Difference between revisions of "2023 SSMO Team Round Problems/Problem 1"
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| − | We can assume because of the symmetry that <math>a=0</math>. Then, the problem is reduced to <math>1+b^c+c^d</math>. Since there are only <math>3</math> possible permutations for <math>b</math>, <math>c</math>, and <math>d</math>, we can try them and find that the maximum possible value is obtained when <math>b=2</math>, <math>c=3</math>, and <math>d=2</math>. Therefore, the answer is <math>1+2^3+3^2=\boxed{18}</math> | + | We can assume because of the symmetry that <math>a=0</math>. Then, the problem is reduced to <math>1+b^c+c^d</math>. Since there are only <math>3</math> possible permutations for <math>b</math>, <math>c</math>, and <math>d</math>, we can try them and find that the maximum possible value is obtained when <math>b=2</math>, <math>c=3</math>, and <math>d=2</math>. Therefore, the answer is <math>1+2^3+3^2=\boxed{18}</math>. |
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| + | ~alexanderruan | ||
Latest revision as of 01:11, 3 January 2024
Problem
Let 
be a permutation of 
. Find the largest possible value of 
Solution
We can assume because of the symmetry that 
. Then, the problem is reduced to 
. Since there are only 
possible permutations for 
, 
, and 
, we can try them and find that the maximum possible value is obtained when 
, 
, and 
. Therefore, the answer is 
.
~alexanderruan
