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(6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink]
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19 Oct 2017, 00:04
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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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(6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink]
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19 Oct 2017, 00:22
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.
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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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19 Oct 2017, 01:28
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.



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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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19 Oct 2017, 02:12



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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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19 Oct 2017, 09:55
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 



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(6^3)a+(6^2)b+6c+d=821, where a, b, c, and d [#permalink]
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20 Oct 2017, 00:12
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 = 173Maximize 36b, with 173 as limit 173/36 = 4.xx b = 4 (36 * 4) = 144 Remaining? 173  144 = 2929 = 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
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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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22 Oct 2017, 21:20
=> 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
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Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d
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