Outline Mathematics
Number Theory

When 3AA1 is divisible by 9?

Here's a problem to tackle:

Solution

Copyright © 1996-2008 Alexander Bogomolny

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 and . 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

As yet we do not know what the missing digit A is. But we do know that 0 is the it can be, whilst its 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):

So we are looking for a number between 4 and 22 ( ) divisible by . There are just two of them: 9 and . Thus, one of the two: either

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:

Answer: A = 7 and the number is .

Check the result: Dividing by 9 we get 3771/9 = - 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

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

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

with no integer solutions. We now check the former:

In other words, B = .

Answer: B = 1 and the number is .

Check the answer: 1413 / 9 = . 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

Copyright © 1996-2008 Alexander Bogomolny