I don't even know how to proceed on this problem.

Let f:**N** \(\rightarrow\) **N** be a function such that f(n+1)>f(n) and f(f(n))=3n \(\forall\) n.
Evaluate f(2010).

Please help.....

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## Comments

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TopNewestUse base systems :)

For any integer \( n \) , let \( n_{[3]} \) denote the base \( 3 \) representation of \( n \) . Let \( n_{[3]} = a_1a_2a_3 \cdots a_k \) Then by using induction we can prove that \[ f(n)_{[3]} = 2a_2a_3 \cdots a_k , \ \text{if} \ a_1 = 1 \] and , \[ f(n)_{[3]} = 1a_2a_3 \cdots a_k0 ,\ \text{if} \ a_1 = 2 \] .

Now using this ,since \( 2010_{[3]} = 2202010 \) we get \( f(n)_{[3]} = 12020100 \) Which gives \( f(n) = \boxed{3816} \) \( \blacksquare \)

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I was quite familar with this technique actually . I had solved a simlar type of problem earlier too and quite familiar with this technique . Base systems are very hand in solving many NT and FE problems .

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No doubt you are familiar with the technique but what I want to know is on what grounds can one use base systems to solve FE ? I mean what are the requirements and also please explain the procedure a bit.

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click here See P20 . You would get the point

See this nice article from IMO compendium group :Log in to reply

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I couldn't get how did u get the idea of using a different base representation and again with a specific 3... when one have to consider this concept of base system ......?..please elaborate....regards....

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hats off to thinking procedure & art of problem solving...

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Even though your soln seems to be concise and simple, but I couldn't get how did you get the idea of using a different base representation and that too only 3 ? Please guide me a bit further. I'd be grateful to you for that.

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Fantastic!

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Here's a suggestion on how to start with the problem:

We know that \(f(f(1))=3\). What are the possible values of \(f(1)\)?

Hint: If \(f(1)=1\) then \(f(f(1))=1\). If \(f(1)=4\) then \(f(4)=3\), but \(f(4)>f(3)>f(2)>f(1)\).

Now think about \(f(n)\) for larger values of \(n\). For example, what is \(f(3)\)?

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Ohhhh....I didn't even think that way. So that way and using mathematical induction, f(n)=n+1. Am I right ?

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no. its simple to obtain:

f(1) =2

f(2) = 3

f(3) = 6

f(4) = 7 (implied by f(6))

f(5) = 8 (implied by f(6))

f(6) = 9 (follows from f(3))

f(7) = 12 (follows from f(4))

f(8) = 15 (follows from f(5))

f(9) = 18 (follows from f(6))

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Comment deleted May 04, 2013

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We are given that \( f : \mathbb{N} \to \mathbb{N} \); that is, \( f \) is a function that maps positive integers to POSITIVE INTEGERS. Your function does not satisfy this criterion.

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