| |||||||||||||||||||||||||||||||||||||||||||
Copyright © 1996-2008 Alexander Bogomolny
Solution
(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
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
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:
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
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
Copyright © 1996-2008 Alexander Bogomolny
|
| ||||||||||||||||||||||||||||||||||||||||||