Difference between revisions of "2023 AMC 8 Problems/Problem 6"
Megaboy6679 (talk | contribs) m (→Solution 1) |
m (→Solution 2) |
||
Line 43: | Line 43: | ||
==Solution 2== | ==Solution 2== | ||
− | The maximum possible value of using the | + | The maximum possible value of using the digits <math>2,0,2,</math> and <math>3.</math>: We can maximize our value by keeping the <math>3</math> and <math>2</math> together in one power (the biggest with the biggest and the smallest with the smallest). This shows <math>3^{2}\times2^{0}=9\times1=9.</math> (We don't want <math>0^{2}</math> because that is <math>0</math>.) It is going to be <math>\boxed{\textbf{(C)}\ 9}.</math> |
~apex304, lpieleanu, (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209, [[User:ILoveMath31415926535|ILoveMath31415926535]] (editing)) | ~apex304, lpieleanu, (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209, [[User:ILoveMath31415926535|ILoveMath31415926535]] (editing)) |
Revision as of 02:01, 5 February 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 digits and : We can maximize our value by keeping the and together in one power (the biggest with the biggest and the smallest with the smallest). This shows (We 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.