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Source: — Data Sufficiency |
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by mendiratta » Tue May 08, 2007 10:35 pm
1) n = 2K + 3
if k=0, n=3 - not divisible.
if k=2, n=7 - divisible.

So we cant say. Therefore A & D are ruled out.
2) It dosn't give anything about n. So B is also ruled out.

We're now left with C & D only.
Combining both the equations we get : -
n = 2K + 3
n - 7 = 2K -4
now eqn 2 says that 2k-4 is divisible by 7 so that means n-7 is also divisible by 7.
Does it mean that n is also divisible by 7?. I think NO.
Suppose n-7 = -7, which is divisible by 7. In this case n=0, that is not divisible by 7.
Suppose n-7 = 7, which is divisible by 7. In this case n=14, that is divisible by 7.

So I think the answer should be E.

Anyone got other ideas?
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by maolivie » Wed May 09, 2007 9:42 am
I'd say C.

Because if you add 7 to both side, you get 2K-4+7. If 2K-4 is divisible by 7, adding 7 to it makes that divisible by 7 also.

Or, since n-3 = 2K, n-3+-4 = -7
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by fushyguru » Wed May 09, 2007 5:43 pm
C is the official answer -

Thanks for the great explanation :D
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by sanju438 » Thu Oct 02, 2008 1:33 pm
from (1) n - 3 = 2K . n = 2K + 3. So n can be 3, 5, 7, 9 etc
insufficient

(2) 2K-4 is divisible by 7. 2K can be 11, 18, 25, 32, 39 etc
Since K is integer, K can be 9, 16, etc
insufficient

Combining (1) and (2)
n - 3 = 2K => n -3 = 7M + 4 (where M is any integer since 2K - 4 is divisible by 7)
This means n = 7M + 7 which is divisible by 7
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