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

Revision as of 19:00, 26 February 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 case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ As $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4,$ the answer is $\boxed{\textbf{(C) }9}.$

~MathFun1000

Solution 2

The maximum possible value of using the digits $2,0,2,$ and $3$ : We can maximize our value by keeping the $3$ and $2$ together in one power (the biggest with the biggest and the smallest with the smallest). This shows $3^{2}\times2^{0}=9\times1=9.$ (We 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

Trying all $12$ distinct 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

Solution 4

There are two 2’s and one 3. To make use of them all, use 2 and 3 as the bases and 2 and 0 as the exponents.

~spacepandamath13

(Creative Thinking) Video Solution

https://youtu.be/HW6TUhQTj0o

~Education the Study of everything

Video Solution by Magic Square

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=EcrktBc8zrM

Video Solution by Interstigation

https://youtu.be/1bA7fD7Lg54?t=331

@@ Line 54: / Line 54: @@
 There are two 2’s and one 3.  To make use of them all, use 2 and 3 as the bases and 2 and 0 as the exponents.
-~spacepandamth13
+~spacepandamath13
 ==(Creative Thinking) Video Solution==

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

Page

Toolbox

Search