If a and n are integers and n > 1, is n odd?
(1) a^n-1 > a^n
(2) a^n > a^3n
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is n odd
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j_shreyans
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Jim@StratusPrep
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The key with both of these is to recognize that a must be negative - its a tricky thing to think about when you have n > 1 making you think about positive numbers.
1) The only way for this to ever be true is if a is negative and n is an odd number, then positive > negative.
2) If a is negative then a^3n will be MORE NEGATIVE, or less than, a^n.
1) The only way for this to ever be true is if a is negative and n is an odd number, then positive > negative.
2) If a is negative then a^3n will be MORE NEGATIVE, or less than, a^n.
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Since n>1, each statement indicates the following:If a and n are integers and n > 1, is n odd?
(1) a^(n-1) > a^n
(2) a^n > a^(3n)
(integer a)^(smaller positive power) > (integer a)^(bigger positive power).
This relationship is valid only if a<0 and the exponent on the lefthand side is EVEN (so that the lefthand side becomes positive), while the exponent on the righthand side is ODD (so that the righthand side stays negative).
In each statement, for the exponent on the righthand side to be ODD, n itself must be odd.
Thus, each statement on its own is SUFFICIENT.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
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nikhilgmat31
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It is very important to note that a is integer.
a can't be 1/2 or so..
otherwise this question is fully reversed and answer will be E
a can't be 1/2 or so..
otherwise this question is fully reversed and answer will be E
