let f ( x , y ) be defined as the remainder when ( x – y ) ! is divided by x . if x = 50 , what is the maximum value of y for which f ( x , y ) = 0 ?
the question is finding y such that ( 50 - y ) ! is a multiple of 50 . that means we need to have 2 * 5 ^ 2 in ( 50 - y ) ! 10 ! is the smallest factorial number with 2 * 5 ^ 2 as a factor . 50 - y = 10 y = 40 the answer is c .
subtract(50, multiply(const_2, divide(divide(50, const_2), add(const_1, const_4))))