Function’s problem (Math Olympiad)

Let $f$ be a function from positive integers to positive integers satisfying $f(n+1) > f(n)$ and $f(f(n)) = 3n$ for all positive integers $n$ . Calculate $f(k)$ for $k = 1, \ldots , 10.$

What is the method to solve this?

#Algebra

Easy Math Editor

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Comments

Consider $f(x)=kx+b$ , $k=\sqrt{3}$ and $b=0$ . But it says positive integers...

Jeff Giff - 9 months, 3 weeks ago

@Jeff Giff Here is their given Solution: To solve this type of problem, one tries to find solutions for n = 0, or n = 1 and/or some other property of f. Then one can conjecture more values, and more general results can usually be proved by induction. In this case, only low-order results are sought, so no inductive proof of general results is needed. We start by showing f(1) > 1, as if f(1) = 1, we would have 3 = f(f(1)) = 1, a contradic- tion. More generally, f(n) > n, for if f(n) ≤ n for some n = k we would have f(k−1) < f(k) ≤ k, so f(k − 1) ≤ k − 1. Repeat this argument k − 1 times and we obtain f(1) ≤ 1, in contra- diction to our result above. Hence f(n) > n for all n. f(1)=2.Supposef(1)=m≥3.Then3=f(f(1))=f(m)>m≥3,acontradiction. So f(1) = 2. So now from the given properties of f we can calculate several values: f(2) = f(f(1)) = 3, f(3) = f(f(2)) = 6, f(6) = f(f(3)) = 9, f(9) = f(f(6)) = 18. Giventhat6=f(3)<f(4)<f(5)<f(6)=9,itfollowsthatf(4)=7, f(5)=8,so f(7) = f(f(4)) = 12 and f(8) = f(f(5)) = 15. Now f(18) = f(f(9)) = 27, and as there are 8 unknown values between f(9) = 18 and f(18) = 27 and 8 numbers between 18 and 27, it follows that f(10) = 19, f(11) = 20, ··· f(17) = 26.

Shevy Doc - 9 months, 3 weeks ago

Oh, yeah! I forgot that it could be random :P

Jeff Giff - 9 months, 3 weeks ago

Here are some things we know:

the constant term is 0
since $f(f(x))=3x$ , the function is linear: or else $f(f(x))$ will have a higher power or $f(x)$ would not be an integer
the question is wrong

Jeff Giff - 9 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$