Difference between revisions of "2023 AMC 8 Problems/Problem 1"
MRENTHUSIASM (talk | contribs) (It is important to center key equations, so the original solution is good enough.) |
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~MathFun1000 | ~MathFun1000 | ||
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+ | ==Solution 3== | ||
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+ | We have | ||
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+ | <cmath>(8 \times 4 + 2) - (8 + 4 \times 2)=34-16=\boxed{\textbf{(D)}\ 18}.</cmath> | ||
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+ | ~AliceDubbleYou | ||
==Video Solution (How to Creatively Think!!!) == | ==Video Solution (How to Creatively Think!!!) == |
Revision as of 20:26, 27 October 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution (How to Creatively Think!!!)
- 6 Video Solution by Magic Square
- 7 Video Solution by SpreadTheMathLove
- 8 Video Solution by Interstigation
- 9 Video Solution by WhyMath
- 10 Video Solution by harungurcan
- 11 Video Solution by Math-X (Smart and Simple)
- 12 Video Explanation by MathTalks_Now
- 13 Video Solution by Dr. David
- 14 See Also
Problem
What is the value of ?
Solution 1
By the order of operations, we have ~apex304, TaeKim, MRENTHUSIASM
Solution 2
We can simplify the expression above in another way:
~MathFun1000
Solution 3
We have
~AliceDubbleYou
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.