# with how many zeros it ends

I have to calculate with how many zeros 1000! ends. This is wat I did, but I am not sure whether its good: I think I have to calculate how many times the product 10 is in 1000!. I found out the factor 10 is 249 times in 1000! using the fact that $s_p(n!)=\sum_{r=1}^{\infty}\lfloor \frac{n}{p^r}\rfloor$. So I think the answer should be that it ends with 249 zeros.

Is this correct/are there other ways to do this? If not how should I do it then?

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Hint: find out the number of 2's and 5's in $\,1,000!\,$..and yes, this is basically what you did, yet the formula you wrote works for primes, so better to separate 2's and 5's. – DonAntonio Feb 5 '13 at 18:49
@DonAntonio yes thats what I did because 10=2*5. there are 249 times 5 and more than 249 times 2. – Badshah Feb 5 '13 at 18:51

First, let's note that $10$ can be factored into $2 \times 5$ so the key is to compute the minimum of the number of $5$'s and $2$'s that appear in $1000!$ as factors of the numbers involved. As an example, consider $5! = 120$, which has one $0$ because there is a single $5$ factor and a trio of $2$ factors in the product, one from $2$ and a pair from $4$.

Thus, the key is to compute the number of $5$'s, $25$'s, $125$'s, and $625$'s in that product as each is contributing a different number of $5$'s to the overall product as these are the powers of $5$ for you to consider as the $2$'s will be much higher and thus not worth computing.

So, while there are 200 times that $5$ will be a factor, there are 40 times for $25$ being a factor, eight for $125$ and one for $625$, which does give the same result as you had of 249, though this is a better explanation.

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actually you should think about factor 5 rather than 10.

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Because for each 5 in $n!$ you'll get (a bit more than) two 2s, and so counting 5s is enough. – vonbrand Feb 5 '13 at 18:52

Also, your sum doesn't have to go to infinity. As you no doubt noticed, the sum itself is finite and the terms after a first few finite terms are all zero. We actually know exactly where we can truncate the sum. We will stop when we hit the largest power of $p$ dividing $n$ which can easily be found by $\left\lfloor \frac{log(n)}{log(p)} \right\rfloor$ because after that the fraction will be less than one so taking its floor will always give zero. So your sum can be written as

$$\sum_{r=1}^{\left\lfloor\frac{\log(n)}{\log(p)}\right\rfloor} \left\lfloor\frac{n}{p^r}\right\rfloor$$

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Actually, what you want to count is the number of $5$'s. If you count how many multiples of $5$ show up (multiples of $25$ should be counted twice, and multiples of $125$ should count for three times), you will get the number of $5$'s in the prime factorization of $1000!$

Similarly, you can count the number of $2$'s in the prime factorization of $1000!$. This number will be much greater.

Since $10=2\times 5$, for each $5$ and $2$ in the prime factorization, you get a power of $10$, ie. a $0$ at the end of the number. So the number of $0$'s will be whichever is less, the number of $5$'s or the number of $2$'s.

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