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If p is an integer such that (−4)2p+6= 49−p, then p =

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If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 05 Feb 2015, 08:18
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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 05 Feb 2015, 08:56
Bunuel wrote:
If p is an integer such that \((-4)^{2p+6}= 4^{9-p}\), then p =

A. -1
B. 0
C. 1
D. 3
E. 5

Kudos for a correct solution.


Plugging in answer C. \((-4)^8=4^8\) this is always true.

Answer C.
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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 05 Feb 2015, 10:52
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Bunuel wrote:
If p is an integer such that \((-4)^{2p+6}= 4^{9-p}\), then p =

A. -1
B. 0
C. 1
D. 3
E. 5

Kudos for a correct solution.



This can be solved easily through a very common funda " when bases are equal powers should be equal"

But there is only one catch we have a "-4" instead of 4 ( makes little sense to ask in GMAT)

Look at the powers 2p+6 => this means the power is always EVEN.

So -4^2p+6 is identical to 4^2p+6.

From here on it is simple math.

P=1 , C is the answer.
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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 05 Feb 2015, 19:15
Answer = C = 1

When bases are same, equate the powers. However, in this case, sign of bases are different. It means the exponent should yield an even value

2p+6 = 9-p

p = 1

By placing p=1, we get \(4^8 = (-4)^8\) which hold true
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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 09 Feb 2015, 04:39
Bunuel wrote:
If p is an integer such that \((-4)^{2p+6}= 4^{9-p}\), then p =

A. -1
B. 0
C. 1
D. 3
E. 5

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: C

Explanation: Any non-zero number raised to an even power (2, 4, 6, etc.) will be a positive number. Because we know that p is an integer, it follows that 2p is even, and that 2p+6 is also even. Therefore we know that (−4)^(2p+6) is the same as 4^(2p+6).

If the bases on both sides of an equation are the same, then the exponents must be equal as well, so: 2p+6 = 9-p;

Adding p and subtracting 6 from both sides gives 3p = 3 => p = 1

The correct answer is C.
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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 30 Mar 2015, 08:52
Another way to look at it would be to separate (-4)^2p+6 into [(-1)^2p+6]*[(4)^2p+6].
We can now safely say that 2p+6 is even since there is no negative sign on the right hand side and equate both the powers of 4.
So, 2p+6 = 9-p => p=1.
A way to validate your answer would be to plug in value of p in (-1)^2p+6 = (-1)^8= 1
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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 30 Mar 2015, 12:10
Hi All,

This question is perfect for a bit of "brute force" and TESTing THE ANSWERS. We just have to plug in the 5 choices until we find the one number that makes the two sides of the equation equal...

(-4)^(2P+6) vs. 4^(9-P)

Answer A: -1
Does (-4)^4 = 4^10?
No. This is NOT the answer

Answer B: 0
Does (-4)^6 = 4^9
No. This is NOT the answer

Answer C: 1
Does (-4)^8 = 4^8
YES. This IS the answer

Final Answer:
[Reveal] Spoiler:
C


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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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New post 31 Mar 2015, 01:27
Bunuel wrote:
If p is an integer such that \((-4)^{2p+6}= 4^{9-p}\), then p =

A. -1
B. 0
C. 1
D. 3
E. 5

Kudos for a correct solution.


2p + 6 is always even.
Hence (-4)^(2p + 6) = 4^(2p + 6)
Now, (-4)^(2p+6) = 4^(9-p)
So, 4^(2p + 6) = 4^(9-p)
So, 2p + 6 = 9 - p
So, p = 5
Hence option (E).

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Re: If p is an integer such that (−4)2p+6= 49−p, then p = [#permalink]

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Re: If p is an integer such that (−4)2p+6= 49−p, then p =   [#permalink] 22 May 2017, 18:54
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