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

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== Solution 4 ==
 
== Solution 4 ==
The minimum term is <math>1 + 1 = 2</math>, and the maximum term is <math>100 + 100 = 200</math>. The average of the <math>100 \cdot 100 = 10{,}000</math> terms is the average of the minimum and maximum terms, which is <math>\frac{2+200}{2}=101</math>. The sum is therefore <math>101 \cdot 10{,}000 = \boxed{\textbf{(E) }1{,}010{,}000}</math>.
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The sum contains <math>100\cdot100=10000</math> terms, and the average value of both <math>i</math> and <math>j</math> is <math>\frac{101}{2}.</math> Therefore, the sum becomes <cmath>10000\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
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~Rejas ~MRENTHUSIASM
  
 
==See Also==
 
==See Also==

Revision as of 13:33, 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 $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ Since the nested summation is symmetric with respect to $i$ and $j,$ it follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\ &= 2\sum^{100}_{i=1} \sum^{100}_{i=1} i \\ &= 2\sum^{100}_{i=1} 5050 \\ &= 2\cdot(5050\cdot100) \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~RandomPieKevin ‎~MRENTHUSIASM

Solution 4

The sum contains $100\cdot100=10000$ terms, and the average value of both $i$ and $j$ is $\frac{101}{2}.$ Therefore, the sum becomes \[10000\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.\] ~Rejas ~MRENTHUSIASM

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 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

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