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

Revision as of 12:55, 20 September 2021

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

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that $\begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ &= \sum^{100}_{i=1} (100i+5050) \\ &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ &= 100\cdot5050+5050\cdot100 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*}$ ~Vfire ~MRENTHUSIASM

Solution 3

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that $\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = 100\cdot(5050\cdot2) = \boxed{\textbf{(E) }1{,}010{,}000}.$

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 $101 \cdot 10{,}000 = \boxed{\textbf{(E) }1{,}010{,}000}$ .

@@ Line 26: / Line 26: @@
 == Solution 2 ==
-We have
+Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that
-<cmath> \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}.</cmath>
+<cmath>\begin{align*}
+\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\
+&= \sum^{100}_{i=1} (100i+5050) \\
+&= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\
+&= 100\cdot5050+5050\cdot100 \\
+&= \boxed{\textbf{(E) }1{,}010{,}000}.
+\end{align*}</cmath>
+~Vfire ~MRENTHUSIASM
 == Solution 3 ==
-We have
+Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that
 <cmath> \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}.</cmath>

2018 AMC 12B (Problems • Answer Key • Resources)
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