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

1
by lanchui

2014-07-18T01:42:38+08:00

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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.
Take time to read. Sa unang basa, nakakahilo talaga, so read again kung di gets :) Don't hesitate to ask kung may part ka na di talaga maintindihan.
okay po, i find it hard to represent....nahihilo po ako kung ano ang gagamitin ko po as tens and ones :(
Kahit anong letter pwede mong gamitin. Ang mahalaga lang ay yung 10 sa tens digit. For example, ang two digit number ko ay 48. So, gamit yung representation sa taas, x=4 and y=8. Kung wala yung 10 sa 10x+y, magiging x+y lang sya, which is equal to 12 and not 48.
Isipin mo na bawat digit ay may katumbas na value. Sa 48 ang value ng 4 ay 40, kaya kailangan imultiply sa 10. Ganon din ang gagawin sa ibang digits in different place values. For example, to represent a 3-digit integer, I will use 100a + 10b + c
ahhhhh okay po, gets ko na yung pag represent..thank you po!!!!