# How many zeros are there at the end of 1000 factorial?

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### How many zeros are there at the end of \(1000! \,?\)

For finding the solution it's convenient to write the factorials of the numbers til 10:

\[\begin{align} 0!&=1 \\ 1!&=1 \\ 2!&=1\times2=2 \\ 3!&=1\times2\times3=6 \\ 4!&=1\times2\times3\times4=24 \\ 5!&=1\times2\times3\times4\times5=120 \qquad \text{(first 0 appearing as a result of a par number, multiple of 2, plus a 5)} \\ 6!&=1\times2\times3\times4\times5\times6=720 \\ 7!&=1\times2\times3\times4\times5\times6\times7=5040 \\ 8!&=1\times2\times3\times4\times5\times6\times7\times8=40320 \\ 9!&=1\times2\times3\times4\times5\times6\times7\times8\times9=362880 \\ 10!&=1\times2\times3\times4\times5\times6\times7\times8\times9\times10 \\ &=3628800. \qquad \qquad \qquad \qquad \quad (\text{second 0 appearing as a result of a second factor } 5 ~(10=2\times5)) \end{align}\]

So, the key to solving this problem is to find the number of 5 factor existing in 1000!.

To find how many numbers from 0 to 1000 has a single factor 5, we can divide 1000 by 5 to obtain \(1000\div 5=200.\)

But there are numbers with two factor 5's like \(25=5\times5, 50=2\times25, 75=3\times25,\) etc. To determine how many numbers betweem 0 and 1000 introduce an extra 0, we have to divide 1000 by 25, so \(1000\div25=40.\)

Likewise, there are numbers with three factor 5's like \(125=5\times5\times5, 250=2\times5\times5\times5,\) etc. So there will be \(1000\div 125= 8.\)

Finally there is only one number with four factor 5's in this set of numbers: \(625=5\times5\times5\times5.\)

Therefore, the number of zeros at the end of \(1000 !\) is \(200+40+8+1=249.\) \(_\square\)

**Cite as:**How many zeros are there at the end of 1000 factorial?.

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