If 73! has 16 zeroes at the end, how many zeroes will 80!

We can solve the question using info given in the stem (about 73!) but I find it easier to use direct approach of trailing zeros.

Trailing zeros:
Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

125000 has 3 trailing zeros;

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!?
\(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

BACK TO THE ORIGINAL QUESTION:

According to above 80! has \(\frac{80}{5}+\frac{80}{25}=16+3=19\) trailing zeros (take only the quotient into account).

Answer: D.

For more on trailing zeros check: everything-about-factorials-on-the-gmat-85592-40.html

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Hope it helps.