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Manager  Joined: 14 Oct 2014
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GMAT 1: 500 Q36 V23 x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 - 2
(2) 2x < x^5
Math Expert V
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x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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x^n = x^(n+2) for any integer n. Is it true that x > 0?

For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1.

(1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient.

(2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient.

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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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viktorija wrote:
x^n=x^(n+2) for any integer n. Is it true that x>0?

(1) x=x^2-2
(2)2x<x^5

$$x^n = x^{n+2}$$

st.1
x=x^2-2

now x^2-2-x=0
x^2-2x+x-2=0
x(x-2)+1(x-2)=0
(x-2)(x+1)=0
x=2 or x=-1

out of these values of x, only x=-1 will satisfy the equation $$x^n = x^{n+2}$$. hence we will reject x=2. thus only possible value of x=-1. which answer the original question as false. Thus st.1 alone is sufficient.

st.2

x^5>2x
x(x^4-2)>0

this equation is satisfied by x=-1, x=2,3,4...

now out of these values of x, only x=-1, satisfy the equation $$x^n = x^{n+2}$$. hence x=-1 . which answer the original question as false. thus st.2 alone is sufficient.

hence answer must be D
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Hi viktorija,

manpreetsingh86's explanation is spot on, so I won't rehash that here.

The information given in the prompt involves a really specific Number Property. Knowing this Number Property would make solving the rest of the question considerably easier:

We're told that x^n=x^(n+2) for ANY integer n. This equation offers a SERIOUS restriction; for it to hold true for ANY integer value of n, the value of X can only be one of 3 things: -1, 0 or 1.

Taking THAT knowledge into Fact 1 and Fact 2 would allow you to work through them quickly and avoid any excess math.

GMAT assassins aren't born, they're made,
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GMAT 1: 630 Q42 V35 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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2
Here x can only be 0 OR 1 OR -1(if n=even)
Statement 1 tells that x=2 or x=-1.
So x=-1. Sufficient.

Statement 2 can holds only when x=-1 and not when x=0 OR x=1.
So x=-1. Sufficient.

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x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 -2
(2) 2x < x^5

When you modify the original condition and the question, it becomes x=-1,0,1. There is 1 variable(x), which should match with the number equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely make D the answer.
For 1), from x^2-x-2=0, (x-2)(x+1)=0, x=2,-1, only x=-1 is possible, which is no and sufficien.t
For 2), what satisfies 2x<x^5 is only x=-1, which is no and sufficient. Therefore, the answer is D.

 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Bunuel wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1.

(1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient.

(2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient.

Hi Bunuel

I have one question for x must be 0,1,-1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat?

I came across similiar question

https://gmatclub.com/forum/what-is-the- ... 51809.html

where 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or -1

Thanks
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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hellosanthosh2k2 wrote:
Bunuel wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1.

(1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient.

(2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient.

Hi Bunuel

I have one question for x must be 0,1,-1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat?

I came across similiar question

https://gmatclub.com/forum/what-is-the- ... 51809.html

where 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or -1

Thanks

0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT.

For this problem, x = n = 0 or x = n + 2 = 0 is technically out because we are given that x^n = x^(n+2) for ANY integer n because for these cases x^n or x^(n + 2) will become undefined and could not be equal to other part.

Apart from these exception x = 0 is perfectly OK.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Bunuel wrote:
hellosanthosh2k2 wrote:
Bunuel wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1.

(1) x = x^2 - 2 --> x^2 - x - 2 = 0 --> (x - 2)(x + 1) = 0 --> x = 2 or x = -1. Since we know that x can only be 0, 1, or -1, then x = -1 < 0. Sufficient.

(2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient.

Hi Bunuel

I have one question for x must be 0,1,-1 for all integers n , if x = 0, and if n= 0, then 0 ^ 0 = 0 ^ 2 , value of 0 ^ 0 = 0 or 0 ^ 0 = 1 or it is out of scope of gmat?

I came across similiar question

https://gmatclub.com/forum/what-is-the- ... 51809.html

where 0 ^ 0 is out of scope, if this were true, x can't take value of 0 right? only 1 or -1

Thanks

0^0, in some sources equals to 1 (not 0), some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT.

For this problem, x = n = 0 or x = n + 2 = 0 is technically out because we are given that x^n = x^(n+2) for ANY integer n because for these cases x^n or x^(n + 2) will become undefined and could not be equal to other part.

Apart from these exception x = 0 is perfectly OK.

Thanks Bunuel for explanation
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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viktorija wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 - 2
(2) 2x < x^5

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

$$x^n = x^{n+2}$$
$$x^{n+2} - x^n = 0$$
$$x^n(x^2-1) = 0$$
$$x^n(x+1)(x-1) = 0$$
$$x -1$$ or $$x = 0$$ or $$x = 1$$
The original condition states $$x -1$$ or $$x = 0$$ or $$x = 1$$.

Condition 1)

$$x = x^2 - 2$$
$$x^2 - x - 2 = 0$$
$$(x+1)(x-2) = 0$$
$$x = -1$$ or $$x = 2$$

From the original condition, we have x = -1.
Then answer is "No"
This is sufficient by CMT(Common Mistake Type) 1, which states "No is also an answer", t

Condition 2)

$$2x < x^5$$
$$x^5 - 2x > 0$$
$$x(x^4 - 2 ) > 0$$
Only $$x = -1$$ from the original condition satisfies this condition.
Then answer is "No"
This is sufficient by CMT(Common Mistake Type) 1, which states "No is also an answer", t

The answer is D.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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viktorija wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

(1) x = x^2 - 2
(2) 2x < x^5

hard problem
from x^n<x^(n+2)
we see that x =0 or 1 or -1

put these number into condition 1 and 2. we see that for both, only x=-1 is good.
so, each condition is ok

D.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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This came up in a GMAT Focus test for me and I selected (B).

Plugging-in numbers and being cognizant of limitations given would have helped me solve this. I think other people fall into this trap as well judging by the difficulty.
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GMAT 1: 560 Q44 V23 Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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I do not know why no body has solved in this way-

$$x^n$$=$$x^{(n+2)}$$
$$x^n$$=$$x^n$$*$$x^2$$

so
$$x^2$$=1
x=-1 or +1

Statement 1: sufficient
put -1 and +1 in the expression
only -1 satisfy the equation
so x=-1

Statement 1: sufficient
put -1 and +1 in the expression
only -1 satisfy the equation
so x=-1

so both the statements gives the solution hence answer is D
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Bunuel wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1.

I know this is a silly doubt, but this is how I solved it.

x^n = x^(n+2)
So, x^n / x^(n+2) =1
x^(n - n - 2) = 1
x^-2 = 1
1/x^2 = 1
x^2 = 1
So, x = 1 or x = -1.

With this approach, I am not getting x = 0.
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Hi Bunuel,

I have gone through your solution but need couple of clarifications in my approach.

I have always committed the same mistake in previous questions so wanted to set the record straight.

So here is what i did
We are given that :
$$x^{n}=x^{n+2}$$
dividing the above equation by$$x^{n}$$ we have
$$x^{2}$$ =1

so x can be +1 or -1

Logically i know that even 0 satisfies the above equation , how and why did i not get algebraically get x= 0 as one of its values .

Second
we have
x= $$x^{2}$$-2
$$x^{2}$$-x= 2
x(x-1)=2
Is this step correct that we say
either x=2 or
(x-1)=2 => x= 3. I know x=3 does not satisfy the equation, but then what is the error i am committing.

Logically i know that either x=2 or x=-1. and since from the stem we have $$x^{2}$$=1 we can say therefor that x=-1

Now in second statement:
we have 2x<$$x^{5}$$
Case (1)
Lets say x>0
then 2<$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is incorrect.
Case (2)
Lets say x<0
then 2>$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is correct.

Need help in above errors and method .

Thanks
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Re: x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Sir Bunuel will even give you more stuff but here is my take
As i said earlier you can’t divide by X^n since x can be Zero. Also you missed the third value
I suggest it should be this way
Simplify the question first
x^n=x^(n+2) => x^n=x^n•x^2 => x^n-x^n•x^2=0 => x^n(1-x^2)=0 , x^n =0 , x=0 ,x^2=1 => x=1 or -1
So you see now the we got the third value
Probus wrote:
Hi Bunuel,

I have gone through your solution but need couple of clarifications in my approach.

I have always committed the same mistake in previous questions so wanted to set the record straight.

So here is what i did
We are given that :
$$x^{n}=x^{n+2}$$
dividing the above equation by$$x^{n}$$ we have
$$x^{2}$$ =1

so x can be +1 or -1

Logically i know that even 0 satisfies the above equation , how and why did i not get algebraically get x= 0 as one of its values .

Second
we have
x= $$x^{2}$$-2
$$x^{2}$$-x= 2
x(x-1)=2
Is this step correct that we say
either x=2 or
(x-1)=2 => x= 3. I know x=3 does not satisfy the equation, but then what is the error i am committing.

Logically i know that either x=2 or x=-1. and since from the stem we have $$x^{2}$$=1 we can say therefor that x=-1

Now in second statement:
we have 2x<$$x^{5}$$
Case (1)
Lets say x>0
then 2<$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is incorrect.
Case (2)
Lets say x<0
then 2>$$x^{4}$$ but since we have $$x^{2}$$=1 then $$x^{4}$$=1 and hence this is correct.

Need help in above errors and method .

Thanks

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x^n = x^(n+2) for any integer n. Is it true that x > 0?  [#permalink]

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Bunuel wrote:
x^n = x^(n+2) for any integer n. Is it true that x > 0?

For x^n = x^(n+2) for ANY integer n, x must be 0, 1, or -1.

(2) 2x < x^5 --> from the possible values of x (0, 1, -1), only -1 satisfies 2x < x^5, thus x = -1 < 0. Sufficient.

Hi Bunuel

For Statement 2 --

1) I tried to solve this algebraically. Just wondering, where is the problem in my procedure ?

We are given in S2 2x < x^5

So moving 2x to the RHS of the equation

-- x^5 - 2x > 0

-- x (x^4 - 2) >0

-- hence there are two possibilities for x with this equation

possibility 1 : x >0
or
possibility 2 : x^4 - 2 >0 or x^4 > 2 or x can be positive or negative

Given we have possibility 1 (Where x has to be positive) or possibility 2(x can be positive or negative) -- hence i marked this as insufficient

I clearly am going wrong somewhere -- just wondering, where is the error in this strategy ?

Thank you !! x^n = x^(n+2) for any integer n. Is it true that x > 0?   [#permalink] 15 Aug 2019, 07:50
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