Difference between revisions of "2023 AIME I Problems/Problem 4"
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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> | 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> | ||
Revision as of 15:46, 8 February 2023
Problem
The sum of all positive integers such that
is a perfect square can be written as
where
and
are positive integers. Find
Solution 1
We first rewrite as a prime factorization, which is
For the fraction to be a square, it needs each prime to be an even power. This means must contain
. Also,
can contain any even power of
up to
, any odd power of
up to
, and any even power of
up to
. The sum of
is
Therefore, the answer is
.
~chem1kall
Solution 2
The prime factorization of is
To get
a perfect square, we must have
, where
,
,
.
Hence, the sum of all feasible is
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)