Question

How many perfect squares are divisors of the product $1!.2!.3!....9!$

• A. $504$
• B. $672$
• C. $864$
• D. $936$
• E. $1008$

Answer 1

The correct option is B $672$

We have
$1!.2!.3!.....9!=\left(1\right)\left(1.2\right)\left(\mathrm{1.2.3}\right)...\left(\mathrm{1.2.3.....9}\right)$
$=1^9{2}^8{3}^7{4}^6{5}^5{6}^4{7}^3{8}^2{9}^1=2^{30}{3}^{13}{5}^5{7}^3$.
The perfect square divisors of that product are the numbers of

the form $2^{2a}{3}^{2b}{5}^{2c}{7}^{2d}$
with $0\le a\le 15,0\le b\le 6,0\le c\le 2,and0\le d\le 1.$Thus there are $\left(16\right)\left(7\right)\left(3\right)\left(2\right)=672$ such numbers.