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# (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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(6^3)a+(6^2)b+6c+d=821, where a, b, c, and d  [#permalink]

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18 Oct 2017, 23:04
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85% (hard)

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54% (03:14) correct 46% (03:01) wrong based on 80 sessions

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[GMAT math practice question]

(6^3)a+(6^2)b+6c+d=821, where a, b, c, and d are integers from 0 to 5, inclusively, c=?

A. 0
B. 1
C. 2
D. 3
E. 4

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior PS Moderator Joined: 26 Feb 2016 Posts: 3253 Location: India GPA: 3.12 (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink] ### Show Tags 18 Oct 2017, 23:22 This expression can be written as 216a + 36b + 6c + d. The value of the expression must be 821 and a,b,c, and d must have integer values, and we have been asked to find the value of c. The value of the expression must be closest to 821 using values of a,b,c,d! If a=3,b=4, the sum of 216a + 36b will be: 216*3 = 648 | 36*4 = 144 | 648+144 = 792. 6c + d must be 27, which is only possible when c=4(Option E) since d cannot be greater than 5. _________________ You've got what it takes, but it will take everything you've got Intern Joined: 29 May 2012 Posts: 33 Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink] ### Show Tags 19 Oct 2017, 00:28 Not sure if my approach is the most appropriate one, but just to share... (6^3)a+(6^2)b+6c+d=216a + 36b + 6c + d=821 Because all a, b, c, and d are integers from 0 to 5, the largest possible value for a is 3; therefore, 648 + 36b + 6c +d=821 and 36b + 6c +d = 173; The largest possible value for b is then 4; therefore, 144 + 6c +d = 173 and 6c + d = 29; Therefore, the largest possible value for c is 4. Answer is E. Math Expert Joined: 02 Sep 2009 Posts: 64174 Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink] ### Show Tags 19 Oct 2017, 01:12 MathRevolution wrote: [GMAT math practice question] (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d are integers from 0 to 5, inclusively, c=? A. 0 B. 1 C. 2 D. 3 E. 4 Similar question: https://gmatclub.com/forum/7-3-a-7-2-b- ... 48834.html _________________ Director Joined: 14 Nov 2014 Posts: 574 Location: India GMAT 1: 700 Q50 V34 GPA: 3.76 Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink] ### Show Tags 19 Oct 2017, 08:55 MathRevolution wrote: [GMAT math practice question] (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d are integers from 0 to 5, inclusively, c=? A. 0 B. 1 C. 2 D. 3 E. 4 As given Value is 821 (6^3)a+(6^2)b+6c---- even and divisible by 6 ------------------1 D must be odd .----------------------------------2 D can be 1 , 3, or 5 let D = 1 = (6^3)a+(6^2)b+6c = 820 ---this is not possible as 820 is not divisible by 6 (as per condition 1) D = 3 =(6^3)a+(6^2)b+6c = 818 ---this is also not possible as 818 is not divisible by 6 D = 5 =(6^3)a+(6^2)b+6c = 816 --yes , only possible value for (6^3)a+(6^2)b+6c 216 * 3 + 36 * 4 + 6 * 4 = 816 C = 4 ---- Senior SC Moderator Joined: 22 May 2016 Posts: 3848 (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink] ### Show Tags 19 Oct 2017, 23:12 MathRevolution wrote: [GMAT math practice question] (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d are integers from 0 to 5, inclusively, c=? A. 0 B. 1 C. 2 D. 3 E. 4 $$(6^3)a + (6^2)b+ 6c+ d=821$$ $$(216)a + (36)b+ 6c + d = 821$$ At most, c = 5, d = 5 At most, 6c + d = 35 So nearly 800 has to be from 216a and 36b. Further: 216 must carry most of the weight. For example, if a = 2, then 216a = 432. Even if b, c, and d = 5, the last three terms cannot sum to 432. So: Maximize 216a, with 821 as limit 821/216 = 3.xx a = 3 (216 * 3) = 648 Remaining? 821 - 648 = 173 Maximize 36b, with 173 as limit 173/36 = 4.xx b = 4 (36 * 4) = 144 Remaining? 173 - 144 = 29 29 = 6c + d d cannot be < 5 If d = 0, 1, 2, 3 or 4, results are 29, 28, 27, 26, and 25. They are not multiples of 6 required by 6c d = 5 (29 - 5) = 24 c = 4 6c = 24 (6c + d) = (24 + 5) = 29 That works. c = 4 Answer E _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr. Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 9018 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink] ### Show Tags 22 Oct 2017, 20:20 => 821 = 6*136 + 5 136 = 6*22 + 4 22 = 6*3 +4 821 = 6*136 + 5 = 6*(6*22+4) + 5 = 6 * ( 6 * (6 *3 + 4) + 4 ) + 5 = 6 * ( 6^2*3 + 6*4 + 4 ) + 5 = 6^3*3 + 6^2*4 + 6*4 + 5 = 6^3*a + 6^2*b + 6*c + d Thus a = 3, b = 4, c = 4 and d = 5. Therefore, the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d  [#permalink]

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26 Mar 2020, 07:12
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Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d   [#permalink] 26 Mar 2020, 07:12

# (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d

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