Difference between revisions of "2023 AMC 8 Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
− | First, let us consider the | + | First, let us consider the case where <math>0</math> is a base: This would result in the entire expression being <math>0.</math> Contrastingly, if <math>0</math> is an exponent, we will get a value greater than <math>0.</math> As <math>3^2\times2^0=9</math> is greater than <math>2^3\times2^0=8</math> and <math>2^2\times3^0=4,</math> the answer is <math>\boxed{\textbf{(C) }9}.</math> |
~MathFun1000 | ~MathFun1000 |
Revision as of 00:16, 27 January 2023
Contents
Problem
The digits and
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 case where is a base: This would result in the entire expression being
Contrastingly, 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
because that is
) It is going to be
~apex304, lpieleanu, (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209, ILoveMath31415926535 (editing))
Solution 3
Trying all distinct orderings, we see that the only possible values are
and
the greatest of which is
~A_MatheMagician
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=5247
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=EcrktBc8zrM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.