olymmath

RS-SMO 2018 Problem 2

    Let n>1n>1 be an integer. Call a number beautiful if its square leaves an odd remainder upon divison by nn . Prove that the number of consecutive beautiful numbers is less or equal to 1+⌊3n⌋1+\lfloor \sqrt{3n} \rfloor .

    Dataset text AI-MO AoPS. Quoted with source.

    Official text: Serbian Mathematical Olympiad 2018 official paper, page 1.

    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 rs-smo/2018/p2 · Tag 4987R89XDP · Edit this page on GitHub · Keys: j next, k previous