Difference between revisions of "2023 SSMO Team Round Problems/Problem 1"
(→Solution) |
(→Solution) |
||
(One intermediate revision by the same user not shown) | |||
Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
− | 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>. |
+ | |||
+ | ~alexanderruan |
Latest revision as of 02:11, 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