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

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\textbf{(E) }1,010,000 \qquad </math>
 
\textbf{(E) }1,010,000 \qquad </math>
  
== Solution #1 ==
+
== 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)
 
<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)
  
  
== Solution #2 ==
+
== Solution 2 ==
  
 
<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>

Revision as of 16:24, 16 February 2018

Problem

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

$\textbf{(A) }100,100 \qquad \textbf{(B) }500,500\qquad \textbf{(C) }505,000 \qquad \textbf{(D) }1,001,00 \qquad \textbf{(E) }1,010,000 \qquad$

Solution 1

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


Solution 2

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


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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