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

m
Line 8: Line 8:
 
What is the value of <math>(5 \, \blacklozenge \, 3) \, \bigstar \, 6?</math>
 
What is the value 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>
  
 
==Solution==
 
==Solution==
  
We get that <math>(5 \, \blacklozenge \, 3) = 5^2 - 3^2 = 16</math>, so <cmath>(5 \, \blacklozenge \, 3) \, \bigstar \, 6 = 16 \, \bigstar \, 6 = (16 - 6)^2 = 10^2 = \boxed{\textbf{(D) }100}.</cmath>  
+
We get that <math>(5 \, \blacklozenge \, 3) = 5^2 - 3^2 = 16</math>, so <cmath>(5 \, \blacklozenge \, 3) \, \bigstar \, 6 = 16 \, \bigstar \, 6 = (16 - 6)^2 = 10^2 = \boxed{\textbf{(D) } 100}.</cmath>  
  
 
<i>pog</i>
 
<i>pog</i>

Revision as of 10:13, 28 January 2022

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 value 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 get that $(5 \, \blacklozenge \, 3) = 5^2 - 3^2 = 16$, so \[(5 \, \blacklozenge \, 3) \, \bigstar \, 6 = 16 \, \bigstar \, 6 = (16 - 6)^2 = 10^2 = \boxed{\textbf{(D) } 100}.\]

pog

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png