Outline Mathematics
Number Theory

When 3AA1 is divisible by 11?

Here's a problem to tackle:

Solution

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Copyright © 1996-2012 Alexander Bogomolny

A is a digit of a 4-digit number divisible by 11. A appears twice in the decimal representation of the number, with two other digits being and . There is a well known criterion for divisibility by 11: a number is divisible by 11 iff the alternating sum of its digits is divisible by 11. The sum in question (for number 3AA1) equals

and is never divisible by 11. We are forced to conclude that the problem has no solution.

May we modify the problem a little to make sure a solution exists? Does the following problem have a solution: find a decimal digit B such that number B4B3 is divisble by 11.

Since it is still a question of divisibility by 11, we are going to compute the alternating sum of the digits, which, in this case, is given by

Let's estimate the possible values of the alternating sum:

There are only two numbers divisible by 11 in this range: 0 and 11. The former leads us nowhere:

with no integer solutions. We now check the latter:

In other words, B = .

Answer: B = 9 and the number is .

Check the answer: 9493 / 11 = . Very good.

You may want to try your hand at a similar problem of divisibility by 9.

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Copyright © 1996-2012 Alexander Bogomolny