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# Difference between revisions of "2018 AMC 12B Problems/Problem 9"

## Problem

What is $$\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?$$

$\textbf{(A) }100{,}100 \qquad \textbf{(B) }500{,}500\qquad \textbf{(C) }505{,}000 \qquad \textbf{(D) }1{,}001{,}000 \qquad \textbf{(E) }1{,}010{,}000 \qquad$

## Solution 1

We can start by writing out the first couple of terms: $$\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\dots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\dots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\dots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\dots &+ &(100+100) \end{array}$$ Looking at the first terms in the parentheses, we can see that $1+2+3+\dots+100$ occurs $100$ times. It goes vertically and exists $100$ times horizontally. Looking at the second terms in the parentheses, we can see that $1+2+3+\dots+100$ occurs $100$ times. It goes horizontally and exists $100$ times vertically.

Thus, we have $$2\left(\dfrac{100\cdot101}{2}\cdot 100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.$$

## Solution 2

We have $$\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} (100i+5050) = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{\textbf{(E) }1{,}010{,}000}.$$

## Solution 3

We have $$\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = 100\cdot(5050\cdot2) = \boxed{\textbf{(E) }1{,}010{,}000}.$$

## Solution 4

The minimum term is $1 + 1 = 2$, and the maximum term is $100 + 100 = 200$. The average of the $100 \cdot 100 = 10{,}000$ terms is the average of the minimum and maximum terms, which is $\frac{2+200}{2}=101$. The sum is therefore $101 \cdot 10{,}000 = \boxed{\textbf{(E) }1{,}010{,}000}$.

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions