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E
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Difficulty:
75%
(hard)
Question Stats:
61% (02:57) correct
39%
(02:56)
wrong
based on 325
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A function V(a, b) is defined for positive integers a, b and satisfies V(a, a) = a, V(a, b) = V(b, a), V(a, a+b) = (1 + a/b) V(a, b). The value represented by V(66, 14) is ?
(A) 364
(B) 231
(C) 455
(D) 472
(E) None of the foregoing
Bunuel,
I know it shouldn't be here but could you explain the solution of this one?
Thanks in advance!
(A) 364
(B) 231
(C) 455
(D) 472
(E) None of the foregoing
Bunuel,
I know it shouldn't be here but could you explain the solution of this one?
Thanks in advance!
11 Dec 2012, 02:57
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felixjkz
Given that:
\(V(a, a) = a\);
\(V(a, b) = V(b, a)\);
\(V(a, a+b) = (1 + \frac{a}{b}) V(a, b)\).
Question asks to find the value of \(V(66, 14)\).
Notice that only the first function gives answer as a simple value rather than another function, thus we should manipulate with \(V(66, 14)\) so that to get \(V(a, a) = a\) in the end.
\(V(66,14 ) = V(14,66) = V(14, 14+52)\);
\(V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{33}{26}*V(14,14+38)\);
\(\frac{33}{26}*V(14,14+38)=\frac{33}{26}*(1+\frac{14}{38})V(14,38)=\frac{33}{19}*V(14, 14+24)\);
\(\frac{33}{19}*V(14,14+24)=\frac{33}{19}*(1+\frac{14}{24})V(14,24)=\frac{33}{12}*V(14,14+10)\);
\(\frac{33}{12}V(14,14+10)=\frac{33}{12}*(1+\frac{14}{10})V(14,10)=\frac{33}{5}*V(10,14)=\frac{33}{5}*V(10, 10+4)\);
\(\frac{33}{5}V(10,10+4)=\frac{33}{5}*(1+\frac{10}{4})V(10,4)=\frac{33*7}{5*2}*V(4,10)=\frac{33*7}{5*2}*V(4, 4+6)\);
\(\frac{33*7}{5*2}*V(4, 4+6)=\frac{33*7}{5*2}*(1+\frac{4}{6})V(4,6)=\frac{33*7}{2*3}*V(4,4+2)\);
\(\frac{33*7}{2*3}*V(4,4+2)=\frac{33*7}{2*3}*(1+\frac{4}{2})V(4,2)=\frac{33*7}{2}*V(2,4)=\frac{33*7}{2}*V(2, 2+2)\);
\(\frac{33*7}{2}*V(2, 2+2)=\frac{33*7}{2}*(1+\frac{2}{2})V(2,2)=\frac{33*7}{2}*2*2=462\).
Answer: E.
General Discussion
ziko
Joined: 28 Feb 2012
Last visit: 29 Jan 2014
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13 Dec 2012, 03:12
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Bunuel
Dear Bunuel,
Your explanation is brilliant. But do you think this is a kind of question that i will face in GMAT because i think the sollution is quite time consuming or there is quiker way?
