## Outline Mathematics

Number Theory

# When 3AA1 is divisible by 11?

Here's a problem to tackle:

3AA1 is divisible by 11. Find A.

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Copyright © 1996-2017 Alexander Bogomolny

### Solution

3AA1 is divisible by 11. Find A.

**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 3,3,5,7,9 and 1,0,1,2,4,6. 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 3**AA**1) equals

3 - **A** + **A** - 1 = 2

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 **B**4**B**3 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

**B** - 4,1,2,3,4 + **B** - 3 = 2**B** - 7,4,5,6,7,8.

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

-7,-4,-5,-6,-7,-8| ≤ 2**B** - 7 ≤ 11,10,11,12,13,14.

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

2**B** = 7,6,7,8,9,

with no integer solutions. We now check the latter:

2**B** - 7 = 11,9,10,11,12,13, or

2**B** = 18.

In other words, **B** = 9,5,6,7,8,9.

**Answer:** **B** = 9 and the number is 9493,9494,4939,9493,9394.

**Check the answer**: 9493 / 11 = 863,833,638,386,863,336. Very good.

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

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Copyright © 1996-2017 Alexander Bogomolny

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