Difference between revisions of "2018 AMC 12B Problems/Problem 9"
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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 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{ | + | Thus, we have: <cmath>2\left(\dfrac{100\cdot101}{2}\cdot 100\right).</cmath> |
This gives us: <cmath>\boxed{\textbf{(E) } 1010000}.</cmath> | This gives us: <cmath>\boxed{\textbf{(E) } 1010000}.</cmath> | ||
Revision as of 02:06, 21 February 2018
Contents
[hide]Problem
What is
![\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]](http://latex.artofproblemsolving.com/8/e/3/8e3d81a14957607d271691b35dd0ebb9fcf8557e.png)
Solution 1
We can start by writing out the first couple of terms:
![\[(1+1) + (1+2) + (1+3) + \dots + (1+100)\]](http://latex.artofproblemsolving.com/b/9/7/b97282365fa09092f8e4b20b29123a7422dc047f.png)
![\[(2+1) + (2+2) + (2+3) + \dots + (2+100)\]](http://latex.artofproblemsolving.com/7/e/c/7ec83b6824059e453146725ccc678f9f78f6b150.png)
![\[(3+1) + (3+2) + (3+3) + \dots + (3+100)\]](http://latex.artofproblemsolving.com/b/5/7/b57eb851bed4faf278c525b5d7105754f781d49f.png)
![\[\vdots\]](http://latex.artofproblemsolving.com/7/6/4/76403adfdd35fb846675aaeaac92daec680a4fba.png)
![\[(100+1) + (100+2) + (100+3) + \dots + (100+100)\]](http://latex.artofproblemsolving.com/1/f/f/1ff9a42d0412c24e7bbd16ad754e71fd7dd35f47.png)
Looking at the second terms in the parentheses, we can see that 
occurs 
times. It goes horizontally and exists times vertically.
Looking at the first terms in the parentheses, we can see that occurs times. It goes vertically and exists times horizontally.
Thus, we have: ![\[2\left(\dfrac{100\cdot101}{2}\cdot 100\right).\]](http://latex.artofproblemsolving.com/6/f/a/6fa6208e316a3026751c1d177c6109103f86fbef.png)
This gives us: ![\[\boxed{\textbf{(E) } 1010000}.\]](http://latex.artofproblemsolving.com/0/5/9/0595e4cc96c6dcc798264e3b99d8a6b86b8b50d9.png)
Solution 2
![\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} 100i+5050 = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{1,010,000}\]](http://latex.artofproblemsolving.com/9/d/a/9da47c708900710786c7a8e297c6578f8365368c.png)
Solution 3
![\[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = (100)*(5050*2) = \boxed{1,010,000}\]](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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
