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

m (Solution 4)
 
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\textbf{(B) }500,500\qquad
 
\textbf{(B) }500,500\qquad
 
\textbf{(C) }505,000 \qquad
 
\textbf{(C) }505,000 \qquad
\textbf{(D) }1,001,00 \qquad
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\textbf{(D) }1,001,000 \qquad
 
\textbf{(E) }1,010,000 \qquad </math>
 
\textbf{(E) }1,010,000 \qquad </math>
  
== Solution #1 ==
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== Solution 1 ==
  
<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> (Vfire)
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We can start by writing out the first couple of terms:
  
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<cmath>(1+1) + (1+2) + (1+3) + \dots + (1+100)</cmath>
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<cmath>(2+1) + (2+2) + (2+3) + \dots + (2+100)</cmath>
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<cmath>(3+1) + (3+2) + (3+3) + \dots + (3+100)</cmath>
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<cmath>\vdots</cmath>
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<cmath>(100+1) + (100+2) + (100+3) + \dots + (100+100)</cmath>
  
== Solution #2 ==
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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.
 +
 
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Thus, we have: <cmath>2\left(\dfrac{100\cdot101}{2}\cdot 100\right).</cmath>
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This gives us: <cmath>\boxed{\textbf{(E) } 1010000}.</cmath>
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== Solution 2 ==
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<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>
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== Solution 3 ==
  
 
<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)*(5050*2) =  \boxed{1,010,000}  </cmath>
  
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== Solution 4 ==
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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>
  
 
==See Also==
 
==See Also==

Latest revision as of 09:21, 4 February 2019

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:

\[(1+1) + (1+2) + (1+3) + \dots + (1+100)\] \[(2+1) + (2+2) + (2+3) + \dots + (2+100)\] \[(3+1) + (3+2) + (3+3) + \dots + (3+100)\] \[\vdots\] \[(100+1) + (100+2) + (100+3) + \dots + (100+100)\]

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. 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.

Thus, we have: \[2\left(\dfrac{100\cdot101}{2}\cdot 100\right).\]

This gives us: \[\boxed{\textbf{(E) } 1010000}.\]

Solution 2

\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} (100i+5050) = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{1,010,000}\]

Solution 3

\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i =  (100)*(5050*2) =  \boxed{1,010,000}\]

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 $101 \cdot 10,000 = \boxed{1,010,000}$

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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