olymmath

IMO 1988 Problem 3

  • Day 1

A function ff defined on the positive integers (and taking positive integers values) is given by:

f(1)=1,f(3)=3f(2⋅n)=f(n)f(4⋅n+1)=2⋅f(2⋅n+1)−f(n)f(4⋅n+3)=3⋅f(2⋅n+1)−2⋅f(n),\begin{matrix} f(1) = 1, f(3) = 3 f(2 \cdot n) = f(n) f(4 \cdot n + 1) = 2 \cdot f(2 \cdot n + 1) - f(n) f(4 \cdot n + 3) = 3 \cdot f(2 \cdot n + 1) - 2 \cdot f(n), \end{matrix}

for all positive integers n.n. Determine with proof the number of positive integers ≤1988\leq 1988 for which f(n)=n.f(n) = n.

Dataset text AI-MO AoPS. Quoted with source.

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Locator imo/1988/p3 · Tag Z746NMBD18 · Edit this page on GitHub · Keys: j next, k previous