2023 AMC 8 Problems/Problem 6: Difference between revisions
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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> | ==Solution 1== | ||
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> | |||
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing) | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209 (editing) | ||
Revision as of 21:30, 24 January 2023
Solution 1
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)
