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sum of consecutive integers

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Source: — Data Sufficiency |
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by ikaplan » Sat Sep 17, 2011 4:52 am
I tried to solve this one by plugging numbers.

(1) n=6; 1+2+3+4+5+6=21 or (-4)+(-3)+(-2)+(-1)+0+1=-9
the sum is ODD, hence Statement (1) is sufficient

(2) n=9, 1+2+3+4+5+6+7+8+9=45 or (-3)+(-2)+(-1)+0+1+2+3+4+5=9, the sum is ODD

hence, Statement (2) is sufficient

IMO, D is the correct answer
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by knight247 » Sat Sep 17, 2011 7:04 am
(1)n=6
Consider different scenarios one beginning with an odd and other beginning with a even number.
1+2+3+4+5+6=21 Not even 8+9+10+11+12+13=63 Not even. So we get the answer that it is not even SUFFICIENT.

(2)The nth term is a multiple of 9. So 1+2+3+4+5+6+7+8+9=45 Not Even
Let the nth term be 18 11+12+13+14+15+16+17+18+19=268 Even. We get conflicting answers so INSUFFICIENT.

Answer is A
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