Difference between revisions of "2022 AMC 10B Problems/Problem 1"
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==Problem== | ==Problem== | ||
Define <math>x\diamondsuit y</math> to be <math>|x-y|</math> for all real numbers <math>x</math> and <math>y.</math> What is the value of <cmath>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)?</cmath> | Define <math>x\diamondsuit y</math> to be <math>|x-y|</math> for all real numbers <math>x</math> and <math>y.</math> What is the value of <cmath>(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)?</cmath> | ||
− | <math>\textbf{(A)}\ {-}2 \qquad\textbf{(B)}\ {-}1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 2</math> | + | |
+ | <math> \textbf{(A)}\ {-}2 \qquad | ||
+ | \textbf{(B)}\ {-}1 \qquad | ||
+ | \textbf{(C)}\ 0 \qquad | ||
+ | \textbf{(D)}\ 1 \qquad | ||
+ | \textbf{(E)}\ 2</math> | ||
== Solution == | == Solution == |
Revision as of 15:26, 24 December 2022
- The following problem is from both the 2022 AMC 10B #1 and 2022 AMC 12B #1, so both problems redirect to this page.
Contents
Problem
Define to be for all real numbers and What is the value of
Solution
We have ~MRENTHUSIASM
Video Solution 1
~Education, the Study of Everything
See Also
2022 AMC 10B (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 AMC 10 Problems and Solutions |
2022 AMC 12B (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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.