olymmath

AUSTRIAN-POLISH 1994 Problem 1

    A function f:R→Rf: R \to R satisfies the conditions: f(x+19)≤f(x)+19f (x + 19) \le f (x) + 19 and f(x+94)≥f(x)+94f (x + 94) \ge f (x) + 94 for all x∈Rx \in R . Prove that f(x+1)=f(x)+1f (x + 1) = f (x) + 1 for all x∈Rx \in R .

    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

    Similar problems

    These problems have the most words in common with this statement.

    Locator austrian-polish/1994/p1 · Tag BBF39CRR6B · Edit this page on GitHub · Keys: j next, k previous