Expert help appreciated: (not sure what the OA is)
Is the sum of a series of n consecutive integers even?
(1) n=6
(2) The n digit number formed by the seriest is a multiple of nine
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sum of consecutive integers
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ikaplan
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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
(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
"Commitment is more than just wishing for the right conditions. Commitment is working with what you have."
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knight247
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(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
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
