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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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==Solution==
 
==Solution==
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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>10 \le \tfrac{1}{d}+\tfrac{1}{c} + \tfrac{7}{cd} \le 9</math>, which is a contradiction. Hence it is impossible to have <math>a\ge 5</math>, as desired.
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~Sedro

Revision as of 23:56, 3 June 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

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 $10 \le \tfrac{1}{d}+\tfrac{1}{c} + \tfrac{7}{cd} \le 9$, which is a contradiction. Hence it is impossible to have $a\ge 5$, as desired.

~Sedro

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