MathDB
2013 HMMT Guts #7: Divisors of 15!

Source:

March 26, 2013
HMMTnumber theoryprime factorization

Problem Statement

Find the number of positive divisors dd of 15!=15⋅14⋅⋯⋅2⋅115!=15\cdot 14\cdot\cdots\cdot 2\cdot 1 such that gcd⁡(d,60)=5\gcd(d,60)=5.
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