Hello, musngiburger!
Find the number of trailing zeros in the ordinary decimal representation of the number: .
We have: .
The numerator is: .
The denominator is: .
The number of trailing zeros depends on the number of factors-of-5
. . that remain in the numerator after reducing the fraction.
The numbers in the numerator which have a factor-of-5 are:
. .
There are ten factors-of-5. .(100 has two.)
The numbers in the denominator which have a factor of 5 are:
. .
There are ten factors-of-5. .(25 has two.)
We find that all the factors-of-5 cancel out.
Therefore, there are no trailing zeros in