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HCF of functions

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HCF of functions

by gmatmachoman » Sat Jul 24, 2010 12:04 pm
A function f(n) is defined for all positive real values of 'n' such that f(n + 2) = f(n + 1) + f(n).

If f(1) = f(2) = 1, then find the highest common factor of f (12) and f (30)


A.f(11)
B. f(4)
C. f(6)
D.f(3)
E.f(5)

OA : C
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by kvcpk » Sat Jul 24, 2010 1:07 pm
gmatmachoman wrote:A function f(n) is defined for all positive real values of 'n' such that f(n + 2) = f(n + 1) + f(n).

If f(1) = f(2) = 1, then find the highest common factor of f (12) and f (30)


A.f(11)
B. f(4)
C. f(6)
D.f(3)
E.f(5)

OA : C
f(n + 2) = f(n + 1) + f(n)
f(3)=f(2)+f(1)
f(3)=2
F(4) = 3
This is a fibonacci series.

1,1,2,3,5,8,13,21,34,55,89,144
F(1) is a factor of f(2)
f(2) is a factor of f(4)
f(3) is a factor of f(6)
f(4) is a factor of f(8)
...
f(6) is a factor of f(12)
..

f(15) is a factor of f(30)

.. Tried lot of things and tired :) Looks very difficult to me.

Would have guessed some answer on the actual test.

I might be missing something.. Will give a try again tomorrow.. :D
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by indiantiger » Sat Jul 24, 2010 6:36 pm
Dude,

Whats the source of the question? I tried similar approach as that of kvcpk, with couple of beers down cannot think properly.
"Single Malt is better than Blended"
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by gmatmachoman » Sat Jul 24, 2010 9:20 pm
kvcpk wrote:
gmatmachoman wrote:A function f(n) is defined for all positive real values of 'n' such that f(n + 2) = f(n + 1) + f(n).

If f(1) = f(2) = 1, then find the highest common factor of f (12) and f (30)


A.f(11)
B. f(4)
C. f(6)
D.f(3)
E.f(5)

OA : C
f(n + 2) = f(n + 1) + f(n)
f(3)=f(2)+f(1)
f(3)=2
F(4) = 3
This is a fibonacci series.

1,1,2,3,5,8,13,21,34,55,89,144
F(1) is a factor of f(2)
f(2) is a factor of f(4)
f(3) is a factor of f(6)
f(4) is a factor of f(8)
...
f(6) is a factor of f(12)
..

f(15) is a factor of f(30)

.. Tried lot of things and tired :) Looks very difficult to me.

Would have guessed some answer on the actual test.

I might be missing something.. Will give a try again tomorrow.. :D

@praveen

Congrats..u have done a good work..Yeah its Fibonacci Series..

So now that u r a Titan..
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by selango » Sun Jul 25, 2010 2:49 am
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610

f(12)=144=2*2*2*2*3*3

f(30)=?

One way I solved it by going thru the answer options.

We can eliminate the following options.

f(11)=89,This cant surely the HCF 89 is not factor of f(12)

f(5)=5,This cant surely the HCF 5 is not factor of f(12)

Tat leaves with f(3)=2,f(4)=3,f(6)=8.
--Anand--
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by selango » Sun Jul 25, 2010 3:17 am
I calculated f(30)=832,040.

f(12)=144

f(30)=832,040

HCF of both terms is 8 which is f(6).

But this is lengthy approach.Any quick approach is appreciated.
--Anand--
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by gmatmachoman » Sun Jul 25, 2010 4:00 am
give me few minutes..will post the OE..

CHeers to the good work Anand..
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by kvcpk » Sun Jul 25, 2010 6:49 am
Finally I was able to crack it.

The formula is:

HCF(f(m), f(n)) = F(HCF(m,n))

HCF(f(12),f(30)) = f(HCF(12,30))

HCF(f(12),f(30)) = f(6)

Feel lot reliefed now!! :)
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by selango » Sun Jul 25, 2010 7:02 am
Praveen,

U proved that ur a TITAN:)
--Anand--
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