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BTG practice Q.. If x, y and k are integers, is xy divisible

Expert replies
by odannyboi » Tue Sep 21, 2010 1:29 pm
If x, y and k are integers, is xy divisible by 3?

(1) y = 2^16 - 1 (EDIT)

(2) The sum of the digits of x equals 6^k

The OA is only S1 is sufficient.

they said (2) isn't sufficient b/c if k=1, xy is div by 3... but if k=0, it isn't.... how can you have the sum of the digits add up to zero and still satisfy the fact that X=6^k??
Last edited by odannyboi on Tue Sep 21, 2010 5:11 pm, edited 1 time in total.
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Source: — Data Sufficiency |

by HPengineer » Tue Sep 21, 2010 4:31 pm
i dont see how one is sufficient and 2 is not sufficient..

what was their reasoning for st1 being sufficient?
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by odannyboi » Tue Sep 21, 2010 5:19 pm
Crap... I fixed it... it should have been 2^16 instead of 216.

They basically factored it out for S1:
=2^8 - 1
=(2^8 + 1)(2^8 - 1)
=(2^8 +1)(2^4 +1) 2^4 -1)
=.... eventually they factored into an expression (2^2-1) which equals to 3... meaning is a multple of 3 and thus xy is div by 3.

I dunno, I didn't factor it out... I guessed that whatever answer you get is gonna tell you if XY is div by 3... so I though S1 was sufficient... WRONG thinking, b/c if y wasnt a multiple of 3, then it would be insufficient.

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by HPengineer » Tue Sep 21, 2010 8:40 pm
ha ok that makes much more sense... but still agree with you on statement two... it should be sufficient as well in my mind....
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