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If p is an integer such that (−4)2p+6= 49−p, then p =
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05 Feb 2015, 09:18
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Re: If p is an integer such that (−4)2p+6= 49−p, then p =
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05 Feb 2015, 09:56
Bunuel wrote: If p is an integer such that \((4)^{2p+6}= 4^{9p}\), 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 =
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05 Feb 2015, 11:52
Bunuel wrote: If p is an integer such that \((4)^{2p+6}= 4^{9p}\), 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 =
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05 Feb 2015, 20:15
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 value2p+6 = 9p 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 =
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09 Feb 2015, 05:39
Bunuel wrote: If p is an integer such that \((4)^{2p+6}= 4^{9p}\), 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 nonzero 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 = 9p; 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 =
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30 Mar 2015, 09:52
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 = 9p => 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 =
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30 Mar 2015, 13:10
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^(9P) 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: GMAT assassins aren't born, they're made, Rich
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Re: If p is an integer such that (−4)2p+6= 49−p, then p =
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31 Mar 2015, 02:27
Bunuel wrote: If p is an integer such that \((4)^{2p+6}= 4^{9p}\), 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^(9p) So, 4^(2p + 6) = 4^(9p) So, 2p + 6 = 9  p So, p = 5 Hence option (E).  Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimusprep.com/gmatondemandcourse



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