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

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==Problem==
 
==Problem==
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The digits <math>2, 0, 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?
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The digits <math>2,0,2,</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>
 
<asy>
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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.
 
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.
  
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~spacepandamath13
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~spacepandamath13 .
  
 
==Solution 5==
 
==Solution 5==
  
 
If 0 is a base, then the whole expression becomes 0. Therefore, 0 must be an exponent. But since <math>2^0</math>=<math>3^0</math>=<math>1</math>, we do not want to waste our 3 on the 0. We will have a 3 and a 2 left, and since <math>3^2</math> is greater than <math>2^3</math>, we get <math>3^2\cdot2^0</math>=<math>9</math>.
 
If 0 is a base, then the whole expression becomes 0. Therefore, 0 must be an exponent. But since <math>2^0</math>=<math>3^0</math>=<math>1</math>, we do not want to waste our 3 on the 0. We will have a 3 and a 2 left, and since <math>3^2</math> is greater than <math>2^3</math>, we get <math>3^2\cdot2^0</math>=<math>9</math>.
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 +
~DrDominic
  
 
==Video Solution by Math-X (Smart and Simple)==
 
==Video Solution by Math-X (Smart and Simple)==
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~harungurcan
 
~harungurcan
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 +
==Video Solution by Dr. David==
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https://youtu.be/5Gw0hMUzp4s
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=5|num-a=7}}
 
{{AMC8 box|year=2023|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:14, 9 November 2024

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 $0.$ $3^2\times2^0=9$ 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

Video Solution by Dr. David

https://youtu.be/5Gw0hMUzp4s

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

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