# 2021 Fall AMC 10B Problems/Problem 22

## Problem

For each integer $n\geq 2$, let $S_n$ be the sum of all products $jk$, where $j$ and $k$ are integers and $1\leq j. What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by $3$?

$\textbf{(A)}\ 196\qquad\textbf{(B)}\ 197\qquad\textbf{(C)}\ 198\qquad\textbf{(D)}\ 199\qquad\textbf{(E)}\ 200$

## Solution 1

To get from $S_n$ to $S_{n+1}$, we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\frac{n(n+1)^2}{2}$.

Now, we can look at the different values of $n$ mod $3$. For $n\equiv 0\pmod{3}$ and $n\equiv 2\pmod{3}$, then we have $\frac{n(n+1)^2}{2}\equiv 0\pmod{3}$. However, for $n\equiv 1\pmod{3}$, we have $$\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.$$

Clearly, $S_2\equiv 2\pmod{3}.$ Using the above result, we have $S_5\equiv 1\pmod{3}$, and $S_8$, $S_9$, and $S_{10}$ are all divisible by $3$. After $3\cdot 3=9$, we have $S_{17}$, $S_{18}$, and $S_{19}$ all divisible by $3$, as well as $S_{26}, S_{27}, S_{28}$, and $S_{35}$. Thus, our answer is $8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{\mathrm{(B)}\ 197} .$ -BorealBear

-minor edit by Yiyj1

## Solution 2 (bash)

Since we have a wonky function, we start by trying out some small cases and see what happens. If $j$ is $1$ and $k$ is $2$, then there is one case. We have $2$ mod $3$ for this case. If $N$ is $3$, we have $1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3$ which is still $2$ mod $3$. If $N$ is $4$, we have to add $1 \cdot 4 + 2 \cdot 4 + 3 \cdot 4$ which is a multiple of $3$, meaning that we are still at $2$ mod $3$. If we try a few more cases, we find that when $N$ is $8$, we get a multiple of $3$. When $N$ is $9$, we are adding $0$ mod $3$, and therefore, we are still at a multiple of $3$.

When $N$ is $10$, then we get $0$ mod $3$ + $10(1+2+3+...+9)$ which is $10$ times a multiple of $3$. Therefore, we have another multiple of $3$. When $N$ is $11$, so we have $2$ mod $3$. So, every time we have $-1$ mod $9$, $0$ mod $9$, and $1$ mod $9$, we always have a multiple of $3$. Think about it: When $N$ is $1$, it will have to be $0 \cdot 1$, so it is a multiple of $3$. Therefore, our numbers are $8, 9, 10, 17, 18, 19, 26, 27, 28, 35$. Adding the numbers up, we get $\boxed{\textbf{(B) 197}}$

~Arcticturn

## Solution 3

Denote $A_{n, <} = \left\{ \left( j , k \right) : 1 \leq j < k \leq n \right\}$, $A_{n, >} = \left\{ \left( j , k \right) : 1 \leq k < j \leq n \right\}$ and $A_{n, =} = \left\{ \left( j , k \right) : 1 \leq j = k \leq n \right\}$.

Hence, $\sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk = S_n$.

Therefore, \begin{align*} S_n & = \frac{1}{2} \left( \sum_{\left( j , k \right) \in A_{n,<}} jk = \sum_{\left( j , k \right) \in A_{n,>}} jk \right) \\ & = \frac{1}{2} \left( \sum_{1 \leq j, k \leq n} jk - {\left( j , k \right) \in A_{n,=}} jk \right) \\ & = \frac{1}{2} \left( \sum_{j=1}^n \sum_{k=1}^n jk - \sum_{j=1}^n j^2 \right) \\ & = \frac{1}{2} \left( \frac{n^2 \left( n + 1 \right)^2}{4} - \frac{n \left( n + 1 \right) \left( 2 n + 1 \right) }{6} \right) \\ & = \frac{\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)}{24} . \end{align*}

Hence, $S_n$ is divisible by 3 if and only if $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is divisible by $24 \cdot 3 = 8 \cdot 9$.

First, $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is always divisible by 8. Otherwise, $S_n$ is not even an integer.

Second, we find conditions for $n$, such that $\left( n - 1 \right) n \left( n + 1 \right) \left( 3 n + 2 \right)$ is divisible by 9.

Because $3 n + 2$ is not divisible by 3, it cannot be divisible by 9.

Hence, we need to find conditions for $n$, such that $\left( n - 1 \right) n \left( n + 1 \right)$ is divisible by 9. This holds of $n \equiv 0, \pm 1 \pmod{9}$.

Therefore, the 10 least values of $n$ such that $\left( n - 1 \right) n \left( n + 1 \right)$ is divisible by 9 (equivalently, $S_n$ is divisible by 3) are 8, 9, 10, 17, 18, 19, 26, 27, 28, 35. Their sum is 197.

Therefore, the answer is $\boxed{\textbf{(B) }197}$.

~Steven Chen (www.professorchenedu.com)

~Interstigation

~savannahsolver