# Find the sum of all positive 2-digit integer that are divisible by each of their digits.

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by lanchui

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by lanchui

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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:

is a positive integer because it is the quotient.

This implies that is a positive integer as well. (a)

Let (b)

Divisible by y:

is also a positive integer because it is the quotient.

But ,

So p should be a positive integer [from (a)] such that 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.**

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:

is a positive integer because it is the quotient.

This implies that is a positive integer as well. (a)

Let (b)

Divisible by y:

is also a positive integer because it is the quotient.

But ,

So p should be a positive integer [from (a)] such that is a positive integer.

The only possible values for p are 1, 2, and 5.

Recall that

List of numbers when

p=1

p=2

p=5

These are all the 2-digit positive integers that satisfy the condition.