Problem:
How many positive cubes divide 3!⋅5!⋅7!?
Answer Choices:
A. 2
B. 3
C. 4
D. 5
E. 6
Solution:
Written as a product of primes, we have
3!⋅5!⋅7!=28⋅34⋅52⋅7
A cube that is a factor has a prime factorization of the form 2p⋅3q⋅5r⋅7s, where p,q,r, and s are all multiples of 3. There are 3 possible values for p, which are 0,3, and 6. There are 2 possible values for q, which are 0 and 3. The only value for r and for s is 0. Hence there are 6=3⋅2⋅1⋅1 distinct cubes that divide 3!⋅5!⋅7!. They are
164=20305070,8=23305070, 27=20335070,=26305070,216=23335070, and 1728=26335070
Answer: E.
The problems on this page are the property of the MAA's American Mathematics Competitions