Difference between revisions of "2022 AMC 8 Problems/Problem 2"
(Created page with "==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 <math>(5 \,...") |
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| Line 2: | Line 2: | ||
Consider these two operations: | Consider these two operations: | ||
| − | \begin{align*} | + | <cmath>\begin{align*} |
a \, \blacklozenge \, b &= a^2 - b^2\\ | a \, \blacklozenge \, b &= a^2 - b^2\\ | ||
a \, \bigstar \, b &= (a - b)^2 | a \, \bigstar \, b &= (a - b)^2 | ||
| − | \end{align*} | + | \end{align*}</cmath> |
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== | ||
| + | |||
| + | 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> | ||
==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}} | ||
Revision as of 10:30, 28 January 2022
Problem
Consider these two operations:
What is the value of 
Solution
We get that 
, so ![\[(5 \, \blacklozenge \, 3) \, \bigstar \, 6 = 16 \, \bigstar \, 6 = (16 - 6)^2 = 10^2 = \boxed{\textbf{(D) }100}.\]](http://latex.artofproblemsolving.com/5/0/1/501ebaa5f45f213f347236883f3bd287d3284e7c.png)
pog
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
