AIME II 2022 (Problem 10)Find the remainder when ((32)2)+((42)2)+⋯+((402)2)\binom{\binom{3}{2}}{2}+\binom{\binom{4}{2}}{2}+\cdots+\binom{\binom{40}{2}}{2}(2(23))+(2(24))+⋯+(2(240)) is divided by 100010001000.Related TopicsToolkit 10 — Difference of squaresToolkit 16 — Hockey Stick IdentityHints (5)Hint 1((k2)2)=(k2)((k2)−1)2=18 k(k−1)(k−2)(k+1)=3(k+14)\binom{\binom{k}{2}}{2} = \frac{\binom{k}{2}\Big(\binom{k}{2}-1\Big)}{2} = \frac{1}{8}\,k(k-1)(k-2)(k+1) = 3\binom{k+1}{4}(2(2k))=2(2k)((2k)−1)=81k(k−1)(k−2)(k+1)=3(4k+1)Hint 2Use 16. Hockey Stick Identity.Hint 33(425)=42×41×40×39×385×4×2=42×41×39×38=(40+2)(40+1)(40−1)(40−2)3\binom{42}{5}=\frac{42\times41\times40\times39\times38}{5\times4\times2} =42\times41\times39\times38=(40+2)(40+1)(40-1)(40-2)3(542)=5×4×242×41×40×39×38=42×41×39×38=(40+2)(40+1)(40−1)(40−2)Hint 4Use 10. Difference of Squares.Hint 5(1600−4)(1600−1)(1600-4)(1600-1)(1600−4)(1600−1)Final Answer004004004