For which value of n ∈ N , n ! has 13 trailing zeros?

Step-by-step solution

We know in n ! exponent of 5 is always less than or equal to exponent of 2 Hence, exponent of 5 is equal to the number of trailing zeroes We know, exponent of 5 in n ! is given as ⇒ 13 = n 5 + n 5 2 + n 5 3 + . . . Using n = 57 from the options, we get = 57 5 + 57 25 + 57 125 = 11 + 2 + 0   . . . So, in 57 ! the power of 5 is 13