Difference between revisions of "2018 AMC 12B Problems/Problem 9"
(Created page with "== Problem == What is <cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath> <math> \textbf{(A) }100,100 \qquad \textbf{(B) }500,500\qquad \textbf{(C) }505,000 \qquad \t...") |
|||
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 #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> | ||
Revision as of 14:56, 16 February 2018
Contents
Problem
What is
Solution #1
(Vfire)
Solution #2
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.