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Difference between revisions of "2024 SSMO Speed Round Problems/Problem 9"
 
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==Solution 1==
 
==Solution 1==
  
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Note that <math>(a,b,c,d) = (4,2,1,1)</math> satisfies the equation. Now, assume for the sake of contradiction that <math>a\geq5</math> has a solution. We have <cmath>a(b+1)cd-a-(b+1)-c-d> abcd-a-b-c-d\implies</cmath><cmath>acd>1,</cmath> clearly true for <math>a\geq5</math> and <math>c,d,\geq1.</math> Clearly, we <math>b=c=d=1</math> doesn't satisfy the equation. So, the smallest value of <math>abcd-a-b-c-d</math> for <math>a\geq5</math> occurs when <math>b=2,c=d=1.</math> This gives <math>abcd-a-b-c-d = a-4\geq1,</math> for all <math>a\geq5,</math> a contradiction. In conclusion, the maximum possible value of <math>a</math> is <math>\boxed{4}.</math>
+
Note that <math>(a,b,c,d) = (4,2,1,1)</math> satisfies the equation. Now, assume for the sake of contradiction that <math>a\geq5</math> has a solution. We have <cmath>a(b+1)cd-a-(b+1)-c-d> abcd-a-b-c-d\implies acd>1,</cmath> clearly true for <math>a\geq5</math> and <math>c,d,\geq1.</math> Clearly, we <math>b=c=d=1</math> doesn't satisfy the equation. So, the smallest value of <math>abcd-a-b-c-d</math> for <math>a\geq5</math> occurs when <math>b=2,c=d=1.</math> This gives <math>abcd-a-b-c-d = a-4\geq1,</math> for all <math>a\geq5,</math> a contradiction. In conclusion, the maximum possible value of <math>a</math> is <math>\boxed{4}.</math>
  
 
~SMO_Team
 
~SMO_Team

Latest revision as of 14:29, 10 September 2025

Problem

Let $a, b, c,$ and $d$ be positive integers such that $abcd = a+b+c+d$. Find the maximum possible value of $a$.

Solution 1

Note that $(a,b,c,d) = (4,2,1,1)$ satisfies the equation. Now, assume for the sake of contradiction that $a\geq5$ has a solution. We have \[a(b+1)cd-a-(b+1)-c-d> abcd-a-b-c-d\implies acd>1,\] clearly true for $a\geq5$ and $c,d,\geq1.$ Clearly, we $b=c=d=1$ doesn't satisfy the equation. So, the smallest value of $abcd-a-b-c-d$ for $a\geq5$ occurs when $b=2,c=d=1.$ This gives $abcd-a-b-c-d = a-4\geq1,$ for all $a\geq5,$ a contradiction. In conclusion, the maximum possible value of $a$ is $\boxed{4}.$

~SMO_Team

Solution 2

We claim that the maximum possible value of $a$ is $\boxed{4}$. This is attainable when $(a,b,c,d)=(4,2,1,1)$. We now show that it is impossible to have $a\ge 5$.

Suppose for the sake of contradiction that $a\ge 5$. WLOG $b = \max\{b,c,d\}$. We have $a = \tfrac{b+c+d}{bcd-1}$, so $b+c+d \ge 5bcd - 5$. Note that $3b \ge b+c+d$ and $5bcd-5 \ge 5b-5$, so $3b \ge 5b-5$, or $5 \ge 2b$. This implies either $b=1$ or $b=2$. If $b=1$, then $c=d=1$, which is absurd because $a\cdot 1\cdot 1\cdot 1 \ne a+1+1+1$. Hence, $b=2$, and $c+d+2\ge 10cd-5$, or $c+d+7 \ge 10cd$. Dividing both sides by $cd$ gives $\tfrac{1}{d}+\tfrac{1}{c} + \tfrac{7}{cd} \ge 10$. However, since $c$, $d$ and $cd$ are all at least $1$, the left hand side of this inequality is at most $9$, which is a contradiction. Hence, it is impossible to have $a\ge 5$, as desired.

~Sedro

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