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# Problem Solving Pack 4, Question 3) If a, b and c are consecutive...

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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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Difficulty:

35% (medium)

Question Stats:

74% (01:55) correct 26% (01:58) wrong based on 117 sessions

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QUANT 4-PACK 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 Window
OA, and explanation will be posted after the 48 hour window closes.

This question is part of the Quant 4-Pack series

Scroll Down For Official Explanation

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# Rich Cohen

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****  EMPOWERgmat Discount Codes e-GMAT Discount Codes Kaplan GMAT Prep Discount Codes Senior Manager Joined: 28 Feb 2014 Posts: 294 Location: United States Concentration: Strategy, General Management Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink] ### Show Tags 19 Nov 2015, 18:28 1 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 Current Student Joined: 20 Mar 2014 Posts: 2633 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink] ### Show Tags 23 Nov 2015, 09:26 1 EMPOWERgmatRichC wrote: QUANT 4-PACK 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 Window Earn 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 4-Pack series Scroll Down For Official Explanation $$\frac {3^{2bc}}{3^{2ab}} = ?$$ ---> $$\frac {3^{2bc}}{3^{2ab}} = 3^{2bc-2ab} = 3^{2b(c-a)}$$and as a,b,c are consecutive integers, c-a will always be = 2. Thus, $$3^{2b(c-a)} = 3^{4b}$$ 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. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12684 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... [#permalink] ### Show Tags 24 Nov 2015, 01:07 1 EMPOWERgmatRichC wrote: QUANT 4-PACK 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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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^[2bc-2ab] or
3^[2(bc-ab)] or
3^[2(b(c-a))]

since a,b,c are consecutive integers, c-a 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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26 Nov 2017, 11:20
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Re: Problem Solving Pack 4, Question 3) If a, b and c are consecutive... &nbs [#permalink] 26 Nov 2017, 11:20
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