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y>=4. But we already saw for y=4, x=45 is a solution. Then y=5,6,7,8... but for higher y, y! is larger, but maybe mod other numbers can rule them out. Alternatively, check modulo 5. Let's see: x² ≡ y! +2001 mod5. 2001 mod5: 2000 is divisible by5, so 2001≡1 mod5. So y! +1 ≡x² mod5. Possible squares mod5: 0,1,4. So y! +1... | {
"page_id": null,
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impossible. But earlier modulo4 already ruled out y=1,2,3. Wait, for y=3: y! =6≡0 mod3, so x²≡0 mod3. Then x must be multiple of3. But y=3: x²=2007, which we saw isn't a square. So maybe combining modulo3 and other moduli. Alternatively, maybe modulo7. Let me try mod7. x² ≡ y! +2001 mod7. Compute 2001 mod7: 2001 divide... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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y=4 might be possible. Which we know works. Therefore, the only solution is y=4 and x=45. Let me check for y=4 again. x²=24 +2001=2025. 45²=2025. Correct. But just to make sure there are no other solutions for higher y. Wait, say y=5. We saw that x²=2121. Let's check if 2121 is a square. Let me compute square roots aro... | {
"page_id": null,
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"title": "from dpo"
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For \(y = 2\): \[ y! = 2 \implies x^2 = 2 + 2001 = 2003. \] Since 2003 is not a perfect square, there is no solution for \(y = 2\). - For \(y = 3\): \[ y! = 6 \implies x^2 = 6 + 2001 = 2007. \] Since 2007 is not a perfect square, there is no solution for \(y = 3\). - For \(y = 4\): \[ y! = 24 \implies x^2 = 24 + 2001 =... | {
"page_id": null,
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"title": "from dpo"
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we have checked \(y = 1, 2, 3, 5, 6\), the only solution is: \[ \boxed{(45, 4)}. \] | cot | | \(\mathbb{Q}\) | has_answer_extracted | aops_c7_college_math | DeepSeek-R1 | 0.96875 | Find all \( x \) in \( \mathbb{R} \) such that the limit \[ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \sum_{k=1}^{n} |\sin (\pi k! x... | {
"page_id": null,
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"title": "from dpo"
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for each term in the sum, when k is fixed and n approaches infinity, |sin(π k! x)|^(1/n) approaches 1 if |sin(π k! x)| ≠ 0, and 0 otherwise. But since we're summing from k=1 to n, when n is going to infinity, the number of terms in the sum is also increasing. So if, for each n, the first n terms are each approaching 1,... | {
"page_id": null,
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"title": "from dpo"
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rational number, say x = p/q where p and q are integers with q > 0. Then for k >= q, k! is divisible by q, so k! x = k! p/q is an integer. Therefore, for k >= q, |sin(π k! x)| = |sin(π * integer)| = 0. So for x rational, when k! x is integer for all k >= q, hence for all k >= q, the terms |sin(π k! x)| = 0, so when k >... | {
"page_id": null,
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"title": "from dpo"
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q divides k! then k! x is integer. So for each prime in q's prime factorization, the prime must be = q. Therefore, for k = q, the sum becomes sum_{k=1}^{q-1} |sin(π k! x)|^(1/n) + sum_{k=q}^{n} 0. So the sum is sum_{k=1}^{q-1} |sin(π k! x)|^(1/n). Now, as n approaches infinity, each term in that sum approaches 1 becaus... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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0 and 1. For each fixed k, as n approaches infinity, |sin(π k! x)|^(1/n) approaches 1. However, when k is varying up to n, as n increases, the number of terms in the sum increases. So we have sum_{k=1}^n a_k(n), where a_k(n) approaches 1 as n approaches infinity for each fixed k. But as n increases, the sum is of n ter... | {
"page_id": null,
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limit is 0, which is bounded. So x=1, a rational number, gives a bounded limit. Another example: x = 1/2. Then q = 2. For k >=2, k! x = k!*(1/2) = (k!)/2. Since k! is divisible by 2 for k >=2, so k! x is integer. Therefore, for k >=2, |sin(π k! x)| = 0. Therefore, the sum from k=1 to n is |sin(π 1! x)|^(1/n) + sum_{k=2... | {
"page_id": null,
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let me check another example where x is rational with a denominator that is not prime. For example, x=1/4. Then q=4. For k >=4, k! x = k!/4, which is integer since 4 divides k! for k >=4. So the sum from k=1 to n is sum_{k=1}^3 |sin(π k! x)|^(1/n) + sum_{k=4}^n 0. So for k=1: |sin(π*(1/4))|=sin(π/4)=sqrt(2)/2≈0.707. Fo... | {
"page_id": null,
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"title": "from dpo"
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is a type of transcendental number that can be very well approximated by rationals. But even then, for each k, k! x may not be integer, so |sin(π k! x)| would still be positive. So even for such x, the sum would still go to infinity. So I don't think there are any other x besides rationals where the limit is bounded. T... | {
"page_id": null,
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\). - For \( k \geq q \), \( k! \) is divisible by \( q \), so \( k! x = k! \frac{p}{q} \) is an integer. - Therefore, \( |\sin (\pi k! x)| = |\sin (\pi \cdot \text{integer})| = 0 \) for \( k \geq q \). - The sum \( \sum_{k=1}^{n} |\sin (\pi k! x)|^{1/n} \) reduces to \( \sum_{k=1}^{q-1} |\sin (\pi k! x)|^{1/n} \) for ... | {
"page_id": null,
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So, this function f has a property where when you multiply the inputs, it adds the outputs. That reminds me of logarithms because log(a*b) = log a + log b. So maybe f is a logarithmic function. But I don't know if that's the case here. Let's explore it step by step. First, the functional equation given is f(xy) = f(x) ... | {
"page_id": null,
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directly. Let's see. Since 256 is a power of 2. Let's write 256 as 2^8. Then f(256) = f(2^8). Now, using the functional equation, how can we break this down? For example, f(2^8) can be written as f(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2). Applying the functional equation multiple times: f(2 * 2) = f(2) + f(2) = 2a. Then f(2 * 2... | {
"page_id": null,
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"title": "from dpo"
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\cdot 1 = k. \] Since \( f(2) = a \), it follows that: \[ k = a. \] Thus, we have: \[ f(x) = a \log_2(x). \] To find \( f(256) \), we substitute \( x = 256 \): \[ f(256) = a \log_2(256). \] Next, we need to determine \( \log_2(256) \). Since \( 256 = 2^8 \), we have: \[ \log_2(256) = \log_2(2^8) = 8. \] Therefore: \[ f... | {
"page_id": null,
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"title": "from dpo"
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The point we want is the closest point on this circle to (-2,11). I remember that the closest point on a circle to an external point lies along the line connecting the center of the circle to that external point. So maybe I can find the line that goes through both the center (6,5) and the point (-2,11), and then find w... | {
"page_id": null,
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and the external point is 10, which is more than the radius 5, so the closest point is the point on the line connecting them, 5 units away from the center towards the external point. So the vector from the center to the external point is (-2 -6, 11 -5) = (-8,6). The direction vector is (-8,6). The unit vector in that d... | {
"page_id": null,
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"title": "from dpo"
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vector in the direction of P multiplied by the radius. Since the distance from C to P is 10, the unit vector is (-8/10,6/10), so scaling by radius 5 gives (-4,3), so adding to C: (6-4,5+3)=(2,8). Yep, same result. Let me check if (2,8) is actually on the circle. Plugging into the equation: (2-6)^2 + (8-5)^2 = (-4)^2 +3... | {
"page_id": null,
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vector from the center to the external point:** The direction vector from the center \((6, 5)\) to the external point \((-2, 11)\) is: \[ (-2 - 6, 11 - 5) = (-8, 6) \] 3. **Find the unit vector in the direction of the external point:** The magnitude of the direction vector \((-8, 6)\) is: \[ \sqrt{(-8)^2 + 6^2} = \sqrt... | {
"page_id": null,
"source": 6842,
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993 | has_answer_extracted | aops_c4_high_school_math | DeepSeek-R1 | 0.96875 | Given the function \( f: \mathbb{R}_{+}^{0} \rightarrow \mathbb{R} \) such that for all \( x > 0 \), \( 2f(x) + 3f\left(\frac{2010}{x}\right) = 5x \), find \( f(6) \). | Okay, so I have this problem where I need to find f(6) given the funct... | {
"page_id": null,
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"title": "from dpo"
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15x - 20100/x So, B = (15x - 20100/x)/5 = 3x - 4020/x But B is f(2010/x). Hmm, so f(2010/x) = 3x - 4020/x. Wait, but I need to express B in terms of x or in terms of 2010/x? Let me think. If I have B = f(2010/x) = 3x - 4020/x, but maybe I need to write this in terms of the argument. Let me set y = 2010/x, so x = 2010/y... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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2 after substitution was 3A + 2B = 5*(2010/x). So 5*(2010/x) is 10050/x. So when I multiplied equation 2 by 2, it should be 6A +4B = 10*(2010/x) = 20100/x. Then when subtracting, 6A +9B - (6A +4B) = 15x -20100/x → 5B =15x -20100/x → B =3x -4020/x. Then B is f(2010/x), so as I did earlier, substituting back to get f(x).... | {
"page_id": null,
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"title": "from dpo"
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\) in the original equation:** \[ 2f\left(\frac{2010}{x}\right) + 3f(x) = 5 \cdot \frac{2010}{x} \] This gives us a second equation: \[ 2f\left(\frac{2010}{x}\right) + 3f(x) = \frac{10050}{x} \] 2. **We now have a system of linear equations:** \[ \begin{cases} 2f(x) + 3f\left(\frac{2010}{x}\right) = 5x \\ 3f(x) + 2f\le... | {
"page_id": null,
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and according to the Goldbach conjecture, every even number greater than 2 can be expressed as the sum of two primes. Well, 126 is even and greater than 2, so there should be some pairs. But I don't need all pairs, just the one with the maximum difference. The strategy here would probably be to find the smallest possib... | {
"page_id": null,
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"title": "from dpo"
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10.6, so we check primes less than 11. 113 is not divisible by 2, 3, 5, 7, or 11. Wait, 11 squared is 121, so up to 11. So yes, 113 is a prime. So that's a valid pair: 13 and 113. The difference here is 113 - 13 = 100. Hmm, that's a large difference. Let's see if there's a smaller prime that pairs with a larger prime. ... | {
"page_id": null,
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check 126-101=25. 25 is not prime. So pair (101,25) invalid. So from the primes up to 63 (since 63+63=126, but primes are odd except 2). Wait, but 126 is even, so primes except 2 are odd. So adding two odd numbers gives an even number. So except for the pair that includes 2, all other primes are odd. So maybe 2 is not ... | {
"page_id": null,
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\( q \) such that \( p + q = 126 \). 1. **Identify the smallest prime \( p \):** - Start with the smallest prime number, which is 2. - Check if \( 126 - 2 = 124 \) is a prime number. Since 124 is even and greater than 2, it is not a prime number. - Next, try the next smallest prime number, which is 3. - Check if \( 126... | {
"page_id": null,
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5). - 113 is not divisible by 7 (since \( 113 \div 7 \approx 16.14 \)). - 113 is not divisible by 11 (since \( 113 \div 11 \approx 10.27 \)). - Since 113 is not divisible by any of these primes, it is a prime number. 2. **Verify the pair:** - The pair \( (13, 113) \) sums to \( 13 + 113 = 126 \). - The difference betwe... | {
"page_id": null,
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| aops_c4_high_school_math | DeepSeek-R1 | 0.96875 | What is the maximum area of a quadrilateral with sides of lengths 1, 4, 7, and 8? | Okay, so I need to find the maximum area of a quadrilateral with sides 1, 4, 7, and 8. Hmm, quadrilaterals can be tricky because there are different types like squares, rectangles, pa... | {
"page_id": null,
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"title": "from dpo"
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+ 4 + 8 = 13 > 7 1 + 7 + 8 = 16 > 4 4 + 7 + 8 = 19 > 1 So yes, the quadrilateral can exist. Now, to be cyclic, another condition must be met. I think for a quadrilateral to be cyclic, the product of the diagonals can be expressed in terms of the sides, but I might be mixing things up. Alternatively, there's Ptolemy's t... | {
"page_id": null,
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the diagonals (Ptolemy's theorem), but without knowing the diagonals, that might not be useful. Alternatively, there's a condition based on the sides. If in a convex quadrilateral, the sum of two opposite angles is 180 degrees, but again, how to check that with just the sides? Wait, another thought: For a quadrilateral... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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gives a real number, the quadrilateral might not be possible. Wait, but the semi-perimeter is greater than each side, which is a necessary condition for a quadrilateral to exist (which we already checked with the triangle inequality). So if the product (s-a)(s-b)(s-c)(s-d) is positive, then the cyclic quadrilateral exi... | {
"page_id": null,
"source": 6842,
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Therefore, if the product (s-a)(s-b)(s-c)(s-d) is non-negative, then the maximum area is sqrt of that product. Since in our case, it's 324, sqrt(324) = 18. Therefore, the maximum area is 18. So I think the answer is 18. But just to check once more: let's compute s = (1+4+7+8)/2 = 20/2 = 10. Then (s-a)(s-b)(s-c)(s-d) = ... | {
"page_id": null,
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of the quadrilateral is \(\boxed{18}\). | cot | | 0 | has_answer_extracted | aops_c4_high_school_math | DeepSeek-R1 | 0.96875 | How many three-digit perfect squares are there such that if each digit is increased by one, the resulting number is also a perfect square? | Alright, let me try to figure out this problem. So,... | {
"page_id": null,
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must be 1-8, and the tens and units places must be 0-8. So, the original number must have digits in the ranges: hundreds (1-8), tens (0-8), units (0-8). That might narrow down the candidates. So, let's first list all three-digit perfect squares without any digits being 9. Let me start by listing all three-digit squares... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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digit by 1, we need to check if the resulting number is also a perfect square. However, when we add 1 to each digit, we have to ensure that no digit was 9 in the original number, which we've already excluded. So, each digit in the original number is at most 8, so adding 1 would make it at most 9, which is a single digi... | {
"page_id": null,
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1 to each digit: 255. 255. Let's see, 15²=225, 16²=256, so 255 is between them. Not a square. 4. 225: Add 1 to each digit: 336. 336. Let's check square roots. 18²=324, 19²=361. 336 is between them. Not a square. 5. 256: Add 1 to each digit: 367. 19²=361, 20²=400. 367 is not a square. 6. 324: Add 1 to each digit: 435. 2... | {
"page_id": null,
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about 225? 2+1=3, 2+1=3, 5+1=6. So 336. 336 is between 18²=324 and 19²=361. Correct. 400: 4+1=5, 0+1=1, 0+1=1. So 511. 511 is not a square. 484: 4+1=5, 8+1=9, 4+1=5. So 595. 595 between 24² and 25². Correct. 576 becomes 687. 26²=676, 27²=729. 687 in between. 625 becomes 736. 27²=729, so 736 is less than 729. Wait, 27² ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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new number can have digits 9, but the original number cannot. So the new number could be 999, but that would require the original number to be 888, which isn't a square. So 999 is 999, which is not a square. Wait, perhaps the problem allows leading zeros? But the original number is a three-digit number, so leading zero... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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equations. Let the original three-digit number be N = ABC, where A, B, C are digits (A ≠ 0). Then, when each digit is increased by 1, the new number is (A+1)(B+1)(C+1), which is a number. Let's express N as 100A + 10B + C. The new number is 100(A+1) + 10(B+1) + (C+1) = 100A + 100 + 10B + 10 + C + 1 = 100A + 10B + C + 1... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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are integers, y - x and y + x must be positive integers and factors of 111. Also, y + x > y - x, and both factors have the same parity since y and x are integers (their sum and difference must both be even or both be odd). Let's check the factor pairs: 1 and 111: 1*111. Then y - x = 1, y + x = 111. Solving these equati... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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+ 111 = 400. But 400 is 4,0,0. How does adding 1 to each digit of 289 become 400? Wait, 2 +1 = 3, 8 +1 =9, 9 +1=10. So, the hundreds digit becomes 3, the tens becomes 9, and the units becomes 10. But 10 is two digits, which would make the number 3 (hundreds), 9 (tens), 10 (units), which is 3*100 + 9*10 +10 = 300 +90 +1... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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a carry-over is different. So this is a problem. Therefore, the previous approach is incorrect because it doesn't account for carry-over when digits are 9. Therefore, we have to discard that method and go back to the original approach of checking each three-digit square without any 9s and see if adding 1 to each digit ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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any 9s, which we had as 15 numbers: 100, 121, 144, 225, 256, 324, 361, 400, 441, 484, 576, 625, 676, 784, 841. Now, for each of these, compute x² + 111 and check if it's a perfect square. Let's go through each: 1. x² = 100. 100 + 111 = 211. Is 211 a square? No. Between 14²=196 and 15²=225. 2. x² = 121. 121 + 111 = 232.... | {
"page_id": null,
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Therefore, the answer is indeed zero? But let me check if there is any other case. Maybe when there are carry-overs, but the problem statement allows that? Wait, the problem says "if each digit is increased by one". If they mean strictly increasing each digit by one regardless of carry-over, then even numbers with 9 di... | {
"page_id": null,
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211. Not a square. N = 625. Increment digits: 736. Not a square. N = 256. Increment digits: 367. Not a square. N = 324. Increment digits: 435. Not a square. N = 361. Increment digits: 472. Not a square. N = 441. Increment digits: 552. Not a square. N = 784. Increment digits: 895. Not a square. N = 841. Increment digits... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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pair for 111? 111 is 3*37. There are no other factors. So, only 1*111 and 3*37. So, no other solutions. Therefore, the only mathematical solution is x=17, y=20, but x²=289 is invalid. So, there are no solutions where N is a three-digit square with no 9s such that N + 111 is also a square. Hence, the answer is zero. The... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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perfect square) 3. \(144 + 111 = 255\) (not a perfect square) 4. \(225 + 111 = 336\) (not a perfect square) 5. \(256 + 111 = 367\) (not a perfect square) 6. \(324 + 111 = 435\) (not a perfect square) 7. \(361 + 111 = 472\) (not a perfect square) 8. \(400 + 111 = 511\) (not a perfect square) 9. \(441 + 111 = 552\) (not ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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means it's proportional to the reciprocal. So combining those, the relationship should be something like \( A^2 = k \times \frac{B}{C^3} \), where \( k \) is the constant of proportionality. That makes sense because directly proportional to B and inversely proportional to C cubed. Alright, the problem gives me that whe... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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25 times 16. Let's compute that. 25 times 10 is 250, 25 times 6 is 150, so 250 + 150 = 400. Wait, is that right? Wait, no, 25*16 is actually 25*(10+6) = 250 + 150 = 400. Yeah, that's correct. So B would be 400. Let me just double-check my steps to make sure I didn't make a mistake. Original equation: \( A^2 = k * (B/C^... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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k \): \[ k = \frac{100}{25} = 4 \] Now that we have \( k = 4 \), the equation becomes: \[ A^2 = 4 \times \frac{B}{C^3} \] Next, we need to find \( B \) when \( A = 5 \) and \( C = 4 \). Substituting these values into the equation, we get: \[ 5^2 = 4 \times \frac{B}{4^3} \] Simplifying the left side: \[ 25 = 4 \times \f... | {
"page_id": null,
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Title: Art of Problem Solving URL Source: Markdown Content: _The following problem is from both the 2018 AMC 12B #18 and 2018 AMC 10B #20, so both problems redirect to this page._ * 1 Problem]( * 3 Solution 2 (Algebra)]( * 5 Solution 4 (Pattern)]( * 7 Solution 6 (Complex Numbers)=f(2)=1$]( and -f(2)=6\cdot336=2016,\]]( from which !Image 28: $f(2018)=f(2)+2016=\boxed{\textbf{(B) } 2017}.$ ------------------------------- Preamble: In this solution, we define the sequence !Image 29: $A$,$]( where !Image 31: $a_n$... | {
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see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence !Image 47: $C$ Now, there are two ways to finish. Finish Method #1: Notice that any six consecutive terms of !Image 54: $B$ } 2017}.$]( Finish Method #2: Notice that !Image 58... | {
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!Image 64: $+2$&=2013 \\ f(2014)&=2016 \\ f(2015)&=2018 \\ f(2016)&=2018 \\ f(2017)&=2017 \\ f(2018)&=\boxed{\textbf{(B) } 2017}. \end{align*}]( Solution 5 (Pattern) -------------------- Writing out the first few values, we get ![Image 73: \[1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\ldots.\]]( We see that eve... | {
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six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now, because they don't follow this pattern. First, we need to know whether or not !Image 104: $2016$,f(12), \ldots,f(6n)$]( all satisfy !Image 106: $6+6(n-1)=n,$ = f(5)$]( and so on for the repeat ... | {
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Resources**Followed by **Problem 21** 1**•**2**•**3**•**4**•**5**•**6**•**7**•**8**•**9**•**10**•**11**•**12**•**13**•**14**•**15**•**16**•**17**•**18**•**19**•**20**•**21**•**22**•**23**•**24**•**25 **All AMC 10 Problems and Solutions** These problems are copyrighted © by the [Mathematical Association of America]( as ... | {
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Title: MaziyarPanahi/OpenMathReasoning_ShareGPT · Datasets at Hugging Face URL Source: Markdown Content: Subset (1) default·2.4M rows Split (1) | expected_answer string lengths 0 1.62k | problem_type string classes 3 values | problem_source string classes 5 values | generation_model string classes 2 values | pass_rate... | {
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"title": "from dpo"
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possible outcomes would be all the ways to pick A balls regardless of color distribution. Let me break it down. The total number of ways to choose A balls from N is the combination of N choose A, which is C(N, A) = N! / (A! (N - A)!). Now, the favorable outcomes. For each color i, we need to choose ai balls from the ni... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
multivariate version extends this to multiple categories, which is exactly the case here with different colors. The formula for the multivariate hypergeometric probability is indeed the product of combinations for each category divided by the combination of the total. Therefore, the probability is the product of combin... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
balls are randomly picked from \( N \) balls, the picked balls consist of \( a_1, a_2, \ldots, a_C \) balls of each color is given by: \[ \boxed{\frac{\prod_{i=1}^{C} \binom{n_i}{a_i}}{\binom{N}{A}}} \] This solution is derived from the multivariate hypergeometric distribution, where the combinations \( \binom{n_i}{a_i... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
balls regardless of color distribution.\n\nLet me break it down. The total number of ways to choose A balls from N is the combination of N choose A, which is C(N, A) = N! / (A! (N - A)!).\n\nNow, the favorable outcomes. For each color i, we need to choose ai balls from the ni available. Since the choices are independen... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
colors. The formula for the multivariate hypergeometric probability is indeed the product of combinations for each category divided by the combination of the total.\n\nTherefore, the probability is the product of combinations C(n_i, a_i) for each color i, divided by the combination C(N, A). So, I think that's the answe... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
color is given by:\n\\[\n\\boxed{\\frac{\\prod_{i=1}^{C} \\binom{n_i}{a_i}}{\\binom{N}{A}}}\n\\]\n\nThis solution is derived from the multivariate hypergeometric distribution, where the combinations \\( \\binom{n_i}{a_i} \\) account for the ways to choose \\( a_i \\) balls from \\( n_i \\) balls of color \\( i \\), and... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
sense that each point has the same distance to the line? That might not even be possible unless all points lie on one side of the line and are arranged in a specific way. Alternatively, maybe equidistant here means that the line is equidistant from each pair of points, but I'm not sure. Wait, the wording is a bit ambig... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
points. So, if we have n points, how many lines can we have that are equidistant from at least two points but not necessarily all. But the question says "equidistant from n given points", which would mean all n points. That's confusing. Wait, maybe the question is asking how many lines can be equidistant from each of t... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
the line, then you can't have a line equidistant to all points unless all points are on the line. If all points are on a line, then that line itself would have all points at distance zero, which is equidistant, but any line parallel to it would have the points at varying distances. Wait, so in that case, there's only o... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
is equidistant from all n points, but that each line is equidistant from the set of n points. But how do you define a line being equidistant from multiple points? The standard definition is that a line is equidistant from two points if it's the perpendicular bisector. For more than two points, it's not a standard term.... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
doesn't specify that the points are arranged in any particular way, so the answer might be that there can be 0, 1, or more lines depending on the configuration. But since the problem asks "How many lines can be drawn...", maybe it's expecting the maximum possible number given n points, regardless of their arrangement. ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
any line that's at a fixed distance from the point. But since the distance can be any non-negative number, there are infinitely many lines (all lines at distance d from the point, for any d ≥ 0). But the problem says "n given points". So, with one point, infinitely many lines. But the problem is asking for n given poin... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
there are zero. So, depending on the configuration, it's 0 or 1. But the problem states "n given points". So, unless the points are in a specific arrangement, the number varies. However, the problem is phrased as a general question, not specifying the configuration. Maybe we need to find the maximum possible number of ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
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diagonal) or some positive distance. So, no, the diagonals are not equidistant. So, in a square, the midlines (the horizontal and vertical through the center) are equidistant to all four points. So, for a square, two lines. Similarly, if points are arranged in a regular hexagon, maybe more lines. Wait, but in regular p... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
in a rectangle, two lines. If arranged in a square, same thing. So, in general, arranging points in two sets of parallel lines (horizontal and vertical) might give two lines. So, maybe the maximum number of such lines is the number of pairs of perpendicular directions in which the points can be arranged. But this seems... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
lines parallel to L, each at distance d on either side. Therefore, for a line L to be equidistant from n points, all n points must lie on the union of two lines parallel to L. So, the number of such lines L is equal to the number of directions such that the n points lie on two parallel lines in that direction. Therefor... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
But in a square, the diagonals are not parallel; they intersect. Wait, if we take a direction of, say, 45 degrees, then two lines parallel to that direction would have to be spaced apart. But in a square, the points (0,0), (1,0), (0,1), (1,1). A line at 45 degrees would have equations y = x + c. To have two parallel li... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
lines c = 0, -2, 1, -1. So, they don't lie on two parallel lines in the direction of 45 degrees. Therefore, no such midline exists for 45 degrees. Therefore, only horizontal and vertical midlines. So, two lines. Therefore, for four points arranged in a rectangle, we have two lines. For four points in general position (... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
points in a configuration where they form several pairs of parallel lines in different directions. For example, a grid. If we have points arranged in a grid with multiple rows and columns, then each row direction and column direction gives a midline. For example, a 3x3 grid has three horizontal lines and three vertical... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
projection coordinates must be exactly two. So, to find the number of such directions, we need to find the number of angles θ such that when the points are projected onto a line making angle θ with the x-axis, there are exactly two distinct projection values. This is similar to problems in computational geometry about ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
point set is a union of two lines in that direction. The number of such directions is the number of ways to split the point set into two lines in some direction. For example, for n=4, if arranged as two lines of two points each, then there's one direction. But if arranged as a rectangle, there are two directions. For n... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
is O(n^2), but here we have a more restrictive condition. But maybe for our problem, the maximum number of equidistant lines is C(n,2), but that's for pairs of points. However, in our case, each line is determined by a direction where the entire point set lies on two lines of that direction. This is similar to the conc... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
there are three pairs of parallel sides, so three midlines. Thus, for a regular 2m-gon, the number of midlines would be m. Therefore, for n=2m points arranged as a regular 2m-gon, the number of equidistant lines is m. So, n/2. But is this the maximum? For example, for n=4 (a square), this would be 2 lines, which matche... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
Hmm, this is getting quite involved. Let's look for references or standard results. Wait, I recall a problem called the "equidistant line problem" or something similar. After a bit of thinking, I remember that for three non-collinear points, there are three lines equidistant from the points. Wait, but earlier I thought... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
all points are co-circular. Wait, if points are concyclic, then the perpendicular bisectors of every pair meet at the center. But equidistant lines would need to have the same distance from all points, which for a circle, any line through the center is a diameter and equidistant to all points? Wait, no. The distance fr... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
have any points on them. But even so, the number would be C(n,2), but the problem states "equidistant from n given points". Wait, maybe the problem is a trick question. If you have n points, the maximum number of lines that are equidistant from all n points is either 0 or 1, depending on the configuration. If the point... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
line has distance zero, and the other two vertices have equal non-zero distances. So, the distances are not the same for all three points, but two points have the same distance, and one has zero. So, the line is not equidistant from all three points. Wait, so even in an equilateral triangle, there is no line that is eq... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
how many directions we can arrange the points into two lines. For example, with n=4, a rectangle gives two directions. With n=6, three directions if arranged as three pairs. In general, if n is even, you can have n/2 directions by arranging the points as n/2 pairs in different directions. But that might not be possible... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
the number of such directions. Therefore, if the points are arranged in a way that they lie on two parallel lines in k different directions, the answer is k. So, the maximum possible k is floor(n/2), if you can pair up the points into floor(n/2) pairs, each in a different direction. But this is possible only if n is ev... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
of parallel lines in three different directions. Hence, three midlines. So, for n=6, three lines. Similarly, in a regular octagon, four pairs of parallel sides, leading to four midlines. Therefore, for a regular 2m-gon, the number of midlines is m. Hence, for n=2m points, m lines. Therefore, the maximum number of lines... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
2. For n=5, arrange as two pairs in two directions and one leftover point, but how? Wait, if n=5, to have two different directions where the points are split into two lines, each direction must accommodate all five points. So, in one direction, split as 3 and 2, and in another direction, another split. For example, if ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
two on a parallel line. Then, there is one midline. If we can rotate the direction and have another split, say three points on one line and two on another in a different direction. For example, arrange the five points in two different directions where they form two pairs of parallel lines. But this would require overla... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
is asking for the number of lines that can be equidistant from n given points. If we interpret this as lines that are equidistant from each pair of points (i.e., their perpendicular bisectors), then the answer would be C(n,2), but each such line is only equidistant to two points, not all n. So that's probably not the c... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
points on a circle. If we take a midline between two opposite sides, then each of those sides is parallel. The midline is equidistant to the two sides. However, the vertices are on the circle, not on the sides. The distance from each vertex to the midline would vary. The vertices on the original sides are at a certain ... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
bisector. Even though these bisectors are only equidistant to two points, if multiple pairs lie on the same bisector, then the line is equidistant to multiple pairs. But not all points. This is really confusing. Given the time I've spent and the lack of progress, I think the problem might have a different interpretatio... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
from multiple points, I think the problem is likely expecting the answer \(\boxed{\frac{n(n-1)}{2}}\) , which is the number of pairs of points and thus the number of perpendicular bisectors. But this is only true if the problem considers lines equidistant from pairs of points. However, the question says "equidistant fr... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
given points. So, if n=1, infinite; n=2, one; n≥3, zero. But the problem says "n given points", so if n≥3, answer is zero. But in the problem statement, n is not specified; it's a general question for any n. Hence, the answer depends on n. For n=1, infinite; n=2, 1; n≥3, 0. But the problem says "n given points", so lik... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
points, i.e., for every pair of points, the line is equidistant to them. In that case, the line must be the perpendicular bisector of every pair of points. Which is only possible if all points are concyclic, and the line is the perpendicular bisector of each pair, which implies all points are on a circle and the line p... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
must have all points at the same distance, which requires them to lie on two parallels. The number of such lines is the number of directions in which the points can be split into two parallels. Therefore, the answer is the number of such directions, which varies from 0 to floor(n/2). But since the problem is general, t... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
defines a unique perpendicular bisector. 2. **Counting Perpendicular Bisectors:** - The number of unique pairs of points from \( n \) points is given by the combination formula \( \binom{n}{2} \). - Therefore, the number of perpendicular bisectors (lines equidistant from each pair of points) is \( \binom{n}{2} = \frac{... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
I think when we say a line is equidistant from two points, it's the perpendicular bisector of the segment connecting those two points. Right? Because the perpendicular bisector is the set of all points equidistant from the two endpoints. So, if a line is equidistant from two points, it's that bisector. So, maybe for ea... | {
"page_id": null,
"source": 6842,
"title": "from dpo"
} |
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