2023 AMC 8 Problems/Problem 1: Difference between revisions
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==Video Explanation by MathTalks_Now== | ==Video Explanation by MathTalks_Now== | ||
https://studio.youtube.com/video/PMOeiGLkDH0/edit | https://studio.youtube.com/video/PMOeiGLkDH0/edit | ||
==Video Solution by Dr. David== | |||
https://youtu.be/GgjbZDHbS1s | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|before=First Problem|num-a=2}} | {{AMC8 box|year=2023|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:11, 12 October 2024
Problem
What is the value of 
?
Solution 1
By the order of operations, we have ![\[(8 \times 4 + 2) - (8 + 4 \times 2) = (32+2) - (8+8) = 34 - 16 = \boxed{\textbf{(D)}\ 18}.\]](http://latex.artofproblemsolving.com/c/a/6/ca65b4deda49fe016bb9a5e17aa20b0b81e87a39.png)
~apex304, TaeKim, MRENTHUSIASM
Solution 2
We can simplify the expression above in another way: ![\[(8 \times 4 + 2) - (8 + 4 \times 2)=8\times4+2-8-4\times2=32+2-8-8=34-16=\boxed{\textbf{(D)}\ 18}.\]](http://latex.artofproblemsolving.com/6/2/3/6234432209092a582bac14d392cbfca5ee640b6e.png)
~MathFun1000
Video Solution (How to Creatively Think!!!)
~Education the Study of everything
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=5746
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=EcrktBc8zrM
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=41
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=35BW7bsm_Cg&t=10s
~harungurcan
Video Solution by Math-X (Smart and Simple)
https://youtu.be/Ku_c1YHnLt0?si=RarnomIDE4gELDM3&t=62 ~Math-X
Video Explanation by MathTalks_Now
https://studio.youtube.com/video/PMOeiGLkDH0/edit
Video Solution by Dr. David
See Also
| 2023 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
