Difference between revisions of "2022 AMC 10B Problems/Problem 1"
MRENTHUSIASM (talk | contribs) (Combined Sols 1 and 2. Now the solution has more steps.) |
MRENTHUSIASM (talk | contribs) (→Solution) |
||
| Line 9: | Line 9: | ||
(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3) &= |1-|2-3|| - ||1-2|-3| \\ | (1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3) &= |1-|2-3|| - ||1-2|-3| \\ | ||
&= |1-1| - |1-3| \\ | &= |1-1| - |1-3| \\ | ||
| − | &= 0-2 | + | &= 0-2 \\ |
&=\boxed{\textbf{(A)}\ {-}2}. | &=\boxed{\textbf{(A)}\ {-}2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 18:20, 18 November 2022
- The following problem is from both the 2022 AMC 10B #1 and 2022 AMC 12B #1, so both problems redirect to this page.
Problem
Define 
to be 
for all real numbers 
and 
What is the value of ![\[(1\diamondsuit(2\diamondsuit3))-((1\diamondsuit2)\diamondsuit3)?\]](http://latex.artofproblemsolving.com/e/4/8/e486594a9c3be11fd5275aa45b53a4aecf616093.png)
Solution
We have 
~MRENTHUSIASM
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. 
