Difference between revisions of "2018 AMC 12B Problems/Problem 9"
| Line 10: | Line 10: | ||
\textbf{(E) }1,010,000 \qquad </math> | \textbf{(E) }1,010,000 \qquad </math> | ||
| − | == Solution | + | == 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 | + | == 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 15:24, 16 February 2018
Contents
Problem
What is
![\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]](http://latex.artofproblemsolving.com/8/e/3/8e3d81a14957607d271691b35dd0ebb9fcf8557e.png)
Solution 1
![\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} 100i+5050 = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{1,010,000}\]](http://latex.artofproblemsolving.com/9/d/a/9da47c708900710786c7a8e297c6578f8365368c.png)
(Vfire)
Solution 2
![\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = (100)*(5050*2) = \boxed{1,010,000}\]](http://latex.artofproblemsolving.com/1/e/4/1e44955d85f7ffa19437e93f63b1ed290fcd84df.png)
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
