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a is a nonzero integer. Is a^a greater than 1?

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a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post Updated on: 14 Aug 2014, 11:35
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a is a nonzero integer. Is a^a greater than 1?

(1) a < -1
(2) a is even

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Originally posted by NoHalfMeasures on 14 Aug 2014, 11:17.
Last edited by Bunuel on 14 Aug 2014, 11:35, edited 1 time in total.
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Re: a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 14 Aug 2014, 11:45
a is a nonzero integer. Is a^a greater than 1?

(1) a < -1. So, a could be -2, -3, -4, ... If a is a negative even integer, then a^a will be a positive fraction less than 1 (for example, if a = -2, then a^a = (-2)^(-2) = 1/4) and if a is a negative odd integer, then a^a will be a negative fraction greater than -1 (for example, if a = -3, then a^a = (-3)^(-3) = -1/27). In any case the result is less than 1. Sufficient.

(2) a is even. If a = 2, then a^a = 4 > 1 but if a = -2, then a^a = 1/4 < 1. Not sufficient.

Answer: A.

Hope it's clear.
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a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 11 Feb 2015, 17:24
Bunuel wrote:
a is a nonzero integer. Is a^a greater than 1?

(1) a < -1. So, a could be -2, -3, -4, ... If a is a negative even integer, then a^a will be a positive fraction less than 1 (for example, if a = -2, then a^a = (-2)^(-2) = 1/4) and if a is a negative odd integer, then a^a will be a negative fraction greater than -1 (for example, if a = -3, then a^a = (-3)^(-3) = -1/27). In any case the result is less than 1. Sufficient.

(2) a is even. If a = 2, then a^a = 4 > 1 but if a = -2, then a^a = 1/4 < 1. Not sufficient.

Answer: A.

Hope it's clear.



Dear Bunuel, I think that (-2)^(-2) which is -1/2^2 should be negative fraction less than 1 and equal to = (- 1/4 )
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Re: a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 11 Feb 2015, 22:57
Hi 23a2012,

Unfortunately, that's NOT how the math "works"

When dealing with a negative exponent, you have to put the entire calculation "under" the 1.

With your example, we have (-2)^(-2). This can be rewritten as....

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

IF....we were dealing with (-3)^(-3) though, we'd have.....

1/[(-3)^3] = 1/-27 = -1/27

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a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 12 Feb 2015, 06:15
EMPOWERgmatRichC wrote:
Hi 23a2012,

Unfortunately, that's NOT how the math "works"

When dealing with a negative exponent, you have to put the entire calculation "under" the 1.

With your example, we have (-2)^(-2). This can be rewritten as....

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

IF....we were dealing with (-3)^(-3) though, we'd have.....

1/[(-3)^3] = 1/-27 = -1/27

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Ok, can you tell me how write -1/4 in the form a^a
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Re: a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 12 Feb 2015, 06:27
23a2012 wrote:
Bunuel wrote:
a is a nonzero integer. Is a^a greater than 1?

(1) a < -1. So, a could be -2, -3, -4, ... If a is a negative even integer, then a^a will be a positive fraction less than 1 (for example, if a = -2, then a^a = (-2)^(-2) = 1/4) and if a is a negative odd integer, then a^a will be a negative fraction greater than -1 (for example, if a = -3, then a^a = (-3)^(-3) = -1/27). In any case the result is less than 1. Sufficient.

(2) a is even. If a = 2, then a^a = 4 > 1 but if a = -2, then a^a = 1/4 < 1. Not sufficient.

Answer: A.

Hope it's clear.



Dear Bunuel, I think that (-2)^(-2) which is -1/2^2 should be negative fraction less than 1 and equal to = (- 1/4 )


hi ,
-2^-2= 1/(-2)^2=1/4....
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Re: a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 12 Feb 2015, 12:20
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Hi 23a2012,

If you have an EVEN exponent, then you CANNOT have a negative outcome (unless the negative "sign" is "outside" of the exponent, and thus unaffected by the exponent).

eg
(2)^2 = 4
(2)^(-2) = 1/4

(-2)^(2) = 4
(-2)^-(2) = 1/4

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

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Re: a is a nonzero integer. Is a^a greater than 1? [#permalink]

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New post 18 Jan 2018, 07:28
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NoHalfMeasures wrote:
a is a nonzero integer. Is a^a greater than 1?

(1) a < -1
(2) a is even


Target question: Is a^a greater than 1?

Given: a is a nonzero integer.

Statement 1: a < -1
Let's start TESTING some values of a and see if we discover a PATTERN

a = -2. Here a^a = (-2)^(-2) = 1/(-2)^2 = 1/4
a = -3. Here a^a = (-3)^(-3) = 1/(-3)^3 = 1/(-27) = - 1/27
a = -4. Here a^a = (-4)^(-4) = 1/(-4)^4 = 1/256
a = -5. Here a^a = (-5)^(-5) = 1/(-5)^5 = 1/(some negative value) = something negative
a = -6. Here a^a = (-6)^(-6) = 1/(-6)^6 = 1/(some positive integer) = a positive fraction that's less than 1
a = -7. Here a^a = (-7)^(-7) = 1/(-7)^7 = 1/(some negative value) = something negative
.
.
.
As we can see, when a is a NEGATIVE EVEN number, a^a = some positive fraction that's LESS THAN 1
When a is a NEGATIVE ODD number, a^a = some negative value that's LESS THAN 1
So, in both possible cases, a^a is definitely LESS THAN 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: a is even
We already saw in statement 1 that if a is a NEGATIVE EVEN integer, then a^a = some positive fraction that's LESS THAN 1
What if a is a POSITIVE EVEN integer?
Let's test some possible values:
a = 2. Here a^a = (2)^(2) = 4. Here, a^a is GREATER THAN 1
So, we'll stop right here.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Re: a is a nonzero integer. Is a^a greater than 1?   [#permalink] 18 Jan 2018, 07:28
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