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 ?
"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 e ."
subtract(36, multiply(const_2, divide(divide(36, const_2), add(const_1, const_4))))