Outline Mathematics
|
| 3AA1 is divisible by 9. Find A. |
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Copyright © 1996-2012 Alexander Bogomolny
Solution
| 3AA1 is divisible by 9. 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 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
| 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
it can be, whilst its
value is 9. It follows that the minimum and maximum values the digital root may have are 4 (assuming
| 4 ≤ 4 + 2A ≤ 22. |
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
|
4 + 2A = 9 or 4 + 2A = 18. |
Assuming the first equation, we arrive at an impossibility:
|
2A =
, or A = . |
Answer: A = 7 and the number is .
Check the result: Dividing by 9 we get
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 + + B + 3 = 2B + . |
As before, we start with estimatng the possible values of the digital root:
| ≤ 2B + 7 ≤ . |
Again there are only two numbers divisible by 9 in this range: 9 and 18. The latter leads us nowhere:
| 2B = , |
with no integer solutions. We now check the former:
|
2B + 7 =
, or 2B = 2. |
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
- G. Lenchner, Math Olympiad Contest Problems For Elementary and Middle Schools, Glenwood Publications, NY, 1997
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