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by 7806 » Mon Sep 06, 2010 12:38 am
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Given that n is an integer, is n - 1 divisible by 3?

(1) (n^2+n ) is not divisible by 3
(2) 3n+5>= K+8 , where k is a positive multiple of 3

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.


A/C...?? i solved it as A.... but OA-C. pl confirm. thanks.
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Source: — Data Sufficiency |

by Rahul@gurome » Mon Sep 06, 2010 4:26 am
(1) n(n + 1) is not divisible by 3 implies n and (n + 1) are not divisible by 3, which implies (n - 1) is always divisible by 3, since every 3rd number is divisible by 3.
So, (1) is SUFFICIENT.

(2) 3n+5 ≥ K+8 6r 3n ≥ K + 3, let us take K = 3M
Then 3n ≥ 3M + 3 or 3n ≥ 3(M + 1) or n ≥ (M + 1), but this is NOT SUFFICIENT to answer whether (n - 1) is divisible by 3.

The correct answer is [spoiler](A)[/spoiler].

Does that help?
Last edited by Rahul@gurome on Mon Sep 06, 2010 5:12 am, edited 1 time in total.
Rahul Lakhani
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by 7806 » Mon Sep 06, 2010 4:51 am
thanks rahul.

but you told ....n - 1 = 0, which is not divisible by 3

(1) n(n + 1) is not divisible by 3.
Let n = 1, then n^2 + n = 2, not divisible by 3. Here n - 1 = 0, which is not divisible by 3.
If n = 4, then n^2 + n = 20, not divisible by 3. But n - 1 = 3, which is divisible by 3.
No unique answer.
So, (1) is NOT SUFFICIENT.



isn't 0 is divisible by all numbers.??

Oa is C itself.
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by Rahul@gurome » Mon Sep 06, 2010 5:14 am
You pointed that right. I have edited the post and did it straight away without taking any values. Statement (1) should be SUFFICIENT and that makes the OA as (A).
Rahul Lakhani
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Gurome, Inc.
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
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