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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 thatthe factors-of-5 cancel out.all

Therefore, there aretrailing zeros inno