Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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| + | ==Problem== | ||
| + | |||
Find the value of the expression | Find the value of the expression | ||
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math> | <cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math> | ||
| − | ==Solution== | + | ==Solutions== |
| + | ===Solution 1=== | ||
We can group each subtracting pair together: | We can group each subtracting pair together: | ||
<cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath> | <cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath> | ||
After subtracting, we have: | After subtracting, we have: | ||
<cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath> | <cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath> | ||
| − | There are <math>50</math> even numbers, therefore there are <math>50 | + | There are <math>50</math> even numbers, therefore there are <math>\dfrac{50}{2}=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math> |
| + | |||
| + | ===Solution 2=== | ||
| + | Since our list does not end with one, we divide every number by 2 and we end up with | ||
| + | <cmath>50-49+48-47+ \ldots +4-3+2-1</cmath> | ||
| + | We can group each subtracting pair together: | ||
| + | <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> | ||
| + | |||
| + | ==Video Solution== | ||
| + | |||
| + | https://youtu.be/iuUwextm334?si=LrUYCG3Cvo0zKqym | ||
| + | |||
| + | A solution so simple a 12-year-old made it! | ||
| + | |||
| + | ~Elijahman~ | ||
| + | |||
| + | ==Video Solution by savannahsolver== | ||
| + | https://youtu.be/TwIwA0XkzoI | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/51K3uCzntWs?t=645 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==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}} | ||
Latest revision as of 10:49, 20 June 2024
Contents
Problem
Find the value of the expression
![\[100-98+96-94+92-90+\cdots+8-6+4-2.\]](http://latex.artofproblemsolving.com/0/2/7/027ea184e9e3fde749ba37eb702069fa8e201387.png)
Solutions
Solution 1
We can group each subtracting pair together:
![\[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\]](http://latex.artofproblemsolving.com/c/f/f/cff0f3dd48ced41da7d2575921b5c0bf5f526e04.png)
After subtracting, we have:
![\[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\]](http://latex.artofproblemsolving.com/1/f/9/1f98cf231e1203e5b694c0eeb37a06c9a99763f5.png)
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
![\[50-49+48-47+ \ldots +4-3+2-1\]](http://latex.artofproblemsolving.com/d/4/2/d42ae896c5dc414cbf054c2400e60d91abdeffd0.png)
We can group each subtracting pair together:
![\[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\]](http://latex.artofproblemsolving.com/7/a/d/7ad418f42d4e4cb58369b973f60676ac04efabcc.png)
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
https://youtu.be/iuUwextm334?si=LrUYCG3Cvo0zKqym
A solution so simple a 12-year-old made it!
~Elijahman~
Video Solution by savannahsolver
~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. 
