2
$\begingroup$

I'm trying to find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$, as part of a basic number systems course.

I'm rather lost with this question. For $10!$ I tried writing the terms out and just extracting powers of $2$ manually, getting $2^8$ as the highest powers of $2$, with $10! = (2^8)(14175)$ as the result.

I'm fairly confident that the answer is correct (although I'm not sure, so confirmation of that would be great!), but this method is rather crude for larger numbers, so I suspect that it isn't the right way to do it.

If anyone can point me in the right direction I would be very grateful.

  • 5
    Use de [Polignac's formula](http://en.wikipedia.org/wiki/De_Polignac%27s_formula). For example, for $40!$ it produces $20+10+5+2+1=38$ and for $10!$ it produces $5+2+1=8$ as answer.2012-10-21
  • 0
    There is a [recent similar](http://math.stackexchange.com/questions/215999/how-many-0s-are-at-the-end-of-20) problem in hich I tried to give the intuition behind the "shortcut" way of computing the highest power of a prime $p$ (in that case $5$) that divides $n!$.2012-10-21

2 Answers 2