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:
Answer (012):
Note that 13!=210⋅35⋅52⋅7⋅11⋅13. The fraction m13! is a perfect square if and only if m is 13! divided by a perfect square divisor. Thus the required sum of all m is
13!(1+(21)2+(21)4+⋯+(21)10)(1+(31)2+(31)4)(1+(51)2)=(210⋅35⋅52⋅7⋅11⋅13)⋅3⋅21046−1⋅3434+32+1⋅5252+1=7⋅11⋅13⋅(7⋅32)⋅(5⋅13)⋅(7⋅13)⋅(2⋅13)=21⋅32⋅51⋅73⋅111⋅134
The requested sum of exponents is 1+2+1+3+1+4=12.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions