Difference between revisions of "2018 AMC 12B Problems/Problem 9"
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<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> | ||
| + | == Solution 4 == | ||
| + | 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{200+2}{2}=101</math>. The sum is therefore <math>101 \cdot 10,000 = \boxed{1,010,000}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 10:20, 4 February 2019
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/4/2/1/4212c5c822717d16ad8e7321127a7ba276a0b7b2.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)
Solution 4
The minimum term is 
, and the maximum term is 
. The average of the 
terms is the average of the minimum and maximum terms, which is 
. The sum is therefore 
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. 
