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Each has one solution except?

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Each has one solution except? [#permalink] New post 01 Oct 2009, 17:45
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Hi folks,
I came across this problem in he MGMAT practice..Can anyone explain the solution more clearly?


Each of the following equations has at least one solution EXCEPT
1. –2n = (–2)-n
2. 2-n = (–2)n
3. 2n = (–2)-n
4. (–2)n = –2n
5. (–2)-n = –2-n

The answer is(1)....
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Re: Each has one solution except? [#permalink] New post 01 Oct 2009, 20:12
I’m assuming that they are all multiples?? That -2-n means (-2)*(-n) and not (-2) subtract (n)?

Let n =1
1) -2(1) = -2(-1) Left = -2 Right = +2. No solution
2) 2(-1) = (-2)(1) Both = -2
3) 2(1) = (-2)(-1) both = +2
4) (-2)1 = -2(1) both = -2
5) (-2)(-1) = (-2)(-1) both = +2
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Re: Each has one solution except? [#permalink] New post 01 Oct 2009, 22:12
vigneshpandi - it is indeed not clear whether the signes should be treated as is or you imply multiplication here. Pls, try to use the math symbols to avoid misunderstanding.

yangsta - in option 1 - what about n=0 - then it does work. Or is it that I cannot have 0 with a minus sign?
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Re: Each has one solution except? [#permalink] New post 01 Oct 2009, 22:46
arkadiyua wrote:
yangsta - in option 1 - what about n=0 - then it does work. Or is it that I cannot have 0 with a minus sign?


well 0 cannot be negative. From what I've seen on most GMAT question stems they will state if a variable can be zero.
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Re: Each has one solution except? [#permalink] New post 02 Oct 2009, 20:25
I am sorry....the question has all the terms 'n' on the powers....the actual question is:

Each of the following equations has at least one solution EXCEPT
1. –2^n = (–2)^-n
2. 2^-n = (–2)^n
3. 2^n = (–2)^-n
4. (–2)^n = –2^n
5. (–2)^-n = –2^-n

The answer is(1)....
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Re: Each has one solution except? [#permalink] New post 02 Oct 2009, 22:00
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vigneshpandi wrote:
I am sorry....the question has all the terms 'n' on the powers....the actual question is:

Each of the following equations has at least one solution EXCEPT
1. –2^n = (–2)^-n
2. 2^-n = (–2)^n
3. 2^n = (–2)^-n
4. (–2)^n = –2^n
5. (–2)^-n = –2^-n

The answer is(1)....
While it is possible to reason out which of these choices must not work, we may not have time or the confidence to do so. However, this problem has variable in its answer choice, and relatively simple math. Therefore, an easy alternative is picking numbers.

Since we're dealing with exponents, we want to keep things as easy as possible. Hence, we'll start with the easiest exponent possible: n = 1. A, B, and C are not solved (x^-n = 1/(x^n), so we're comparing integers to fractions), but choices D and E both end up valid, eliminating them from contention.

In the process of doing this, however, we've uncovered a major clue to our next step: A, B, and C, all compared integers to fractions, and the only integer equal to it's reciprocal is 1, which is equal to 1/1. This, in turn, tells us the we need to pick n = 0. Remember, for all non-zero x, x^0 = 1.

If we plug n = 0 into choices B and C, we end up with 1 = 1 both times. Choice A, however, results in the false 1 = -1. Thus, we conclude that the first choice has no valid solutions, and is therefore the correct answer.
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Re: Each has one solution except? [#permalink] New post 03 Oct 2009, 09:29
Thank you for the explanation....
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Re: Each has one solution except? [#permalink] New post 03 May 2011, 07:43
1 has no solution.
others have 0,1 and 2.
hence A
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Re: Each has one solution except? [#permalink] New post 25 Jun 2011, 04:37
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I have a quick answer, using elimination.
If n1 is the solution of the second equation, then (-n1) must be the solution of the third question. Therefore 2nd and 3rd question must have a solution or must have no solution at the same time.
The same situation with the 4th and 5th equation.
Then we left the 1st equation which is the correct answer :)
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Re: Each has one solution except? [#permalink] New post 25 Jun 2011, 18:42
1 has no solution , others has at least one solution.
Re: Each has one solution except?   [#permalink] 25 Jun 2011, 18:42
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