Difference between revisions of "2022 AMC 10B Problems/Problem 1"
MRENTHUSIASM (talk | contribs) m (→Problem) |
MRENTHUSIASM (talk | contribs) |
||
Line 2: | Line 2: | ||
==Problem== | ==Problem== | ||
− | Define <math>x\ | + | Define <math>x\diamond 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\diamond(2\diamond3))-((1\diamond2)\diamond3)?</cmath> |
<math> \textbf{(A)}\ {-}2 \qquad | <math> \textbf{(A)}\ {-}2 \qquad | ||
Line 12: | Line 12: | ||
== Solution == | == Solution == | ||
We have <cmath>\begin{align*} | We have <cmath>\begin{align*} | ||
− | (1\ | + | (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ |
&= |1-1| - |1-3| \\ | &= |1-1| - |1-3| \\ | ||
&= 0-2 \\ | &= 0-2 \\ |
Revision as of 14:56, 6 March 2023
- 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.