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If n is a prime number greater than 2, is 1/x > 1?

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emmak
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If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   Updated on: 25 Nov 2018, 10:25
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If n is a prime number greater than 2, is 1/x > 1?


(1) \(x^n < x < x^{\frac{1}{n}}\)

(2) \(x^{n-1} > x^{2n-2}\)
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C

Originally posted by emmak on 08 May 2013, 22:03.
Last edited by Bunuel on 25 Nov 2018, 10:25, edited 3 times in total.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   09 May 2013, 00:41
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emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?
(1) \(x^n < x < x^{\frac{1}{n}}\)
(2)\(x^{n-1} > x^{2n-2}\)


The question is asking, is \(\frac{1}{x}\)>1 --> Is 0<x<1

From F.S 1, for n=3, we have\(x^3<x<x^{\frac{1}{3}}\) . As \(x>x^3,\) we have \(x(1-x^2)>0\) --> x(x+1)(x-1)<0 .

Hence 0<x<1 OR x<-1. Again, from \(x<x^{\frac{1}{3}}\), we can cube on both sides and get, \(x^3<x\), which is the same as above. Thus, from F.S 1, we get :

0<x<1 OR x<-1. Insufficient.

From F.S 2, we know that for n=3, \(x^2 > x^4 --> x^2<1\) --> -1<x<1. Clearly Insufficient.

Taking both together, we get 0<x<1 as the common range and a YES for the question stem. Sufficient.

C.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   30 Oct 2014, 08:52
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emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) \(x^n < x < x^{\frac{1}{n}}\)

(2)\(x^{n-1} > x^{2n-2}\)


Statement 1 -> will be true for negative integers and positive fractions. So X can either be a negative integer or positive fraction. If X is a negative integer 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 1 alone is insufficient.


Statement 2 -> Will be true for negative and positive fractions. If X is a negative fraction 1/X will not be greater than 1 but if X is a positive fraction 1/X will be greater than 1. So we have two possibilities.

Hence Statement 2 alone is insufficient.

But for both statement 1 and 2 to be true, X can only be a positive fraction. Here 1/X will be greater than 1.

Hence both statements together are sufficient.

Answer -> C
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   01 Nov 2014, 08:12
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emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) \(x^n < x < x^{\frac{1}{n}}\)

(2)\(x^{n-1} > x^{2n-2}\)


I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   01 Nov 2014, 08:38
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thefibonacci wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) \(x^n < x < x^{\frac{1}{n}}\)

(2)\(x^{n-1} > x^{2n-2}\)


I got A for this one.

1) x^n < x < x^(1/n)
n = odd (since it is prime and > 2)
so the inequality will hold true only for 0 < x < 1
it will not hold true for -ve x as in that case x^(1/n) would be an imaginary number.
so this is sufficient.

2) x^(n-1) > x^(2n-2)
=> x^(n-1) > x^2(n-1)
n=odd, so n-1=even=2k(say)
x^2k > x^4k
this inequality holds true for -1<x<1
so insufficient.

Is the OA correct? Bunuel can you please have a look at this one.


Odd root from negative number is negative, not imaginary. For, example, \(\sqrt[3]{-8}=-2\).

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   11 May 2016, 13:45
I believe the answer should be B

Statement 1: Insufficient [0<x<1] or [x<-1] (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct ([1/4] is greater than [1/16])
When plugging in x=(-1/2) and n=3, statement 2 is incorrect ([-1/4] is NOT greater than [-1/16])
Therefore: [0<x<1] only ; [1/x]>1 in all cases ; Statement 2 Sufficient

Answer is B.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   11 May 2016, 15:40
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mattyahn wrote:
I believe the answer should be B

Statement 1: Insufficient [0<x<1] or [x<-1] (refer to others)

Statement 2: Fractions get closer to zero when raised to a power greater than 1. I think of it like this - when fractions are raised to a power greater than 1, the fraction gets closer to zero. A positive fraction would decrease, while a negative fraction would increase.
When plugging in x= (1/2) and n=3, statement 2 is correct ([1/4] is greater than [1/16])
When plugging in x=(-1/2) and n=3, statement 2 is incorrect ([-1/4] is NOT greater than [-1/16])
Therefore: [0<x<1] only ; [1/x]>1 in all cases ; Statement 2 Sufficient

Answer is B.


The OA is given under the spoiler and it's C, not B.

If x=-1/2 and n=3, \(x^{n-1} > x^{2n-2}\) holds true: \([(-\frac{1}{2})^{(3-1)} =\frac{1}{4}] > [\frac{1}{16}=(-\frac{1}{2})^{(2*3-2)}]\)
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If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   06 Nov 2019, 05:32
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?

(1) \(x^n < x < x^{\frac{1}{n}}\)
(2) \(x^{n-1} > x^{2n-2}\)


\(n=[3,5,7…odd.prime]\)
\(1/x>1…when:x=positive.proper.fraction…0<x<1\)

(1) \(x^n < x < x^{\frac{1}{n}}\) insufic.

\(n=3:x^n < x…x^3-x<0…x(x^2-1)<0…x(x-1)(x+1)<0…(less.than=inside.rng)\)
\(0<x<1…x<-1\)

\(n=3:x<x{\frac{1}{n}}…x<x{\frac{1}{3}}…x^3<x…0<x<1…x<-1\)

(2) \(x^{n-1} > x^{2n-2}\) insufic.

\(n=3:x^{n-1} > x^{2n-2}…x^2>x^4…x=|proper.fraction|…0<x<1…-1<x<0\)

(1 & 2) sufic.

(1) x<-1 or 0<x<1
(2) 0<x<1 or -1<x<0
(1&2) combined 0<x<1

Answer (C)
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If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   20 Jun 2020, 12:12
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?


(1) \(x^n < x < x^{\frac{1}{n}}\)

(2) \(x^{n-1} > x^{2n-2}\)
Statement 1: \(x^n\) < x < \(x^\frac{1}{n}\)

This is only valid for 0 < x <1, since if x<0 then \(x^\frac{1}{n}\) is not defined.

Thus \(\frac{1}{x}\)>1

Statement 1 is Sufficient.

Statement 2: \(x^{n-1} > x^{2n-2}\)

\(x^{n-1}(1 - x^{n-1})>0 \)

\(x^{n-1}>0 and (1 - x^{n-1})>0 \) or
\(x^{n-1}<0 and (1 - x^{n-1})<0 \)


-1 < x < 1, \(x\neq{0} \) or
Not Possible, since n is a prime number greater than 2 and thus n is odd, and n-1 is even


Thus \(x^{n-1}\) cannot be negative.


Statement 2 is Not Sufficient

Answer A

@Bunel please check
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   20 Jun 2020, 12:31
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GMATStudyStudent wrote:
emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?


(1) \(x^n < x < x^{\frac{1}{n}}\)

(2) \(x^{n-1} > x^{2n-2}\)
Statement 1: \(x^n\) < x < \(x^\frac{1}{n}\)

This is only valid for 0 < x <1, since if x<0 then \(x^\frac{1}{n}\) is not defined.

Thus \(\frac{1}{x}\)>1

Statement 1 is Sufficient.

Statement 2: \(x^{n-1} > x^{2n-2}\)

\(x^{n-1}(1 - x^{n-1})>0 \)

\(x^{n-1}>0 and (1 - x^{n-1})>0 \) or
\(x^{n-1}<0 and (1 - x^{n-1})<0 \)


-1 < x < 1, \(x\neq{0} \) or
Not Possible, since n is a prime number greater than 2 and thus n is odd, and n-1 is even


Thus \(x^{n-1}\) cannot be negative.


Statement 2 is Not Sufficient

Answer A

@Bunel please check


The red part is not correct. For example, \(\sqrt[3]{-8}=-2\). BTW, this is explained a above.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   09 Feb 2021, 10:14
hi Bunuel, VeritasKarishma mikemcgarry

I generally understand the solutions above, except, I don't understand the following: we use n=3 to work out the individual statements. However, how are we sure that the statements work out the same way for every single odd prime?

Thank you in advance
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   10 Feb 2021, 00:18
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sidship21 wrote:
hi Bunuel, VeritasKarishma mikemcgarry

I generally understand the solutions above, except, I don't understand the following: we use n=3 to work out the individual statements. However, how are we sure that the statements work out the same way for every single odd prime?

Thank you in advance


You are given that statement 1 holds. This means that for whatever values n can take,

x^n < x < x^(1/n)

So this will be true for n = 3, 5, 7, 11 etc all values of n.

You can plug in any one of these values and it will hold for that.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   23 May 2021, 20:43
One off topic question related to this concept....

If x^(n-1) > x^2(n-1), then let's say by plugging n=3, is the below solution acceptable?

x^2 > x^4
x^2 > x^(2+2)
x^2 > x^2 * x^2
1 > x^2

But the final solution doesn't hold good for the equation we started with in the first place. I don't think x can be equal to 0. Can you please help me understand this better?

Bunuel VeritasKarishma mikemcgarry
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If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   Updated on: 28 Nov 2023, 00:04
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amoghhlgr wrote:
One off topic question related to this concept....

If x^(n-1) > x^2(n-1), then let's say by plugging n=3, is the below solution acceptable?

x^2 > x^4
x^2 > x^(2+2)
x^2 > x^2 * x^2
1 > x^2

But the final solution doesn't hold good for the equation we started with in the first place. I don't think x can be equal to 0. Can you please help me understand this better?

Bunuel VeritasKarishma mikemcgarry


This solution is incorrect. You cannot cancel off x^2 until and unless you know that it is not 0.
\(x^2 > x^4\)
\(x^4 - x^2 < 0\)
\(x^2 ( x^2 - 1) < 0\)
\(x^2 (x + 1)(x - 1) < 0\)

Since x^2 cannot be negative, it will not change the sign of the expression so ignore x^2 in transition points (-1 and 1) but it will need to be considered while writing the answer later.

-1 < x < 1 and x cannot be 0.

Check here:
https://youtu.be/PWsUOe77__E

Originally posted by KarishmaB on 24 May 2021, 01:13.
Last edited by KarishmaB on 28 Nov 2023, 00:04, edited 1 time in total.
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   10 Aug 2021, 01:56
value of n can be 3,5,7..
let n = 3
#1
\(x^n < x < x^{\frac{1}{n}}\)
this is possible when
0>x<1 and x<-1
insufficient
#2
\(x^{n-1} > x^{2n-2}\)[/quote]
possible at
fraction +ve and -ve -1<x<0 and 0>x<1
insufficient
from 1 &2
0>x<1
sufficient
option C :)

emmak wrote:
If n is a prime number greater than 2, is 1/x > 1?


(1) \(x^n < x < x^{\frac{1}{n}}\)

(2) \(x^{n-1} > x^{2n-2}\)
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink] New post   11 Nov 2023, 09:50
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Re: If n is a prime number greater than 2, is 1/x > 1? [#permalink]
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