Difference between revisions of "2018 AMC 12B Problems/Problem 9"
MRENTHUSIASM (talk | contribs) (Reformatted answer choices and writing in this page.) |
MRENTHUSIASM (talk | contribs) (Improved Sol 2 significantly.) |
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== Solution 2 == | == Solution 2 == | ||
− | + | Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that | |
− | <cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) | + | <cmath>\begin{align*} |
+ | \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ | ||
+ | &= \sum^{100}_{i=1} (100i+5050) \\ | ||
+ | &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ | ||
+ | &= 100\cdot5050+5050\cdot100 \\ | ||
+ | &= \boxed{\textbf{(E) }1{,}010{,}000}. | ||
+ | \end{align*}</cmath> | ||
+ | ~Vfire ~MRENTHUSIASM | ||
== Solution 3 == | == Solution 3 == | ||
− | + | Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that | |
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = 100\cdot(5050\cdot2) = \boxed{\textbf{(E) }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\cdot(5050\cdot2) = \boxed{\textbf{(E) }1{,}010{,}000}.</cmath> | ||
Revision as of 13:55, 20 September 2021
Problem
What is
Solution 1
We can start by writing out the first couple of terms: Looking at the first terms in the parentheses, we can see that occurs times. It goes vertically and exists times horizontally. Looking at the second terms in the parentheses, we can see that occurs times. It goes horizontally and exists times vertically.
Thus, we have
Solution 2
Recall that the sum of the first positive integers is It follows that ~Vfire ~MRENTHUSIASM
Solution 3
Recall that the sum of the first positive integers is It follows that
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.