| Manifesto | Bookstore | Contents | Amazon store | Term index | What changed? | Contact | Recommend |
Outline Mathematics
|
| 3AA1 is divisible by 11. Find A. |
Copyright © 1996-2009 Alexander Bogomolny
Solution
| 3AA1 is divisible by 11. Find A. |
(In the text, some words are omitted. These have been underlined. Click just above the line. See what happens.)
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
| 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 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
| B - + B - 3 = 2B - . |
Let's estimate the possible values of the alternating sum:
| ≤ 2B - 7 ≤ . |
There are only two numbers divisible by 11 in this range: 0 and 11. The former leads us nowhere:
| 2B = , |
with no integer solutions. We now check the latter:
|
2B - 7 =
, or 2B = 18. |
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.
Copyright © 1996-2009 Alexander Bogomolny
| 34385524 |  |
