Difference between revisions of "2023 AMC 8 Problems/Problem 1"
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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 24</math> | <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 24</math> | ||
| − | ==Solution== | + | ==Solution (PEMDAS)== |
By the order of operations, we have <cmath>(8 \times 4 + 2) - (8 + 4 \times 2) = (32+2) - (8+8) = 34 - 16 = \boxed{\textbf{(D)}\ 18}.</cmath> | By the order of operations, we have <cmath>(8 \times 4 + 2) - (8 + 4 \times 2) = (32+2) - (8+8) = 34 - 16 = \boxed{\textbf{(D)}\ 18}.</cmath> | ||
Revision as of 02:15, 25 January 2023
Problem 1
What is the value of 
?
Solution (PEMDAS)
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, peelybonehead, MRENTHUSIASM
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. 
