Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=36, what is the maximum value of y for which f(x,y)=0?  Choose the correct choice.
A) 9  B) 12  C) 18  D) 20  E) 30
The question is finding y such that (36-y)! is a multiple of 36. That means we need to have 2^2*3^2 in (36-y)!   6! is the smallest factorial number with 2^2*3^2 as a factor.   36-y = 6   y = 30   The answer is 30.
<result>E</result>
Let f(x,y) be defined as the remainder when (x–y)! is divided by x. If x=36, what is the maximum value of y for which f(x,y)=0?