Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 8 Problems/Problem 6"

(→‎Solution 2)
Line 54: Line 54:
 
==Video Solution by Magic Square==
 
==Video Solution by Magic Square==
 
https://youtu.be/-N46BeEKaCQ?t=5247
 
https://youtu.be/-N46BeEKaCQ?t=5247
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=EcrktBc8zrM
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=5|num-a=7}}
 
{{AMC8 box|year=2023|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:18, 26 January 2023

Problem

The digits $2, 0, 2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?

[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path bsql = J--O; path lsqb = shift((1.2,0.75))*scale(0.5)*bsqb; path lsqr = shift((1.2,0.75))*scale(0.5)*bsqr; path lsqt = shift((1.2,0.75))*scale(0.5)*bsqt; path lsql = shift((1.2,0.75))*scale(0.5)*bsql; draw(bsqb,dashed); draw(bsqr,dashed); draw(bsqt,dashed); draw(bsql,dashed); draw(lsqb,dashed); draw(lsqr,dashed); draw(lsqt,dashed); draw(lsql,dashed); label(scale(3)*"$\times$",(w,1/3)); draw(shift(1.3w,0)*bsqb,dashed); draw(shift(1.3w,0)*bsqr,dashed); draw(shift(1.3w,0)*bsqt,dashed); draw(shift(1.3w,0)*bsql,dashed); draw(shift(1.3w,0)*lsqb,dashed); draw(shift(1.3w,0)*lsqr,dashed); draw(shift(1.3w,0)*lsqt,dashed); draw(shift(1.3w,0)*lsql,dashed); [/asy]

$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$

Solution 1

First, let us consider the cases where $0$ is a base. This would result in the entire expression being $0.$ However, if $0$ is an exponent, we will get a value greater than $0.$ As $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4,$ the answer is $\boxed{\textbf{(C) }9}.$

~MathFun1000

Solution 2

The maximum possible value of using the digit $2,0,2,3.$ We can maximize our value by keeping the $3$ and $2$ together in one power. (Biggest with biggest and smallest with smallest) This shows $3^{2}\times2^{0}=9\times1=9.$ (Don't want $0^{2}$ because that is $0$) It is going to be $\boxed{\textbf{(C)}\ 9}.$

~apex304, lpieleanu, (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209, ILoveMath31415926535 (editing))

Solution 3

Trying all $24$ possible orderings, we see that the only possible values are $0,4,8,$ and $9,$ the greatest of which is $\boxed{\textbf{(C)}\ 9}.$

~A_MatheMagician

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5247

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=EcrktBc8zrM

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_6&oldid=188127"