2024 SSMO Speed Round Problems/Problem 9

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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2024 SSMO Speed Round Problems/Problem 9

Problem

Solution 1

Solution 2