How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)? Options:
A) 504 B) 672 C) 864 D) 936 E) 1008
N=(1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)   1!=1   2!=1*2   3!=1*2*3   4!=1*2*3*4   5!=1*2*3*4*5   6!=1*2*3*4*5*6   7!=1*2*3*4*5*6*7   8!=1*2*3*4*5*6*7*8   9!=1*2*3*4*5*6*7*8*9   N=(2^8)(3^7)(4^6)(5^5)(6^4)(7^3)(8^2)(9^1)   =(2^30)(3^13)(5^5)(7^3)   To find the number of perfect squares each exponent in the prime factorization must be even .   Comparing to the form (2^p)(3^q)(5^r)(7^s).   In this case,   p can take values from 0 to 30   q can take values from 0 to 13   r can take values from 0 to 5   s can take values from 0 to 3   Number of even exponents   p=16   q=7   r=3   s=2   Number of perfect squares= 16*7*3*2= 
<gadget id="calculator">16*7*3*2</gadget>
<output>672</output>
672   Answer 672
<result>B</result>
How many perfect squares are divisors of the product (1!)(2!)(3!)(4!)(5!)(6!)(7!)(8!)(9!)?