Difference between revisions of "2023 AIME I Problems/Problem 4"
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==Problem 4== | ==Problem 4== | ||
Unofficial problem: The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f</math>, where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math> | Unofficial problem: The sum of all positive integers <math>m</math> such that <math>\frac{13!}{m}</math> is a perfect square can be written as <math>2^a3^b5^c7^d11^e13^f</math>, where <math>a,b,c,d,e,</math> and <math>f</math> are positive integers. Find <math>a+b+c+d+e+f.</math> | ||
− | ===Solution=== | + | ==Solution== |
+ | ===Solution 1=== | ||
+ | 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\cdot\13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)</math>, which is <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 13:23, 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 $(7\cdot11\cdot\13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)$ (Error compiling LaTeX. Unknown error_msg), which is .