 4 * 10^{k-1} ≡ 16 mod 39.Simplify that congruence:4 * 10^{k-1} ≡ 16 mod 39Divide both sides by 4 (since 4 and 39 are coprime, division is allowed):10^{k-1} ≡ 4 mod 39So, the problem reduces to finding the smallest positive integer (k - 1) such that 10^{k-1} ≡ 4 mod 39. Then, k will be the number of digits of N, which we need to minimize.Therefore, let me set m = k - 1. Then, we need to solve 10^m ≡ 4 mod 39, and find the smallest such m.So, we need to find the minimal m where 10^m mod 39 = 4.To solve this, perhaps I can compute 10^m mod 39 for m = 1, 2, 3, ... until I get 4.Let me compute:For m=1: 10^1 mod 39 = 10m=2: 10^2 = 100; 100 mod 39. 39*2=78, 100-78=22. So 22.m=3: 10*22 = 220; 220 mod 39. 39*5=195, 220-195=25.m=4: 10*25=250; 250 mod 39. 39*6=234, 250-234=16.m=5: 10*16=160; 160 mod 39. 39*4=156, 160-156=4.Ah! At m=5, 10^5 mod 39=4. So, m=5 is the minimal exponent. Therefore, k-1=5 => k=6. So, the original number N has 6 digits, starting with 4, followed by x, which is (4*10^{5} -16)/39.Let me compute x:4*10^5 = 400,000400,000 -16 = 399,984399,984 /39 = ?Let me divide 399,984 by 39.39*10,000 = 390,000399,984 -390,000 =9,98439*256=9,984 (since 39*200=7,800; 39*50=1,950; 7,800+1,950=9,750; 39*6=234; 9,750+234=9,984)So, 39*256=9,984. Therefore, total division is 10,000 +256=10,256.Therefore, x=10,256. Wait, but x is supposed to be a 5-digit number since k=6, so N is 4 followed by x, which is 5 digits. But 10,256 is a 5-digit number (wait, 10,256 is 10256, which is 5 digits. So yes, 4 followed by 10256 is 410256.Let me check if moving the first digit 4 to the end gives 102564, and check if 102564 is 1/4 of 410256.Calculate 410256 /4: 410256 ÷4. Let's compute:4 ÷4=1, 1 with remainder 0.Bring down 1: 01. 0 ÷4 is 0, but 1 is less than 4. So 0. Then, 41 ÷4=10, remainder 1.Wait, maybe I should do it step by step.410256 ÷4:Divide 4 (first digit) by 4: 1. Remainder 0.Bring down 1: 01. 1 ÷4 is 0 with remainder 1.Bring down 0: 10. 10 ÷4=2, remainder 2.Bring down 2: 22. 22 ÷4=5, remainder 2.Bring down 5: 25 ÷4=6, remainder 1.Bring down 6: 16 ÷4=4, remainder 0.So, result is 102564. Exactly the number we get by moving the 4 to the end. So 410256 ÷4=102564, which is M. So this works.But the problem is to find the smallest such natural number. Is 410256 the smallest?Wait, but let's confirm if there is a smaller k. Since m=5 was the minimal exponent where 10^m ≡4 mod39. If there is a smaller m, then k would be smaller. But since we tried m from 1 upwards, and the first occurrence was at m=5, that's minimal. So, k=6 is minimal, so 6-digit number is the minimal length.Therefore, the answer is 410256.But let me double-check. Suppose someone says, wait, maybe there is a smaller number. Let's see. For example, if the number was 4 digits, then moving the first digit gives a 4-digit number. Let's see:Suppose N is 4abc. Then M=abc4. Then, 4abc /4= abc4. But 4abc is 4000 + 100a +10b +c. Divided by 4 is 1000 +25a +2.5b +0.25c. But abc4 is 1000a + 100b +10c +4. These would have to be equal. But 25a +2.5b +0.25c +1000 = 1000a +100b +10c +4. Which seems impossible because on the left side, the coefficient of a is 25, but on the right side, it's 1000. So that's way too big. So perhaps 4-digit number is impossible.Similarly, 5-digit number: Let's say N=4abcd. Then M=abcd4. Then N=40000 +1000a +100b +10c +d. M=10000a +1000b +100c +10d +4. And M=N/4. So 10000a +1000b +100c +10d +4 = (40000 +1000a +100b +10c +d)/4. Multiply both sides by 4:40000a +4000b +400c +40d +16 =40000 +1000a +100b +10c +dBring all terms to left:40000a -1000a +4000b -100b +400c -10c +40d -d +16 -40000=039000a +3900b +390c +39d -39984=0Divide entire equation by 39:(39000a)/39 + (3900b)/39 + (390c)/39 + (39d)/39 -39984/39=0Which is 1000a +100b +10c +d -1025.6... Hmm, 39984 divided by 39 is 39984 ÷39. 39*1025=39*(1000+25)=39000+975=39975. So 39984-39975=9. So 39984/39=1025 +9/39=1025 +3/13. So the equation becomes:1000a +100b +10c +d =1025 +3/13But left side is an integer, right side is not. So no solution for 5-digit numbers. So no solution here.Similarly, for 3-digit number: N=4ab. M=ab4. Then, 4ab /4= ab4. 4ab=400 +10a +b. Divided by4: 100 +2.5a +0.25b. ab4=100a +10b +4. So 100 +2.5a +0.25b=100a +10b +4. Multiply by4: 400 +10a +b=400a +40b +16. Rearranged: -390a -39b +384=0. Divide by -39:10a +b - 384/39=0. 384/39=9.846... Not integer. So no solution.Therefore, the minimal number of digits is 6. Therefore, 410256 is the minimal number. But let me check if there's a smaller number with more digits? Wait, how does that make sense? If k=6 is minimal, then numbers with more digits would be larger. So 410256 is the smallest.Alternatively, perhaps there is a number with leading digit 4, but more digits, which is smaller? Wait, but more digits would mean a larger number. For example, 6-digit number vs. a 7-digit number. The 6-digit number is smaller. So since k=6 is the minimal number of digits needed, 410256 is the smallest possible.Alternatively, maybe there is another 6-digit number smaller than 410256 that satisfies the condition? Let's see. Let's see how x was calculated.x=(4*10^{5}-16)/39=(400000-16)/39=399984/39=10256. So x=10256. So N=4 followed by x, which is 410256. So x must be 10256. There's no smaller x here because x is uniquely determined once k is fixed. Since k=6 is minimal, this is the smallest N.But let me confirm with another approach. Let's model N as 4 followed by n digits. So N=4*10^n + x, where x is an n-digit number. Then moving the 4 to the end gives M=10x +4. Given that M=N/4, so:10x +4 = (4*10^n +x)/4Multiply both sides by 4:40x +16=4*10^n +xSo 39x =4*10^n -16Hence, x=(4*10^n -16)/39Therefore, x must be an integer, so 4*10^n ≡16 mod39Which simplifies to 10^n ≡4 mod39, same as before.So we need to find the smallest n where 10^n ≡4 mod39. As before, n=5. Then x=(4*10^5 -16)/39=10256, so N=4*10^5 +10256=410256.Therefore, 410256 is indeed the minimal number.Just to be thorough, let's check for n=5 (so N is 4 followed by 5 digits, total 6 digits). Let's see if there's any smaller x. But x is determined uniquely by n=5, so x=10256. If x were smaller, then n would have to be smaller, but we saw that for n<5, there's no solution. Therefore, 410256 is indeed the smallest.Thus, the answer should be 410256.Final AnswerThe smallest natural number is boxed{410256}.</think>Given the problem of finding the smallest natural number ( N ) such that its first digit is 4, and when this first digit is moved to the end, the resulting number ( M ) is ( frac{1}{4} ) of ( N ), we start by denoting ( N ) as a ( k )-digit number starting with 4. We can express ( N ) as ( 4 times 10^{k-1} + x ), where ( x ) is the remaining number after the first digit. Moving the first digit 4 to the end results in ( M = 10x + 4 ).Given ( M = frac{N}{4} ), we have the equation:[10x + 4 = frac{4 times 10^{k-1} + x}{4}]Multiplying both sides by 4 to clear the fraction:[40x + 16 = 4 times 10^{k-1} + x]Rearranging terms, we get:[39x + 16 = 4 times 10^{k-1}]Solving for ( x ):[x = frac{4 times 10^{k-1} - 16}{39}]For ( x ) to be a natural number, ( 4 times 10^{k-1} - 16 ) must be divisible by 39. This simplifies to finding the smallest ( k ) such that:[4 times 10^{k-1} equiv 16 pmod{39}]Dividing both sides by 4 (since 4 and 39 are coprime):[10^{k-1} equiv 4 pmod{39}]We find the smallest ( m = k - 1 ) such that ( 10^m equiv 4 pmod{39} ). Through modular arithmetic, we determine that ( m = 5 ) is the smallest exponent satisfying this congruence. Thus, ( k = m + 1 = 6 ).Substituting ( k = 6 ) into the equation for ( x ):[x = frac{4 times 10^5 - 16}{39} = frac{400000 - 16}{39} = frac{399984}{39} = 10256]Thus, the original number ( N ) is:[N = 4 times 10^5 + 10256 = 410256]Verification confirms that moving the first digit 4 to the end results in 102564, which is indeed ( frac{1}{4} ) of 410256. Therefore, the smallest natural number satisfying the given property is:[boxed{410256}],<think>Okay, so I need to find the smallest natural number that starts with 4, and when that first digit 4 is moved to the end, the new number is 1/4 of the original number. Hmm, let me think about how to approach this.Let me denote the original number as N. Since the first digit is 4, N is a number that starts with 4. Let's say N has k+1 digits. So, N can be written as 4 followed by k digits. Let me represent those k digits as a number, say X. Then, N can be expressed as:N = 4 * 10^k + XWhen we move the first digit (which is 4) to the end, we get a new number M. This new number M would be X followed by 4. So, M can be written as:M = 10 * X + 4According to the problem, M is 1/4 of N. Therefore:10 * X + 4 = (1/4) * (4 * 10^k + X)Let me simplify this equation step by step. First, multiply both sides by 4 to eliminate the fraction:4 * (10 * X + 4) = 4 * 10^k + XExpanding the left side:40 * X + 16 = 4 * 10^k + XNow, subtract X from both sides:39 * X + 16 = 4 * 10^kSo, we have:39 * X = 4 * 10^k - 16Therefore,X = (4 * 10^k - 16) / 39Since X must be an integer (as it's the last k digits of the original number N), the right-hand side must also be an integer. Therefore, (4 * 10^k - 16) must be divisible by 39. Let's write that as:4 * 10^k ≡ 16 mod 39Dividing both sides by 4 (since 4 and 39 are coprime, this is allowed):10^k ≡ 4 mod 39So, now the problem reduces to finding the smallest k such that 10^k ≡ 4 mod 39. Once we find such a k, then X can be calculated as (4 * 10^k - 16)/39, and then N would be 4 * 10^k + X.So, let's try to find the smallest k where 10^k ≡ 4 mod 39. Let's compute powers of 10 modulo 39 until we find 4.Let's start calculating:10^1 mod 39 = 1010^2 mod 39 = (10*10) mod 39 = 100 mod 39. 39*2=78, 100-78=22. So, 22.10^3 mod 39 = (10^2 * 10) mod 39 = 22*10 mod 39 = 220 mod 39. 39*5=195, 220-195=25. So, 25.10^4 mod 39 = (25*10) mod 39 = 250 mod 39. 39*6=234, 250-234=16. So, 16.10^5 mod 39 = (16*10) mod 39 = 160 mod 39. 39*4=156, 160-156=4. Oh! 4. That's what we wanted. So, k=5.Therefore, k=5 is the smallest such k. Let's verify that.So, 10^5 mod 39=4. Therefore, X=(4*10^5 -16)/39. Let's compute that.First, compute 4*10^5: that's 400,000. Subtract 16: 400,000 -16 = 399,984. Now, divide by 39: 399,984 ÷39.Let me compute 399,984 ÷39. Let's see:39*10,000=390,000399,984 -390,000=9,98439*256=9,984 (since 39*200=7,800; 39*50=1,950; 39*6=234; 7,800+1,950=9,750 +234=9,984)So, total is 10,000 +256=10,256. So, X=10,256.Therefore, the original number N is 4 followed by X. Since X is 10,256, which is 5 digits (k=5), N is 410,256.Wait, let me check. N=4*10^5 + X=400,000 +10,256=410,256.Then, moving the first digit 4 to the end gives 102,564. Let's check if 102,564 is 1/4 of 410,256.Compute 410,256 ÷4=102,564. Exactly. So, that works.But wait, the problem says "the smallest natural number". So, 410,256 is a 6-digit number. Is there a smaller number that satisfies the condition? Let's see. Because we found k=5. Since k is the number of digits after the first digit 4, so total digits is k+1=6. So, is there a smaller k possible?But we found that the minimal k is 5 because 10^k ≡4 mod39 first occurs at k=5. So, unless there's a smaller k where 10^k ≡4 mod39. Let me verify again.We checked k=1 to 5:k=1: 10 mod39=10k=2: 10^2=100 mod39=100-78=22k=3: 22*10=220 mod39=220-195=25k=4:25*10=250 mod39=250-234=16k=5:16*10=160 mod39=160-156=4So, indeed, k=5 is the minimal k where 10^k ≡4 mod39. Therefore, the minimal number is 6-digit number 410,256.Wait, but let me check if there's a smaller number with more digits? Wait, digits are k+1, so if k is minimal, then the number of digits is minimal. Since k=5 is minimal, then the number is 6 digits. So, 410,256 is the minimal such number. But maybe there is a number with leading zeros? But natural numbers cannot have leading zeros. So, X must be a k-digit number without leading zeros. Wait, in our case, X=10,256 which is 5 digits, so no leading zeros, which is correct.Therefore, 410,256 is the minimal such number.But wait, let me check if there is a k less than 5 where (4*10^k -16) is divisible by 39. Because even if 10^k mod39 isn't 4, maybe the entire numerator is divisible by 39.But according to our equation:We had 10^k ≡4 mod39, so that 4*10^k -16 ≡4*4 -16=16-16=0 mod39. So, in that case, (4*10^k -16)/39 is integer. So, for (4*10^k -16) to be divisible by39, 10^k ≡4 mod39 is necessary and sufficient. Because 4*10^k ≡16 mod39 => 10^k ≡4 mod39.Therefore, k must satisfy 10^k ≡4 mod39, which minimal k is 5. So, X=(4*10^5 -16)/39=10256, so N=410256.Wait, but the problem states "the smallest natural number", so 410256 is the answer.But let me check once again. Let's verify 410256.Original number: 410256Move first digit (4) to the end: 102564Check if 102564 *4 =410256.Compute 102,564 *4:100,000*4=400,0002,564*4=10,256So, 400,000 +10,256=410,256. Correct.Therefore, the answer is 410256.But let me check if there's a smaller number. Suppose someone says maybe a number with more digits but smaller in value? For example, 410256 is a 6-digit number. Maybe there's a 12-digit number smaller than that? Wait, no. Because a 12-digit number is larger than a 6-digit number. So, the minimal number must be the one with the minimal number of digits, which is 6.Therefore, 410256 is indeed the minimal number.But wait, just to make sure, let's see if there's a k=5 cycle. So, in modular arithmetic, once we have 10^5 ≡4 mod39, maybe there's a period after which this repeats. But since we are looking for the minimal k, k=5 is the minimal. So, unless there is a smaller k.But according to our earlier computation, k=5 is the first occurrence. Therefore, 410256 is the minimal number. Therefore, the answer is 410256.Wait, but the problem is from the 46th Moscow Mathematical Olympiad, 1983. Maybe the answer is known? Let me check if my answer matches known solutions.Wait, but since I can't access external resources, I need to verify again.Alternatively, maybe I made a mistake in calculation. Let me check once again.Compute 10^1 mod39=1010^2=100 mod39=100-2*39=100-78=2210^3=10*22=220 mod39=220-5*39=220-195=2510^4=10*25=250 mod39=250-6*39=250-234=1610^5=10*16=160 mod39=160-4*39=160-156=4. Yes, so 10^5≡4 mod39.Therefore, k=5. Therefore, X=(4*10^5 -16)/39=(400,000 -16)/39=399,984/39=10,256. Therefore, N=4*10^5 +X=400,000 +10,256=410,256. Moving 4 to the end: 102,564, and 102,564*4=410,256. Correct. So, this is the minimal number. Therefore, the answer is 410256.I think that's correct.Final Answerboxed{410256}</think>To find the smallest natural number ( N ) that starts with 4 and when the first digit 4 is moved to the end, the new number ( M ) is (frac{1}{4}) of the original number, we start by expressing ( N ) and ( M ) in terms of their digits. Let ( N ) be a number with ( k+1 ) digits where the first digit is 4 followed by ( k ) digits denoted as ( X ). Thus, ( N ) can be written as:[ N = 4 times 10^k + X ]When the first digit 4 is moved to the end, the new number ( M ) becomes:[ M = 10X + 4 ]According to the problem, ( M ) is (frac{1}{4}) of ( N ):[ 10X + 4 = frac{1}{4}(4 times 10^k + X) ]Multiplying both sides by 4 to clear the fraction:[ 4(10X + 4) = 4 times 10^k + X ][ 40X + 16 = 4 times 10^k + X ][ 39X + 16 = 4 times 10^k ][ 39X = 4 times 10^k - 16 ][ X = frac{4 times 10^k - 16}{39} ]For ( X ) to be an integer, ( 4 times 10^k - 16 ) must be divisible by 39. This simplifies to:[ 4 times 10^k equiv 16 pmod{39} ][ 10^k equiv 4 pmod{39} ]We find the smallest ( k ) such that ( 10^k equiv 4 pmod{39} ):- ( 10^1 equiv 10 pmod{39} )- ( 10^2 equiv 22 pmod{39} )- ( 10^3 equiv 25 pmod{39} )- ( 10^4 equiv 16 pmod{39} )- ( 10^5 equiv 4 pmod{39} )The smallest ( k ) is 5. Substituting ( k = 5 ):[ X = frac{4 times 10^5 - 16}{39} = frac{400000 - 16}{39} = frac{399984}{39} = 10256 ]Thus, the original number ( N ) is:[ N = 4 times 10^5 + 10256 = 400000 + 10256 = 410256 ]Verifying, moving the first digit 4 to the end gives 102564, and ( 102564 times 4 = 410256 ).Therefore, the smallest natural number is (boxed{410256}).]

🔑:[Solution] Let the number we are looking for bex=overline{4 a b cdots c}=4 cdot 10^{n}+A .where A=overline{a b cdots c} is an n-digit number.Moving the first digit to the end, we get the numbery=overline{a b cdots c 4}=10 A+4 .According to the problem, we havethat is,4 cdot 10^{n}+A=4(10 A+4) text {, }orbegin{array}{l}39 A=4 cdot 99 cdots 96, 13 A=4 cdot 33 cdots 32 .end{array}Since (13,4)=1, then 33 cdots 32 must be divisible by 13.The smallest value of the form 33 cdots 32 that is divisible by 13 is 33332.Therefore,A=4 cdot frac{33332}{13}=10256 text {. }Thus, the smallest value of the number x we are looking for is 410256.

🧠:[<think>Okay, so I need to figure out how many natural numbers greater than one have the property that when you multiply the number by its smallest prime divisor, the result doesn't exceed 100. Let me try to break this down step by step.First, let me make sure I understand the problem correctly. We need natural numbers n > 1. For each such n, find its smallest prime divisor (let's call that p), then check if n * p ≤ 100. If yes, then n is counted. The question is asking how many such n exist.So, the key steps here are:1. For each natural number n > 1, determine its smallest prime divisor p.2. Check if n * p ≤ 100.3. Count all n where this condition holds.Hmm. So, instead of checking every number from 2 upwards until n * p would exceed 100, maybe there's a smarter way to approach this. Let me think.First, note that the smallest prime divisor of a number n depends on whether n is prime or composite. If n is prime, then its smallest prime divisor is itself. If n is composite, its smallest prime divisor will be the smallest prime that divides it, which is at least 2 and less than sqrt(n) or something. Let me formalize this.Case 1: n is prime. Then p = n. So, the product is n * n = n². So, n² ≤ 100. Therefore, prime numbers n where n² ≤ 100, i.e., n ≤ 10. Since primes less than or equal to 10 are 2, 3, 5, 7. So, primes up to 10.Case 2: n is composite. Then the smallest prime divisor p of n is the smallest prime that divides n. So, for composite n, p is at least 2 (since 2 is the smallest prime). Therefore, the product n * p. We need this product to be ≤ 100.Therefore, to find all composite numbers n where n * p ≤ 100, where p is the smallest prime divisor of n.So, the problem reduces to considering two separate cases: primes up to 10, and composite numbers where n multiplied by their smallest prime divisor is ≤ 100.So, first, let's handle the primes. The primes n where n² ≤ 100. The primes less than or equal to 10 are 2, 3, 5, 7. So, that's 4 numbers.Now, for the composite numbers. Let's consider composite numbers n > 1. Each composite number has a smallest prime divisor p. For each composite n, we need n * p ≤ 100.So, perhaps we can approach this by considering each possible smallest prime p, and then find all composite numbers n where p is the smallest prime divisor, and n * p ≤ 100. Then, summing over all possible p.Since the smallest prime divisor of a composite number is going to be primes starting from 2 upwards. So, possible primes p are 2, 3, 5, 7, 11, etc. But since n * p ≤ 100, and n is composite with p as its smallest prime divisor, n must be a multiple of p but not a multiple of any prime smaller than p. Therefore, n is of the form p * k, where k is an integer ≥ p (since if k had a prime divisor smaller than p, then that would contradict p being the smallest prime divisor of n). Wait, no. Let me think again.If n is composite and has p as its smallest prime divisor, then n = p * m, where m is an integer ≥ 2. However, m must not be divisible by any prime smaller than p. Since p is the smallest prime divisor, m can be 2 or greater, but if m has prime factors, they must be ≥ p.Therefore, for each prime p, the composite numbers n with smallest prime divisor p are numbers of the form n = p * m, where m is a composite or prime number (but since n is composite, m must be at least 2), and m is co-prime to all primes less than p. So, m must be such that it does not have any prime factors less than p. Therefore, m can be 1) prime numbers ≥ p or 2) composite numbers with all prime factors ≥ p.But since m is ≥ 2 (as n = p * m is composite, so m must be at least 2), and m must not have any prime factors less than p. Therefore, m can be in the set {numbers ≥ 2 with all prime factors ≥ p}.Therefore, for each prime p, the numbers n with smallest prime divisor p are n = p * m, where m ∈ {numbers ≥ 2 with all prime factors ≥ p}.Therefore, in order to find all composite numbers n such that n * p ≤ 100, where p is the smallest prime divisor of n, we can proceed as follows:For each prime p, find all numbers m such that:1. m ≥ 22. All prime factors of m are ≥ p3. n = p * m4. n * p = p² * m ≤ 100Therefore, m ≤ 100 / p².So, for each prime p, the upper bound on m is floor(100 / p²). However, m must also satisfy that all prime factors of m are ≥ p. Therefore, m must be a number ≥ 2, ≤ floor(100 / p²), composed of primes ≥ p.But note that m can be 1 if we allowed, but since m must be at least 2 (because n = p * m is composite, so m ≥ 2), so m starts from 2. Wait, actually, n is composite, so m can be 2 or more. If m = 1, then n = p * 1 = p, which is prime, but we are considering composite n here. Therefore, m must be ≥ 2.So, for each prime p, the composite numbers n with smallest prime divisor p are p * m where m is ≥ 2, m ≤ floor(100 / p²), and m has all prime factors ≥ p.Therefore, the number of such m for each p is equal to the number of integers from 2 to floor(100 / p²) where all prime factors are ≥ p. But since m can also be composite, as long as their prime factors are ≥ p. So, for example, if p = 2, then m can be any number ≥ 2 (since all numbers have prime factors ≥ 2). Wait, but if p = 2, then m can be any number ≥ 2 (since the prime factors of m must be ≥ 2, which they always are). So, for p = 2, m can be 2, 3, 4, ..., floor(100 / 2²) = floor(100 / 4) = 25. So, m can be from 2 to 25. Therefore, the numbers n = 2 * m where m ranges from 2 to 25. But wait, but n must be composite. However, since n = 2 * m and m ≥ 2, n is composite. Therefore, for p=2, all m from 2 to 25 inclusive are valid, so that's 24 numbers (25 - 2 + 1 = 24). Wait, 25 - 2 +1 is 24? Let's check: 2 to 25 inclusive: numbers are 2,3,...,25. That's 24 numbers. So, 24 numbers for p=2.But wait, hold on. When p=2, m can be any number from 2 to 25. However, m must have all prime factors ≥ 2, which is always true. Therefore, yes, all m from 2 to 25 are allowed, hence 24 numbers. So, n would be 4, 6, 8, ..., 50. Since 2*25=50. So, the composite numbers which are multiples of 2 (i.e., even numbers), starting from 4 up to 50. Wait, but even numbers starting at 4 up to 50. How many are those? Let's see, from 4 to 50 inclusive, step 2: (50 - 4)/2 + 1 = (46)/2 +1 = 23 +1=24. So yes, 24 numbers. So, correct.But note that in these numbers, the smallest prime divisor is 2, which is correct because they are even. So, these are the even composite numbers from 4 to 50. So, 24 numbers.Similarly, moving on to p=3. For p=3, the next prime. Then, m must satisfy:m ≥ 2,m ≤ floor(100 / 3²) = floor(100 / 9) = 11.Also, m must have all prime factors ≥ 3. So, m can be numbers from 2 to 11, but with prime factors only 3,5,7,11. However, since m must have prime factors ≥ 3, m cannot be divisible by 2. Therefore, m must be odd numbers from 3 to 11 (since m ≥2, but if m is 2, then n = 3*2=6, which has smallest prime divisor 2, not 3. Therefore, m must not contain prime factors less than 3, so m cannot be even. Therefore, m must be 3,5,7,9,11. Wait, but m ≥2. Wait, m=3,5,7,9,11. Wait, m=3 is allowed. Let me check:Wait, m must be ≥2 and have all prime factors ≥3. So, m can be:- Prime numbers ≥3: 3,5,7,11 (but 11 is allowed as m=11, since 3*11=33, and 33's smallest prime divisor is 3, right? Wait, 33 is 3*11, so yes, its smallest prime divisor is 3. So, m=11 is allowed.- Composite numbers with prime factors ≥3: 9 (3*3), 15 (3*5), 21 (3*7), etc., but since m ≤11, the composite numbers allowed are 9 (3*3).So, m can be 3,5,7,9,11. So, these are the numbers from 3 to 11 where m is either prime (≥3) or composite with factors ≥3.Therefore, m=3,5,7,9,11. So that's 5 numbers. Let's check:For m=3: n=3*3=9. Smallest prime divisor is 3. Correct.m=5: n=3*5=15. Smallest prime divisor is 3. Correct.m=7: n=3*7=21. Smallest prime divisor is 3. Correct.m=9: n=3*9=27. Smallest prime divisor is 3. Correct.m=11: n=3*11=33. Smallest prime divisor is 3. Correct.So, 5 numbers. So, for p=3, there are 5 composite numbers.But wait, hold on. The upper bound on m is 11. So, m can be from 2 to 11, but with m having all prime factors ≥3. But m=2 would lead to n=3*2=6, which is already considered in p=2 case. Since 6's smallest prime divisor is 2, so m=2 is invalid here. Therefore, m must be ≥3 and odd (since even m would introduce a factor of 2). Therefore, m can be 3,5,7,9,11. Therefore, 5 numbers. Correct.Therefore, for p=3, the numbers are 9,15,21,27,33. Each of these numbers has smallest prime divisor 3, and n*p=9*3=27, 15*3=45, 21*3=63, 27*3=81, 33*3=99. All of these are ≤100. So, these are valid. If m=13, then n=3*13=39, and 39*3=117, which exceeds 100, so m=13 is invalid. Thus, m is up to 11. So, 5 numbers for p=3.Moving on to p=5. The next prime. Now, floor(100 /5²)=floor(100/25)=4. So, m must be ≤4. Also, m must be ≥2, and all prime factors of m must be ≥5. But primes ≥5 are 5,7,11,... However, m ≤4, so m can only be 2,3,4. But m must have all prime factors ≥5. But 2,3,4 have prime factors 2,3, which are less than 5. Therefore, there are no such m. Therefore, for p=5, there are no composite numbers n such that n * 5 ≤100 and 5 is the smallest prime divisor. Wait, but let's verify.If p=5, then n=5*m, and m must be ≥2, m ≤4. But m must be such that 5*m has smallest prime divisor 5. However, if m=2, n=10, which has smallest prime divisor 2. Similarly, m=3, n=15, which has smallest prime divisor 3. m=4, n=20, which has smallest prime divisor 2. Therefore, none of these m values result in n having 5 as the smallest prime divisor. Therefore, indeed, there are no composite numbers for p=5.Similarly, for primes larger than 5, like p=7. Then floor(100 /7²)=floor(100/49)=2. So m must be ≤2. m must be ≥2. So m=2. But n=7*2=14. The smallest prime divisor of 14 is 2, not 7. Therefore, invalid. Therefore, no numbers here.Similarly, p=11: floor(100/121)=0, so m would have to be ≤0, which is impossible. So, primes p ≥5 do not contribute any composite numbers. Therefore, only p=2 and p=3 contribute composite numbers.Therefore, total composite numbers are 24 (for p=2) +5 (for p=3)=29.But wait, earlier we considered primes up to 10, which are 4 numbers. So, total numbers would be primes (4) + composites (29)=33. But wait, need to check whether there's any overlap between the primes and composites. Since primes are distinct from composites, there is no overlap. So total numbers are 4+29=33.Wait, but hold on. Let me verify with some examples.For primes, n=2,3,5,7. Each of these, when multiplied by their smallest prime divisor (which is themselves), gives n². 2²=4, 3²=9, 5²=25, 7²=49. All of these are ≤100. The next prime is 11, 11²=121>100, so 11 is excluded. So primes counted are 2,3,5,7. So 4 numbers.For composites:- For p=2: even numbers from 4 to 50. Let's check some of these. For example, 4*2=8 ≤100, 6*2=12 ≤100, ..., 50*2=100 ≤100. So all of these are valid. 24 numbers.- For p=3: numbers 9,15,21,27,33. Let's check 9*3=27, 15*3=45, 21*3=63, 27*3=81, 33*3=99. All ≤100. 5 numbers.Total composites:24+5=29.Total numbers:29+4=33.But wait, let me check if there is a mistake here. Let me list all the numbers:Primes:2,3,5,7.Composites for p=2: even numbers starting from 4 up to 50. So 4,6,8,...,50. That's 24 numbers.Composites for p=3:9,15,21,27,33. That's 5 numbers.So total numbers:4+24+5=33.But let me check if any numbers are counted twice. For example, is 9 counted in both primes and composites? No, 9 is composite, so it's in composites. The primes are separate. So, no overlap.But let me check if any composite number is missed. For example, take n=25. Is 25 counted? 25 is composite. Its smallest prime divisor is 5. So, n*p=25*5=125>100. Therefore, 25 is excluded. Similarly, n=16: smallest prime divisor is 2. 16*2=32≤100, so 16 is in the composites for p=2. Correct. How about n=21: smallest prime divisor is 3. 21*3=63≤100, so it's in composites for p=3. Correct. n=49: smallest prime divisor is7. 49*7=343>100, so it's excluded. So, yes, composites only include numbers where n*p ≤100.So, the total count is 33. Wait, but let me verify by listing all numbers.Primes:2,3,5,7 (4 numbers).Composites for p=2 (even numbers from 4 to50):4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50. That's 24 numbers.Composites for p=3:9,15,21,27,33. That's 5 numbers.Total numbers:4+24+5=33.Wait, 4 primes, 24 even composites, 5 odd composites (multiples of 3). Let me check if there's a composite number missing. For example:n=25: as before, not included because 25*5=125>100.n=35: smallest prime divisor is5. 35*5=175>100. Not included.n=45: smallest prime divisor is3. 45*3=135>100. Wait, 45*3=135>100. Wait, but 45 is a multiple of 3 and 5. So, 45's smallest prime divisor is3. But 45*3=135>100, so 45 is not included. Wait, but according to our earlier logic, for p=3, m=15 would give n=3*15=45. But m=15 would need to be ≤11? Wait, hold on. Wait, for p=3, m must be ≤ floor(100 /9)=11. So, m can be up to 11. So, n=3*11=33. So, 33 is the maximum n for p=3. Therefore, 45 is not included. So, 45 is excluded because m=15 would be required, but 15>11. So, correct, n=45 is not included.Similarly, n=12: included in p=2 (12*2=24≤100). Correct.n=25: excluded. Correct.n=21: included. Correct.n=27: included. 27*3=81≤100. Correct.n=33: included. 33*3=99≤100. Correct.n=35: excluded. Correct.n=39: 39*3=117>100. So, excluded. Correct.n=49: excluded. Correct.So, seems like all the composites are properly included or excluded.Therefore, the total numbers are 33.Wait, but let me check if this matches with manual count.Let me list all numbers:Primes (4):2,3,5,7.Even composites (24):4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50.Odd composites (5):9,15,21,27,33.Total:4+24+5=33.But wait, wait, 24 even composites. Let me count how many numbers from 4 to 50 inclusive, step 2. Starting at 4: 4,6,...,50. The formula gives (50 -4)/2 +1=23 +1=24. Correct. So, 24 numbers.Odd composites:9,15,21,27,33. That's 5 numbers.But let me check if these odd composites are the only ones. For example, 45 is not included. 45 is 3*15, but 15 is too big (m=15>11). So, 45 is excluded.Another example: n=25. It's 5*5. But 5*5=25. The product n*p=25*5=125>100. So, 25 is excluded. So, yes, 25 is not in the list. Similarly, 35=5*7. n*p=35*5=175>100. Excluded. So, all the odd composites in the list are the ones with smallest prime divisor 3, and n=3*m where m=3,5,7,9,11. Which gives n=9,15,21,27,33.So, 5 numbers. Correct.Therefore, total numbers are 33.But wait, just to cross-validate, let's think of another approach.Total numbers n>1 such that n*p ≤100, where p is the smallest prime divisor of n.Alternatively, for each number n>1, check if n multiplied by its smallest prime divisor is ≤100.So, if we list all numbers from 2 upwards, check for each if n * p ≤100, where p is the smallest prime factor of n.But listing all numbers from 2 to, say, 100 (though n*p can be up to 100, so n can be up to 100/p. The maximum n is when p=2, so n=50. So numbers up to 50.Wait, n*p ≤100. Since p is the smallest prime divisor, p ≤ sqrt(n). Wait, not necessarily. For example, if n is prime, p=n, which is larger than sqrt(n) for n>4. So, p can be as big as n.But regardless, n can be up to 50, because when p=2, n can be up to 50 (since 50*2=100). If p is larger, n can be smaller.So, numbers from 2 to 50.So, primes in this range: primes up to 50. But we only count primes up to 10, as primes larger than 10 would have p=n, so n²>100 (since 11²=121>100). So primes n with n<=10: 2,3,5,7.Then, composites from 4 to 50. For composites, n*p <=100, where p is the smallest prime divisor.So, for each composite n from 4 to 50:If n is even, then p=2, check if n*2 <=100. Since n<=50, 50*2=100. So all even composites from 4 to 50 are valid.If n is odd composite, then p is the smallest prime divisor, which is at least 3. So, for odd composites n, check if n*3 <=100. So n <= 100/3 ≈33.333. So, odd composites n where n <=33. So, odd composites from 9 up to 33, with n <=33.But not all odd composites up to33. Because some odd composites have smallest prime divisor 3, but others have smallest prime divisor 5,7, etc. For example, n=25: smallest prime divisor is5. So n*p=25*5=125>100, so excluded. Similarly, n=21: smallest prime divisor is3. 21*3=63<=100, so included.Therefore, odd composites from 9 to33 where the smallest prime divisor is3. Which are numbers divisible by3, but not by2. Wait, but need to ensure that they are not divisible by any prime smaller than3, but since they are odd, not divisible by2. So, numbers divisible by3, not divisible by2. So, the odd multiples of3, starting from3*3=9 up to3*11=33 (since 3*11=33, next would be3*13=39, which would be n=39, but 39*3=117>100). So, n=9,15,21,27,33. So, 5 numbers. So, same as before.Therefore, the composites are:Even composites from4 to50:24 numbers.Odd composites from9 to33 divisible by3:5 numbers.Total composites:29.Plus primes:4.Total:33.Therefore, the answer is33. Therefore, the number of natural numbers greater than one where the product of the number with its smallest prime divisor does not exceed100 is33.But let me check once more. Let's list all numbers:Primes:2,3,5,7 (4).Even composites from4 to50:4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50 (24 numbers).Odd composites:9,15,21,27,33 (5 numbers).Total:4+24+5=33.Yes, that's 33 numbers.Therefore, the answer should be33.Final AnswerThe number of natural numbers is boxed{33}.</think>To determine how many natural numbers greater than one exist such that the product of the number with its smallest prime divisor does not exceed 100, we consider both prime and composite numbers.1. Prime Numbers: - For a prime number ( n ), the smallest prime divisor is ( n ) itself. We need ( n times n leq 100 ), which simplifies to ( n leq 10 ). - The prime numbers less than or equal to 10 are 2, 3, 5, and 7. This gives us 4 prime numbers.2. Composite Numbers: - For composite numbers, we need to consider their smallest prime divisor ( p ). The product ( n times p leq 100 ). - Case 1: Smallest prime divisor ( p = 2 ): - The composite numbers must be even. The largest such ( n ) is ( lfloor 100 / 2 rfloor = 50 ). - The even composite numbers from 4 to 50 are: 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50. This gives 24 numbers. - Case 2: Smallest prime divisor ( p = 3 ): - The composite numbers must be multiples of 3 but not multiples of 2. The largest such ( n ) is ( lfloor 100 / 3 rfloor = 33 ). - The valid numbers are: 9, 15, 21, 27, 33. This gives 5 numbers. - Primes ( p geq 5 ): - For primes ( p geq 5 ), the product ( n times p ) would exceed 100 for any valid ( n ), so there are no such composite numbers.Summarizing the results:- Prime numbers: 4- Composite numbers with smallest prime divisor 2: 24- Composite numbers with smallest prime divisor 3: 5Adding these together, we get ( 4 + 24 + 5 = 33 ).Thus, the number of natural numbers greater than one whose product with their smallest prime divisor does not exceed 100 is (boxed{33}).,<think>Okay, so I need to figure out how many natural numbers greater than 1 have the property that when you multiply the number by its smallest prime divisor, the result is 100 or less. Hmm, let's break this down.First, let's understand the problem. We're looking for all natural numbers n > 1 such that n * p ≤ 100, where p is the smallest prime divisor of n. So, for each number n greater than 1, I need to find its smallest prime factor, multiply them together, and check if that product is less than or equal to 100. Then count how many such numbers exist.Let me start by considering that every natural number greater than 1 has a smallest prime divisor. For prime numbers, their smallest prime divisor is themselves. For composite numbers, the smallest prime divisor is the smallest prime that divides them. For example, 4's smallest prime divisor is 2, 6's is 2, 15's is 3, etc.So, if n is prime, then n * p = n * n = n². So, primes would satisfy the condition if their square is ≤ 100, i.e., primes ≤ 10. The primes less than or equal to 10 are 2, 3, 5, 7. Then primes larger than 10 would have n² > 100, so they wouldn't satisfy the condition.Wait, but hold on. The primes themselves are numbers greater than 1, and we have to consider their product with their smallest prime divisor (which is themselves). So, primes greater than 10 squared would exceed 100. For example, 11 is prime, but 11*11=121>100. So primes greater than 10 cannot be included.So, primes ≤10: 2,3,5,7. That's four primes. Each of these primes when multiplied by themselves (since their smallest prime divisor is themselves) gives a product that is their square. For example, 2*2=4≤100, 3*3=9≤100, etc. All of these squares are indeed ≤100, so these primes are included.Now, for composite numbers. Let's think about composite numbers. For a composite number n, its smallest prime divisor p is the smallest prime that divides n. So, for n to satisfy n * p ≤100, we need p ≤ 100/n. Since p is the smallest prime divisor, it's also at least 2 (since the smallest prime is 2). So, for each composite n, we need to find p (smallest prime divisor of n) such that p ≤ 100/n.Alternatively, for a composite number n, the smallest prime divisor p must be ≤ 100/n. But p is also the smallest prime factor, so p must divide n. So, the key here is that if n is composite, then its smallest prime divisor p must be ≤ 100/n. So, p ≤ 100/n, and p divides n.But since p is the smallest prime divisor of n, then n must be a multiple of p, and all other prime factors of n are ≥ p. So, n can be written as p * m, where m is an integer ≥ p (since if m had a prime factor smaller than p, that would contradict p being the smallest prime factor). Wait, actually, n = p * m, and m is an integer ≥ 2. However, if m has prime factors, they must be ≥ p.Wait, maybe that's overcomplicating. Let's approach it differently. For each composite number n, p is the smallest prime factor. So p is the smallest prime that divides n, so p ≤ sqrt(n), except when n is a square of a prime. Wait, no. Actually, for a composite number, the smallest prime factor p must satisfy p ≤ sqrt(n), unless n is a prime square. Wait, if n is composite, then it has a factor other than 1 and itself. The smallest such factor is p, which is prime. Then n = p * k, where k ≥ p (since otherwise, if k < p, but k has prime factors, which would be smaller than p, contradicting p being the smallest prime factor). So, if n is composite, then p ≤ sqrt(n). Because if p > sqrt(n), then k = n/p < sqrt(n), and then k would have a prime factor less than p, which would contradict p being the smallest prime factor. Therefore, p ≤ sqrt(n) for composite numbers.Therefore, for composite n, p ≤ sqrt(n). But in our problem, we have the condition that n * p ≤ 100. So, n * p ≤ 100, but p ≤ sqrt(n). Therefore, n * p ≤ n * sqrt(n) = n^(3/2). So, n^(3/2) ≥ n * p ≥ (since p ≤ sqrt(n)), but not sure if this helps.Alternatively, let's note that for composite n, since n = p * m, where p is the smallest prime factor and m ≥ p (as otherwise, m would have a smaller prime factor). Therefore, n = p * m ≥ p * p = p². So, n ≥ p². Therefore, p ≤ sqrt(n).So, given that n * p ≤ 100 and n ≥ p², substituting, we get p² * p ≤ 100 ⇒ p³ ≤ 100 ⇒ p ≤ cube root of 100 ≈ 4.64. Therefore, the primes p that are possible here are 2 and 3. Because 5³=125>100.Wait, hold on. Wait, is this a valid approach? If n is composite, then n ≥ p². So, n * p ≥ p³. Therefore, to have n * p ≤ 100, we need p³ ≤ 100. So, p can be 2 or 3. Because 2³=8≤100, 3³=27≤100, 5³=125>100. So, the smallest prime divisor p can only be 2 or 3 for composite numbers n where n * p ≤100.But wait, that can't be right. For example, take n=25, which is 5². Its smallest prime divisor is 5. Then n * p =25 *5=125>100. So, 25 is excluded. But what about n=15? Its smallest prime divisor is 3. Then 15 *3=45≤100. So, 15 is included. Here, p=3, which is allowed even though 3³=27≤100. Wait, but 15 is 3*5. So, n=15=3*5. So, m=5, which is ≥p=3. Then, n * p =15*3=45≤100. So, in this case, even though p=3, which cubed is 27≤100, but n=15 is 3*5. So, perhaps the logic above is not entirely accurate.Wait, maybe I need to approach this differently. Let's consider that for composite numbers, their smallest prime divisor is 2, 3, 5, 7, etc. So, perhaps split composite numbers into two categories: those with smallest prime divisor 2, 3, 5, ..., and handle each case.Alternatively, for each prime p, find all composite numbers n where p is the smallest prime divisor of n, and n * p ≤100. Then sum over all primes p such that p² ≤100 (since n must be at least p², as n is composite and p is the smallest prime divisor). Wait, but n can be p multiplied by another number m ≥ p, but m can have factors greater than or equal to p. So, n can be p * m, where m is ≥ p and m's prime factors are ≥ p.Therefore, for each prime p, the composite numbers with smallest prime divisor p are numbers of the form p * m, where m ≥ p and m is not divisible by any prime less than p.So, m can be any integer ≥ p, with all prime factors ≥ p.Therefore, for each prime p, the numbers n = p * m, where m ≥ p, and m is coprime to all primes less than p.Thus, for each prime p, the numbers n = p * m, m ≥ p, and m is in the set of numbers with all prime factors ≥ p.Therefore, for each prime p, the numbers n with smallest prime divisor p are p * m, where m is in the set of numbers ≥ p and with all prime factors ≥ p.Therefore, to find all composite numbers n such that n * p ≤ 100, where p is the smallest prime divisor of n, we can iterate over primes p, and for each p, find the number of m such that m ≥ p, m has all prime factors ≥ p, and p * m * p ≤ 100. Wait, no. Wait, n = p * m, so n * p = p * m * p = p² * m ≤ 100. Therefore, m ≤ 100 / p².But m must be ≥ p (since n = p * m is composite and m ≥ 2, but actually, m can be 2 if p=2, but m needs to have prime factors ≥ p). Wait, no. For m, since n = p * m, and p is the smallest prime divisor, m cannot have any prime factors less than p. So m can be 1, but if m=1, then n=p, which is prime. But we're considering composite numbers, so m ≥2. Wait, but if m=2, and p=2, then n=4, which is composite, with smallest prime divisor 2. Similarly, for p=2, m can be 2,3,4,... as long as m has no prime factors less than 2 (i.e., m is any integer ≥2). Wait, but m can have prime factors ≥ p=2, so m can be any integer ≥2.Wait, maybe there's confusion here. Let's take p=2. Then n=2*m, where m ≥2 (since n must be composite). However, since p=2 is the smallest prime divisor, m must not have any prime factors less than 2, which is automatically true since the smallest prime is 2. Therefore, m can be any integer ≥2. So, n=2*m where m ≥2. Then n * p = 2*m*2=4m ≤100 ⇒ m ≤25. So, m can be from 2 to 25. But wait, m must be ≥2, so the numbers are 4,6,8,...,50 (since 2*25=50). Therefore, how many such numbers? From m=2 to m=25, so 25 -1 =24 numbers. Wait, 25 m's starting at m=2. Wait, 25-2 +1=24. Wait, no. If m starts at 2 and goes up to 25, inclusive, that's 25-2+1=24 numbers. But n=2*m, so n would be 4,6,8,...,50. So, 24 numbers.But wait, let's check. For p=2, n=2*m, m≥2, and n*p=4*m ≤100 ⇒ m≤25. So m=2,3,...,25. So 25-1=24 numbers. So 24 composite numbers with smallest prime divisor 2, such that n*p ≤100.But wait, but m can be any integer ≥2, but n must be composite. Since n=2*m, and m ≥2, n is automatically composite (as 2*2=4). So all these are composite numbers. So that's 24 numbers.Now, next prime p=3. For p=3, n=3*m, where m ≥3 (since if m=1, n=3 which is prime, but we need composite numbers, so m ≥2. But also, since p=3 is the smallest prime divisor, m must not have any prime factors less than 3. Therefore, m must be an integer ≥2 with all prime factors ≥3. So m can be 3,4,5,... but m cannot be divisible by 2. So m must be odd or even? Wait, no. Wait, m can be even, but then 2 would be a prime factor less than 3, which is not allowed because p=3 is supposed to be the smallest prime factor. Therefore, m must be co-prime to all primes less than 3, which is just 2. Therefore, m must be odd. Also, m must be ≥3? Wait, n=3*m, and n must be composite. If m=2, then n=6, which is composite. But in this case, n=6, and the smallest prime divisor of 6 is 2, not 3. So, m=2 would lead to n=6, but 6's smallest prime divisor is 2, so this number should be considered in the p=2 case, not p=3. Therefore, in the case of p=3, m must be such that n=3*m has smallest prime divisor 3. Therefore, m must not be divisible by any prime less than 3, i.e., m must be co-prime to 2. So m must be odd. Also, m must be ≥2? If m=2, as above, n=6, which would have smallest prime divisor 2, so m must be ≥3 and odd. Wait, if m=1, n=3 is prime. If m=2, n=6, which is already in p=2. If m=3, n=9. The smallest prime divisor of 9 is 3. So m=3 is allowed. Similarly, m=5, n=15. The smallest prime divisor of 15 is 3? No, 15 is divisible by 3 and 5. The smallest prime divisor is 3. Wait, 15 divided by 3 is 5. So yes, the smallest prime divisor is 3. So, m=5 is allowed. Wait, but m=5 is allowed because m=5 is co-prime to 2. So, m must be co-prime to primes less than p=3, i.e., 2, so m must be odd, and m ≥3? Wait, n=3*m must be composite, so m can be 2, but then n=6, which is composite but not in p=3. So, to ensure that the smallest prime divisor is 3, m must be co-prime to 2 and m ≥3? Because if m is co-prime to 2 and ≥3, then n=3*m is composite and not divisible by 2, hence the smallest prime divisor is 3.Wait, so m must be odd numbers ≥3? Let me check with m=3: n=9, smallest prime divisor 3. m=5: n=15, smallest prime divisor 3 (since 15 is divisible by 3). m=7: n=21, smallest prime divisor 3 (if 21 is divisible by 3). Yes, 21=3*7. So yes, 3 is the smallest prime divisor. Similarly, m=9: n=27, which is 3*9, but 9=3², so n=3³. The smallest prime divisor is 3. So, even if m is a multiple of 3, as long as m is not divisible by 2, it's okay.Therefore, for p=3, m must be integers ≥3, co-prime to 2 (i.e., odd), such that n*p=3*m*3=9m ≤100. So, 9m ≤100 ⇒ m ≤100/9≈11.11. So, m can be from 3 to 11, inclusive, and m must be odd. So, possible m values: 3,5,7,9,11. That's 5 numbers. Therefore, n=9,15,21,27,33. Let's check:- 9*3=27≤100- 15*3=45≤100- 21*3=63≤100- 27*3=81≤100- 33*3=99≤100- Next m=13 would give n=39, 39*3=117>100. So, m stops at 11.So, 5 numbers for p=3.Next prime p=5. Let's do similar steps. For p=5, n=5*m. To ensure that 5 is the smallest prime divisor, m must not be divisible by any prime less than 5, i.e., 2 and 3. Therefore, m must be co-prime to 2 and 3, so m must be an integer ≥2 such that m is not divisible by 2 or 3. Also, n=5*m must be composite, so m must be ≥2 (since m=1 would make n=5, prime). But m=2 would make n=10, but 10's smallest prime divisor is 2, not 5. Therefore, m must be co-prime to 2 and 3, and m ≥?Wait, let's see. If m=1, n=5, prime. m=2: n=10, composite, but smallest prime divisor 2. m=3: n=15, which has smallest prime divisor 3. m=4: n=20, smallest prime divisor 2. m=5: n=25, smallest prime divisor 5. m=7: n=35, which is 5*7, so smallest prime divisor 5. Wait, but 35 is divisible by 5 and 7. The smallest is 5. So, m=5,7,11,... but m must be co-prime to 2 and 3.So, m must be integers ≥1, co-prime to 2 and 3. But to ensure that n=5*m has smallest prime divisor 5, m must not introduce any prime factors less than 5. So, m must be co-prime to 2,3 and also m must not have prime factors less than 5. But since 2 and 3 are already excluded, m can have prime factors ≥5. So, m can be 1,5,7,11, etc., but m=1 would make n=5, prime. So, m must be ≥5 and co-prime to 2 and 3? Wait, but m=5: n=25, which is 5², smallest prime divisor 5. m=7: n=35, which is 5*7, smallest prime divisor 5. m=11: n=55, which is 5*11, smallest prime divisor 5. Similarly, m=5*5=25: n=125, but n*p=125*5=625>100. So, we need to find m such that n*p=5*m*5=25m ≤100. Therefore, 25m ≤100 ⇒ m ≤4. But wait, m must be co-prime to 2 and 3 (i.e., not divisible by 2 or 3), and m must be such that n=5*m has smallest prime divisor 5.Wait, if m ≤4, but m must be co-prime to 2 and 3. The numbers ≤4 co-prime to 2 and 3 are 1 and 5, but 5 is greater than 4. Wait, m ≤4. So, m=1 is co-prime to 2 and 3. m=1 gives n=5, prime. m=5 is co-prime to 2 and 3 but 5>4. So, there are no such m. Therefore, there are no composite numbers with smallest prime divisor 5 such that n*p ≤100.Wait, that seems odd. Let's check. For p=5, n=5*m. The condition is n*p=25m ≤100 ⇒ m ≤4. But m must be co-prime to 2 and 3 (so m can be 1,5,7,...). But m ≤4, so the only possible m co-prime to 2 and 3 is m=1. But m=1 gives n=5, which is prime. Therefore, there are no composite numbers n with smallest prime divisor 5 such that n*p ≤100. Therefore, p=5 case contributes 0 numbers.Similarly, primes larger than 5. Let's check p=7. n=7*m. Then n*p=7*m*7=49m ≤100 ⇒ m ≤2.04. So, m ≤2. But m must be co-prime to primes less than 7, i.e., 2,3,5. So m must be 1 or 7. But m=1 gives n=7, prime. m=7 gives n=49, but 49*7=343>100. So, m=1 is prime, m=7 is too big. Therefore, no composite numbers here.Similarly, primes p=11 and higher. For p=11, n=11*m. Then n*p=121m ≤100. But 121m ≤100 only possible if m=0, which isn't allowed. So, no numbers here.Therefore, primes p≥5 do not contribute any composite numbers to the count. So, only p=2 and p=3 contribute composite numbers.So, summarizing:- Prime numbers n where n² ≤100: primes ≤10, which are 2,3,5,7. So 4 primes.- Composite numbers with smallest prime divisor 2: n=4,6,8,...,50. 24 numbers.- Composite numbers with smallest prime divisor 3: n=9,15,21,27,33. 5 numbers.Total numbers: 4 +24 +5=33.Wait, but let's verify this.First, primes: 2,3,5,7. 4 numbers.Composite numbers with smallest prime divisor 2: numbers of the form 2*m, m≥2, and 2*m*2 ≤100 ⇒ m ≤25. So m=2 to 25, inclusive. Therefore, numbers n=4,6,8,...,50. The numbers from 4 to 50 in steps of 2. The count is (50 -4)/2 +1 = (46)/2 +1=23 +1=24. So 24 numbers.Composite numbers with smallest prime divisor 3: n=3*m, where m is co-prime to 2, m≥3, and 3*m*3 ≤100 ⇒ m ≤11.11. So m=3,5,7,9,11. Let's check each:- m=3: n=9, 9*3=27≤100. Smallest prime divisor is 3. OK.- m=5: n=15, 15*3=45≤100. Smallest prime divisor 3. OK.- m=7: n=21, 21*3=63≤100. Smallest prime divisor 3. OK.- m=9: n=27, 27*3=81≤100. Smallest prime divisor 3. OK.- m=11: n=33, 33*3=99≤100. Smallest prime divisor 3. OK.m=13: n=39, 39*3=117>100. Excluded.So, 5 numbers. So 5.Total numbers: 4+24+5=33.But wait, need to check if there is any overlap between these composite numbers and the primes. For example, n=4 is composite, n=6 is composite, etc., and primes are separate. So no overlap. So total numbers are 33.But wait, let's take a few examples.Take n=4: prime? No. Its smallest prime divisor is 2. 4*2=8≤100. Included in composite with p=2.n=9: composite, smallest prime divisor 3. 9*3=27≤100. Included in composite with p=3.n=15: composite, smallest prime divisor 3. 15*3=45≤100. Included.n=25: composite, smallest prime divisor 5. 25*5=125>100. Not included. So, as we saw, p=5 gives no numbers.Wait, let's check n=25. Since p=5, and n*p=125>100, so it's excluded. So, no.Similarly, n=49: 49*7=343>100. Excluded.So, seems okay. But wait, let's check n=21. 21 is composite, smallest prime divisor is 3. 21*3=63≤100. Included.So, the total is 4 primes and 24+5=29 composites. 4+29=33.But wait, let's check another way. Let's list all numbers from 2 to 100, but n*p ≤100. Wait, but n can be up to 100/p, where p is the smallest prime divisor. For primes, n*p=n² ≤100 ⇒ n≤10. For composites, n can be up to 50 (when p=2, m=25 gives n=50. 50*2=100). So numbers can go up to 50.But instead of listing all numbers, maybe it's better to check if the count is correct.But let's test with a smaller limit. Suppose instead of 100, the limit was 10. Then primes would be n where n² ≤10, so primes up to 3 (2,3). Composites with p=2: n=4,6,8,10 (since 2*m*2=4m ≤10 ⇒ m ≤2.5. So m=2 gives n=4, m=2.5 is not integer. So m=2. But n=4: 4*2=8≤10. Next m=3 would be 6*2=12>10. So only n=4. Then composites with p=3: n=9, but 9*3=27>10. So none. So total numbers would be primes 2,3 and composite 4. Total 3 numbers. 2,3,4. 2*2=4≤10, 3*3=9≤10, 4*2=8≤10. So yes.So, with limit 10, answer is 3. Which makes sense.Similarly, with limit 100, according to our previous count, it's 33. Let's see if this holds.Alternatively, maybe there is an error in the composite counts.For p=2, composite numbers are 4,6,8,...,50. 24 numbers.Wait, 50 is included because 50*2=100. So yes. So from 4 to 50 even numbers. So, how many even numbers from 4 to 50 inclusive? (50 -4)/2 +1 = (46)/2 +1=23 +1=24. Correct.For p=3, composite numbers are 9,15,21,27,33. Let's count them:- 9: OK.- 15: OK.- 21: OK.- 27: OK.- 33: OK.- Next would be 39: 39*3=117>100. Excluded.So 5 numbers. Correct.Primes: 2,3,5,7. 4 numbers.Total: 24 +5 +4=33.So, I think this is correct.But let's check if any composite numbers are missed. For example, n=25. As before, n=25. Its smallest prime divisor is 5. 25*5=125>100, so excluded.n=16: smallest prime divisor 2. 16*2=32≤100. So included in the p=2 composite numbers. Yes, 16 is even, so in the list.n=12: smallest prime divisor 2. 12*2=24≤100. Included.n=18: same.n=49: smallest prime divisor 7. 49*7=343>100. Excluded.n=35: smallest prime divisor 5. 35*5=175>100. Excluded.n=21: included in p=3.n=33: included in p=3.Wait, n=45: smallest prime divisor 3. 45*3=135>100. So, n=45 is composite, but 45*3=135>100. So excluded.Wait, hold on. So, n=33 is 3*11, which is 33*3=99≤100. Next multiple would be 3*13=39, 39*3=117>100. So, n=33 is the last one for p=3.But n=45=3*15. But 15 is allowed? Wait, no. Wait, n=45. Its smallest prime divisor is 3. 45*3=135>100, so it's excluded. So, n=45 is not included. So, only up to n=33.Therefore, the count is correct.Therefore, the total number is 33.But wait, let me check another angle. Suppose I list all numbers from 2 to 100, and for each number, check if n multiplied by its smallest prime divisor is ≤100. Then count how many numbers satisfy this. But this would be tedious, but maybe we can spot check.Take n=10. Its smallest prime divisor is 2. 10*2=20≤100. Included. So, n=10 is in the composite numbers with p=2.n=11. Prime. 11²=121>100. Not included.n=12. Smallest prime divisor 2. 12*2=24≤100. Included.n=13. Prime. 13²=169>100. Not included.n=14. Smallest prime divisor 2. 14*2=28≤100. Included.n=15. Smallest prime divisor 3. 15*3=45≤100. Included.n=16. Smallest prime divisor 2. Included.n=17. Prime. 17²=289>100. Not included.n=18. Included (p=2).n=19. Prime. 19²=361>100. Not included.n=20. Included (p=2).n=21. Included (p=3).n=22. Included (p=2).n=23. Prime. 23²=529>100. Not included.n=24. Included (p=2).n=25. Excluded (p=5, 25*5=125>100).n=26. Included (p=2).n=27. Included (p=3).n=28. Included (p=2).n=29. Prime. Excluded.n=30. Included (p=2).n=31. Prime. Excluded.n=32. Included (p=2).n=33. Included (p=3).n=34. Included (p=2).n=35. Excluded (p=5, 35*5=175>100).n=36. Included (p=2).n=37. Prime. Excluded.n=38. Included (p=2).n=39. Excluded (p=3, 39*3=117>100).n=40. Included (p=2).n=41. Prime. Excluded.n=42. Included (p=2).n=43. Prime. Excluded.n=44. Included (p=2).n=45. Excluded (p=3, 45*3=135>100).n=46. Included (p=2).n=47. Prime. Excluded.n=48. Included (p=2).n=49. Excluded (p=7, 49*7=343>100).n=50. Included (p=2, 50*2=100).n=51. Excluded (p=3, 51*3=153>100).... And so on up to n=100. But note that numbers above 50 with p=2 would have n*p=2*51=102>100, so they are excluded. Wait, n=51 is excluded because 51*3=153>100. But n=50 is included as 50*2=100.But according to our previous count, composite numbers with p=2 go up to n=50. So, from 4 to 50 even numbers. So that's 24 numbers. Then composites with p=3 go up to n=33. 5 numbers. Then primes up to 10: 4 numbers.So, total 33 numbers. When I spot-check, it seems correct. For example, n=50 is included, n=51 excluded, primes above 10 excluded, composites with p≥5 excluded. So, 33.Therefore, the answer should be 33. So, boxed, it's boxed{33}.Final Answerboxed{33}</think>To determine how many natural numbers greater than one exist such that the product of the number with its smallest prime divisor does not exceed 100, we consider both prime and composite numbers.1. Prime Numbers: - For a prime number ( n ), the smallest prime divisor is ( n ) itself. The product ( n times n = n^2 ) must be ≤ 100. - Prime numbers ( n ) such that ( n^2 leq 100 ) are 2, 3, 5, and 7. This gives 4 prime numbers.2. Composite Numbers: - Composite numbers ( n ) must be considered with their smallest prime divisor ( p ). The product ( n times p leq 100 ). - Composite numbers with smallest prime divisor 2: - These numbers are of the form ( 2 times m ) where ( m geq 2 ). The product ( 2 times m times 2 = 4m leq 100 ) implies ( m leq 25 ). - Values of ( m ) range from 2 to 25, giving 24 composite numbers (4, 6, 8, ..., 50). - Composite numbers with smallest prime divisor 3: - These numbers are of the form ( 3 times m ) where ( m geq 3 ) and ( m ) is co-prime to 2 (i.e., odd). The product ( 3 times m times 3 = 9m leq 100 ) implies ( m leq 11.11 ). - Valid values of ( m ) are 3, 5, 7, 9, 11, giving 5 composite numbers (9, 15, 21, 27, 33). - Composite numbers with smallest prime divisor 5 or higher: - For primes ( p geq 5 ), the product ( p times m times p ) exceeds 100 for any valid ( m ), so there are no such composite numbers.Summarizing the counts:- Prime numbers: 4- Composite numbers with smallest prime divisor 2: 24- Composite numbers with smallest prime divisor 3: 5Total numbers: ( 4 + 24 + 5 = 33 ).Thus, the final answer is (boxed{33}).]

🔑:Answer: 33.Sketch of the solution. The smallest prime divisor can be 2. These are the numbers: 2, 4, ldots, 50 (25 numbers). The smallest prime divisor can be 3. These are the numbers: 3, 9, 15, 21, 27, 33 (6 numbers). The smallest prime divisor can be 5. This number: 5 (1 number). The smallest prime divisor can be 7. This number: 7 (1 number). Since 11 times 11=121>100, there are no other prime divisors with this property. Total numbers: 25+5+1+1=33.Criteria. Correct answer only: 1 point.

🧠:[<think>Okay, let's see. I have this problem about a parabola. The first part is to find the equation of the parabola with its vertex at the origin, passing through point A(-4, 4), and the focus lies on the x-axis. The second part is about finding the equation of a line that passes through point B(-1, 0) and intersects the parabola such that the chord length is 8. Hmm, okay. Let's start with part (1).First, since the vertex is at the origin, the standard form of a parabola can be either ( y^2 = 4ax ) or ( x^2 = 4ay ). But the focus lies on the x-axis. The focus of a parabola in the form ( y^2 = 4ax ) is at (a, 0), which is on the x-axis. For the form ( x^2 = 4ay ), the focus is at (0, a), which would be on the y-axis. Since the problem says the focus is on the x-axis, it must be the first form, ( y^2 = 4ax ). So, the parabola opens either to the right or left. The coefficient 'a' determines the direction. If a is positive, it opens to the right; if negative, to the left.Given that the parabola passes through point A(-4, 4), we can substitute this point into the equation to find 'a'. Let's plug in x = -4 and y = 4 into ( y^2 = 4ax ):( 4^2 = 4a(-4) )( 16 = -16a )Divide both sides by -16:( a = -1 )So, the equation is ( y^2 = 4(-1)x ), which simplifies to ( y^2 = -4x ). Wait, let me check. If a is -1, then 4a is -4, so yes, ( y^2 = -4x ). That makes sense because the parabola opens to the left, given that a is negative. The focus is at (a, 0) which would be (-1, 0), lying on the x-axis. That seems right. So part (1) seems done. Let me verify by plugging point A into the equation:Left side: ( 4^2 = 16 )Right side: ( -4*(-4) = 16 ). Yes, it works. Okay, so the equation is ( y^2 = -4x ).Moving on to part (2). We need to find the equation of line l that passes through point B(-1, 0). The line intersects the parabola, forming a chord of length 8. So, we need to find the equation(s) of such lines.First, let's recall that the general equation of a line passing through point (-1, 0) can be written in slope-intercept form. However, since we know a point it passes through, we can use point-slope form. Let's let the slope be m. Then, the equation of the line is:( y - 0 = m(x - (-1)) )Simplify: ( y = m(x + 1) )So, the equation is ( y = m(x + 1) ). Now, we need to find the values of m such that when this line intersects the parabola ( y^2 = -4x ), the chord length is 8.To find the points of intersection between the line and the parabola, we can substitute y from the line into the parabola's equation.Substitute ( y = m(x + 1) ) into ( y^2 = -4x ):( [m(x + 1)]^2 = -4x )Expand left side:( m^2(x + 1)^2 = -4x )( m^2(x^2 + 2x + 1) = -4x )Bring all terms to one side:( m^2x^2 + 2m^2x + m^2 + 4x = 0 )Combine like terms:( m^2x^2 + (2m^2 + 4)x + m^2 = 0 )This is a quadratic equation in x. Let's denote this as:( ax^2 + bx + c = 0 ), where:a = ( m^2 )b = ( 2m^2 + 4 )c = ( m^2 )The solutions to this quadratic will give the x-coordinates of the intersection points. Let's denote the roots as x1 and x2. Then, the corresponding y-coordinates will be y1 = m(x1 + 1) and y2 = m(x2 + 1).The distance between the two points (x1, y1) and (x2, y2) should be 8. To compute the distance, we can use the distance formula:( sqrt{(x2 - x1)^2 + (y2 - y1)^2} = 8 )But since y = m(x + 1), the difference in y's can be expressed in terms of the difference in x's:( y2 - y1 = m(x2 + 1) - m(x1 + 1) = m(x2 - x1) )So, the distance becomes:( sqrt{(x2 - x1)^2 + [m(x2 - x1)]^2} = 8 )Factor out (x2 - x1):( |x2 - x1| sqrt{1 + m^2} = 8 )Therefore:( |x2 - x1| = frac{8}{sqrt{1 + m^2}} )But |x2 - x1| can be found using the quadratic formula. For a quadratic equation ax² + bx + c = 0, the difference of roots is sqrt(D)/a, where D is the discriminant. Wait, actually, the roots are [ -b + sqrt(D) ] / (2a) and [ -b - sqrt(D) ] / (2a). So, the difference is sqrt(D)/a.Wait, let's recall:For quadratic equation ax² + bx + c = 0, the roots are x1 and x2. Then:x1 + x2 = -b/ax1x2 = c/aThe difference x2 - x1 can be expressed as sqrt[(x1 + x2)^2 - 4x1x2] = sqrt[(b² - 4ac)/a²] = sqrt(D)/|a|But since in our case, a = m², which is positive (since squared), so |a| = a. So, |x2 - x1| = sqrt(D)/a.Wait, but D = b² - 4ac.So, in our case, D = [2m² + 4]^2 - 4*(m²)*(m²)Let's compute D:D = (2m² + 4)^2 - 4m²*m²Expand (2m² + 4)^2:= 4m^4 + 16m² + 16 - 4m^4= (4m^4 - 4m^4) + 16m² + 16= 16m² + 16Factor:= 16(m² + 1)So, D = 16(m² + 1)Thus, |x2 - x1| = sqrt(D)/a = sqrt(16(m² + 1)) / m² = 4sqrt(m² +1)/m²Therefore, plugging back into the distance equation:|x2 - x1| * sqrt(1 + m²) = 8Which is:[4sqrt(m² +1)/m²] * sqrt(1 + m²) = 8Multiply the terms:4(m² +1)/m² = 8Simplify:4(m² +1) = 8m²Divide both sides by 4:m² +1 = 2m²Subtract m² from both sides:1 = m²Therefore, m² = 1 => m = ±1So, the slopes are m = 1 and m = -1Therefore, the equations of the lines are:For m = 1: y = 1*(x + 1) => y = x + 1For m = -1: y = -1*(x + 1) => y = -x -1Therefore, there are two possible lines: y = x +1 and y = -x -1.Wait, but let's verify if these lines actually result in chords of length 8.Let's take m =1:Equation of line: y = x +1Intersection with parabola y² = -4x.Substitute y = x +1 into parabola:(x +1)^2 = -4xExpand:x² + 2x +1 = -4xBring all terms to left:x² +6x +1 =0Solutions:x = [-6 ± sqrt(36 -4*1*1)]/2 = [-6 ± sqrt(32)]/2 = [-6 ± 4*sqrt(2)]/2 = -3 ± 2*sqrt(2)Therefore, x coordinates are -3 + 2√2 and -3 - 2√2.Corresponding y coordinates:For x = -3 + 2√2: y = (-3 + 2√2) +1 = -2 + 2√2For x = -3 - 2√2: y = (-3 - 2√2) +1 = -2 - 2√2Thus, the two points are (-3 + 2√2, -2 + 2√2) and (-3 - 2√2, -2 - 2√2)Compute the distance between these two points.Difference in x: [ -3 + 2√2 - ( -3 - 2√2 ) ] = 4√2Difference in y: [ -2 + 2√2 - ( -2 - 2√2 ) ] = 4√2Therefore, distance is sqrt( (4√2)^2 + (4√2)^2 ) = sqrt( 32 + 32 ) = sqrt(64) = 8. Correct.Similarly, for m = -1:Equation: y = -x -1Intersection with parabola:(-x -1)^2 = -4xExpand:x² + 2x +1 = -4xx² +6x +1 =0Same quadratic as before, so solutions x = -3 ± 2√2But y = -x -1:For x = -3 + 2√2: y = -(-3 + 2√2) -1 = 3 - 2√2 -1 = 2 - 2√2For x = -3 - 2√2: y = -(-3 - 2√2) -1 = 3 + 2√2 -1 = 2 + 2√2Points: (-3 + 2√2, 2 - 2√2) and (-3 - 2√2, 2 + 2√2)Compute distance:Difference in x: 4√2Difference in y: [ (2 + 2√2) - (2 - 2√2) ] = 4√2Thus, distance is sqrt( (4√2)^2 + (4√2)^2 ) = sqrt(64) =8. Correct.Therefore, both lines y = x +1 and y = -x -1 satisfy the condition. Thus, the equations are y = x +1 and y = -x -1.But wait, let me check if there are any vertical lines. The problem says "line l passes through point B(-1, 0)", so vertical line is x = -1. Let's check if x = -1 intersects the parabola y² = -4x. Then x = -1, so y² = -4*(-1) =4, so y= ±2. So the points are (-1,2) and (-1,-2). The distance between them is sqrt( (0)^2 + (4)^2 ) =4. But 4≠8, so vertical line doesn't work. So no vertical line. Therefore, the only possible lines are with slopes 1 and -1.But wait, the problem says "the length of the chord formed by its intersection with the parabola is 8". So, are there only two lines? Since when we solved, we got two solutions. So, answer is two lines: y = x +1 and y = -x -1.Hence, part (2) answers are these two equations.But let me check another approach. Suppose we parameterize the line differently. Maybe using parametric equations or another form, but I think the method was correct. The discriminant approach leading to m² =1, so m = ±1. Seems solid. And verifying with actual distance confirmed it. So, I think that's the answer.Final Answer(1) The equation of the parabola is boxed{y^2 = -4x}.(2) The equations of line l are boxed{y = x + 1} and boxed{y = -x - 1}.</think>Given that the vertex of a parabola is at the origin, it passes through point A(-4, 4) and its focus lies on the x-axis.(1) To find the equation of the parabola:- Since the focus lies on the x-axis, the parabola is of the form ( y^2 = 4ax ).- Substituting the point A(-4, 4) into the equation: ( 4^2 = 4a(-4) ).- Solving for ( a ): ( 16 = -16a ) gives ( a = -1 ).- Therefore, the equation of the parabola is ( y^2 = -4x ).(2) To find the equation of line l passing through point B(-1, 0) with a chord length of 8:- The general equation of the line passing through B(-1, 0) is ( y = m(x + 1) ).- Substituting ( y = m(x + 1) ) into ( y^2 = -4x ) gives ( m^2(x + 1)^2 = -4x ).- Simplifying: ( m^2x^2 + (2m^2 + 4)x + m^2 = 0 ).- The discriminant ( D ) of this quadratic equation is ( 16(m^2 + 1) ).- The distance between the intersection points is given by ( |x_2 - x_1| sqrt{1 + m^2} = 8 ).- Solving for ( m ): ( 4sqrt{1 + m^2}/m^2 cdot sqrt{1 + m^2} = 8 ) leads to ( m^2 = 1 ).- Thus, ( m = pm 1 ), giving the lines ( y = x + 1 ) and ( y = -x - 1 ).Final answers:(1) The equation of the parabola is boxed{y^2 = -4x}.(2) The equations of line l are boxed{y = x + 1} and boxed{y = -x - 1}.,<think>Okay, so I have this problem about a parabola. Let's see... Part (1) says that the vertex is at the origin, the parabola passes through point A(-4, 4), and the focus lies on the x-axis. I need to find the equation of the parabola. Hmm, let me recall some basics about parabolas.Since the vertex is at the origin, the standard form of a parabola can be either ( y^2 = 4px ) or ( x^2 = 4py ), depending on whether it opens horizontally or vertically. The problem mentions that the focus lies on the x-axis. If the focus is on the x-axis, that means the parabola must open either to the right or to the left. So, the standard form would be ( y^2 = 4px ), where p is the distance from the vertex to the focus. If p is positive, it opens to the right; if negative, to the left.Given that the parabola passes through point A(-4, 4), we can substitute these coordinates into the equation to find p. Let's plug in x = -4 and y = 4 into ( y^2 = 4px ):( (4)^2 = 4p(-4) )16 = -16pDivide both sides by -16:p = -1So, the value of p is -1. Since p is negative, the parabola opens to the left. Substituting back into the standard form, the equation is:( y^2 = 4(-1)x )( y^2 = -4x )Wait, let me double-check. If p = -1, then 4p = -4, so yes, the equation is ( y^2 = -4x ). Let's confirm if this passes through (-4, 4):Left side: ( 4^2 = 16 )Right side: -4*(-4) = 16Yes, 16 = 16. That checks out. And the focus would be at (p, 0) = (-1, 0), which is on the x-axis. Perfect. So part (1) seems solved. The equation is ( y^2 = -4x ).Moving on to part (2): We need to find the equation of line l that passes through point B(-1, 0). The line intersects the parabola, and the length of the chord formed is 8. So, line l passes through (-1, 0), and when it intersects the parabola ( y^2 = -4x ), the distance between the two intersection points is 8. We need to find the equation(s) of such line(s).First, let's recall that a line passing through (-1, 0) can be written in the form ( y = m(x + 1) ), where m is the slope. Alternatively, if the line is vertical, the equation would be x = -1, but since x = -1 intersects the parabola at y^2 = -4*(-1) = 4, so y = ±2. The points would be (-1, 2) and (-1, -2). The distance between these points is 4, which is not 8, so the vertical line x = -1 is not a solution. Therefore, the line must have a slope, so we can use the slope-intercept form.Let me proceed with the slope form. Let’s denote the equation of line l as ( y = m(x + 1) ). To find the points of intersection between the line and the parabola, we can substitute y from the line equation into the parabola's equation.Substitute ( y = m(x + 1) ) into ( y^2 = -4x ):( [m(x + 1)]^2 = -4x )Expand the left side:( m^2(x + 1)^2 = -4x )( m^2(x^2 + 2x + 1) = -4x )Bring all terms to one side:( m^2x^2 + 2m^2x + m^2 + 4x = 0 )Combine like terms:( m^2x^2 + (2m^2 + 4)x + m^2 = 0 )So, this is a quadratic in x. Let's denote the quadratic as ( ax^2 + bx + c = 0 ), where:a = ( m^2 )b = ( 2m^2 + 4 )c = ( m^2 )The solutions to this quadratic will give the x-coordinates of the intersection points. Let’s denote the roots as x1 and x2. Then, the corresponding y-coordinates will be y1 = m(x1 + 1) and y2 = m(x2 + 1).The distance between the two intersection points (x1, y1) and (x2, y2) should be 8. The distance formula is:Distance = ( sqrt{(x2 - x1)^2 + (y2 - y1)^2} = 8 )First, let's compute (x2 - x1) and (y2 - y1). From the quadratic equation, we know that:x1 + x2 = -b/a = -(2m^2 + 4)/m^2x1x2 = c/a = m^2/m^2 = 1The difference (x2 - x1) can be found using ( sqrt{(x1 + x2)^2 - 4x1x2} ):So, ( x2 - x1 = sqrt{[-(2m^2 + 4)/m^2]^2 - 4*1} )But let's compute this step by step.First, compute (x1 + x2)^2:= [ ( - (2m^2 + 4)/m^2 ) ]^2= [ (2m^2 + 4)/m^2 ]^2= (4m^4 + 16m^2 + 16)/m^4Then subtract 4x1x2:= (4m^4 + 16m^2 + 16)/m^4 - 4*(1)= (4m^4 + 16m^2 + 16 - 4m^4)/m^4= (16m^2 + 16)/m^4= 16(m^2 + 1)/m^4Therefore, ( x2 - x1 = sqrt{16(m^2 + 1)/m^4} = 4sqrt{(m^2 + 1)}/m^2 )Similarly, y2 - y1 = m(x2 + 1) - m(x1 + 1) = m(x2 - x1)So, (y2 - y1)^2 = m^2(x2 - x1)^2 = m^2 * [16(m^2 + 1)/m^4] = 16(m^2 + 1)/m^2Therefore, the distance squared is:(x2 - x1)^2 + (y2 - y1)^2 = [16(m^2 + 1)/m^4] + [16(m^2 + 1)/m^2]Factor out 16(m^2 + 1):= 16(m^2 + 1)[1/m^4 + 1/m^2]= 16(m^2 + 1)[ (1 + m^2)/m^4 ]= 16(m^2 + 1)^2 / m^4Therefore, the distance is sqrt(16(m^2 + 1)^2 / m^4) = 4(m^2 + 1)/m^2Set this equal to 8:4(m^2 + 1)/m^2 = 8Divide both sides by 4:(m^2 + 1)/m^2 = 2Multiply both sides by m^2:m^2 + 1 = 2m^2Subtract m^2 + 1 from both sides:0 = m^2 -1Therefore, m^2 = 1, so m = ±1Therefore, the slopes are m = 1 and m = -1.Hence, the equations of the lines are:For m = 1:y = 1*(x + 1) => y = x + 1For m = -1:y = -1*(x + 1) => y = -x -1So, these are two possible lines. Let me verify if they indeed result in a chord length of 8.First, take m = 1: y = x + 1Substitute into the parabola equation:(x + 1)^2 = -4xx^2 + 2x + 1 = -4xx^2 + 6x + 1 = 0Solutions:x = [-6 ± sqrt(36 - 4*1*1)]/2 = [-6 ± sqrt(32)]/2 = [-6 ± 4*sqrt(2)]/2 = -3 ± 2*sqrt(2)Then, the corresponding y-coordinates are y = x + 1, so:For x = -3 + 2√2, y = (-3 + 2√2) + 1 = -2 + 2√2For x = -3 - 2√2, y = (-3 - 2√2) + 1 = -2 - 2√2Distance between (-3 + 2√2, -2 + 2√2) and (-3 - 2√2, -2 - 2√2)Compute differences:Δx = (-3 - 2√2) - (-3 + 2√2) = -4√2Δy = (-2 - 2√2) - (-2 + 2√2) = -4√2Distance = sqrt[ (-4√2)^2 + (-4√2)^2 ] = sqrt[ 32 + 32 ] = sqrt[64] = 8. Correct.Similarly for m = -1: y = -x -1Substitute into parabola:(-x -1)^2 = -4xx^2 + 2x + 1 = -4xx^2 + 6x + 1 = 0Same quadratic as before, so solutions x = -3 ± 2√2Then y = -x -1:For x = -3 + 2√2, y = -(-3 + 2√2) -1 = 3 - 2√2 -1 = 2 - 2√2For x = -3 - 2√2, y = -(-3 - 2√2) -1 = 3 + 2√2 -1 = 2 + 2√2Distance between (-3 + 2√2, 2 - 2√2) and (-3 - 2√2, 2 + 2√2):Δx = (-3 - 2√2) - (-3 + 2√2) = -4√2Δy = (2 + 2√2) - (2 - 2√2) = 4√2Distance = sqrt[ (-4√2)^2 + (4√2)^2 ] = sqrt[32 + 32] = sqrt[64] = 8. Correct.Therefore, both lines y = x + 1 and y = -x -1 satisfy the condition.But wait, the problem says "the equation of line l". Since there are two solutions, we need to present both. However, sometimes problems expect all possible equations. Let me check if there are any other possibilities. We considered lines with slope m, but what if the line is vertical? Earlier, we saw that the vertical line x = -1 gives a chord length of 4, which is too short. Any other vertical line? Wait, no. The line has to pass through B(-1, 0). If the line is vertical, it's x = -1. We already checked that, and it's not 8. So, all other lines passing through (-1, 0) must have a slope, so they can be written as y = m(x + 1). We found m = ±1. Therefore, the two lines are y = x + 1 and y = -x -1.Hence, the answer for part (2) should be these two equations.But let me confirm once again.Alternatively, maybe we can approach this problem using parametric equations or other methods. Let's see.Another approach: The length of the chord in a parabola can also be related to the focal length. But I think the method we did is solid. The key was to set up the distance formula based on the roots of the quadratic equation, which we did by using Vieta's formula. Since the algebra checked out and the verification with actual points gave the correct distance, it's confident that these are the correct lines.So, summarizing:1) The equation of the parabola is ( y^2 = -4x ).2) The equations of the lines are ( y = x + 1 ) and ( y = -x -1 ).Final Answer(1) The equation of the parabola is boxed{y^2 = -4x}.(2) The equations of line l are boxed{y = x + 1} and boxed{y = -x - 1}.</think>Given that the vertex of a parabola is at the origin, it passes through point A(-4, 4), and its focus lies on the x-axis:1. Finding the equation of the parabola: - Since the focus lies on the x-axis, the parabola opens horizontally. The standard form is ( y^2 = 4px ). - Substituting the coordinates of point A(-4, 4) into the equation: [ (4)^2 = 4p(-4) implies 16 = -16p implies p = -1 ] - The equation of the parabola is ( y^2 = -4x ).2. Finding the equation of line l: - Line l passes through point B(-1, 0) and intersects the parabola such that the chord length is 8. - The line can be written as ( y = m(x + 1) ). Substituting this into the parabola's equation: [ [m(x + 1)]^2 = -4x implies m^2(x^2 + 2x + 1) = -4x implies m^2x^2 + (2m^2 + 4)x + m^2 = 0 ] - Using the quadratic roots, the distance between the intersection points is calculated and set to 8: [ sqrt{(x2 - x1)^2 + (y2 - y1)^2} = 8 ] - Solving for ( m ), we find ( m = pm 1 ). - The equations of the lines are ( y = x + 1 ) and ( y = -x - 1 ). Final Answer(1) The equation of the parabola is boxed{y^2 = -4x}.(2) The equations of line l are boxed{y = x + 1} and boxed{y = -x - 1}.]

🔑:(1) Let the equation of the parabola be y^2 = -2px. Since the parabola passes through point A(-4, 4), we substitute the coordinates into the equation:4^2 = -2p(-4)This simplifies to p = 2. Thus, the equation of the parabola is y^2 = -4x.(2) We consider two cases:Case 1: When the slope of line l does not exist, the equation of line l is x = -1. Intersecting this line with the parabola yields points (-1, -2) and (-1, 2). The length of the chord formed is 4, which is not consistent with the given information.Case 2: When the slope of line l exists, let the slope be k. The equation of line l is y = k(x + 1). Eliminating y by substituting this into the equation of the parabola, we obtain:begin{cases} y = k(x + 1) y^2 = -4x end{cases}Solving for y gives us k^2x^2 + (2k^2 + 4)x + k^2 = 0. The sum and product of the roots are given by:x_1 + x_2 = -frac{2k^2 + 4}{k^2}x_1x_2 = 1The length of the chord is given by the formula:text{Chord length} = sqrt{1 + k^2} times frac{sqrt{(2k^2 + 4)^2 - 4k^2}}{k^2}Setting this equal to 8 and solving for k, we find k^2 = 1. Hence, k = pm 1.Therefore, the equations of line l are boxed{y = x + 1} or boxed{y = -x - 1}.

🧠:[<think>Okay, let's see. The problem is about classifying natural numbers as "good" or "bad" based on certain rules. We know that 1 is good, and we need to find the nearest good number to 2015. Hmm. Let me try to understand the rules first.Rule a) says that from any bad number, you can subtract some natural number not exceeding its half, resulting in a good number. So if a number is bad, there's at least one number you can subtract from it (up to half its value) to get a good number.Rule b) states that from a good number, you can't subtract any number up to half of it and still have a good number. In other words, any subtraction of a number up to half of a good number will result in a bad number.So, good numbers are sort of "foundational" in the sense that once you have a good number, any number you can reach by subtracting up to half of it is bad. Conversely, bad numbers can reach a good number by such a subtraction.Given that 1 is good, let's try to see if we can figure out a pattern here.Starting from 1, since it's good, any number that can reach 1 by subtracting up to half of itself must be bad. Let's think: if a number n is such that n - k = 1, where k ≤ n/2, then n - k = 1 implies k = n - 1. So n - 1 ≤ n/2. Solving for n: n - 1 ≤ n/2 → n/2 ≤ 1 → n ≤ 2. So for n=2, k=1 (since 2-1=1). Therefore, 2 is bad because you can subtract 1 (which is ≤ 2/2=1) and get 1, which is good. Similarly, n=3: can we subtract k ≤ 1.5 (so k=1) to get 2. But 2 is bad. So we need to check if all possible subtractions from a good number lead to bad numbers. Wait, but 3: if 3 is good, then subtracting any k ≤ 1.5 (k=1) would result in 2, which is bad. So that fits rule b). Wait, but is 3 good or bad?Wait, let's re-examine. Let's start building the sequence.1 is good.Now, 2: from 2, you can subtract 1 (which is half of 2) to get 1 (good). Therefore, by rule a), 2 is bad.3: Let's see. If 3 were bad, then there must be some k ≤ 1.5 (so k=1) such that 3 - 1 = 2, which is bad. But rule a) requires that subtracting k from a bad number results in a good number. But 3 - 1 = 2, which is bad. Therefore, 3 cannot be bad because there is no k ≤ 1.5 that gives a good number. Wait, this is confusing.Wait, perhaps my approach is incorrect. Let's formalize this.Let G denote good numbers and B denote bad numbers.Given:1 is G.For any number n:If n is in B, then there exists k (1 ≤ k ≤ floor(n/2)) such that n - k is in G.If n is in G, then for all k (1 ≤ k ≤ floor(n/2)), n - k is in B.So starting with 1 (G), let's check numbers sequentially.n=1: Gn=2: Check if it's G or B. If it were G, then subtracting any k=1 (since floor(2/2)=1) should result in B. But 2 - 1 =1, which is G. Therefore, 2 cannot be G. Hence, 2 must be B.n=3: If it were G, then subtracting k=1 (floor(3/2)=1) would have to result in B. 3 -1=2, which is B. So that works. So 3 is G.n=4: If it were G, then subtracting k=1 or 2 (floor(4/2)=2) must result in B. Let's check:4 -1=3 (G), which violates the rule. Therefore, 4 cannot be G. Therefore, 4 must be B. Now, since it's B, there must exist some k (1≤k≤2) such that 4 -k is G. Indeed, k=1: 4-1=3 (G), so that works. So 4 is B.n=5: If G, then subtracting k=1 or 2 (floor(5/2)=2) must result in B. Let's check:5-1=4 (B), 5-2=3 (G). Since one of them is G, 5 cannot be G. Therefore, 5 must be B. Then, check if there's a k such that 5 -k is G. 5-2=3 (G). So yes, k=2. Hence, 5 is B.n=6: If G, then subtracting k=1,2,3 (floor(6/2)=3) must all be B.6-1=5 (B), 6-2=4 (B), 6-3=3 (G). Since 6-3=3 (G), 6 cannot be G. Therefore, B. Check for B: there must exist some k (1-3) such that 6 -k is G. 6-3=3 (G). So yes, k=3. So 6 is B.n=7: If G, then subtracting k=1,2,3 must result in B.7-1=6 (B), 7-2=5 (B), 7-3=4 (B). All are B. Therefore, 7 is G.Wait, so 7 is G. Hmm. So here we have G numbers at 1,3,7.Hmm. Let's see if there's a pattern here. 1,3,7. The differences are 2,4. Maybe the next one is 15? Let's check.n=8: If G, subtract k=1-4, must all be B.8-1=7 (G). So 8 cannot be G. Therefore, B. Check if B: 8 -k=7 (G) when k=1. So 8 is B.n=9: If G, subtract k=1-4 (floor(9/2)=4). 9-1=8 (B), 9-2=7 (G). Since 9-2=7 (G), 9 cannot be G. Therefore, B. Check B: 9 -2=7 (G). So yes, 9 is B.n=10: If G, subtract k=1-5. 10-1=9 (B), 10-2=8 (B), 10-3=7 (G). So 10-3=7 (G), hence 10 is B. Check B: 10-3=7 (G). So yes, B.n=11: If G, subtract k=1-5. 11-1=10 (B), 11-2=9 (B), 11-3=8 (B), 11-4=7 (G), 11-5=6 (B). So 11-4=7 (G). Hence, 11 cannot be G. Therefore, B. Check B: 11-4=7 (G). So B.n=12: If G, subtract k=1-6. 12-1=11 (B), 12-2=10 (B), 12-3=9 (B), 12-4=8 (B), 12-5=7 (G), 12-6=6 (B). So 12-5=7 (G). Therefore, 12 is B. Check B: 12-5=7 (G). So yes.n=13: If G, subtract k=1-6. 13-1=12 (B), 13-2=11 (B), 13-3=10 (B), 13-4=9 (B), 13-5=8 (B), 13-6=7 (G). So 13-6=7 (G). Hence, 13 cannot be G. Therefore, B. Check B: 13-6=7 (G). So B.n=14: If G, subtract k=1-7. 14-1=13 (B), 14-2=12 (B), 14-3=11 (B), 14-4=10 (B), 14-5=9 (B), 14-6=8 (B), 14-7=7 (G). So 14-7=7 (G). Therefore, 14 is B. Check B: 14-7=7 (G). So B.n=15: If G, subtract k=1-7. 15-1=14 (B), 15-2=13 (B), 15-3=12 (B), 15-4=11 (B), 15-5=10 (B), 15-6=9 (B), 15-7=8 (B). All subtractions result in B. Therefore, 15 is G.So 15 is G. So the G numbers so far are 1,3,7,15. Ah, this is 2^n -1. 2^1 -1=1, 2^2 -1=3, 2^3 -1=7, 2^4 -1=15. So the next would be 31, 63, 127, 255, 511, 1023, 2047... But wait, 2047 is 2^11 -1. But 2015 is between 1023 (2^10 -1) and 2047. Let's check.Wait, but let's confirm if the pattern holds. Let's check n=16.n=16: If G, subtract k=1-8. 16-1=15 (G). So 16 cannot be G. Therefore, B. Check B: 16 -1=15 (G). So B.n=17: If G, subtract k=1-8. 17-1=16 (B), 17-2=15 (G). So 17 cannot be G. Therefore, B. Check B: 17-2=15 (G). So B.Continuing up to n=31. Let's jump.n=31: If G, subtract k=1-15. All subtractions from 31 must lead to B. Let's check 31-15=16 (B), which is B. 31-14=17 (B), 31-13=18 (B)... all the way down. Since 31 is 2^5 -1, the previous G number is 15. So subtracting any k from 1 to 15 gives numbers from 30 down to 16, all of which are B. Therefore, 31 is G.Similarly, 63 would be G, 127, 255, 511, 1023, 2047... So the pattern seems like all numbers of the form 2^n -1 are G. Then the numbers in between are B.Therefore, if the "good" numbers are Mersenne numbers (2^n -1), then the nearest good number to 2015 would be either 2047 (2^11 -1) or 1023 (2^10 -1). Since 2015 is closer to 2047 (difference of 32) than to 1023 (difference of 992), the nearest good number is 2047. But wait, let's confirm.Wait, 2015 to 2047 is 2047 -2015 =32. 2015 -1023=992. So yes, 2047 is closer. But is 2047 indeed a good number?But wait, according to our earlier pattern, 2^n -1 are good. So 2047=2^11 -1. However, 2047 is actually a composite number, known as a Mersenne number but not a prime (23*89). But in our problem, the classification is based on the rules, not primes. So even if it's composite, according to the rules, if 2047 is a good number, then it should follow the rules. Let's check.n=2047: If it's G, then subtracting any k up to floor(2047/2)=1023 must result in B. Since 2047 is a G number, all numbers from 2047 -1=2046 down to 2047 -1023=1024 must be B. Since the previous G number is 1023 (2^10 -1), subtracting from 2047 would land in numbers above 1023, which are all B. Therefore, 2047 is G.Similarly, 1023 is G. So the nearest G number to 2015 is 2047. But wait, 2015 is between 1023 and 2047. However, perhaps there is another G number between them? Wait, the pattern is 1,3,7,15,31,63,127,255,511,1023,2047,... So each subsequent G number is double the previous plus 1. So each G number is 2^(n) -1. Therefore, the next after 1023 is 2047. Therefore, between 1023 and 2047, there are no other G numbers. Therefore, the nearest is 2047.But wait, the problem says "the nearest good number to 2015". Since 2015 is 32 less than 2047 and 992 more than 1023, so 2047 is indeed closer. Therefore, the answer should be 2047.But let me check for possible errors. Suppose there is a G number between 1023 and 2047. Is that possible? For example, if there is a number that is not of the form 2^n -1 but still G. Let's test n=127:127 is G. Then 255 is G, 511 is G, 1023 is G, 2047 is G. So the pattern holds. Let's verify for n=15 (G), next is 31 (G). 15*2 +1=31. Similarly, 31*2 +1=63, etc. So the sequence is each time doubling and adding 1. So 2^(n+1)-1 = 2*(2^n -1)+1. So the pattern is consistent.Therefore, unless there's a different pattern, the good numbers are exactly the numbers one less than a power of two. Therefore, the nearest good number to 2015 is 2047.Wait, but 2015 is 2047 - 32 =2015. So 32 less. But is there a good number between 2015 and 2047? If not, then 2047 is the closest. But according to the pattern, the next good number after 1023 is 2047. So between 1023 and 2047, all numbers are bad. Therefore, 2047 is the closest.But let me check n=2015 itself. Is 2015 a good number? If we follow the pattern, since 2047 is good, 2015 is 2047 -32. So 2015 is in the range 1024-2047, all bad. Therefore, 2015 is bad. So the closest good number is 2047. However, wait, 2015 is closer to 2047 than to 1023.Alternatively, maybe my initial assumption is wrong. Let's check a number in between, say 1535. Wait, 1535 is 1023 + 512 =1535. Wait, 1535=2^11 - 1 - 512=2047 -512=1535. But 1535 is not of the form 2^n -1. Let's check if 1535 is good. To do that, according to the rules, if 1535 is good, then subtracting any k up to 767 (half of 1535) must result in a bad number. Let's take k=512. 1535-512=1023, which is good. Therefore, 1535 cannot be good. So 1535 is bad.Similarly, take another number, say 2015. If 2015 is bad, which it is, then there exists some k ≤1007 such that 2015 -k is good. The nearest good number would be 2047, but 2047 is larger than 2015, so k would have to be negative, which isn't allowed. Wait, no. To get from 2015 to a good number, we subtract k ≤1007. So 2015 -k must be good. The closest good number less than 2015 is 1023, but 2015 -1023=992. So k=992. But 992 is less than 1007. So 2015 -992=1023, which is good. Therefore, 2015 is bad. But the question is not about whether 2015 is good or bad, but the nearest good number to 2015. So since 2015 is between 1023 and 2047, and 2047 is closer, the answer is 2047.Therefore, the final answer is boxed{2047}.Final Answerboxed{2047}</think>Given the problem of classifying natural numbers as "good" or "bad" based on specific rules, we need to find the nearest "good" number to 2015. The rules are:a) From any bad number, subtracting a natural number not exceeding its half results in a good number.b) From a good number, subtracting any number up to half of it results in a bad number.Starting with the knowledge that 1 is good, we analyzed the sequence of numbers and observed the pattern. Through detailed analysis, we found that the good numbers are of the form (2^n - 1) (Mersenne numbers). These numbers are good because subtracting any number up to half of them results in a bad number, and they cannot be further reduced to another good number.The sequence of good numbers is: 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, ...Given 2015, the closest good numbers are 1023 (2^10 - 1) and 2047 (2^11 - 1). Calculating the differences:- (2015 - 1023 = 992)- (2047 - 2015 = 32)Since 2047 is closer to 2015 than 1023, the nearest good number to 2015 is (boxed{2047}).,<think>Okay, so I need to figure out the nearest "good" number to 2015 based on the given rules. Let's start by understanding the problem step by step.First, the rules:a) From any bad number, you can subtract some natural number not exceeding its half, and the result is good.b) From a good number, no matter what you subtract (as long as it's a natural number not exceeding half of it), the result will be bad.Given that 1 is good. So starting from 1, we need to determine which numbers are good and which are bad, and then find the closest good number to 2015.Hmm. This reminds me of games where positions are either winning or losing. Maybe it's similar to the concept of P-positions and N-positions in combinatorial game theory. In such games, a position is a P-position (good) if every move from it leads to an N-position (bad), and a N-position (bad) if there exists at least one move to a P-position (good).So translating that here:- A number is good (P-position) if all possible subtractions (of up to half its value) result in bad numbers.- A number is bad (N-position) if there exists at least one subtraction (of up to half its value) that results in a good number.Given that 1 is good. Let's start building the list of good and bad numbers step by step.Starting with n=1:1 is good.Now, n=2:If we can subtract up to half of 2, which is 1. So from 2, we can subtract 1, resulting in 1, which is good. So according to rule a), since from a bad number (if 2 were bad), you can subtract 1 to get a good number. But wait, actually, we need to check if 2 is good or bad.Wait, the definitions are:A number is good if, from it, you cannot subtract any number up to half of it and remain good. Wait, rule b) says: From a good number, you cannot subtract no more than half of it so that it remains good. Wait, maybe my initial translation to game theory terms isn't perfect.Wait, let's parse the rules again:a) From any bad number, there exists a subtraction (≤ half of it) leading to a good number.b) From any good number, any subtraction (≤ half of it) leads to a bad number. So for a good number, all possible subtractions (as per the rules) result in bad numbers.So yes, exactly like P-positions and N-positions. So a good number is a P-position: every move from it leads to an N-position (bad). A bad number is an N-position: there exists at least one move to a P-position (good).Therefore, starting with 1 as good (P-position). Let's list the numbers one by one and determine their status.n=1: Good (given).n=2: Let's check. From 2, the possible subtractions are numbers from 1 to 1 (since half of 2 is 1). So we can subtract 1, resulting in 1. Since 1 is good, according to rule a), if 2 were bad, this would satisfy the condition. But we need to determine if 2 is good or bad. To check if 2 is good, we need to see if all possible subtractions (which is only 1) result in bad numbers. But 1 is good, so if we subtract 1 from 2, we get a good number. But if 2 were good, rule b) says that you cannot subtract any number (up to half) and remain good. However, subtracting 1 from 2 gives 1, which is good. This would violate rule b) if 2 were good. Therefore, 2 must be bad. Because from 2, you can subtract 1 to get a good number. So 2 is bad.Wait, maybe I need to formalize this.To determine if a number is good or bad:- If all moves from n lead to bad numbers, then n is good.- If there exists at least one move from n leading to a good number, then n is bad.Therefore, starting from n=1:n=1: Good (base case).n=2: Check moves. Subtract 1, get 1 (good). Therefore, there exists a move leading to good, so n=2 is bad.n=3: Possible subtractions are 1 (since half of 3 is 1.5, so natural numbers up to 1). So subtract 1, resulting in 2 (bad). Since all possible subtractions (only 1 in this case) lead to bad numbers, n=3 is good.Wait, wait. For n=3, half of 3 is 1.5, so we can subtract 1 at most. So subtracting 1 gives 2, which is bad. Since from n=3, the only possible subtraction leads to bad, so n=3 is good.n=4: Half of 4 is 2, so we can subtract 1 or 2.Subtract 1: 4-1=3 (good). Subtract 2: 4-2=2 (bad). So there exists a move (subtract 1) leading to good, so n=4 is bad.n=5: Half of 5 is 2.5, so subtract up to 2.Subtract 1: 5-1=4 (bad). Subtract 2: 5-2=3 (good). Therefore, there exists a move (subtract 2) leading to good, so n=5 is bad.n=6: Half of 6 is 3. So subtract 1, 2, or 3.Subtract 1: 6-1=5 (bad). Subtract 2: 6-2=4 (bad). Subtract 3: 6-3=3 (good). So there exists a move (subtract 3) leading to good. Therefore, n=6 is bad.n=7: Half of 7 is 3.5, so subtract up to 3.Subtract 1: 7-1=6 (bad). Subtract 2: 7-2=5 (bad). Subtract 3: 7-3=4 (bad). All subtractions lead to bad numbers. Therefore, n=7 is good.So far, the good numbers are 1, 3, 7.n=8: Half of 8 is 4. Subtract 1,2,3,4.Subtract 1: 7 (good). Therefore, exists a move leading to good. So n=8 is bad.n=9: Half of 9 is 4.5, subtract up to 4.Subtract 1: 8 (bad). Subtract 2: 7 (good). Therefore, exists a move. So n=9 is bad.n=10: Half of 10 is 5. Subtract up to 5.Subtract 1: 9 (bad). Subtract 2: 8 (bad). Subtract 3: 7 (good). Therefore, exists a move. n=10 is bad.n=11: Half of 11 is 5.5, subtract up to 5.Subtract 1: 10 (bad). Subtract 2: 9 (bad). Subtract 3: 8 (bad). Subtract 4: 7 (good). So subtracting 4 gives 7 (good). Therefore, n=11 is bad.n=12: Half of 12 is 6. Subtract up to 6.Subtract 1: 11 (bad). Subtract 2: 10 (bad). Subtract 3: 9 (bad). Subtract 4: 8 (bad). Subtract 5: 7 (good). Subtract 6: 6 (bad). So subtract 5 gives good. Therefore, n=12 is bad.n=13: Half of 13 is 6.5, subtract up to 6.Subtract 1: 12 (bad). Subtract 2: 11 (bad). Subtract 3: 10 (bad). Subtract 4: 9 (bad). Subtract 5: 8 (bad). Subtract 6: 7 (good). Therefore, subtract 6 leads to good. Therefore, n=13 is bad.n=14: Half of 14 is 7. Subtract up to 7.Subtract 1: 13 (bad). Subtract 2: 12 (bad). Subtract 3: 11 (bad). Subtract 4: 10 (bad). Subtract 5: 9 (bad). Subtract 6: 8 (bad). Subtract 7: 7 (good). So subtracting 7 leads to good. Therefore, n=14 is bad.n=15: Half of 15 is 7.5, subtract up to 7.Subtract 1: 14 (bad). Subtract 2: 13 (bad). Subtract 3: 12 (bad). Subtract 4: 11 (bad). Subtract 5: 10 (bad). Subtract 6: 9 (bad). Subtract 7: 8 (bad). All subtractions lead to bad numbers. Therefore, n=15 is good.So now, good numbers are 1, 3, 7, 15.Wait, that seems like 2^n -1. 1=2^1-1, 3=2^2-1, 7=2^3-1, 15=2^4-1. So perhaps the good numbers are Mersenne numbers, numbers of the form 2^k -1. Let me check the next one.n=16: Half is 8. Subtract up to 8.Subtract 1: 15 (good). Therefore, n=16 is bad.n=17: Half is 8.5, subtract up to 8.Subtract 1: 16 (bad). Subtract 2: 15 (good). Therefore, n=17 is bad.n=18: Half is 9. Subtract up to 9.Subtract 1: 17 (bad). Subtract 2: 16 (bad). Subtract 3: 15 (good). Therefore, n=18 is bad.Continuing this pattern, n=15 is good. Next, n=31? Let's check n=16 to n=31.n=15 is good. Let's see n=16 to 31.For n=16, subtracting 1 gives 15 (good). So bad.Similarly, n=17, subtract 2 gives 15 (good). So bad.n=18: subtract 3 gives 15. So bad.Similarly up to n=30.n=31: Half is 15.5, subtract up to 15.Subtract 1: 30 (bad). Subtract 2: 29 (bad). ..., Subtract 15: 16 (bad). So all subtractions lead to bad numbers. Therefore, n=31 is good.Yes, so the pattern seems to be that the good numbers are 2^k -1. Each time, when we reach a number that is one less than a power of two, it is good. Then, the next good number is double that plus one. For example, 1=2^1-1, 3=2^2-1, 7=2^3-1, 15=2^4-1, 31=2^5-1, etc. So the general formula for the nth good number is 2^n -1.If that's the case, then to find the nearest good number to 2015, we need to find the Mersenne number closest to 2015.So let's find the exponents k such that 2^k -1 is closest to 2015.Compute log2(2015 +1) = log2(2016). Since 2^10=1024, 2^11=2048. 2016 is between 2^11 and 2^10. Let's see:2^11=2048, which is 2048-2015=33 more than 2015.2^10=1024, which is 2015-1024=991 less than 2015.So the nearest Mersenne number would be 2047 (2^11 -1) and 1023 (2^10 -1). Wait, but 2048 is 2^11, so 2^11 -1=2047, which is 2047-2015=32 away. While 2015-1023=992. Therefore, 2047 is closer.But wait, wait. Wait, 2^11 -1=2047, which is 2047-2015=32. 2015 is between 1023 and 2047. But 2015 is much closer to 2047.But hold on, maybe there is another Mersenne number between 1023 and 2047? Let's check. The Mersenne numbers are 1,3,7,15,31,63,127,255,511,1023,2047,4095,... So between 1023 and 2047, the next is 2047. So yes, 2047 is the next good number after 1023. So the nearest good number to 2015 is 2047, which is 32 greater than 2015. But wait, maybe 2015 is a bad number, but perhaps there's a lower good number closer than 2047? Wait, but the previous good number is 1023, which is 2015-1023=992 away, which is much further. So the closest is 2047.But hold on, 2047 is 2047-2015=32. But wait, maybe there's a closer number? Let me check if there's any other good numbers between 2015 and 2047. But according to the pattern, the next good number after 1023 is 2047, so between 1023 and 2047, all numbers are bad. Therefore, 2047 is the next good number.But wait, let me confirm this pattern. If the good numbers are indeed Mersenne numbers, then the nearest one to 2015 is 2047. But let me verify for a number between, say, 15 and 31. Let's take n=23. According to the rules, is 23 a good or bad number?Wait, n=15 is good. Then n=16 to 30 are bad. n=31 is good. So between 15 and 31, all numbers are bad except 31. Similarly, between 31 and 63, all numbers are bad except 63. Therefore, the pattern holds. Therefore, the good numbers are indeed 2^k -1. Therefore, the closest good number to 2015 is 2047.Wait, but 2015 is 2015, 2047 is 32 more. Let me check if 2015 is 2047 - 32. Yes, 2047 - 2015 = 32. So 2015 + 32 = 2047. The difference is 32. Since the previous good number is 1023, which is 992 less than 2015, so indeed 2047 is the closest.Wait, but perhaps I made a wrong assumption here. Because maybe the good numbers are not exactly the Mersenne numbers. Let's verify with the earlier numbers.Wait, n=1 is good, 3 is good, 7 is good, 15 is good, 31 is good, 63, 127, 255, 511, 1023, 2047, etc. These are all Mersenne numbers. Since when we checked up to n=15, the good numbers were exactly 2^k -1, so perhaps this continues.Alternatively, maybe there's a different pattern, but given the initial numbers, 1,3,7,15,31,... it's consistent with 2^n -1.Therefore, if this pattern continues, then the closest good number to 2015 is 2047.But let me check n=2047. Let's verify if 2047 is indeed a good number. According to the rules, if 2047 is good, then any subtraction of up to half of it (which is 1023.5, so up to 1023) would lead to a bad number.But 2047 - x, where x is 1 to 1023, should all be bad numbers. Since all numbers between 1024 and 2046 are bad, as per the previous pattern. Thus, subtracting any x from 1 to 1023 from 2047 would result in numbers from 2046 down to 1024, which are all bad. Hence, 2047 is good.Similarly, 1023 is good. So yes, the pattern holds. Therefore, 2047 is good, and the next one is 4095, which is way higher.Therefore, the closest good number to 2015 is 2047.But wait, let me check if there's a smaller good number closer. For example, is 2015 itself a good number? If 2015 is good, then the answer is 2015. But according to the pattern, it's not a Mersenne number. 2^11 -1=2047, 2^10 -1=1023. So 2015 is not a Mersenne number, so it should be bad. Therefore, the closest good number is 2047.But let me check 2015 manually. Let's suppose 2015 is bad, so there exists some x ≤ 1007 (half of 2015) such that 2015 - x is good. If the good numbers are 2^k -1, then 2015 - x = 2^k -1. Therefore, x = 2016 - 2^k. Since x must be ≤ 1007, then 2016 -2^k ≤ 1007 → 2^k ≥ 2016 -1007 = 1009. The smallest k where 2^k ≥ 1009 is k=10, since 2^10=1024. Therefore, x=2016 -1024=992. So subtract 992 from 2015 gives 2015-992=1023, which is good. Therefore, 2015 is bad because we can subtract 992 (which is ≤1007) to get 1023 (good). Therefore, 2015 is bad. Therefore, the nearest good number is 2047, which is 32 above, or 1023, which is 992 below. Since 32 < 992, the closest is 2047.Therefore, the answer should be boxed{2047}.Final AnswerThe nearest "good" number to 2015 is boxed{2047}.</think>Given the problem of determining the nearest "good" number to 2015 based on the provided rules, we start by understanding the definitions of "good" and "bad" numbers:1. A number is "good" if every possible subtraction (of a natural number not exceeding half its value) results in a "bad" number.2. A number is "bad" if there exists at least one subtraction (of a natural number not exceeding half its value) that results in a "good" number.Starting with the given information that 1 is "good", we observe the pattern of "good" numbers. Through analysis, we find that the "good" numbers follow the sequence of Mersenne numbers, which are of the form (2^k - 1). These numbers are 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, etc.To find the nearest "good" number to 2015, we identify the closest Mersenne numbers. The Mersenne numbers around 2015 are 1023 (which is (2^{10} - 1)) and 2047 (which is (2^{11} - 1)). Calculating the distances:- (2015 - 1023 = 992)- (2047 - 2015 = 32)Since 32 is smaller than 992, the nearest "good" number to 2015 is 2047.Thus, the answer is (boxed{2047}).]

🔑:Answer: 2047. Solution. Considering the first few natural numbers, we notice that good numbers have the form 2^{n}-1 (while 2^{n}, ldots, 2 n+1-2 are bad).We will prove this by mathematical induction. For n=1, the statement is given in the condition. Suppose the statement is proven for n-1. Consider a number of the form M=2^{n}+k, where k=0,1, ldots, 2^{n}-2. Then from such a number, we can subtract k+1 leqslant frac{1}{2}left(2^{n}+kright)=frac{M}{2}.On the other hand, consider a number of the form N=2^{n}-1. From it, we need to subtract at least 2^{n-1}, which exceeds frac{N}{2}=2^{n-1}-frac{1}{2}.Thus, the nearest good number to 2015 will be the number 2047=2^{11}-1.

🧠:[<think>Okay, let me try to figure out this problem. So, there's an alloy that's 3.5 kilograms and it's 75% silver. Then they melt it and mix it with another alloy, and they end up with 10.5 kilograms of a new alloy that's 84% silver. We need to find the percentage of silver in the second alloy. Hmm.Alright, let me start by recalling some basic concepts. When you mix two alloys, the total mass should be the sum of the masses of the two original alloys. Also, the amount of silver in the final alloy should be the sum of the silver from both original alloys. So maybe I can set up some equations here.First, let's note down the given information:1. First alloy: mass = 3.5 kg, 75% silver.2. Second alloy: mass = unknown, let's call it m kg, and percentage of silver = p% (which we need to find).3. Resulting alloy: mass = 10.5 kg, 84% silver.Since the total mass after mixing is 10.5 kg, the mass of the second alloy must be 10.5 kg minus the mass of the first alloy. Wait, but the first alloy is 3.5 kg. So, mass of second alloy would be 10.5 kg - 3.5 kg = 7 kg? Wait, is that correct?Hold on, when you melt and mix two alloys, the total mass is the sum. So, first alloy is 3.5 kg, second alloy is m kg, so total mass should be 3.5 + m = 10.5 kg. Therefore, m = 10.5 - 3.5 = 7 kg. So the second alloy is 7 kg. That seems straightforward.So, the second alloy has a mass of 7 kg. Now, we need to find the percentage of silver in this 7 kg alloy.Now, let's think about the amount of silver. The first alloy is 3.5 kg with 75% silver. So the amount of silver in the first alloy is 0.75 * 3.5 kg. Let me calculate that. 3.5 times 0.75. Let's see, 3 * 0.75 is 2.25, and 0.5 * 0.75 is 0.375, so total is 2.25 + 0.375 = 2.625 kg of silver from the first alloy.The second alloy is 7 kg with p% silver, so the amount of silver from the second alloy is (p/100) * 7 kg. Let me write that as 0.01p * 7 kg = 0.07p kg.The resulting alloy is 10.5 kg with 84% silver, so the total silver is 0.84 * 10.5 kg. Let me compute that. 10 * 0.84 = 8.4, and 0.5 * 0.84 = 0.42, so total is 8.4 + 0.42 = 8.82 kg of silver in the final alloy.Now, since the total silver from both alloys should equal the silver in the final alloy, we can set up the equation:Silver from first alloy + Silver from second alloy = Silver in final alloyWhich translates to:2.625 + 0.07p = 8.82Now, we can solve for p.Subtract 2.625 from both sides:0.07p = 8.82 - 2.625Let me compute that. 8.82 minus 2.625. Hmm, 8.82 minus 2 is 6.82, minus 0.625 more is 6.82 - 0.625 = 6.195.So, 0.07p = 6.195Therefore, p = 6.195 / 0.07Let me calculate that. Dividing 6.195 by 0.07. Well, 6 divided by 0.07 is 85.714..., and 0.195 divided by 0.07 is 2.7857..., so total is approximately 85.714 + 2.7857 ≈ 88.5. Wait, but let me do it more accurately.6.195 divided by 0.07. Multiply numerator and denominator by 1000 to eliminate decimals:6195 / 70. Let's compute that.70 goes into 6195 how many times?70*88 = 6160. 6195 - 6160 = 35. So 88 + 35/70 = 88.5. So p = 88.5.So the percentage of silver in the second alloy is 88.5%.Wait, that seems straightforward, but let me check my steps again to make sure I didn't make a mistake.First, mass of the second alloy: 10.5 - 3.5 = 7 kg. That seems right.Silver from first alloy: 3.5 * 0.75 = 2.625 kg. Correct.Silver from second alloy: 7 * (p/100) = 0.07p kg. Correct.Total silver in final alloy: 10.5 * 0.84 = 8.82 kg. Correct.Equation: 2.625 + 0.07p = 8.82. Then solving for p: subtract 2.625, get 6.195 = 0.07p. Then p = 6.195 / 0.07 = 88.5%. Yes, that seems correct.So the answer is 88.5%. Therefore, the percentage of silver in the second alloy is 88.5%.But wait, just to double-check, let's verify the calculation:If the second alloy is 7 kg with 88.5% silver, then the silver is 7 * 0.885 = 6.195 kg.First alloy has 2.625 kg silver. Total silver is 2.625 + 6.195 = 8.82 kg, which is indeed 84% of 10.5 kg (since 10.5 * 0.84 = 8.82). So all the numbers check out.Therefore, the answer is 88.5%.Final AnswerThe percentage of silver in the second alloy is boxed{88.5}.</think>Given an alloy of zinc and silver with a mass of 3.5 kilograms containing 75% silver, and another alloy of unknown silver percentage, when melted together, they form a third alloy of 10.5 kilograms with 84% silver. We need to find the percentage of silver in the second alloy.1. Mass of the second alloy: [ 10.5 , text{kg} - 3.5 , text{kg} = 7 , text{kg} ]2. Amount of silver in the first alloy: [ 0.75 times 3.5 , text{kg} = 2.625 , text{kg} ]3. Amount of silver in the resulting alloy: [ 0.84 times 10.5 , text{kg} = 8.82 , text{kg} ]4. Let ( p ) be the percentage of silver in the second alloy. The amount of silver in the second alloy is: [ 0.01p times 7 , text{kg} = 0.07p , text{kg} ]5. Setting up the equation for total silver: [ 2.625 + 0.07p = 8.82 ]6. Solving for ( p ): [ 0.07p = 8.82 - 2.625 ] [ 0.07p = 6.195 ] [ p = frac{6.195}{0.07} = 88.5 ]Thus, the percentage of silver in the second alloy is boxed{88.5}.,<think>Okay, let me try to figure out this problem. So, we have two alloys: the first one is 3.5 kilograms and it's 75% silver. Then, they melt this and mix it with another alloy to get a third alloy that's 10.5 kilograms and 84% silver. We need to find the percentage of silver in the second alloy. Hmm, seems like a mixture problem where we have to find the concentration of one component.First, let me note down the given information. The first alloy is 3.5 kg with 75% silver. The second alloy's mass we don't know yet, but when mixed with the first, the total becomes 10.5 kg. So, the mass of the second alloy must be 10.5 kg minus 3.5 kg. Let me check that: 10.5 - 3.5 = 7.0 kg. So, the second alloy is 7.0 kilograms. Okay, that makes sense.Now, the problem is about the percentage of silver in this second alloy. Let's denote the percentage of silver in the second alloy as x%. So, the total amount of silver in the second alloy would be 7.0 kg times x%, right?Similarly, the amount of silver in the first alloy is 75% of 3.5 kg. Let me compute that. 75% is 0.75 in decimal, so 0.75 * 3.5. Let me do that multiplication. 3.5 * 0.75. Well, 3 * 0.75 is 2.25, and 0.5 * 0.75 is 0.375. Adding those together: 2.25 + 0.375 = 2.625 kg. So, the first alloy has 2.625 kg of silver.Then, the second alloy has 7.0 kg * x% silver. So, that's 0.01x * 7.0 kg. Let's write that as 0.07x kg of silver.When we mix these two, the total silver in the third alloy would be 2.625 + 0.07x kg. The total mass of the third alloy is 10.5 kg, and it's 84% silver, so the amount of silver in the third alloy is 0.84 * 10.5 kg. Let me calculate that. 10.5 * 0.84. Hmm, 10 * 0.84 is 8.4, and 0.5 * 0.84 is 0.42. Adding those together gives 8.4 + 0.42 = 8.82 kg. So, the third alloy has 8.82 kg of silver.Therefore, we can set up an equation: the silver from the first alloy plus the silver from the second alloy equals the silver in the third alloy. So:2.625 + 0.07x = 8.82Now, let's solve for x. Subtract 2.625 from both sides:0.07x = 8.82 - 2.625Calculating the right side: 8.82 - 2.625. Let's subtract 2 from 8.82 first, which is 6.82, then subtract 0.625 from that. 6.82 - 0.625 = 6.195. So, 0.07x = 6.195Now, divide both sides by 0.07:x = 6.195 / 0.07Let me compute that. 6.195 divided by 0.07. Well, 6 divided by 0.07 is 85.714..., since 0.07 times 85 is 5.95, and 0.07 times 86 is 6.02. But we have 6.195. Let me do it more accurately.6.195 / 0.07Multiply numerator and denominator by 1000 to eliminate decimals:6195 / 70Now, divide 6195 by 70.70 goes into 6195 how many times?70*88=6160, so 6195 - 6160=35So, 70*88 + 35 = 6195So, 6195 /70=88.5So, x=88.5Therefore, the percentage of silver in the second alloy is 88.5%Wait, let me verify this. So, 7.0 kg of 88.5% silver is 7 * 0.885 = 6.195 kg. Then, adding that to 2.625 kg from the first alloy gives 6.195 + 2.625 = 8.82 kg, which matches the third alloy's silver content. So, that seems correct.But let me check if my calculation was right. 0.07x = 6.195. So x = 6.195 / 0.07. Let's compute 6.195 divided by 0.07. 0.07 is 7 cents. So, 6.195 divided by 0.07. Let's move the decimal two places: 619.5 divided by 7. 7 into 619.5: 7*80=560, 619.5-560=59.5. 7*8=56, 59.5-56=3.5. 7*0.5=3.5. So, 80 +8 +0.5=88.5. Yes, that's 88.5. So, correct.So, the percentage of silver in the second alloy is 88.5%. Therefore, the answer should be 88.5%, which in boxed form would be boxed{88.5}But wait, in some countries, they use comma as decimal separator. But in the problem, percentages are given with a dot, like 75%, 84%, so likely expects a dot. So, 88.5% is correct.But let me just check again if I made any mistake in setting up the equations. So, mass of first alloy: 3.5 kg, 75% Ag. Mass of second alloy: 10.5 -3.5=7 kg, x% Ag. Total silver: 3.5*0.75 +7*x =10.5*0.84. Yes, exactly. Then, 3.5*0.75=2.625, 10.5*0.84=8.82. So, 2.625 +7x=8.82. So, 7x=8.82-2.625=6.195. So, x=6.195/7=0.885, so 88.5%. Yup, that's correct. So, answer is 88.5%, so boxed boxed{88.5}Final AnswerThe percentage of silver in the second alloy is boxed{88.5}.</think>Given an alloy of zinc and silver with a mass of 3.5 kilograms containing 75% silver, and another alloy of unknown silver percentage, when melted and mixed together, they form a third alloy of 10.5 kilograms with 84% silver. We need to find the percentage of silver in the second alloy.1. Calculate the mass of the second alloy: [ 10.5 , text{kg} - 3.5 , text{kg} = 7.0 , text{kg} ]2. Calculate the amount of silver in the first alloy: [ 3.5 , text{kg} times 0.75 = 2.625 , text{kg} ]3. Let ( x ) be the percentage of silver in the second alloy. The amount of silver in the second alloy is: [ 7.0 , text{kg} times frac{x}{100} = 0.07x , text{kg} ]4. Calculate the amount of silver in the third alloy: [ 10.5 , text{kg} times 0.84 = 8.82 , text{kg} ]5. Set up the equation for the total silver in the mixture: [ 2.625 + 0.07x = 8.82 ]6. Solve for ( x ): [ 0.07x = 8.82 - 2.625 ] [ 0.07x = 6.195 ] [ x = frac{6.195}{0.07} ] [ x = 88.5 ]The percentage of silver in the second alloy is boxed{88.5}.]

🔑:Solution. In the first alloy, there was 3.5 cdot 0.75 mathrm{~kg}=2.625 mathrm{~kg} of silver. In the third alloy, there was 10.5 cdot 0.84 mathrm{~kg}=8.82 mathrm{~kg} of silver. To the first alloy, (10.5-3.5) mathrm{kg}=7 mathrm{~kg} of alloy was added, which contained (8.82-2.66) mathrm{kg}=6.16 mathrm{~kg} of silver. Therefore, in the second alloy, there was frac{6.195 cdot 100}{7}=88 % silver.3AB. Igor, Jovan, Kire, and Lazar invited their girlfriends to a dance. During one dance, Betty danced with Igor, Ana with the boy of Vера, Gordana with the boy of Ana, Jovan with the girl of Kire, and Kire with the girl of Igor. Determine the couples in love.