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If a, b, c and d are positive integers less than 4, and 4^a+3^b+2

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If a, b, c and d are positive integers less than 4, and 4^a+3^b+2  [#permalink]

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New post 16 Sep 2014, 04:40
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A
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C
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E

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Question Stats:

73% (01:54) correct 27% (01:49) wrong based on 162 sessions

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If a, b, c and d are positive integers less than 4, and \(4^a+3^b+2^c+1^d=78\) then what is the value of b/c?

A. 3
B. 2
C. 1
D. 1/2
E. 1/3

This problem took me considerable time for solving. Please suggest quick ways to approach this problem.

Many Thanks.
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If a, b, c and d are positive integers less than 4, and 4^a+3^b+2  [#permalink]

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New post Updated on: 17 Sep 2014, 00:41
1
gabriel87 wrote:
If a, b, c and d are positive integers less than 4, and , \(4^a+3^b+2^c+1^d\)=78 then what is the value of b/c?

A) 3
B) 2
C) 1
D)1/2
E)1/3

This problem took me considerable time for solving. Please suggest quick ways to approach this problem.

Many Thanks.


Not sure it's the quickest way to solve it, but this is how I solved it (under 30s).

\(4^a+3^b+2^c+1^d = 78\)---- >\(4^a+3^b+2^c = 77\)

Since a, b and c are positive integers less than 4 we have the following possibilities:

\(4^a = 4, 16, or 64\)
\(3^b = 3, 9, or 27\)
\(2^c = 2, 4, or 8\)

Trial and error gives us quite quick the solution of

\(4^a = 64\)
\(3^b = 9\)
\(2^c = 4\)

\(64 + 9+ 4 = 77\)

i.e. \(c = 2\) and \(b = 2\) ----> \(b/c = 1/1 = 1\)

The correct answer is C
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Originally posted by TehMoUsE on 16 Sep 2014, 10:02.
Last edited by TehMoUsE on 17 Sep 2014, 00:41, edited 2 times in total.
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Re: If a, b, c and d are positive integers less than 4, and 4^a+3^b+2  [#permalink]

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New post 16 Sep 2014, 22:53
gabriel87 wrote:
If a, b, c and d are positive integers less than 4, and \(4^a+3^b+2^c+1^d=78\) then what is the value of b/c?

A. 3
B. 2
C. 1
D. 1/2
E. 1/3

This problem took me considerable time for solving. Please suggest quick ways to approach this problem.

Many Thanks.


It does involve a bit of hit and trial.

\(4^a+3^b+2^c = 77\)
\(2^{2a}+3^b+2^c = 77\)

So we are adding two powers of 2 and a power of 3 to get 77.

Let's try to figure out the power of 3.
Say b = 1, then \(2^{2a}+2^c = 74\)
Can 74 be the sum of two powers of 2? No. The largest power of 2 less than 74 is 64 but 64+10 doesn't work. So 64 doesn't work. The power smaller than that is 32. There is no other power smaller than 64 which could add up with 32 to make 74. So the power of 3 is not 1.

Say b = 2, then \(2^{2a}+2^c = 68\)
This is simply 64 + 4.

Since all powers are less than 4, a must be 3 to get 4^3 = 64 and c must be 2 to get 2^2 = 4.

b = 2, c = 2 so b/c = 1

Answer (C)
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Re: If a, b, c and d are positive integers less than 4, and 4^a+3^b+2  [#permalink]

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New post 22 Sep 2014, 02:30
1
\(4^a + 3^b + 2^c + 1^d=78\)

For any positive value of d, \(1^d = 1\)

\(4^a + 3^b + 2^c = 77\)

Using plug-in method (Reverse Order)

E: \(\frac{b}{c} = \frac{1}{3}\)

\(4^a = 77 - 3 - 8 = 66\) ......... Ignore

D: \(\frac{b}{c} =\frac{1}{2}\)

\(4^a = 77 - 3 - 4 = 70\) ............... Ignore

C: \(\frac{b}{c} = \frac{1}{1} = \frac{2}{2} = \frac{3}{3}\)

\(4^a = 77 - 3 - 2 = 72\) ................ Ignore OR

\(4^a = 77 - 9 - 4 = 77-13 = 64\) .......... Possible

Answer = C
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Re: If a, b, c and d are positive integers less than 4, and 4^a+3^b+2  [#permalink]

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Re: If a, b, c and d are positive integers less than 4, and 4^a+3^b+2 &nbs [#permalink] 11 Oct 2017, 02:42
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