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

 
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&= \boxed{\textbf{(D) } 100}.
 
&= \boxed{\textbf{(D) } 100}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
<i>~pog</i> ~MathFun1000 (Minor Edits)
 
 
\end{align*}<math></math>
 
 
<i>~pog</i> ~MathFun1000 (Minor Edits)
 
<i>~pog</i> ~MathFun1000 (Minor Edits)
  
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<cmath> =((a^2-b^2) \, \bigstar \, c) </cmath>
 
<cmath> =((a^2-b^2) \, \bigstar \, c) </cmath>
 
<cmath> =(a^2-b^2-c)^2 </cmath>
 
<cmath> =(a^2-b^2-c)^2 </cmath>
<cmath> =a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+(c^2) </cmath>
+
<cmath> =a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2 </cmath>
<cmath> =5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+(6^2) </cmath>
+
<cmath> =5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2 </cmath>
 
<cmath> =625+81-225-300-225+108+36 </cmath>
 
<cmath> =625+81-225-300-225+108+36 </cmath>
 
<cmath> =\boxed{\textbf{(D) } 100} </cmath>
 
<cmath> =\boxed{\textbf{(D) } 100} </cmath>
  
~[[megaboy6679]]
+
~megaboy6679
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
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~harungurcan
 
~harungurcan
 +
 +
==Video Solution 7 by Dr. David==
 +
 +
https://youtu.be/ItPIuHdgyNk
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=1|num-a=3}}
 
{{AMC8 box|year=2022|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:47, 19 November 2024

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 1

We can substitute $5$, $3$, and $6$ into the functions' 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)

Solution 2

We can find a general solution to any $((a \, \blacklozenge \, b) \, \bigstar \, c)$. \[((a \, \blacklozenge \, b) \, \bigstar \, c)\] \[=((a^2-b^2) \, \bigstar \, c)\] \[=(a^2-b^2-c)^2\] \[=a^4+b^4-(a^2)(b^2)-2(a^2)(c)-(b^2)(a^2)+2(b^2)(c)+c^2\] \[=5^4+3^4-(5^2)(3^2)-2(5^2)(6)-(3^2)(5^2)+2(3^2)(6)+6^2\] \[=625+81-225-300-225+108+36\] \[=\boxed{\textbf{(D) } 100}\]

~megaboy6679

Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=JlQdlwkdbIYFNJXj&t=123

~Math-X

Video Solution 2 (HOW TO THINK CREATIVELY!!!)

https://youtu.be/ytDV0GNc9Mw

~Education, the Study of Everything

Video Solution 3

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

Video Solution 4

https://youtu.be/YYuEBGoEK1Y

~savannahsolver

Video Solution 5

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

~STEMbreezy

Video Solution 6

https://www.youtube.com/watch?v=c123kPqd11I

~harungurcan

Video Solution 7 by Dr. David

https://youtu.be/ItPIuHdgyNk

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
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Problem 1
Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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