consecutive 0’s

We can phrase this question as

"How many 10s can be found in the factorization of 60! ?"

or, since 10 = 5 * 2 and 5s are less common than 2s,

"How many 5s can be found in the factorization of 60! ?"

Then we just count the factors of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.

The numbers in red count twice, since they contain 5*5. Counting those twice, we have a total of FOURTEEN 5's in the list above, and we're set!

jain2016 wrote:If 60! is written out as an integer, with how many consecutive 0’s will that integer end?

A) 6

B) 12

C) 14

D) 42

E) 56

OAC

Hi Experts ,

Please explain.

Many thanks in advance

SJ

Zeros as the end of a number means we are multiplying the number by powers of 10
We know that 10 = 5*2, this means a 10 has 1 power of 5 and one power of 2 in it.

Therefore if we have to calculate the number of 10s at the end, we need to calculate the number of powers of 5 in the number.

In the given number 60!,
Total powers of 5 = [60/5] + [60/25] = 12 + 2 = 14
Total powers of 2 = [60/2] + [60/4] + [60/8] + [60/16] + [60/32] = 20 + 15 + 7 + 3 + 1 = 46

Clearly the powers of 5 are less than the powers of 2. We need not calculate the powers of 2 as we know for sure that the powers of 2 would be higher than the powers of 5 in 60!

Hence the total number of 10s = 14
Number of zeors at the end = 14

Correct Option: C