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

Revision as of 20:37, 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))

Solution 3 (Bash)

Trying all $24$ possible 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

@@ Line 47: / Line 47: @@
 ~apex304 (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209, [[User:ILoveMath31415926535|ILoveMath31415926535]] (editing))
+==Solution 3 (Bash)==
+Trying all <math>24</math> possible orderings, we see that the only possible values are <math>0</math>, <math>4</math>, <math>8</math>, and <math>9</math>, the greatest of which is <math>\boxed{\textbf{(C)}\ 9}</math>.
+~A_MatheMagician
 ==Video Solution by Magic Square==

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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