Difference between revisions of "2023 AIME I Problems/Problem 4"
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We first rewrite 13! as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | We first rewrite 13! as a prime factorization, which is <math>2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.</math> | ||
− | For the fraction to be a square, it needs each prime to be an even power. <math>m</math> must contain <math>7\cdot\cdot11\cdot13</math>. <math>m</math> can contain any even power of 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of <math>m</math> is <math>(7\cdot11\ | + | For the fraction to be a square, it needs each prime to be an even power. <math>m</math> must contain <math>7\cdot\cdot11\cdot13</math>. <math>m</math> can contain any even power of 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of <math>m</math> is <math>(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)</math>, which simplifies to <math>1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4</math>. <cmath>1+2+1+3+1+4=\boxed{12}</cmath> |
Revision as of 13:25, 8 February 2023
Problem 4
Unofficial problem: The sum of all positive integers such that is a perfect square can be written as , where and are positive integers. Find
Solution
Solution 1
We first rewrite 13! as a prime factorization, which is For the fraction to be a square, it needs each prime to be an even power. must contain . can contain any even power of 2 up to 10, any odd power of 3 up to 5, and any even power of 5 up to 2. The sum of is , which simplifies to .