How do I find sum of digits of a given factorial with missing digits?

Question

Suppose its given that

21!=5109094x17170y440000

How do I find

x+y

I know any factorial bigger than 6! will be divisible by 9. So I can apply that rule to find out it should be

52+x+y divisible by 9

. Which will imply that

x+y = 2

But looking at the answer

21! = 51090942171709440000

so

x+y = 11

Is there any rule that I am missing which will tell me the actual sum, not the digit sum?

Answer 1

You can use the divisibility test for $11$. Add the digits in even places and subtract the digits in odd places. The result must be a multiple of $11$. That will distinguish your two cases.

Answer 2

One way to proceed is to note the excess factors of $2$ over factors of $5$. You may want to check out Legendre's Formula, which is relevant to what follows.

Write down the numbers from $1$ to $21$. You see four of them with factors of $5$, each without a second factor of $5$. So the factorial has four factors of $5$. But for factors of $2$ first there are ten of those from $2$ up to $20$, then five of those ten are multiples of $4$ and thus have an additional factor of $2$, and you tack on still more factors of $2$ for the multiples of $8$ and finally $16$. You add all those up and there are $18$ factors of $2$.

This difference between factors of $2$ and $5$ implies that there are only four terminal zeroes but the preceding digits must be a multiple of $2^{14}$. This means, for instance, the last six digits before the terminal zeroes must be a multiple of $64$. Why do I choose six digits in this example? As we will see, going three digits before the final $y$ will lead to a unique solution.

Divide $170y44$ by $64$ by first putting in $y=0$, getting a remainder $R$, then interpreting the remainder as $100y+R$ for any $y$. Knowing $R$, you are to match $y$ with $R$ so that the combined remainder $100y+R$ is a multiple of $64$. Here $R=60$, and only $y=9$ among single positive digits matches with a multiple of $64$. Having found $x+y=11$ with the divisibility tests for $9$ and $11$ (Ross Milliken), you then conclude $x=2$.