The answer should be E.
This is not a tough question,if you used numbers rather than tedious algebra.
Prompt says a ,when divided by 5,leaves remainder 3.
So a could be = 8,13,18,23... and so on
1) a when divided by 6 leaves remainder 1.
Think of no's that leave remainder 1:
7,13 [match with 13 in prompt].Go on to see whether its sufficient 19,25,31,37,43 [another match]
Divide 43 by 5 remainder = 3
Insufiicient
2) Same procedure 4:
5,9,13 [match],17,21,25,29,33 [match]
Insufficient.
1 and 2
6 -> 7,13 [Match],19,25,31,37,43,49,55,61,67,73 [match]
4 -> 5,9,13 [Match],17,21,25,29,33,37,41,45,49,53,57,61,65,69,73[Match]
a could be either 13 or 73.likewise there will be numerous other possible values of a.
You don't have to write down every value like I did above.you could have don the entire question mentally:just write down the matches on scratch paper so that you don't forget them.
What is the OA?
This is the main reason why I say algebra can be tedious.I solved this in under 45 seconds using this approach.GMAT doesn't check what you did on scratch paper.
Disagree with OA [GMAT-Hacks Challenge number properties]
- rishab1988
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