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Difference between revisions of "2023 AMC 8 Problems/Problem 6"
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==Problem==
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The digits 2, 0, 2, and 3 are placed in the expression below, one digit per box. What is the maximum
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possible value of the expression?
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[[File:2023 AMC 8-6.png|200px|thumb|center]]
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<math>\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18</math>
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==Solution 1==
 
==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\cdot2^0=9</math> is greater than <math>2^3\cdot2^0=8</math> and <math>2^2\cdot3^0=4</math>, the answer is <math>\boxed{\textbf{(C) }9}</math>.
 
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\cdot2^0=9</math> is greater than <math>2^3\cdot2^0=8</math> and <math>2^2\cdot3^0=4</math>, the answer is <math>\boxed{\textbf{(C) }9}</math>.

Revision as of 00:24, 25 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?

2023 AMC 8-6.png

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

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 $3^2\cdot2^0=9$ is greater than $2^3\cdot2^0=8$ and $2^2\cdot3^0=4$, 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 $3^{2}*0^{2}$=$9*1$=$9$. (Don't want $2^{0}$ cause that's $0$) It is going to be $\boxed{\text{(C)}9}$

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

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