10
$\begingroup$

Is it possible to find the digit sum of $n!$ ($n \in \mathbb{N} \text{ and } n \le100$) without actually computing the factorial?

I faced this problem in quantitative aptitude test which asks to find the sum of the digits of $25!$, I was wondering if there is any paper pencil approach to deduce the sum from counting the number of each digits in the factorial without actually computing the whole thing.

  • 2
    Related: http://stackoverflow.com/questions/1469529/sum-of-digits-of-a-factorial2011-11-07
  • 1
    In the present form of the question, the answer is "yes, obviously"; there are lots of ways of doing that -- for instance you can take out all factors of $2$ and $5$ and use only the excess of one of them over the other; then you've chopped off the terminal $0$s and never compute $n!$. So to make this an interesting question you'll have to make some quantitative specification how much easier the method should be than actually computing the factorial.2011-11-07
  • 2
    You can find the digit sum mod nine very easily: any natural number $m$ is congruent to its digit sum mod 9. If $n\geq 6$ then $9 | n!$ so the digit sum has to be zero mod 9...2011-11-07
  • 0
    @ mt_:I think it's called digital root, but I don't want that here.2011-11-07

1 Answers 1