2023 AMC 8 Problems/Problem 6: Difference between revisions

Revision as of 19:31, 4 April 2024

Problem

The digits $2,1,2,$ 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 $0.$ $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4.$ Therefore, the answer is $\boxed{\textbf{(C) }9}.$

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 $\boxed{\textbf{(C)}\ 9}.$

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

Solution 5

If 0 is a base, then the whole expression becomes 0. Therefore, 0 must be an exponent. But since $2^0$ = $3^0$ = $1$ , we do not want to waste our 3 on the 0. We will have a 3 and a 2 left, and since $3^2$ is greater than $2^3$ , we get $3^2\cdot2^0$ = $9$ .

~DrDominic

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=TGWfKGTgBNwzH3xf&t=763 ~Math-X

Video Solution (HOW TO CREATIVELY THINK!!!)

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/DBqko2xATxs&t=439

Video Solution by WhyMath

https://youtu.be/vKdWbtXYgz4

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=35BW7bsm_Cg&t=679s

~harungurcan

@@ Line 1: / Line 1: @@
 ==Problem==
-The digits <math>2, ,1 2,</math> and <math>3</math> are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
+The digits <math>2,1,2,</math> and <math>3</math> are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
 <asy>

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AJHSME/AMC 8 Problems and Solutions

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