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 $2$.
[Math] Sum of integers divisible by their digits
elementary-number-theory
elementary-number-theory
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 $2$.
Best Answer
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