If I expand 25*24*23* ... *12*11*10, how many zeros are there at the end of the number I get?
+185 We can use the trailing zeroes trick.
We can divide 25! by each power of 5.
25/5=5. But since we don't include the first multiple of 5, there would be four fives. So this would get 4 zeroes.
Now we can divide by 25, or 5^2. This gets an additional one more zero.
This then gets 5 trailing zeroes.
