Peter wrote: The function $ f: \mathbb{N}\to\mathbb{N}_{0}$ satisfies for all $ m,n\in\mathbb{N}$: \[ f(m + n) - f(m) - f(n) = 0\text{ or }1, \; f(2) = 0, \; f(3) > 0, \; \text{ and }f(9999) = 3333. \] Determine $ f(1982)$. We show that $ f(n) = [\frac{n}{3}]$ for $ n \leq 9999$, where$ [ ]$ denotes the integral part. We show first that $ f(3) = 1. f(1)$ must be $ 0$, otherwise $ f(2) - f(1) - f(1)$ would be negative. Hence $ f(3) = f(2) + f(1) + 0$ or $ 1 = 0$ or $ 1$. But we are told $ f(3) > 0$, so $ f(3) = 1$. It follows by induction that $ f(3n) \geq n$. For $ f(3n+3) = f(3) + f(3n) + 0$ or $ 1 = f(3n) + 1$ or $ 2$. Moreover if we ever get $ f(3n) > n$, then the same argument shows that $ f(3m) > m$ for all $ m > n$. But $ f(3.3333) = 3333$, so $ f(3n) = n$ for all $ n \leq 3333$. Now $ f(3n+1) = f(3n) + f(1) + 0$ or $ 1 = n$ or $ n + 1$. But $ 3n+1 = f(9n+3) ≥ f(6n+2) + f(3n+1) ≥ 3f(3n+1)$, so $ f(3n+1) < n+1$. Hence $ f(3n+1) = n$. Similarly, $ f(3n+2) = n$. In particular $ f(1982) = 660$