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If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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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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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]

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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]

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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)$$x^{n-1} > x^{2n-2}$$

Couldn't the answer be [A],

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

In such a case, 1/x>1 or x belongs to {0,1} -ve values again will not satisfy the equation.

Statement 2: x^(n-1) > x^2(n-1), for n being odd, n-1 will be even. Hence for the above equation the solution set will be {-1,1}
As mentioned above, the negative values of x will not hold the 1/x>1 equation. Hence the range is again {0,1}.

Am I doing sumthing wrong?

Regards,
Arpan
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a +ve proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X is a +ve proper fraction ?????

So, We have to find if X is a +ve Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a +ve proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a +ve proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

It can be written as ... .... $$x^{n+1} > x^{2n}$$

Therefore.. InSufficient ....

Hence, ...... A.......
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Originally posted by manishuol on 08 May 2013, 23:56.
Last edited by manishuol on 09 May 2013, 01:42, edited 3 times in total.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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1
manishuol wrote:
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X a proper fraction ?????

So, We have to find if X is a Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...$$x^2$$ > $$x^4$$

& when n = 5 .. we will get $$x^4$$ > $$x^8$$ ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation:
1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards,
Arpan
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Originally posted by arpanpatnaik on 09 May 2013, 00:14.
Last edited by arpanpatnaik on 09 May 2013, 00:35, edited 1 time in total.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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arpanpatnaik 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)$$x^{n-1} > x^{2n-2}$$

Couldn't the answer be [A],

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As x>2 and is a prime number, x will always be an odd integer. For x = -8,$$x^{1/3}$$ = -2.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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arpanpatnaik wrote:
manishuol wrote:
emmak wrote:
If n is a prime number greater than 2, is $$\frac{1}{x}$$ > 1?
(1) $$x^n < x < x^{\frac{1}{n}}$$
(2)$$x^{n-1} > x^{2n-2}$$

The Question asks ......Is $$\frac{1}{x}>1$$ ...... & this is only possible only if x is a proper fraction .. where

Denominator is greater than the numerator. So in other words.. Question Asks ... Is X a proper fraction ?????

So, We have to find if X is a Proper fraction or not ......

Given, N is a prime Number greater than 2.....

Statement :: 1 Says ..... $$x^n < x < x^{\frac{1}{n}}$$ ....

This is only possible if x is a proper fraction .. if you want to check .. plugin the value of n as 3 or 5 .... Therefore, from this ....

we can say that x is a proper fraction.. Therefore. Sufficient ..

Statement :: 2 says ... $$x^{n-1} > x^{2n-2}$$ ...

$$x^{n-1} > x^{2n-2}$$

In this one also if we plugin in the value of n as 3 or 5 ... we will get when n = 3 ...$$x^2$$ > $$x^4$$

& when n = 5 .. we will get $$x^4$$ > $$x^8$$ ....... Both of these states that x is a proper fraction .....

Therefore.. Sufficient ....

Hence, ...... D.......

I agree with Statement 1 being sufficient, but when it comes to Statement 2,

For example, x = -0.1, in such a case, assuming n is Odd and equal to 3, the inequality can be reduced to x^2>x^4. The value x = -0.1 satisfies.

Since the power term (n-1) will always be even, the inequality will hold good for even -ve numbers between {-1,0}. But when it comes to our parent equation:
1/-0.1 = -10 < 1. Hence the inequality doesn't satisfy. Therefore, I feel [A] should be the answer. Please look into my method and let me know if I am doing anything wrong! Also please refer to my post above for a general approach to the prob.

Regards,
Arpan

My answer is still A ...
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Originally posted by manishuol on 09 May 2013, 00:59.
Last edited by manishuol on 09 May 2013, 01:34, edited 2 times in total.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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vinaymimani wrote:
arpanpatnaik 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)$$x^{n-1} > x^{2n-2}$$

Couldn't the answer be [A],

Statement 1: for all n where n is prime and > 2, x^n < x < x^1/n.
Since n>2, n must be odd. Hence the above values of x belongs to {0,1} will hold for the above equation. For -ve values of x, x^1/n will not be defined.

Not true. As n>2 and is a prime number, will n always be an odd integer. For x = -8,$$x^{1/3}$$ = -2.

My Bad!! I missed that... Thanks for correcting me Vinay! I got the answer now! [C] it is Can't believe I used the odd principle in Statement 2 and failed to use it in Statement 1. The range x < -1 will be valid and the solution set needs to be an intersection of Statement 1 and 2 i.e. x belongs to {0,1}.

Regards,
Arpan
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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the question asks if 1/x > 1
or, 1/x - 1 > 0
or, (1-x) / x > 0
again for the above condition to be true;
(1-x) > 0 and x > 0 or (1-x) < 0 and x<0 (for x<0, 1-x can never be less than 0, hence ignore this)
basically the question asks if 0 < x < 1 <----------- this is what we need to find, does x lie between 0 and 1??

statement 1 is clearly insufficient

i've got a doubt here with statement 2.

x^(n-1) > x^(2n-2)
=> (x^(n-1)) - (x^(2n-2)) > 0
=> (x^n/x) - (x^2n/x^2) > 0
=> x^n/x (1-x^n/x) > 0 taking x^n/x common outside the bracket
=> for the above condition to be true
either or
x^n/x > 0 and (1- x^n/x) > 0 x^n/x < 0 and (1 - x^n/x) < 0
when x^n/x > 0, when x^n/x < 0, (1 - x^n/x) can never be less than 0
it means for (1 - x^n/x) > 0 hence, we can ignore this option
x^n/x has to be a fraction between 0 and 1
i.e. 0< x^n/x < 1
or, 0 < x^(n-1) < 1

when x^(n-1) is between 0 and 1, and n is a prime number greater than 2,
it implies that x is a fraction between 0 and 1 and x^(n-1) will always be positive and true for any value between 0 and 1.
in that case Statement 2 is satisfying what we are asked to find out. statement 2 should be sufficient.
correct me if i am wrong.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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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.

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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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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]

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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, $$\sqrt{-8}=-2$$.

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]

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Bunuel wrote:
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, $$\sqrt{-8}=-2$$.

The question asks whether 0 < x < 1. For (1) check x = -8 and n = 3 prime.

Is this case GMAT specific?

Please have a look at this:

http://www.wolframalpha.com/input/?i=%2 ... 281%2F3%29
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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thefibonacci wrote:
Is this case GMAT specific?

Please have a look at this:

http://www.wolframalpha.com/input/?i=%2 ... 281%2F3%29

No, that's generally true.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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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

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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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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

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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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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Bunuel, thanks - I overlooked the fact that the exponent will always be even given that n is prime greater than 2.
Man, this a tricky one.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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both statements lead to either x is negative or x belongs to (0;1)
only C can settle the problem => x is not negative.
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Re: If n is a prime number greater than 2, is 1/x > 1?  [#permalink]

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Vetrik 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 -> 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.

I kinda followed the same approach. Just I actually took the value of n as 3 and tried to work calculations
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