Increasing Function

A bit of playing around identifies the pattern for this function, and then it is possible to prove by induction that $f\big(3^n+j\big) \; = \; 2\times3^n + j \qquad \qquad f\big(2\times3^n + j\big) \; = \; 3^{n+1} + 3j$ for $1 \le j \le 3^n$ and $n \ge 0$ . Since $955 = 3^6 + 226$ , we have $f(955) = 2\times3^6 + 226 = 1684$ .

First, notice that by definition we have: $f(3x) = f(f(f(x))) = 3f(x)$ for all positive integers $x$ . Let's call this equation $(1)$ .

Second, we can show by induction that if $a \ge b$ with $a,b \in \mathbb{N}$ , then $f(a) \ge f(b) + (a-b)$ . Let's call this inequality $(2)$ .

Now, we must have $f(1) \ge 1$ , and by (ii), $f(f(1)) = 3$ . Thus, using $(2)$ with $a = f(1)$ and $b = 1$ , we have: $3 = f(f(1)) \ge f(1) + (f(1)-1) = 2f(1)-1$ Thus $f(1) \le 2$ . However $f(1)$ cannot equal 1, because if it did, we would have a contradiction: $f(f(1)) = f(1) = 1$ , which does not equal $3*1$ .

Therefore $f(1) = 2$ .

Therefore, by equation $(1)$ , it's easy to show by induction that for any $n \in \mathbb{N}$ , we have $f(3^n) = 2*3^n$ . Therefore, $f(2*3^n) = f(f(3^n)) = 3^{n+1}$ .

Therefore, $f(729) = 1458$ and $f(1458) = 2187$ .

By $(2)$ with $a = 955, b = 729$ , we have $f(955) \ge f(729) + (955 - 729) = 1458 + 226 = 1684$ .

Again by $(2)$ with $a = 1458, b = 955$ , we have $f(1458) \ge f(955) + (1458 - 955)$ , so that $f(955) \le 2187 + (955 - 1458) = 1684$ .

Hence $1684 \le f(955) \le 1684$ . Therefore $f(955) = \boxed{1684}$

One function that works is as follows: if the base-3 expansion of $x$ starts with a $1$ , then $f(x)$ is the same expansion, with the $1$ replaced by a $2$ . And if the base-3 expansion of $x$ starts with a $2$ , then $f(x)$ replaces the $2$ with a $1$ and adds a $0$ on the far right.

This $f$ clearly satisfies the conditions of the problem. I believe it is unique, but the problem doesn't ask us to prove that.

In any case, $955$ 's ternary expansion starts with a $1$ , since $729$ is the next lowest power of $3$ , and so $f(955) = 955 + 729 = \fbox{1684}$ .

In my solution, I have used a notation, [a,b]->[c,d]. It means that for a<=x<=b, c<=f(x)<=d. If f(1)=n, f(f(1))=f(n)=3. Then f(f(n))=f(3)=3n. Note if n=1, we have f(f(1))=3=f(1)=1(not possible), and for n>2, 4<=f(2)<=8, but in that case we won't have a fixed value of f(x) for some bigger values of x. Thus we conclude n=2. Then in the last interval [730,1457]->[1459,2186], we note that 1457-730=2186-1459=727, it means that no elements in domain and domain set are equal, hence the mapping of f in that interval will be linear, thus f(955)=955+727=1684

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