MathCounts 2025 Chapter — Sprint Round (Problem 27)Let 2n2^n2n be the greatest power of 222 that divides 1×2×3×4+2×3×4×5+3×4×5×6+⋯+25×26×27×281\times2\times3\times4+2\times3\times4\times5+3\times4\times5\times6+\cdots+25\times26\times27\times281×2×3×4+2×3×4×5+3×4×5×6+⋯+25×26×27×28. What is the value of nnn?Related TopicsCoreToolkit 16 — Hockey Stick IdentityHints (4)Hint 1a(a+1)(a+2)(a+3)=4!(a+34)a(a+1)(a+2)(a+3)=4!\binom{a+3}{4}a(a+1)(a+2)(a+3)=4!(4a+3)Hint 2By Hint 1,1×2×3×4+2×3×4×5+3×4×5×6+⋯+25×26×27×281\times2\times3\times4+2\times3\times4\times5+3\times4\times5\times6+\cdots+25\times26\times27\times281×2×3×4+2×3×4×5+3×4×5×6+⋯+25×26×27×28=4!(44)+4!(54)+4!(64)+⋯+4!(284)=4!\binom{4}{4}+4!\binom{5}{4}+4!\binom{6}{4}+\cdots+4!\binom{28}{4}=4!(44)+4!(45)+4!(46)+⋯+4!(428)=4!((44)+(54)+(64)+⋯+(284))=4!\left(\binom{4}{4}+\binom{5}{4}+\binom{6}{4}+\cdots+\binom{28}{4}\right)=4!((44)+(45)+(46)+⋯+(428))Hint 3By Toolkit 16 — Hockey Stick Identity,4!((44)+(54)+(64)+⋯+(284))4!\left(\binom{4}{4}+\binom{5}{4}+\binom{6}{4}+\cdots+\binom{28}{4}\right)4!((44)+(45)+(46)+⋯+(428))=4!(295)=4!\binom{29}{5}=4!(529)Hint 44!(295)4!\binom{29}{5}4!(529)=4!⋅29⋅28⋅27⋅26⋅255!=4!\cdot\frac{29\cdot28\cdot27\cdot26\cdot25}{5!}=4!⋅5!29⋅28⋅27⋅26⋅25=29⋅28⋅27⋅26⋅5=29\cdot28\cdot27\cdot26\cdot5=29⋅28⋅27⋅26⋅5=29⋅22⋅7⋅27⋅2⋅13⋅5=29\cdot2^2\cdot7\cdot27\cdot2\cdot13\cdot5=29⋅22⋅7⋅27⋅2⋅13⋅5=23⋅(an odd integer)=2^3\cdot\text{(an odd integer)}=23⋅(an odd integer)Therefore,n=3n=3n=3Final Answer333Related Problems (1)AIME II 2022 (Problem 10)