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2023 AMC 8 Problems/Problem 6: Difference between revisions

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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
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?
possible value of the expression?
[[File:2023 AMC 8-6.png|200px|thumb|center]]
 
<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>


<math>\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18</math>
<math>\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18</math>

Revision as of 15:16, 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?

[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 $3^2\cdot2^0=9$ 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 (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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