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) |
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(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 19:20, 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
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.