Problem:
How many perfect squares are divisors of the product 1!⋅2!⋅3!⋯9! ?
Answer Choices:
A. 504
B. 672
C. 864
D. 936
E. 1008
Solution:
We have
1!⋅2!⋅3!⋯9!=(1)(1⋅2)(1⋅2⋅3)⋯(1⋅2⋯9)=192837465564738291=2303135573
The perfect square divisors of that product are the numbers of the form
22a32b52c72d
with 0≤a≤15,0≤b≤6,0≤c≤2, and 0≤d≤1. Thus there are (16)(7)(3)(2)=672 such numbers.
The problems on this page are the property of the MAA's American Mathematics Competitions