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

Revision as of 23:16, 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 case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, 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 $12$ distinct 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

@@ Line 38: / Line 38: @@
 ==Solution 1==
-First, let us consider the cases where <math>0</math> is a base. This would result in the entire expression being <math>0.</math> However, if <math>0</math> is an exponent, we will get a value greater than <math>0.</math> As <math>3^2\times2^0=9</math> is greater than <math>2^3\times2^0=8</math> and <math>2^2\times3^0=4,</math> the answer is <math>\boxed{\textbf{(C) }9}.</math>
+First, let us consider the case where <math>0</math> is a base: This would result in the entire expression being <math>0.</math> Contrastingly, if <math>0</math> is an exponent, we will get a value greater than <math>0.</math> As <math>3^2\times2^0=9</math> is greater than <math>2^3\times2^0=8</math> and <math>2^2\times3^0=4,</math> the answer is <math>\boxed{\textbf{(C) }9}.</math>
 ~MathFun1000

2023 AMC 8 (Problems • Answer Key • Resources)
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