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amijags
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we have 100a+10b+c
from (1) we have a=b, thus
100a+10a+с
from (2) we have c=6+a, so
100a+10a+a+6 = 111a+6
plug any figure instead of a, you will see it divisible by 3.
(C)
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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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