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# If a and n are integers and n > 1, is n odd?

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Math Expert
Joined: 02 Sep 2009
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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21 May 2016, 05:06
sa18 wrote:
Can someone explain why my way of solving A is incorrect?

a^(n-1)>a^n
=> a^n/a >a^n
=> a^n/a - a^n > 0
=> a^n((1/a)-a) > 0
=> ((1-a)/a) a^n > 0

Both the terms can either be positive or negative
Taking the first case i.e. positive
Say ((1-a)/a) = 1
=> a=1/2
Substituting the value of a in a^n
For the second term i.e. a^n to be positive
(1/2)^n can take the value of n as either positive or negative
Why is then A sufficient?

Notice that we are told that a and n are integers.
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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26 Dec 2016, 12:57
Nighmare question to get on the real thing, wonder if there's any educated guessing that would make this quit
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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16 Apr 2017, 11:14
Theoretical method - way too LONG:
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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16 Apr 2017, 11:16
a |= 0, 1, (-1)
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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16 Apr 2017, 11:19
Yesterday when i got this question in exam this is what i did:
This took me > 2 min.
Please let me know if anyone knows shorter/faster method!!!
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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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16 Apr 2017, 11:57
EvaJager wrote:
conty911 wrote:
If a and n are integers and n > 1, is n odd?

(1) a^(n–1) > a^n
(2) a^n > a^(3n)

Please provide me pointers to any similar questions, also if possible

Can some one explain steps to tackle such problems with least amount of time.For eg. how to select the values for plugging etc.
THanks

From the conditions in both statements, we can deduce that $$a$$ cannot be neither $$0$$ nor $$1.$$
A little algebraic manipulation can help in understanding the inequalities and also in choosing values for $$a$$.
Usually it is easier to compare a product of numbers to $$0$$, so it is worth rearranging the inequalities such that there is a comparison to $$0$$ involved.

(1) $$a^{n-1}>a^n$$ can be rewritten as $$a^{n-1}-a^n>0$$ or $$a^{n-1}(1-a)>0$$.
If $$a>1$$, doesn't matter if $$n$$ is even or odd, the given inequality cannot hold.
If $$a<0,$$ then necessarily $$n$$ must be odd.
Sufficient.

(2)$$a^n>a^{3n}$$ can be rewritten as $$a^n-a^{3n}$$ or $$a^n(1-a^{2n})>0$$.
Since $$a$$ cannot be neither $$0$$ nor $$1,$$ $$\,\,1-a^{2n}$$ is certainly negative. (Now $$a$$ cannot be $$-1$$ either.)
Then $$a^n$$ is negative if $$a$$ is negative and $$n$$ is odd, and this is the only way the given inequality can hold.
If $$a$$ is positive, doesn't matter the value of $$n$$, the inequality cannot hold.
Sufficient.

hello, can you explain more about statement 2, why a is positive, the inequality never holds? if 0<a<1, it still holds right?

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Re: If a and n are integers and n > 1, is n odd? [#permalink]

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28 Aug 2017, 12:14
This can be done with algebra if you're adept at factoring inequalities.

Statement 1)

$$a^{n-1}>a^n$$
$$a^{n-1}-a^n>0$$
$$a^{n-1}(1-a)>0$$
$$a^{n-1}(-1)(1-a)>0(-1) \rightarrow a^{n-1}(a-1)<0$$

If $$n$$ is even then $$a^{n-1}$$ is odd, and the solution to the inequality would be $$0<a<1$$.
Since it's given that $$a$$ is an integer, $$0<a<1$$ can't be the solution, and $$n$$ must be odd.

Sufficient

Statement 2)

$$a^n>a^{3n}$$
$$a^n-a^{3n}>0$$
$$a^n(1-a^{2n})>0$$
$$a^n(-1)(1-a^{2n})>0(-1) \rightarrow a^n(a^{2n}-1)<0$$
$$a^n(a^n+1)(a^n-1)<0$$

The solution to this inequality is $$a^n<-1$$ and $$0<a^n<1$$
Again, it's given that $$a$$ is an integer, so the second case isn't possible.
Therefore $$a^n<-1$$, and $$n$$ must be odd

Sufficient

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Re: If a and n are integers and n > 1, is n odd?   [#permalink] 28 Aug 2017, 12:14

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# If a and n are integers and n > 1, is n odd?

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