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Suppose x=-1 and k is a positive number. If n is a positive root of t

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4908
GPA: 3.82
Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink]

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09 Jan 2018, 02:54
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91% (01:16) correct 9% (01:15) wrong based on 56 sessions

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[GMAT math practice question]

Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$?

A. $$-2$$
B. $$-1$$
C. $$0$$
D. $$1$$
E. $$2$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Aug 2009
Posts: 5663
Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink]

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09 Jan 2018, 06:11
MathRevolution wrote:
[GMAT math practice question]

Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$?

A. $$-2$$
B. $$-1$$
C. $$0$$
D. $$1$$
E. $$2$$

FIRST step - solve for 'n'...
$$n^4 – 2^{4k} = 0..........(n^2-2^{2k})(n^2+2^{2k})=0...............(n-2^{k})(n+2^k)(n^2+2^{2k})=0$$
so what all can be the value of n... $$2^k$$ or $$-(2^k)$$ or $$\sqrt{-(2^{2k})}$$, which is not possible
positive root of n is $$2^k$$
and since k is positive integer, $$2^k$$ or n will also be EVEN..

$$x+x^n+x^{n+1}+x^{n+2} = (-1)+(-1)^{even}+(-1)^{even+1}+(-1)^{even+2}=(-1)+1+(-1)+1=0$$
C
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Re: Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink]

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09 Jan 2018, 10:21
MathRevolution wrote:
[GMAT math practice question]

Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$?

A. $$-2$$
B. $$-1$$
C. $$0$$
D. $$1$$
E. $$2$$

$$n^4-2^{4k}=0=>n^4=2^{4k}$$. Taking fourth root of both sides we have

$$n=2^k=Even$$ as $$k$$ is a positive integer(Note as $$n$$ is a positive root, so $$n=-2^k$$ is not possible here)

Hence $$x+x^n+x^{n+1}+x^{n+2}=-1+(-1)^{even}+(-1)^{odd}+(-1)^{even}=0$$

Option C
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4908
GPA: 3.82
Re: Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink]

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11 Jan 2018, 00:44
=>

$$n^4 – 2^{4k} = 0$$
$$⇔ (n^2-2^{2k})(n^2+2^{2k}) = 0$$
$$⇔ (n^2-2^{2k}) = 0, since n^2+2^{2k} ≠ 0$$
$$⇔ (n+2^k)(n-2^k) = 0$$
$$⇔ n = 2^k$$
Since$$k > 0, n$$ must be even.

Thus $$n = 2m$$ for some integer $$m$$. It follows that
$$x+x^n+x^{n+1}+x^{n+2}$$
$$= (-1)+(-1)^n+(-1)^{n+1}+(-1)^{n+2}$$
$$= (-1)+(-1)^{2m}+(-1)^{2m} +1+(-1)^{2m+2}$$
$$= (-1) + 1 + (-1) + 1 = 0$$

Therefore, the answer is C.
Answer: C
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Joined: 02 Jan 2016
Posts: 19
Re: Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink]

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25 Jan 2018, 22:41
MathRevolution wrote:
=>

$$n^4 – 2^{4k} = 0$$
$$⇔ (n^2-2^{2k})(n^2+2^{2k}) = 0$$
$$⇔ (n^2-2^{2k}) = 0, since n^2+2^{2k} ≠ 0$$
$$⇔ (n+2^k)(n-2^k) = 0$$
$$⇔ n = 2^k$$
Since$$k > 0, n$$ must be even.

Thus $$n = 2m$$ for some integer $$m$$. It follows that
$$x+x^n+x^{n+1}+x^{n+2}$$
$$= (-1)+(-1)^n+(-1)^{n+1}+(-1)^{n+2}$$
$$= (-1)+(-1)^{2m}+(-1)^{2m} +1+(-1)^{2m+2}$$
$$= (-1) + 1 + (-1) + 1 = 0$$

Therefore, the answer is C.
Answer: C

I do not understand why do we need this complex calculation, when we know it for a fact that any No. multiplied by 2 is always an Even number and as "K" is positive, so 2^4k has to be an even no. and thus n^4 is even too. and that makes the process much easier.
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Joined: 25 Feb 2013
Posts: 948
Location: India
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Re: Suppose x=-1 and k is a positive number. If n is a positive root of t [#permalink]

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25 Jan 2018, 23:34
MathRevolution wrote:
[GMAT math practice question]

Suppose $$x=-1$$ and $$k$$ is a positive number. If $$n$$ is a positive root of the equation $$n^4 – 2^{4k} = 0$$, what is the value of $$x+x^n+x^{n+1}+x^{n+2}$$?

A. $$-2$$
B. $$-1$$
C. $$0$$
D. $$1$$
E. $$2$$

Hi MathRevolution

shouldn't it be better to mentioned "k" as positive integer rather than positive number because positive number can be a decimal value which will create ambiguity.

we all have solved the question assuming "k" to be a positive integer
Re: Suppose x=-1 and k is a positive number. If n is a positive root of t   [#permalink] 25 Jan 2018, 23:34
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