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Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink]
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19 Nov 2015, 17:43
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QUANT 4PACK SERIES Problem Solving Pack 4 Question 3 If a, b and c are consecutive...If a, b and c are consecutive positive integers and a < b < c, then what is the minimum possible value of \(\frac{3^{2bc}}{3^{2ab}}\) A) 81 B) 2,187 C) 3,300 D) 6,561 E) 19,683 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation
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Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink]
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19 Nov 2015, 18:28
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If a, b and c are consecutive positive integers and a < b < c, then what is the minimum possible value of
\(3^{2bc}\) / \(3^{2ab}\)
A) 81 B) 2,187 C) 3,300 D) 6,561 E) 19,683
assume a=1, b=2, c=3 for minimum possible value
the equation then becomes \(3^{12}\) / \(3^{4}\) =\(3^{8}\) calculating only for the units digit to save time, we can then see that 1 is the last units digit of \(3^{8}\)
Answer: D



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Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink]
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23 Nov 2015, 09:26
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EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 4 Question 3 If a, b and c are consecutive...If a, b and c are consecutive positive integers and a < b < c, then what is the minimum possible value of \(3^{2bc}\) / \(3^{2ab}\) A) 81 B) 2,187 C) 3,300 D) 6,561 E) 19,683 48 Hour Window Answer & Explanation WindowEarn KUDOS! Post your answer and explanation. OA, and explanation will be posted after the 48 hour window closes. This question is part of the Quant 4Pack seriesScroll Down For Official Explanation \(\frac {3^{2bc}}{3^{2ab}} = ?\) > \(\frac {3^{2bc}}{3^{2ab}} = 3^{2bc2ab} = 3^{2b(ca)}\)and as a,b,c are consecutive integers, ca will always be = 2. Thus, \(3^{2b(ca)} = 3^{4b}\) and as all 3 a,b,c are POSITIVE integers, minimum value of a=1, making b =2 > \(3^{4b} = 3^8 = 65\)61. D is the correct answer.



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Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink]
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24 Nov 2015, 01:07
EMPOWERgmatRichC wrote: QUANT 4PACK SERIES Problem Solving Pack 4 Question 3 If a, b and c are consecutive...
If a, b and c are consecutive positive integers and a < b < c, then what is the minimum possible value of
\(3^{2bc}\) / \(3^{2ab}\)
A) 81 B) 2,187 C) 3,300 D) 6,561 E) 19,683
Hi All, To start, the answer choices to this question are rather 'spread out', so we might be able to use that spread to our advantage (and avoid some calculations). We're told that a, b and c are consecutive positive integers and a < b < c. We're asked for the MINIMUM value of \(3^{2bc}\) / \(3^{2ab}\)... Since the prompt asks for the MINIMUM value, it's likely that we'll have to make the 3 variables as small as possible, but we'll have to check to make sure that that's what is required to get to the correct answer. The smallest 3 numbers that 'fit' are... a = 1 b = 2 c = 3 This makes the calculation... \(3^{12}\) / \(3^{4}\) = \(3^{8}\) \(3^{8}\)= \(9^{4}\) = \(81^{2}\) = ABOUT \(80^{2}\) = ABOUT 6400, so the answer appears to be D. We just need to confirm that that's the case.... IF.... a = 2 b = 3 c = 4 This makes the calculation... \(3^{24}\) / \(3^{12}\) = \(3^{12}\) \(3^{12}\) is clearly bigger than \(3^{8}\), meaning that increasing the values of a, b and c will lead to a LARGER end result. This proves that D is actually the smallest possible result. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink]
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28 Nov 2015, 10:57
since a, b, and c are consecutive integers, and positive: we can rewrite the fraction as
3^[2bc2ab] or 3^[2(bcab)] or 3^[2(b(ca))]
since a,b,c are consecutive integers, ca is always = 2 regardless of the values they take.
now, we are left with 3^4b to minimize the value, we need to take the minimal value of b, while taking into consideration that a,b,c  must be positive and numbers are consecutive numbers. minimal value of b is 2.
now, 3^8 = 9^4 or 81^2. don't need to solve for the exact value. the correct answer choice must have 1 at the end, and since A cannot be the answer, the only correct answer that it can be is D.



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Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink]
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