Thread: Factorial

Hi everyone.

Show that if $\displaystyle n$ is a integer, then is impossible that $\displaystyle n!$ finish with $\displaystyle 153, 154$ or $\displaystyle 155$ zeros.

Thanks so much.

20! has 4 zeros because it has factors including 5,10,15 and 20. That gives us the factor 5 four times. There will always be enough 2s to pair up with the 5s.

25! has 6 zeros because you have factors 5, 10, 15, 20 and 25. That gives us the factor 5 six times.

Another example: 600! has 5,10,15....600 and 25,50,75...600 and 125,250,375,500. So how many zeros?

600/5=120
600/25=24
int(600/125)=4
I think that gives you 148 zeros.

The answer lies near here.