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Difference between revisions of "2023 AMC 8 Problems/Problem 6"
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The maximum possible value of using the digit <math>2,0,2,3</math> is going to be <math>\boxed{\text{(C)}9}</math> because we can maximize our value by keeping the <math>3</math> and <math>2</math> together in one power.
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The maximum possible value of using the digit <math>2,0,2,3</math>. We can maximize our value by keeping the <math>3</math> and <math>2</math> together in one power. (Biggest with biggest and smallest with smallest) This shows <math>3^{2}*0^{2}</math>=<math>9*1</math>=<math>9</math>. (Don't want <math>2^{0}</math> cause that's <math>0</math>!) It is going to be <math>\boxed{\text{(C)}9}</math>
  
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing)

Revision as of 21:30, 24 January 2023

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 $3^{2}*0^{2}$=$9*1$=$9$. (Don't want $2^{0}$ cause that's $0$!) It is going to be $\boxed{\text{(C)}9}$

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing)

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