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is n — 1 divisible by 3?

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Source: — Data Sufficiency |
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by thephoenix » Mon Apr 05, 2010 9:43 am
imo a

s1) =n(n+1) any value of n for which n(n+1) is not div by 3 n-1 will be div by 3
suff
s2)i tried alot and think its in suff
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by eaakbari » Mon Apr 05, 2010 11:58 am
Statement one
Since n and n+1 are not divisible then obviously n-1 will be divisible by 3. Suff

Statement two

3n + 5 >= k + 8
3n>= 3.c +3 (let k = 3.c)

n>= c+ 1
which does not tell us anything

Hence Insuff

IMO A
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by dxgamez » Mon Apr 05, 2010 7:34 pm
see this post by Stuart. very useful to see why statement 1 is insufficient in one glance.

https://www.beatthegmat.com/divisibility ... tml#240208
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by eaakbari » Mon Apr 05, 2010 9:26 pm
dxgamez wrote:see this post by Stuart. very useful to see why statement 1 is insufficient in one glance.

https://www.beatthegmat.com/divisibility ... tml#240208

The question is different and its the same logic that it applied. When you have n , n-1 and n+ 1 , one of them have to be a multiple of 3. Plug in any number and you will see thats the case. Since we are given n and n+1 are not divisible by 3 then definitely n -1 is
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by thephoenix » Mon Apr 05, 2010 9:28 pm
dxgamez wrote:see this post by Stuart. very useful to see why statement 1 is insufficient in one glance.

https://www.beatthegmat.com/divisibility ... tml#240208
yes and for the very same reason here s1 is suff....
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by Stuart@KaplanGMAT » Mon Apr 05, 2010 9:42 pm
thephoenix wrote:
dxgamez wrote:see this post by Stuart. very useful to see why statement 1 is insufficient in one glance.

https://www.beatthegmat.com/divisibility ... tml#240208
yes and for the very same reason here s1 is suff....
Correct - the logic applied in that thread proves that (1) is sufficient, not insufficient.

(1) breaks down to n(n+1) is NOT a multiple of 3, which means that neither n nor (n+1) is a multiple of 3.

Since one of (n-1), n and (n+1) must be a multiple of 3, by default (n-1) must be divisible by 3.
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by istuti » Tue Apr 06, 2010 2:40 am
Just wish to cross-check, we'd okay 0 as div by 3 right?
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by Stuart@KaplanGMAT » Tue Apr 06, 2010 8:09 am
istuti wrote:Just wish to cross-check, we'd okay 0 as div by 3 right?
0 is divisible by (i.e. is a multiple of) everything except 0.
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