If 40! × 40!/40^n is an integer, what is the largest possible value of n?	Answers:
A) 24	B) 10	C) 7	D) 3	E) 1
40! × 40!/40^n is an integer.   Or 40! × 40!/(4^2n*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 40! = [40/5] + [40/25] = 5   Since there are 2 20!'s, power of 5 in the numerator = 2*5= 
<gadget id="calculator">2*5</gadget>
<output>10</output>
10   Hence the largest value of b for which 40! × 40!/40^n is integer= 10   Correct Option : 10
<result>B</result>
If 40! × 40!/40^n is an integer, what is the largest possible value of n?