Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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<cmath>(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).</cmath> | <cmath>(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).</cmath> | ||
There are now <math>25</math> pairs of numbers, and the value of each pair is <math>1</math>. This sum is <math>25</math>. However, we divided by <math>2</math> originally so we will multiply <math>2*25</math> to get the final answer of <math>\boxed{\textbf{(C) }50}</math> | There are now <math>25</math> pairs of numbers, and the value of each pair is <math>1</math>. This sum is <math>25</math>. However, we divided by <math>2</math> originally so we will multiply <math>2*25</math> to get the final answer of <math>\boxed{\textbf{(C) }50}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/TwIwA0XkzoI | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2016|num-b=7|num-a=9}} | {{AMC8 box|year=2016|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:55, 18 March 2022
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
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.