Digit Product Equal to Half

Write the integer as $10a+b$ with digits $a\in\{1,\dots,9\}$ and $b\in\{0,\dots,9\}$. The condition $ab=\tfrac12(10a+b)$ becomes

Since $a\ge 1$, we have $2a-1\ge 1$, so we may divide:

For $b$ to be an integer, the odd positive number $2a-1$ must divide $5$, so $2a-1\in\{1,5\}$.

• If $2a-1=1$, then $a=1$ and $b=10$. This is not a digit.
• If $2a-1=5$, then $a=3$ and $b=6$.

Check: $3\cdot 6=18=\tfrac12\cdot 36$.