# Sum of integers divisible by their digits

Determine the sum of : all two-digit positive integers that are divisible by each of their digits.

For example :

$12$ is divisible by $1$ and $2$.

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ALL CAPS, abbreviations like "Thx", and commands like "Determine the sum of..." are not very polite. It's also good form to include your thoughts on the problem (what you've tried) and maybe why you want to know the answer. Finally, you mean "divisible by each of its digits", not "divisible to its each digit". –  Alex Kruckman Mar 22 '13 at 7:59
Have you proceeded with the question? –  hjpotter92 Mar 22 '13 at 7:59

Let such a number be $10x+y$. It is divisible by $x$ and $y$.

So $(10x+y)/x$ = $10+ y/x$ = $10 + m$ should be a natural number where $m=y/x$.

Similarly $(10x+y)/y$ = $10(x/y) + 1$ = $10/m + 1$ should be a natural number.

This is only possible for $m = 1, 2$ and $5$. So the possible numbers are

For $m=1$: No.s = $\{11,22,33,...99\}$

For $m=2$: No.s = $\{12,24,36,48\}$

For $m=5$: No.s = $\{15\}$.

Hence there are 14 such numbers

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And their sum is 630? –  user47805 Mar 22 '13 at 17:15

Hint: work them out and add them up

Check: I think there are 14 cases: nine of one type, four of another type, and one more which would give you an answer of the form $$a \sum_{i=1}^9 i + b \sum_{j=1}^4 j + c$$

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