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

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==Solution==
 
==Solution==
We have
+
We can substitute <math>5</math>, <math>3</math>, and <math>6</math> into the function definitions:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
(5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\
 
(5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\
 +
(5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(25-9\right) \, \bigstar \, 6 \\
 
&= 16 \, \bigstar \, 6 \\
 
&= 16 \, \bigstar \, 6 \\
 
&= (16-6)^2 \\
 
&= (16-6)^2 \\
 
&= \boxed{\textbf{(D) } 100}.
 
&= \boxed{\textbf{(D) } 100}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
<i>~pog</i>
+
<i>~pog</i> ~MathFun1000 (Minor Edits)
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:06, 16 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 can substitute $5$, $3$, and $6$ into the function definitions: \begin{align*} (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\ (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(25-9\right) \, \bigstar \, 6 \\ &= 16 \, \bigstar \, 6 \\ &= (16-6)^2 \\ &= \boxed{\textbf{(D) } 100}. \end{align*} ~pog ~MathFun1000 (Minor Edits)

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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