2023 AMC 8 Problems/Problem 6: Difference between revisions

Revision as of 08:21, 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?

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

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))

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 ~apex304 (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing))
+==See Also==
+{{AMC8 box|year=2023|num-b=5|num-a=7}}
+{{MAA Notice}}

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2023 AMC 8 Problems/Problem 6: Difference between revisions

Revision as of 08:21, 25 January 2023

Contents

Problem

Solution 1

Solution 2

See Also