First, we need to find all numbers that satisfy the condition.
Represent each 2-digit number by 10x+y (x is the tens digit, y is the units digit)
Condition: the number is divisible by each of its digits.
Divisible by x:
[tex] \frac{10x+y}{x}= 10 + \frac{y}{x} [/tex]
[tex]10 +\frac{y}{x}[/tex] is a positive integer because it is the quotient.
This implies that [tex]\frac{y}{x}[/tex] is a positive integer as well. (a)
Let [tex]\frac{y}{x}=p[/tex] (b)
Divisible by y:
[tex] \frac{10x+y}{y}= \frac{10x}{y}+ 1 [/tex]
[tex]\frac{10x}{y}+ 1 [/tex] is also a positive integer because it is the quotient.
But [tex]\frac{10x}{y}=\frac{10}{p}[/tex] ,
So p should be a positive integer [from (a)] such that [tex]\frac{10}{p}[/tex] is a positive integer.
The only possible values for p are 1, 2, and 5.
Recall that p is the ratio between the ones and the tens digit [from (b)].
List of numbers when
p=1 {11, 22, 33, 44, 55, 66, 77, 88, 99}
p=2 {12, 24, 36, 48}
p=5 {15}
These are all the 2-digit positive integers that satisfy the condition.
The sum of all those numbers is 630.
