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# A function V(a, b) is defined for positive integers a, b and

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A function V(a, b) is defined for positive integers a, b and [#permalink]

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10 Dec 2012, 12:25
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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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Dec 2012, 01:14, edited 1 time in total.
Edited the question and added the OA.
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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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11 Dec 2012, 01:57
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Expert's post
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?

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.

$$V(66,14 ) = V(14,66) = V(14, 14+52)$$;

$$V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{33}{26}*V(14,14+38)$$;

$$\frac{33}{26}*V(14,14+38)=\frac{33}{26}*(1+\frac{14}{38})V(14,38)=\frac{33}{19}*V(14, 14+24)$$;

$$\frac{33}{19}*V(14,14+24)=\frac{33}{19}*(1+\frac{14}{24})V(14,24)=\frac{33}{12}*V(14,14+10)$$;

$$\frac{33}{12}V(14,14+10)=\frac{33}{12}*(1+\frac{14}{10})V(14,10)=\frac{33}{5}*V(10,14)=\frac{33}{5}*V(10, 10+4)$$;

$$\frac{33}{5}V(10,10+4)=\frac{33}{5}*(1+\frac{10}{4})V(10,4)=\frac{33*7}{5*2}*V(4,10)=\frac{33*7}{5*2}*V(4, 4+6)$$;

$$\frac{33*7}{5*2}*V(4, 4+6)=\frac{33*7}{5*2}*(1+\frac{4}{6})V(4,6)=\frac{33*7}{2*3}*V(4,4+2)$$;

$$\frac{33*7}{2*3}*V(4,4+2)=\frac{33*7}{2*3}*(1+\frac{4}{2})V(4,2)=\frac{33*7}{2}*V(2,4)=\frac{33*7}{2}*V(2, 2+2)$$;

$$\frac{33*7}{2}*V(2, 2+2)=\frac{33*7}{2}*(1+\frac{2}{2})V(2,2)=\frac{33*7}{2}*2*2=462$$.

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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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13 Dec 2012, 02:12
Bunuel wrote:
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?

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.

$$V(66,14 ) = V(14,66) = V(14, 14+52)$$;

$$V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{33}{26}*V(14,14+38)$$;

$$\frac{33}{26}*V(14,14+38)=\frac{33}{26}*(1+\frac{14}{38})V(14,38)=\frac{33}{19}*V(14, 14+24)$$;

$$\frac{33}{19}*V(14,14+24)=\frac{33}{19}*(1+\frac{14}{24})V(14,24)=\frac{33}{12}*V(14,14+10)$$;

$$\frac{33}{12}V(14,14+10)=\frac{33}{12}*(1+\frac{14}{10})V(14,10)=\frac{33}{5}*V(10,14)=\frac{33}{5}*V(10, 10+4)$$;

$$\frac{33}{5}V(10,10+4)=\frac{33}{5}*(1+\frac{10}{4})V(10,4)=\frac{33*7}{5*2}*V(4,10)=\frac{33*7}{5*2}*V(4, 4+6)$$;

$$\frac{33*7}{5*2}*V(4, 4+6)=\frac{33*7}{5*2}*(1+\frac{4}{6})V(4,6)=\frac{33*7}{2*3}*V(4,4+2)$$;

$$\frac{33*7}{2*3}*V(4,4+2)=\frac{33*7}{2*3}*(1+\frac{4}{2})V(4,2)=\frac{33*7}{2}*V(2,4)=\frac{33*7}{2}*V(2, 2+2)$$;

$$\frac{33*7}{2}*V(2, 2+2)=\frac{33*7}{2}*(1+\frac{2}{2})V(2,2)=\frac{33*7}{2}*2*2=462$$.

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?
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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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13 Dec 2012, 02:15
ziko wrote:
Bunuel wrote:
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?

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.

$$V(66,14 ) = V(14,66) = V(14, 14+52)$$;

$$V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{33}{26}*V(14,14+38)$$;

$$\frac{33}{26}*V(14,14+38)=\frac{33}{26}*(1+\frac{14}{38})V(14,38)=\frac{33}{19}*V(14, 14+24)$$;

$$\frac{33}{19}*V(14,14+24)=\frac{33}{19}*(1+\frac{14}{24})V(14,24)=\frac{33}{12}*V(14,14+10)$$;

$$\frac{33}{12}V(14,14+10)=\frac{33}{12}*(1+\frac{14}{10})V(14,10)=\frac{33}{5}*V(10,14)=\frac{33}{5}*V(10, 10+4)$$;

$$\frac{33}{5}V(10,10+4)=\frac{33}{5}*(1+\frac{10}{4})V(10,4)=\frac{33*7}{5*2}*V(4,10)=\frac{33*7}{5*2}*V(4, 4+6)$$;

$$\frac{33*7}{5*2}*V(4, 4+6)=\frac{33*7}{5*2}*(1+\frac{4}{6})V(4,6)=\frac{33*7}{2*3}*V(4,4+2)$$;

$$\frac{33*7}{2*3}*V(4,4+2)=\frac{33*7}{2*3}*(1+\frac{4}{2})V(4,2)=\frac{33*7}{2}*V(2,4)=\frac{33*7}{2}*V(2, 2+2)$$;

$$\frac{33*7}{2}*V(2, 2+2)=\frac{33*7}{2}*(1+\frac{2}{2})V(2,2)=\frac{33*7}{2}*2*2=462$$.

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?

I doubt that this is a GMAT question. So, I wouldn't worry about it at all.
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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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31 Jan 2014, 13:43
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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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20 Sep 2015, 08:40
Hello from the GMAT Club BumpBot!

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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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21 Sep 2015, 19:15
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?

Once you have figured out the way to solve this (see Bunnel's explanation above), it becomes clear that the expansion of the function continues till you arrive at the greatest common factor between a and b. As the multiplication to arrive at the solution starts with 66, the solution itself should be divisible by 66. Without any further calculations it becomes clear that none of the options here satisfy that condition.
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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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17 Jun 2016, 03:21
Bunuel wrote:
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?

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.

$$V(66,14 ) = V(14,66) = V(14, 14+52)$$;

$$V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{33}{26}*V(14,14+38)$$;

$$\frac{33}{26}*V(14,14+38)=\frac{33}{26}*(1+\frac{14}{38})V(14,38)=\frac{33}{19}*V(14, 14+24)$$;

$$\frac{33}{19}*V(14,14+24)=\frac{33}{19}*(1+\frac{14}{24})V(14,24)=\frac{33}{12}*V(14,14+10)$$;

$$\frac{33}{12}V(14,14+10)=\frac{33}{12}*(1+\frac{14}{10})V(14,10)=\frac{33}{5}*V(10,14)=\frac{33}{5}*V(10, 10+4)$$;

$$\frac{33}{5}V(10,10+4)=\frac{33}{5}*(1+\frac{10}{4})V(10,4)=\frac{33*7}{5*2}*V(4,10)=\frac{33*7}{5*2}*V(4, 4+6)$$;

$$\frac{33*7}{5*2}*V(4, 4+6)=\frac{33*7}{5*2}*(1+\frac{4}{6})V(4,6)=\frac{33*7}{2*3}*V(4,4+2)$$;

$$\frac{33*7}{2*3}*V(4,4+2)=\frac{33*7}{2*3}*(1+\frac{4}{2})V(4,2)=\frac{33*7}{2}*V(2,4)=\frac{33*7}{2}*V(2, 2+2)$$;

$$\frac{33*7}{2}*V(2, 2+2)=\frac{33*7}{2}*(1+\frac{2}{2})V(2,2)=\frac{33*7}{2}*2*2=462$$.

Is there any shorter method to solve this sort of question as it will take a lot of time in the exam to solve it
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Re: A function V(a, b) is defined for positive integers a, b and [#permalink]

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17 Jun 2016, 04:08
rhine29388 wrote:
Bunuel wrote:
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?

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.

$$V(66,14 ) = V(14,66) = V(14, 14+52)$$;

$$V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{33}{26}*V(14,14+38)$$;

$$\frac{33}{26}*V(14,14+38)=\frac{33}{26}*(1+\frac{14}{38})V(14,38)=\frac{33}{19}*V(14, 14+24)$$;

$$\frac{33}{19}*V(14,14+24)=\frac{33}{19}*(1+\frac{14}{24})V(14,24)=\frac{33}{12}*V(14,14+10)$$;

$$\frac{33}{12}V(14,14+10)=\frac{33}{12}*(1+\frac{14}{10})V(14,10)=\frac{33}{5}*V(10,14)=\frac{33}{5}*V(10, 10+4)$$;

$$\frac{33}{5}V(10,10+4)=\frac{33}{5}*(1+\frac{10}{4})V(10,4)=\frac{33*7}{5*2}*V(4,10)=\frac{33*7}{5*2}*V(4, 4+6)$$;

$$\frac{33*7}{5*2}*V(4, 4+6)=\frac{33*7}{5*2}*(1+\frac{4}{6})V(4,6)=\frac{33*7}{2*3}*V(4,4+2)$$;

$$\frac{33*7}{2*3}*V(4,4+2)=\frac{33*7}{2*3}*(1+\frac{4}{2})V(4,2)=\frac{33*7}{2}*V(2,4)=\frac{33*7}{2}*V(2, 2+2)$$;

$$\frac{33*7}{2}*V(2, 2+2)=\frac{33*7}{2}*(1+\frac{2}{2})V(2,2)=\frac{33*7}{2}*2*2=462$$.

Is there any shorter method to solve this sort of question as it will take a lot of time in the exam to solve it

Hi,

the method by Bunuel is the best and the way to do the Q...
Just a small shortcut can be used to the method..

$$V(a, a+b) = (1 + \frac{a}{b}) V(a, b) = \frac{(a+b)}{b} V(a,b)$$....

IMP STEP -In the NEXT step b = a+c.. so V(a,c) will become $$\frac{(a+c)}{c} V(a,c)$$..... and $$\frac{(a+b)}{b}* V(a,b)=\frac{(a+b)}{b}*\frac{(a+c)}{c} V(a,c) = \frac{(a+b)}{c} V(a,c)$$.. and will continue till the right portion , b or c, becomes lesser than a, and at that POINT the right portion will become the REMAINDER when b is divided by a...

so $$V(14, 14+52)=(1+\frac{14}{52})V(14,52)=\frac{66}{38}*V(14,14+38) = \frac{66}{38}*\frac{38}{38-14}*V(14,14+24) = \frac{66}{10}V(14,10)$$.... 10 since 66 div by 14 gives 10 as remainder....
Reverse a and b and continue....
$$\frac{66}{10}V(14,10)=\frac{66}{10}V(10,14) = \frac{66}{10}*\frac{14}{4}V(10,4)$$
reverse a and b and same steps
$$\frac{66}{10}*\frac{14}{4}V(10,14)=\frac{66}{10}*\frac{14}{4}V(4,10)=\frac{66}{10}*\frac{14}{4}*\frac{10}{2}V(4,2)=\frac{66}{10}*\frac{14}{4}*\frac{10}{2}V(2,4)=\frac{66}{10}*\frac{14}{4}*\frac{10}{2}*\frac{4}{2}V(2,2)= \frac{66}{10}*\frac{14}{4}*\frac{10}{2}*\frac{4}{2}*2 = 66*7 = 462$$
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Re: A function V(a, b) is defined for positive integers a, b and   [#permalink] 17 Jun 2016, 04:08
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