2023 AMC 8 Problems/Problem 6

Revision as of 15:32, 25 January 2023 by Mrthinker (talk | contribs) (Solution 2)

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 $3^{2}*2^{0}$=$9*1$=$9$. (Don't want $0^{2}$ because that is $0$) It is going to be $\boxed{\textbf{(C)}\ 9}$

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

Video Solution by Magic Square

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

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
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

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