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Difference between revisions of "2024 SSMO Speed Round Problems/Problem 9"

(Created page with "==Problem== Let <math>a, b, c,</math> and <math>d</math> be positive integers such that <math>abcd = a+b+c+d</math>. Find the maximum possible value of <math>a</math>. ==Sol...")
 
 
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Let <math>a, b, c,</math> and <math>d</math> be positive integers such that <math>abcd = a+b+c+d</math>. Find the maximum possible value of <math>a</math>.
 
Let <math>a, b, c,</math> and <math>d</math> be positive integers such that <math>abcd = a+b+c+d</math>. Find the maximum possible value of <math>a</math>.
  
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==Solution==
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==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 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>
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~SMO_Team
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==Solution 2==
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We claim that the maximum possible value of <math>a</math> is <math>\boxed{4}</math>. This is attainable when <math>(a,b,c,d)=(4,2,1,1)</math>. We now show that it is impossible to have <math>a\ge 5</math>.
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Suppose for the sake of contradiction that <math>a\ge 5</math>. WLOG <math>b = \max\{b,c,d\}</math>. We have <math>a = \tfrac{b+c+d}{bcd-1}</math>, so <math>b+c+d \ge 5bcd - 5</math>. Note that <math>3b \ge b+c+d</math> and <math>5bcd-5 \ge 5b-5</math>, so <math>3b \ge 5b-5</math>, or <math>5 \ge 2b</math>. This implies either <math>b=1</math> or <math>b=2</math>. If <math>b=1</math>, then <math>c=d=1</math>, which is absurd because <math>a\cdot 1\cdot 1\cdot 1 \ne a+1+1+1</math>. Hence, <math>b=2</math>, and <math>c+d+2\ge 10cd-5</math>, or <math>c+d+7 \ge 10cd</math>. Dividing both sides by <math>cd</math> gives <math>\tfrac{1}{d}+\tfrac{1}{c} + \tfrac{7}{cd} \ge 10</math>. However, since <math>c</math>, <math>d</math> and <math>cd</math> are all at least <math>1</math>, the left hand side of this inequality is at most <math>9</math>, which is a contradiction. Hence, it is impossible to have <math>a\ge 5</math>, as desired.
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~Sedro

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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