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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
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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
[Reveal] Spoiler: OA

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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.

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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
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
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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 = 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

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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
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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.

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Re: (6^3)a+(6^2)b+6c+d=821, where a, b, c, and d   [#permalink] 22 Oct 2017, 21:20
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