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When k=odd, f(k)=2k and when k=even, f(k)=3k, f(210)=?

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When k=odd, f(k)=2k and when k=even, f(k)=3k, f(210)=?

by Max@Math Revolution » Wed Jan 13, 2016 6:32 pm
When k=odd, f(k)=2k and when k=even, f(k)=3k, f(210)=?


A. 9^(2^9 ) B. 9^(2^10 ) C. 9^(3^9 ) D. 910 E. 99


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by Brent@GMATPrepNow » Wed Jan 13, 2016 8:39 pm
Max@Math Revolution wrote:When k=odd, f(k)=2k and when k=even, f(k)=3k, f(210)=?
A. 9^(2^9 )
B. 9^(2^10 )
C. 9^(3^9 )
D. 910
E. 99
Is this question transcribed correctly?
210 is even, so f(210) = (3)(210) = 630

Answer: ???

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by chetan.sharma » Thu Jan 14, 2016 3:54 am
Max@Math Revolution wrote:When k=odd, f(k)=2k and when k=even, f(k)=3k, f(210)=?


A. 9^(2^9 ) B. 9^(2^10 ) C. 9^(3^9 ) D. 910 E. 99


* A solution will be posted in two days.
Hi,

looking at the Q and the answer choices, I feel some signs are missing.
The Q should be something like this..

Q. When k is odd, f(k)=2^k and when k is even, f(k)=3^k, f(2^10)=?

A. 9^(2^9 )
B. 9^(2^10 )
C. 9^(3^9 )
D. 9^10
E. 9^9
So , the answer in this case will be..
here k is 2^10 and so even..
so f(2^10)= 3^( 2^10)..
now to get this to base 9, we will square 3 and half the power..

3^(2^10)=(3^2)^{(2^10)/2}..
= 9^(2^9)..
A
ans A
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by Max@Math Revolution » Mon Jan 18, 2016 8:55 pm
When k=odd, f(k)=2k and when k=even, f(k)=3^k, f(2^10)=?


A. 9^(2^9 )
B. 9^(2^10 )
C. 9^(3^9 )
D. 9^10
E. 9^9


-> f(210)=3^(2^10 )=3^((2)(2^9))=(〖3^2)〗^(2^9 )=9^(2^9 )
Therefore, the answer is A.
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by Brent@GMATPrepNow » Tue Jan 19, 2016 10:22 am
Max@Math Revolution wrote:When k=odd, f(k)=2k and when k=even, f(k)=3^k, f(2^10)=?


A. 9^(2^9 )
B. 9^(2^10 )
C. 9^(3^9 )
D. 9^10
E. 9^9
When posting questions in the future, please take a few seconds to ensure that your question is displayed properly. As you can see from the original wording...
Max@Math Revolution wrote:When k=odd, f(k)=2k and when k=even, f(k)=3k, f(210)=?
A. 9^(2^9 ) B. 9^(2^10 ) C. 9^(3^9 ) D. 910 E. 99
..... the question you answered differs from the question you posted.

Cheers,
Brent
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