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# If x is a positive even integer, and n and m are consecutive integers,

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Math Expert
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If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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03 Nov 2015, 13:35
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85% (01:02) correct 15% (01:08) wrong based on 122 sessions

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If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

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

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If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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03 Nov 2015, 19:57
x= +ve even integer
Since,n and m are consecutive integers , their difference will be 1

((n-m)^x)/((m-n)^x)= ((n-m)/(m-n))^x = (-1)^x
Since we are raising the difference of n and m to power x , which is even , the answer will be 1 .

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Re: If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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04 Nov 2015, 03:49
Bunuel wrote:
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

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

Let,
x = 2 (Positive EVEN Integer)
m = 3 and n = 2

Now,
(n - m)^x/(m - n)^x = (2-3)^2 / (3-2)^2 = 1

Key: When the question gives open ended information like Positive Even Integer etc. then it's best to assume values and solve the expression. The answer will remain constant for any value assumed as per information unless there is an option like 'Can NOT be determined'
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If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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05 Nov 2015, 15:05
Since n and m are consecutive integers, the difference will always be either 1 or -1.
E.g.: n=5, m=6

$$\frac{(5-6)^x}{(6-5)^x}$$= $$\frac{(-1)^x}{(1)^x}$$

And since we have the even exponent the resolved exponent will always be 1. Thus result 1.
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Re: If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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06 Nov 2015, 13:07
Bunuel wrote:
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

Lets plug in some values -

x = 2
m = 4 ; n = 5

$$\frac{(n - m)^x}{(m - n)^x}$$ = $$\frac{(5 - 4)^2}{(4 - 5)^2}$$ =$$\frac{1^2}{-1^2}$$ = 1

Hence among the given options IMHO answer is D. 1
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Re: If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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06 Nov 2015, 20:08
X is even
N and M are consecutive, thus their difference will always be -1 or 1.
Anything raised to even power is positive and 1 to power anything is 1
Thus OME : D
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Re: If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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18 Jan 2018, 14:28
Bunuel wrote:
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

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

Notice that (n - m)^x / (m - n)^x = [(n - m)/(m - n)]^x = (-1)^x

Since x is a positive even integer, (-1)^x will always be equal to 1.

Alternate solution:

Let’s use some strategic numbers for n, m, and x.

n = 4, m = 3, and x = 2, thus:

[(4 - 3)^2]/(3 - 4)^2 = 1^2/(-1)^2 = 1

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Re: If x is a positive even integer, and n and m are consecutive integers,  [#permalink]

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19 Mar 2018, 09:18
Bunuel wrote:
If x is a positive even integer, and n and m are consecutive integers, then (n - m)^x/(m - n)^x =

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

If power is even, then base can never be negative. Therefore, A & B are out.

Try n = 1, m = 2, x = 2

$$\frac{(1 - 2)^2}{(2 - 1)^2}$$

$$\frac{(-1)^2}{(1)^2}$$

1

(D)
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Re: If x is a positive even integer, and n and m are consecutive integers,   [#permalink] 19 Mar 2018, 09:18
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