How many terminal zeros are at the end of (20!)?
How many numbers are divisible by 5 between 1 and 20?
20,15,10, and 5 are all divisible by 5. so 4. Is that how you solve for the amount of terminal zero's?
uh oh my mistake. I missed a key part of the question. the question is..... How many terminal zeros are at the end of (20!)^2?
The way we can get a terminal zero in a product depends on two factors (no pun intended), 2 and 5.
There are plenty of two's between 1 and 20, namely we get three of them in 8=2^3, and 16=2^4.
Therefore the governing factor is the number of 5's, which when combined with 2 gives a terminal zero.
If there are four 5's between 1 and 20, then 20! will have 4 terminal zeroes.
As you probably figured it out, squaring doubles the number of terminal zeroes.
So if 20! has 4, then (20!)^2 should have???
Note: watch out if and when you come to 25!
25=5^2, so counts as 2 factors of 5.