2023 AIME I Problems/Problem 4

Problem 4

Unofficial problem: The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f$ , where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$

Solution

Solution 1

We first rewrite 13! as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$ For the fraction to be a square, it needs each prime to be an even power. $m$ must contain $7\cdot\cdot11\cdot13$ . $m$ 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 $m$ is $(7\cdot11\cdot13)(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)$ , which simplifies to $1365\cdot26\cdot270\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4$ . $1+2+1+3+1+4=\boxed{12}$

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2023 AIME I Problems/Problem 4

Problem 4

Solution

Solution 1