Fall 2021 AMC 10 B

$$U_n=(1+2+3+\cdots +(n-1))\cdot n=\frac{(n-1)n^2}{2}.$$ Then $S_n=S_{n-1}+U_n$ for $n\ge 3$. Note that $U_n$ is divisible by 3 if $n\equiv 0\text{ or }1(\mathrm{mod}3)$; and if $n\equiv 2(\mathrm{mod}3)$, then $(n-1)n^2\equiv 1(\mathrm{mod}3)$ and is even, so $U_n\equiv 2(\mathrm{mod}3)$. Hence $S_{n+3}\equiv S_n+2(\mathrm{mod}3)$ for $n\ge 2$. It is readily verified that $S_2\equiv S_3\equiv S_4\equiv 2(\mathrm{mod}3)$, so $S_5\equiv S_6\equiv S_7\equiv 1(\mathrm{mod}3)$ and $S_8\equiv S_9\equiv S_{10}\equiv 0(\mathrm{mod}3)$, and it follows that $S_n$ is divisible by 3 if and only if $n\equiv 0\text{ or }\pm 1(\mathrm{mod}9)$. Thus the sum of the 10 least values of $n$ that satisfy the required condition is $$8+9+10+17+18+19+26+27+28+35=197.$$

The sum of the products $jk$ as $j$ and $k$ run independently from 1 to $n$ is $$(1+2+\cdots +n{)}^2={\left(\frac{n(n+1)}{2}\right)}^2.$$ To eliminate the cases in which $j=k$, subtract $$1^2+2^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}.$$ Thus $$\sum _{\begin{array}{c}1\le j\le n\\ 1\le k\le n\\ j\ne k\end{array}}jk=\frac{n^2(n+1{)}^2}{4}-\frac{n(n+1)(2n+1)}{6}=\frac{(n-1)n(n+1)(3n+2)}{12}.$$ For a given pair $j,k$ with $j\ne k$ either $j<k$ or $j>k$, but their product is the same in either order. To impose the condition $j<k$, it suffices to divide by 2. Thus $$S_n=\frac{(n-1)n(n+1)(3n+2)}{24}.$$ There is one factor of 3 in the denominator. For any $n$, exactly one of $n+1,n,n-1$ is divisible by 3, and $3n+2$ is not divisible by 3. In order that $S_n$ be divisible by 3 it is necessary and sufficient that the factor that is divisible by 3 should in fact be divisible by 9. That is, $n\equiv 0\text{ or }\pm 1(\mathrm{mod}9)$, and the answer can be calculated as above.