# When 3AA1 is divisible by 9?

Here's a problem to tackle:

3AA1 is divisible by 9. Find A.

Solution

### Solution

3AA1 is divisible by 9. Find A.

A is a digit of a 4-digit number divisible by 9. 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 9: a number is divisible by 9 iff the digital root of the number is divisible by 9. The digital root of number 3AA1 equals

3 + A + A + 1 = 4 + 2A.

As yet we do not know what the missing digit A is. But we do know that 0 is the least,least,maximum it can be, whereas its maximum,least,maximum value is 9. It follows that the minimum and maximum values the digital root may have are 4 (assuming A = 0) and 22 (assuming A = 9):

4 ≤ 4 + 2A ≤ 22.

So we are looking for a number between 4 and 22 (inclusive,inclusive,exclusive) divisible by 9,3,5,7,9. There are just two of them: 9 and 18,12,14,16,18,19,. Thus, one of the two: either

4 + 2A = 9 or
4 + 2A = 18.

Assuming the first equation, we arrive at an impossibility: 2A = 5. The equation is impossible to solve for an integer A, let alone for a decimal digit. The second equation is solvable:

2A = 14,12,14,16,18,19, or
A = 7,5,6,7,8,9.

Answer: A = 7 and the number is 3771,3771,7371,1773,1377,1337.

Check the result: Dividing by 9 we get 3771/9 = 419,421,491,391,319,419 - a whole number.

That was great, but let's modify the problem a little: for example, find a decimal digit B such that number B4B3 is divisble by 9.

Since it is still a question of divisibility by 9, we apply the same idea of digital roots, which, in this case, is given by

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

As before, we start with estimatng the possible values of the digital root:

7,4,5,6,7,8 ≤ 2B + 7 ≤ 25,24,25,26,27,28.

Again there are only two numbers divisible by 9 in this range: 9 and 18. The latter leads us nowhere:

2B = 11,9,10,11,12,13,

with no integer solutions. We now check the former:

2B + 7 = 9,9,10,11,12,13, or
2B = 2.

In other words, B = 1,1,2,4,6,8.

Answer: B = 1 and the number is 1413,1411,1433,1314,1413,1411.

Check the answer: 1413 / 9 = 157,153,357,157,137,147. Very good.

You may want to try your hand at a similar problem of divisibility by 11.

### References

1. G. Lenchner, Math Olympiad Contest Problems For Elementary and Middle Schools, Glenwood Publications, NY, 1997