Number Theory
Number Theory
Level
3
Find the sum of all positive two-digit integers that are divisible by each of their digits.
Source: AIME 2001.
The answer is 630.
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1 solution
Assume that the number is AB.
It can be written as (10×A)+B
According to the question, both A & B must divide (10×A)+B
Then we can conclude, [(10×A)+B] is divisible by A if B is divisible by A & [(10×A)+B] is divisible by B if 10×A is divisible by B.
The above two conditions must be satisfied simultaneously.
Mathematically,
B = KA [where K is a constant]
10A = LB [where L is a constant]
Now, all possibilities for first condition in tabulated form:
& all possibilities for second condition in tabulated form:
As, both conditions must be satisfied simultaneously so we have to choose only those combinations of A & B that are common in both the tables.
So, the numbers [AB not BA] are:
11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99
Sum of these numbers is: 630