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divisibility by 3

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divisibility by 3

by mehrasa » Thu Sep 29, 2011 8:19 pm
If X and Y represent digits of a two digit number divisible by 3, is the two digit number less than 50?
1. Sum of the digits is a multiple of 18
2. Product of the digits is a multiple of 9

[spoiler]OA: A[/spoiler]
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by shankar.ashwin » Thu Sep 29, 2011 8:44 pm
Statement 1:

Sum should be 18,36 etc..

For sum to be 18, minimum 2 digit integer is 99 (9+9=18) > 50. Hence sufficient

Statement 2:

Product is multiple of 9 (9,18,27..)

Consider 19 & 91 (product is 9, but one is lesser and the other is greater than 50)

Hence A
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by gmatboost » Mon Oct 03, 2011 10:53 am
Shankar:

19 and 91 don't count because they aren't multiples of 3, as stated in the question.

We need two-digit numbers that are multiples of 3 AND whose digits multiply to a multiple of 9.

The only way the product of digits is a multiple of 9 is if either
1. Both digits are multiples of 3
2. One of the digits is 9

So, we are talking about numbers whose digits are both 3, 6, or 9: 33, 36, 39, 63, 66, 69, 93, 96, 99
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