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## Wednesday, 17 August 2011

### Math GRE - #16

For how many positive integers $k$ does the ordinary decimal representation of the integer $k!$ end in exactly 99 zeros?

• None
• One
• Four
• Five
• Twenty-Four

Solution :

There are five positive integer $k$ for which this is true.

Note that the number of zeros depend directly on how many factors of 10 are present in $k!$. However, the number of 10s depend on the number of factors of 2s and 5s. Since there are an abundance of 2 in the factorization of $k!$, we just have to count the number of times a factor of 5 appears. Since one (or more) factor of 5 appear every five numbers, every five numbers have the same number of zeros at the end.

There is a simple flaw with the argument above, do you see it?
We haven't proven the existence of a single integer $k$ such that $k!$ ends in 99 zeros! It turns out that 400! has 99 zeros (and thus 401! to 404! do as well). I believe that we were supposed to assume the existence of such an integer $k$ and that the question is also a bit poorly worded. However, I could be wrong.

Can anybody see why there must be an integer $k$ such that $k!$ ends in 99 zeros?

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