2023 AMC 8 Problems/Problem 6
Contents
[hide]Problem
The digits 2, 0, 2, and 3 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 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 because that is ) It is going to be
~apex304 (SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, stevens0209, ILoveMath31415926535 (editing))
Solution 3 (Bash)
Trying all possible 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
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.