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When 3-digit integer 6a3 plus 3-digit integer 3b5 is divisib

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by Max@Math Revolution » Mon Mar 07, 2016 4:00 pm
When 3-digit integer 6a3 plus 3-digit integer 3b5 is divisible by 9, is a+b=1?

1) The units of all the digits' sum of 6a3+3b5 is 8
2) a+b<10


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Source: — Data Sufficiency |
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by Max@Math Revolution » Fri Mar 11, 2016 4:59 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

When 3-digit integer 6a3 plus 3-digit integer 3b5 is divisible by 9, is a+b=1?

1) The units of all the digits' sum of 6a3+3b5 is 8
2) a+b<10


In the original condition, in order to satisfy 6a3+3b5=9integer, 6+a+3+3+b+5=17+a+b should be divided by 9 as well. Then, a+b should be 1, 10. Since 1)=2) a+b=1, it is yes and sufficient.
Therefore, the answer is D.
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