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

Sagot :

AnneC
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.