I want to know how to find the last non-zero digit of $n$.

For example $n = 100!$

my try:

First i have to know how much Zeros $100!$ has so i did this:

$$E_{5}100 = \sum _{1\leq k <n} \Bigg[\frac{100}{5^{k}}\Bigg] =\Bigg[\frac{100}{5}\Bigg] + \Bigg[\frac{100}{25}\Bigg] = 24$$

So $100!$ has $24$ zeros which means that the last digit of $\quad\frac{100!}{10^{24}}\quad$ is the number that i´m looking for.

so if $x = \frac{100!}{10^{24}}$ i need to find $x (mod 10)$ to get it but here is where i got stuck...

  • $\begingroup$ First multiply digits 1 to 9 and note the last non zero number this might be your answer which will be 4. This is how you will get the last number. $\endgroup$ – Jasser Sep 22 '14 at 7:50
  • $\begingroup$ @user291957 actually, $9! = 362880$, whose last non-zero number is 8. $\endgroup$ – symmetricuser Sep 22 '14 at 8:42
  • $\begingroup$ Yes yes it's 8 @user125084 . Apologies. $\endgroup$ – Jasser Sep 22 '14 at 8:45
    int digit=1;
    int tmp;
    int cnt_2=0;
        tmp =i;
            tmp = (tmp>>1);
            tmp = tmp/5;
        digit = (digit*tmp)%10;
            digit = ((digit<<1)%10);

This is the code in C which I wrote to find the last nonzero digit of n!

  • $\begingroup$ Can you provide formal mathematical description of your algorithm in addition to the programming implementation? $\endgroup$ – Vlad Aug 20 '15 at 17:14
  • $\begingroup$ sorry ! I am not so good at mathematics. But I can try to explain what I did here. $\endgroup$ – nhimran Aug 22 '15 at 4:57

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