Brazilian Math Olympiad

First note that if $n$ is written in the Fibonacci basis (as a sum of distinct Fibonacci numbers containing no neighbors), then the representation of $f(n)$ is just that of $n$ with a 0 in the end. The proof goes by induction: it is true for $n=0$. Suppose $n>0$. Then $r$ will be such that $f(r)=n$, if the last digit of the Fibonacci of $n$ is zero, or $f(r)=n-1$, if it is one. Then $f(n)=n+r+n-f(r)$. If the last digit of the Fibonacci of $n$ is zero, then $r$ is obtained by deleting the last digit from $n$. So $f(n)=n+r$ is the sum of the Fibonacci numbers used in the representation of $n$ and its respective antecessors in the Fibonacci sequence, resulting in the subsequent Fibonacci numbers, which is exactly $n$ with a zero at its right. If the last digit of the Fibonacci of $n$ is zero, then $r$ is still obtained by deleting the last digit from $n$, but $f(n)=n+1+r$. Note that $n-1$ ends with a zero, so summing $r$ will do the same thing as the preceding case; and we're substituting the rightmost digit one with 2, which corresponds to a rightmost 10.

Now, express $n$ and $f(n)$ by using the closed form for Fibonacci numbers $F_k=\frac{1}{\sqrt{5}}({\varphi }^k-(-\varphi {)}^{-k})$, and note that we cannot use two consecutive Fibonacci numbers in base Fibonacci, so $|f(n)-\varphi \cdot nZ|<{\varphi }^{-1}(1+{\varphi }^{-2}+{\varphi }^{-4}+\dots )=\frac{{\varphi }^{-1}}{1-{\varphi }^{-2}}=1$, that is, $f(n)-1<\varphi n<f(n)+1$. Thus if $m\le \varphi n$ then $m<f(n)+1⟺m\le f(n)$ and the match is tough; if $m\ge \varphi n+1$ then $m>f(n)-1+1=f(n)$ and the match is not tough.