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16 Sep 2014, 04:40
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:08) correct
26%
(02:27)
wrong
based on 272
sessions
History
Date
Time
Result
Not Attempted Yet
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.
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.
gabriel87
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
16 Sep 2014, 22:53
Kudos
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gabriel87
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)
