Question

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

Answer 1

14

Explanation:

Ok, 60! is a really big number

`x!=x(x-1)(x-2)...1`

So we have 60*59*58...*1

2 times 5 is 10 which makes a terminal zero (ending zero)

We need to count how many fives and twos are in 60!

There are 60/5=12 5^1's

There are 60/25=2 5^2's

So 12+2=14

There are way more 2's than fives so we take the lesser number

So the answer is 14

*Note that 60/25 is not 2, we need an integer rounded down