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TR-MO-R1 2008 Problem 26

    A=22+3⋅2+13!⋅4!+32+3⋅3+14!⋅5!+42+3⋅4+15!⋅6!+⋯+102+3⋅10+111!⋅12!A = \frac{2^2+3\cdot 2+1}{3!\cdot 4!} + \frac{3^2+3\cdot 3+1}{4!\cdot 5!} + \frac{4^2+3\cdot 4+1}{5!\cdot 6!} + \cdots + \frac{10^2+3\cdot 10+1}{11!\cdot 12!}

    toplamı için 11!⋅12!⋅A11!\cdot 12!\cdot A sayısını 11 e bölünce kalan nedir?

    a) 0 b) 1 c) 5 d) 8 e) 10

    Official text Turkish Mathematical Olympiad, first round 2008 official paper, page 8. 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.

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