# Advent calendar 2017

## 10 December

How many zeros does 1000! (ie 1000 × 999 × 998 × ... × 1) end with?

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The number of 0s at the end of 1000! will be equal to the number of times 10 is a factor of it.
For each 10 that is a factor, there will be a 5 and a 2 that are factors. Therfore the number of 0s will be equal to the number of times
5 is a factors of 100! (because 5 is larger than 2).

Therefore the number of 0s at the end of 1000! will be equal to
\(\left\lfloor\frac{1000}{5}\right\rfloor+\left\lfloor\frac{1000}{25}\right\rfloor+\left\lfloor\frac{1000}{125}\right\rfloor+\left\lfloor\frac{1000}{625}\right\rfloor\).
This is **249**.