# Functional Relation

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).

Note by Nishant Sharma
5 years, 3 months ago

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Use 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 $$f(n) = \boxed{3816}$$ $$\blacksquare$$

- 5 years, 3 months ago

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 .

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

See this nice article from IMO compendium group : click here See P20 . You would get the point

- 5 years, 3 months ago

It was good but since I am not much familiar with these type of techniques I was not able to grasp it completely. Though indirectly you gave a nice link to the general methods of solving FE. Great !!!

- 5 years, 3 months ago

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....

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

Fantastic!

- 5 years, 3 months ago

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 $$f(3)$$?

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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))

- 5 years, 3 months ago

ya got your point. I'm working on it....

- 5 years, 3 months ago

ya I was wrong......By the way how did u obtain values f(4) onwards ?

- 5 years, 3 months ago

I edited the previous post to show the flow of logic... it does look like induction will work, but you need to notice the pattern.

- 5 years, 3 months ago

Comment deleted May 04, 2013

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.

- 5 years, 3 months ago