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
Problem
What is
Solution 1
We can start by writing out the first couple of terms:
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:
This gives us:
Solution 2
Solution 3
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 |
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