x is the product of each integer from 1 to 50, inclusive and y=100^k, where k is an integer . What is the greatest value of k for which y is a factor of x? Choose the correct option.
A) 0
B) 5
C) 6
D) 10
E) 12

Rationale: x = 1 * 2 * 3 * 4 * 5 * . . . . . . * 49 * 50 = 50 ! y = 100 ^ k 100 = 10 ^ 2 = 2 ^ 2 * 5 ^ 2 on prime factorization of x , power of 5 will always be greater than power of 2 concept : in every factorial value ( x ! ) , on prime factorization , the power of bigger prime number > the power of smaller prime number also , power of any prime number in any factorial can be calculated by following understanding power of prime x in n ! = [ n / x ] + [ n / x ^ 2 ] + [ n / x ^ 3 ] + [ n / x ^ 4 ] + . . . and so on where , [ x ] = greatest integer less than or equal to x and the explanation of terms is as follows [ n / x ] = no . of integers that are multiple of x from 1 to n [ n / x ^ 2 ] = no . of integers that are multiple of x ^ 2 from 1 to n whose first power has been counted in previous step and second is being counted at this step [ n / x ^ 3 ] = no . of integers that are multiple of x ^ 3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step and so on . . . . . where [ n / x ] is greatest integer value of ( n / x ) less than or equal to ( n / x ) i . e . [ 100 / 3 ] = [ 33.33 ] = 33 i . e . [ 100 / 9 ] = [ 11.11 ] = 11 etc . power of 5 in x = [ 50 / 5 ] + [ 50 / 5 ^ 2 ] + [ 50 / 5 ^ 3 ] . . . = 10 + 2 + 0 = 12 power of 2 in x = [ 50 / 2 ] + [ 50 / 2 ^ 2 ] + [ 50 / 2 ^ 3 ] . . . = 25 + 12 + 6 + 3 + 1 = 47 i . e . x = 50 ! = 2 ^ 47 * 5 ^ 12 * . . . = 10 ^ 12 * . . . = 100 ^ 6 * . . . i . e . x is divisible by a 6 th power of 100 at the most i . e . y = 100 ^ k = 100 ^ 6 i . e . k = 6 answer : option c
Answer: C