Problem:
The sum of all positive integers m such that m13! is a perfect square can be written as 2a3b5c7d11e13f, where a,b,c,d,e, and f are positive integers. Find a+b+c+d+e+f.
Solution:
We first rewrite 13! as a prime factorization, which is 210⋅35⋅52⋅7⋅11⋅13. For the fraction to be a square, it needs each prime to be an even power. This means m must contain 7⋅11⋅13. Also, m can contain any even power of 2 up to 210, any odd power of 3 up to 35, and any even power of 5 up to 52. The sum of m is
(20+22+24+26+28+210)(31+33+35)(50+52)(71)(111)(131)=1365⋅273⋅26⋅7⋅11⋅13=2⋅32⋅5⋅73⋅11⋅134
Therefore, the answer is 1+2+1+3+1+4=12.
The problems on this page are the property of the MAA's American Mathematics Competitions