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

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<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath>
 
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath>
  
−
<math> \textbf{(A) }100,100 \qquad
+
<math> \textbf{(A) }100{,}100 \qquad
−
\textbf{(B) }500,500\qquad
+
\textbf{(B) }500{,}500\qquad
−
\textbf{(C) }505,000 \qquad
+
\textbf{(C) }505{,}000 \qquad
−
\textbf{(D) }1,001,000 \qquad
+
\textbf{(D) }1{,}001{,}000 \qquad
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\textbf{(E) }1,010,000 \qquad </math>
+
\textbf{(E) }1{,}010{,}000 \qquad </math>
  
 
== Solution 1 ==
 
== Solution 1 ==
  
 
We can start by writing out the first couple of terms:
 
We can start by writing out the first couple of terms:
−
 
+
<cmath>\begin{array}{ccccccccc}
−
<cmath>(1+1) + (1+2) + (1+3) + \dots + (1+100)</cmath>
+
(1+1) &+ &(1+2) &+ &(1+3) &+ &\dots &+ &(1+100) \\
−
<cmath>(2+1) + (2+2) + (2+3) + \dots + (2+100)</cmath>
+
(2+1) &+ &(2+2) &+ &(2+3) &+ &\dots &+ &(2+100) \\
−
<cmath>(3+1) + (3+2) + (3+3) + \dots + (3+100)</cmath>
+
(3+1) &+ &(3+2) &+ &(3+3) &+ &\dots &+ &(3+100) \\ [-1ex]
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<cmath>\vdots</cmath>
+
&&&&\vdots&&&& \\
−
<cmath>(100+1) + (100+2) + (100+3) + \dots + (100+100)</cmath>
+
(100+1) &+ &(100+2) &+ &(100+3) &+ &\dots &+ &(100+100)
−
 
+
\end{array}</cmath>
 +
Looking at the first terms in the parentheses, we can see that <math>1+2+3+\dots+100</math> occurs <math>100</math> times. It goes vertically and exists <math>100</math> times horizontally.
 
Looking at the second terms in the parentheses, we can see that <math>1+2+3+\dots+100</math> occurs <math>100</math> times. It goes horizontally and exists <math>100</math> times vertically.
 
Looking at the second terms in the parentheses, we can see that <math>1+2+3+\dots+100</math> occurs <math>100</math> times. It goes horizontally and exists <math>100</math> times vertically.
−
Looking at the first terms in the parentheses, we can see that <math>1+2+3+\dots+100</math> occurs <math>100</math> times. It goes vertically and exists <math>100</math> times horizontally.
 
  
−
Thus, we have: <cmath>2\left(\dfrac{100\cdot101}{2}\cdot 100\right).</cmath>
+
Thus, we have <cmath>2\left(\dfrac{100\cdot101}{2}\cdot 100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
−
 
 
−
This gives us: <cmath>\boxed{\textbf{(E) } 1010000}.</cmath>
 
  
 
== Solution 2 ==
 
== Solution 2 ==
−
 
+
We have
−
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} (100i+5050) = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{1,010,000} </cmath>
+
<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>
  
 
== Solution 3 ==
 
== Solution 3 ==
−
 
+
We have
−
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i =  (100)*(5050*2) =  \boxed{1,010,000}   </cmath>
+
<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>
  
 
== 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{1,010,000}</math>
+
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>.
  
 
==See Also==
 
==See Also==

Revision as of 11:25, 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

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

See Also

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