Difference between revisions of "2023 AMC 8 Problems/Problem 6"
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==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>. | ||
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+ | ~MathFun1000 | ||
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+ | ==Solution 2== | ||
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> | 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 22:55, 24 January 2023
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)