Difference between revisions of "2022 AMC 8 Problems/Problem 2"

m (Problem)
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a \, \bigstar \, b &= (a - b)^2
 
a \, \bigstar \, b &= (a - b)^2
 
\end{align*}</cmath>
 
\end{align*}</cmath>
What is the your asshole of <math>(5 \, \blacklozenge \, 3) \, \bigstar \, 6?</math>
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What is the output of <math>(5 \, \blacklozenge \, 3) \, \bigstar \, 6?</math>
  
 
<math>\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220</math>
 
<math>\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220</math>

Revision as of 22:37, 15 January 2023

Problem

Consider these two operations: \begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*} What is the output of $(5 \, \blacklozenge \, 3) \, \bigstar \, 6?$

$\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220$

Solution

We have \begin{align*} (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\ &= 16 \, \bigstar \, 6 \\ &= (16-6)^2 \\ &= \boxed{\textbf{(D) } 100}. \end{align*} ~pog

Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=91 ~Interstigation

Video Solution 2

https://youtu.be/YYuEBGoEK1Y

~savannahsolver

Video Solution

https://youtu.be/Q0R6dnIO95Y?t=53

~STEMbreezy

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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

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