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Difference between revisions of "2023 SSMO Team Round Problems/Problem 1"

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Let <math>(a, b, c, d)</math> be a permutation of <math>(2, 0, 2, 3)</math>. Find the largest possible value of <math>a^b + b^c + c^d + d^a</math>
 
Let <math>(a, b, c, d)</math> be a permutation of <math>(2, 0, 2, 3)</math>. Find the largest possible value of <math>a^b + b^c + c^d + d^a</math>
  
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==Solution==
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==Solution 1==
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WLOG, assume that <math>a = 0</math>. Therefore, we have <math>(a,b,c,d) = (0,2,2,3),(0,2,3,2),</math> or <math>(0,3,2,2).</math> The value of <math>a^b+b^c+c^d+d^a</math> for these three permutations are <math>13,18,</math> and <math>14,</math> respectively, meaning the greatest possible sum is <math>\boxed{18}.</math>
  
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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>
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~SMO_Team
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==Solution 2==
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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>.
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~alexanderruan

Latest revision as of 21:14, 9 September 2025

Problem

Let $(a, b, c, d)$ be a permutation of $(2, 0, 2, 3)$. Find the largest possible value of $a^b + b^c + c^d + d^a$

Solution 1

WLOG, assume that $a = 0$. Therefore, we have $(a,b,c,d) = (0,2,2,3),(0,2,3,2),$ or $(0,3,2,2).$ The value of $a^b+b^c+c^d+d^a$ for these three permutations are $13,18,$ and $14,$ respectively, meaning the greatest possible sum is $\boxed{18}.$

~SMO_Team

Solution 2

We can assume because of the symmetry that $a=0$. Then, the problem is reduced to $1+b^c+c^d$. Since there are only $3$ possible permutations for $b$, $c$, and $d$, we can try them and find that the maximum possible value is obtained when $b=2$, $c=3$, and $d=2$. Therefore, the answer is $1+2^3+3^2=\boxed{18}$.

~alexanderruan

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