olymmath

HMMT 2013 U7

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

Dataset text NuminaMath olympiads. Quoted with source.

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