olymmath

TR-MO-R1 2013 Problem 22

    n4+2n3−20n2+2n−21n^4 + 2n^3 - 20n^2 + 2n - 21 sayısı, 0≤n<20130 \le n < 2013 koşulunu sağlayan kaç nn tam sayısı için, 2013 ile bölünür?

    a) 6 b) 8 c) 12 d) 16 e) Hiçbiri

    Official text Turkish Mathematical Olympiad, first round 2013 official paper, page 7. Quoted with source.

    A model read this text from the scan of the paper by OCR, with vision/claude@2.1.318 (claude-opus-5-5). No person has checked it against the paper yet.

    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 tr-mo-r1/2013/p22 · Tag DMSESG6QH8 · Edit this page on GitHub · Keys: j next, k previous