A number theory problem by Worranat Pakornrat

For simplicity, assume that $a \ge b \ge c$ .

The equation can be rewritten as $\frac{1}{10} = \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$ .

If $c \ge 6$ then $\tfrac{1}{ab} + \tfrac{1}{ac} + \tfrac{1}{bc} \le \tfrac{3}{36} < \tfrac{1}{10}$ . If $a,b \ge 6$ and $c=5$ then $\tfrac{1}{ab} + \tfrac{1}{ac} + \tfrac{1}{bc} \leq \tfrac{2}{30} + \tfrac{1}{36} < \tfrac{1}{10}$ . If $a=b=c=5$ then $\tfrac{1}{ab} + \tfrac{1}{ac} + \tfrac{1}{bc} = \tfrac{3}{25} \neq \tfrac{1}{10}$ . Thus we can assume that $c \le 4$ .

• If $c=1$ the equation can be written as $(a-10)(b-10) = 110$ , and so we can consult the following table to find four solutions $\begin{array}{|cc|ccc|} \hline a-10 & b-10 & a & b & c \\ \hline 110 & 1 & 120 & 11 & 1 \\ 55 & 2 & 65 & 12 & 1 \\ 22 & 5 & 32 &15 & 1 \\ 11 & 10 & 21 & 20 & 1 \\ \hline \end{array}$
• If $c=2$ the equation can be written as $(a-5)(b-5) = 35$ , and we find two solutions $\begin{array}{|cc|ccc|} \hline a-5 & b-5 & a & b & c \\ \hline 35 & 1 & 40 &6 & 2 \\ 7 & 5 & 12 & 10 & 2 \\ \hline \end{array}$
• If $c=3$ the equation can be written as $(3a-10)(3b-10) = 190$ , and we find two solutions $\begin{array}{|cc|ccc|} \hline 3a-10 & 3b-10 & a & b & c \\ \hline 190 & 1 & \\ 95 & 2 & 35 & 4 & 3 \\ 38 & 5 & 16 & 5 & 3 \\ 19 & 10 & \\ \hline \end{array}$
• If $c=4$ the equation can be written as $(2a-5)(2b-5) = 65$ , and we find one more solution $\begin{array}{|cc|ccc|} \hline 2a-5 & 2b-5 & a & b & c \\ \hline 65 & 1 \\ 13 & 5 & 9 & 5 & 4 \\ \hline \end{array}$ Note that we have discarded the $2a-5=65,2b-5=1$ case because it gives a value of $b$ less than $4$ (in effect, it repeats the $35,4,3$ solution already found).

The solution with the smallest value of $ab+ac+bc$ is $(9,5,4)$ , so the answer is $\boxed{101}$ .