2016 AMC 8 Problems/Problem 8
Contents
Problem
Find the value of the expression
Solutions
Solution 1
We can group each subtracting pair together: After subtracting, we have: There are even numbers, therefore there are even pairs. Therefore the sum is
Solution 2
Since our list does not end with one, we divide every number by 2 and we end up with We can group each subtracting pair together: There are now pairs of numbers, and the value of each pair is . This sum is . However, we divided by originally so we will multiply to get the final answer of
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=645
~ pi_is_3.14
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.