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Number Properties

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Number Properties

by Aman verma » Fri May 21, 2010 8:38 am
Q: ( 3^2n ) - 1 is divisible by 2^(n+3) for n =

a)1

b)2

c)3

d)4

e)None of these

Regret there was an error in the original post. It's 2^(n+3). Now corrected !
Last edited by Aman verma on Fri May 21, 2010 9:04 am, edited 1 time in total.
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by liferocks » Fri May 21, 2010 8:49 am
( 3^2n ) - 1=k2^(n+2)
or (9^n)-1=4*2^n
or (9^n)=4*2^n+1

so for n=1,LHR=RHS

Ans option A
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by STEVEN SPIELBERG » Sat May 22, 2010 7:17 am
What's the OA ?
I want to win an OSCAR on the GMAT !!!
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by kstv » Sat May 22, 2010 9:47 am
Aman verma wrote:Q: ( 3^2n ) - 1 is divisible by 2^(n+3) for n =
a)1 b)2 c)3 d)4 e)None of these
(3^2n) - 1 / 2^(n+3)

IMO E
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by liferocks » Sat May 22, 2010 9:20 pm
Aman verma wrote: It's 2^(n+3). Now corrected !
Now it should be
( 3^2n ) - 1=k2^(n+3)
or (9^n)-1=k8*2^n
or (9^n)=k8*2^n+1

For n=1 ,(9^n)=9..or 8=k*8*2^n..this is possible for n=0 and k=1..not possible
For n=2 ,(9^n)=81..or 80=k*8*4..or k=5/2..not possible
For n=3 ,(9^n)=729..or 728=k*8*8..or k=81/8..not possible
For n=4 ,(9^n)=6561..or 6560=k*8*16..or k=205/4..not possible

Ans E
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by Aman verma » Sun May 23, 2010 8:26 am
OAE. Thanks to all . Regret the error !
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by odod » Mon May 24, 2010 5:47 am
Hello - thanks for the post and answer. I'm having troubles understanding the alegbra. Can someone help explain how k2^(n+3) becomes k8*2^n+1 ?


i would appreciate the help. thanks
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by gmatjedi » Fri May 28, 2010 2:45 pm
k2^(n+3)=K8*2^n

principle is a^b * a^c = a^(b+c)

k2^n * 2^3=8k2^n

1 is added to the right side since it was on the left side

hope that helpbs
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by odod » Fri May 28, 2010 2:54 pm
thanks. I think i get it.. but did you mean: k2^n * 2^3=8k2^n to read k2^n * 2k^3=8k2^n


in other words, was the K missing in your post above?
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by odod » Fri May 28, 2010 3:00 pm
ohhh nevermind I get it..thanks again.

cheers
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by odod » Fri May 28, 2010 3:09 pm
ohhh nevermind I get it..thanks again.

cheers
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