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

Solution :

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

**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?

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

Solution :

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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Let's say div(i,j) is the integer div operation.

For a given k, let's say that i=div(k,5). Then k!, when prime-factorised, will contain 5 to the power of

i + div(i,5) + div(i,25) + div(i,125) + ...

All we need to show is that there exists an i such that this sum equals 99. As it turns out, i=80 satisfies this, as

99 = 80 + div(80,5) + div(80,25) = 80 + 16 + 3

Therefore, if i=80 and i=div(k,5), then k! will figure 5 to the power of 99 in its prime-factorisation. Solving div(k,5)=80 gives k=400,401,402,403,404.

Yes, but spotting i = 80 is a bit difficult for those who are not good with mental arithmetic (myself included).

I haven't found any quick tests of determining of a specific number of zeros exist at OEIS either, though there are a few computation intensive formulae that give the exact number of trailing zeros in k! for any k. My guess is that there is no easy way (i.e. without much computation) to find if 99 zeros are possible.

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