AMC 10B Fall 2021 (Problem 22)For each integer n≥2n\ge2n≥2, let SnS_nSn be the sum of all products jkjkjk, where jjj and kkk are integers and 1≤j<k≤n1\le j<k\le n1≤j<k≤n. What is the sum of the 101010 least values of nnn such that SnS_nSn is divisible by 333?(A) 196\text{(A)}\;196(A)196(B) 197\text{(B)}\;197(B)197(C) 198\text{(C)}\;198(C)198(D) 199\text{(D)}\;199(D)199(E) 200\text{(E)}\;200(E)200Related TopicsCoreToolkit 47 — Modular Arithmetic: Definition and PropertiesMajorToolkit 1 — Sum of the first n integersToolkit 24 — Square of a sum (n terms)MinorToolkit 46 — Divisibility: Definition and PropertiesHints (8)Hint 1Sn=1⋅2+1⋅3+⋯+1⋅n+2⋅3+⋯+(n−1)nS_n=1\cdot2+1\cdot3+\cdots+1\cdot n+2\cdot3+\cdots+(n-1)nSn=1⋅2+1⋅3+⋯+1⋅n+2⋅3+⋯+(n−1)n=(1+2+3+⋯+n)2−(12+22+32+⋯+n2)2=\frac{(1+2+3+\cdots+n)^2-(1^2+2^2+3^2+\cdots+n^2)}{2}=2(1+2+3+⋯+n)2−(12+22+32+⋯+n2)Hint 2Sn=(n(n+1)2)2−n(n+1)(2n+1)62S_n=\frac{\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2n+1)}{6}}{2}Sn=2(2n(n+1))2−6n(n+1)(2n+1)=n2(n+1)24−n(n+1)(2n+1)62=\frac{\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{6}}{2}=24n2(n+1)2−6n(n+1)(2n+1)=3n2(n+1)2−2n(n+1)(2n+1)24=\frac{3n^2(n+1)^2-2n(n+1)(2n+1)}{24}=243n2(n+1)2−2n(n+1)(2n+1)=n(n+1)(3n(n+1)−2(2n+1))24=\frac{n(n+1)\left(3n(n+1)-2(2n+1)\right)}{24}=24n(n+1)(3n(n+1)−2(2n+1))=n(n+1)(3n2−n−2)24=\frac{n(n+1)(3n^2-n-2)}{24}=24n(n+1)(3n2−n−2)=n(n+1)(n−1)(3n+2)24=\frac{n(n+1)(n-1)(3n+2)}{24}=24n(n+1)(n−1)(3n+2)Hint 33∣Sn3\mid S_n3∣Sn⟹9∣n(n+1)(n−1)(3n+2)\Longrightarrow 9\mid n(n+1)(n-1)(3n+2)⟹9∣n(n+1)(n−1)(3n+2)Hint 4Consider n(mod9)n\pmod 9n(mod9).Hint 5n(mod9)n(n+1)(n−1)(3n+2)(mod9)001022⋅3⋅1⋅8≢033⋅4⋅2⋅2≢044⋅5⋅3⋅5≢055⋅6⋅4⋅8≢066⋅7⋅5⋅2≢077⋅8⋅6⋅5≢080\begin{array}{c|c}n\pmod 9&n(n+1)(n-1)(3n+2)\pmod 9\\ \hline0&0\\1&0\\2&2\cdot3\cdot1\cdot8\not\equiv0\\3&3\cdot4\cdot2\cdot2\not\equiv0\\4&4\cdot5\cdot3\cdot5\not\equiv0\\5&5\cdot6\cdot4\cdot8\not\equiv0\\6&6\cdot7\cdot5\cdot2\not\equiv0\\7&7\cdot8\cdot6\cdot5\not\equiv0\\8&0\end{array}n(mod9)012345678n(n+1)(n−1)(3n+2)(mod9)002⋅3⋅1⋅8≡03⋅4⋅2⋅2≡04⋅5⋅3⋅5≡05⋅6⋅4⋅8≡06⋅7⋅5⋅2≡07⋅8⋅6⋅5≡00Hint 6n≡0,1,8(mod9)n\equiv0,1,8\pmod 9n≡0,1,8(mod9)Hint 7The 101010 least possible values of nnn are8,9,10,17,18,19,26,27,28,35.8,9,10,17,18,19,26,27,28,35.8,9,10,17,18,19,26,27,28,35.Hint 8Ans=8+9+10+17+18+19+26+27+28+35=197\text{Ans}=8+9+10+17+18+19+26+27+28+35=197Ans=8+9+10+17+18+19+26+27+28+35=197Final Answer(B) 197197197Related Problems (6)AIME I 2026 (Problem 8)AMC 10A 2020 (Problem 8)USAJMO 2023 (Problem 1)USAMO 2018 (Problem 1)AMC 12A 2024 (Problem 15)AMC 12A 2022 (Problem 16)