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Turkey NMO 2008 1st Round - P26 (Number Theory)

Source:

August 26, 2012
factorial

Problem Statement

Let 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!} + \dots + \frac{10^2+3\cdot 10 + 1}{11! \cdot 12!}. What is the remainder when 11!⋅12!⋅A11!\cdot 12! \cdot A is divided by 1111?
<spanclass=′latex−bold′>(A)</span> 0<spanclass=′latex−bold′>(B)</span> 1<spanclass=′latex−bold′>(C)</span> 5<spanclass=′latex−bold′>(D)</span> 8<spanclass=′latex−bold′>(E)</span> 10 <span class='latex-bold'>(A)</span>\ 0 \qquad<span class='latex-bold'>(B)</span>\ 1 \qquad<span class='latex-bold'>(C)</span>\ 5 \qquad<span class='latex-bold'>(D)</span>\ 8 \qquad<span class='latex-bold'>(E)</span>\ 10
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