felixjkz wrote:
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!
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.
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?