2023 AMC 8 Problems/Problem 6
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Video Solution by Math-X (Smart and Simple)
- 8 Video Solution (HOW TO CREATIVELY THINK!!!)
- 9 Video Solution by Magic Square
- 10 Video Solution by SpreadTheMathLove
- 11 Video Solution by Interstigation
- 12 Video Solution by WhyMath
- 13 Video Solution by harungurcan
- 14 See Also
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 is greater than and Therefore, the answer is
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
Solution 3
Trying all distinct orderings, we see that the only possible values are and the greatest of which is
~A_MatheMagician
Solution 4
There are two 2’s and one 3. To make use of them all, use 2 and 3 as the bases and 2 and 0 as the exponents.
~spacepandamath13
Solution 5
If 0 is a base, then the whole expression becomes 0. Therefore, 0 must be an exponent. But since ==, we do not want to waste our 3 on the 0. We will have a 3 and a 2 left, and since is greater than , we get =.
~DrDominic
Video Solution by Math-X (Smart and Simple)
https://youtu.be/Ku_c1YHnLt0?si=TGWfKGTgBNwzH3xf&t=763 ~Math-X
Video Solution (HOW TO CREATIVELY THINK!!!)
~Education the Study of everything
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=5247
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=EcrktBc8zrM
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=439
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=35BW7bsm_Cg&t=679s
~harungurcan
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.