If 20! × 20!/20^n is an integer, what is the largest possible value of n?  Choose the correct answer:	A) 20	B) 16	C) 8	D) 4	E) 1
20! × 20!/20^n is an integer.   Or 20! × 20!/(2^2n*5^n)   This means we need to find the power of 5 in the numerator. We can safely say that the number of powers of 5 will be lower than the number of powers of 4 in the numerator.   Largest power of 5 in 20! = [20/5] + [20/25] = 4   Since there are 2 20!'s, power of 4 in the numerator = 2*4= 
<gadget id="calculator">2*4</gadget>
<output>8</output>
8   Hence the largest value of b for which 20! × 20!/20^n is integer= 8   ANSWER:8
<result>C</result>
If 20! × 20!/20^n is an integer, what is the largest possible value of n?