2023 AMC 8 Problems/Problem 6: Difference between revisions
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==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? | |||
[[File:2023 AMC 8-6.png|200px|thumb|center]] | |||
<math>\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18</math> | |||
==Solution 1== | ==Solution 1== | ||
First, let us consider the cases where <math>0</math> is a base. This would result in the entire expression being <math>0</math>. However, if <math>0</math> is an exponent, we will get a value greater than <math>0</math>. As <math>3^2\cdot2^0=9</math> is greater than <math>2^3\cdot2^0=8</math> and <math>2^2\cdot3^0=4</math>, the answer is <math>\boxed{\textbf{(C) }9}</math>. | First, let us consider the cases where <math>0</math> is a base. This would result in the entire expression being <math>0</math>. However, if <math>0</math> is an exponent, we will get a value greater than <math>0</math>. As <math>3^2\cdot2^0=9</math> is greater than <math>2^3\cdot2^0=8</math> and <math>2^2\cdot3^0=4</math>, the answer is <math>\boxed{\textbf{(C) }9}</math>. | ||
Revision as of 00:24, 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?
Solution 1
First, let us consider the cases where 
is a base. This would result in the entire expression being . However, if is an exponent, we will get a value greater than . As 
is greater than 
and 
, the answer is 
.
~MathFun1000
Solution 2
The maximum possible value of using the digit 
. We can maximize our value by keeping the 
and 
together in one power. (Biggest with biggest and smallest with smallest) This shows 
=
=
. (Don't want 
cause that's ) It is going to be 
~apex304 (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing))
