If an ≠0 and n is a positive integer, is n odd?
(1) a^n + a^n+1 < 0
(2) a is an integer.
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n is odd?
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j_shreyans
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Statement 1: a^n + a^(n+1) < 0j_shreyans wrote:If an ≠0 and n is a positive integer, is n odd?
(1) a^n + a^n+1 < 0
(2) a is an integer.
This inequality holds true only if a<0.
Test one case that also satisfies statement 2 and one case that does not.
It's possible that a=-2 and n=2, since (-2)² + (-2)³ = -4.
It's possible that a=-1/2 and n=1, since (-1/2)¹ + (-1/2)² = -1/4.
Since n is EVEN in the first case but ODD in the second case, INSUFFICIENT.
Statement 2: a is an integer
No information about n.
INSUFFICIENT.
Statements combined:
a must be a negative integer.
If n is odd, then a^n + a^(n+1) = nonnegative.
To illustrate:
If a=-1 and n=1, then a^n + a^(n+1) = (-1)¹ + (-1)² = 0.
If a=-2 and n=3, then a^n + a^(n+1) = (-2)³ + (-2)� = 8.
If a=-3 and n=3, then a^n + a^(n+1) = (-3)³ + (-3)� = 54.
Thus, to satisfy the constraint that a^n + a^(n+1) < 0, n CANNOT be odd.
SUFFICIENT.
The correct answer is C.
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If an ≠0 and n is a positive integer, is n odd?
(1) a^n + a^n+1 < 0
(2) a is an integer
Think if the logic this way: This is basically testing even/odd powers of a negative number concept.
1) a^n (1 + a) <0..now can a>0? NO as a^n >0 and (+a)>0. So, a <0. There is a reason why they are asking whether n is odd. We want to find out whether a^n is positive or negative (give that a<0).
now, since we don't know what a is (integer or not), (1+a) can be <0 or >0, giving us no clue whether a^n is positive or negative (or n even or odd, given that a<0).
Statement 2 tells us exactly that a integer,,,,we know a<0 so, (1+a) <0 => (-ve)^n needs to be positive. So, we know n has to be even.
(1) a^n + a^n+1 < 0
(2) a is an integer
Think if the logic this way: This is basically testing even/odd powers of a negative number concept.
1) a^n (1 + a) <0..now can a>0? NO as a^n >0 and (+a)>0. So, a <0. There is a reason why they are asking whether n is odd. We want to find out whether a^n is positive or negative (give that a<0).
now, since we don't know what a is (integer or not), (1+a) can be <0 or >0, giving us no clue whether a^n is positive or negative (or n even or odd, given that a<0).
Statement 2 tells us exactly that a integer,,,,we know a<0 so, (1+a) <0 => (-ve)^n needs to be positive. So, we know n has to be even.
