You just got to test all the possible products

Two more solutions exist $15\times 23 =345 , \ 12 \times 29 = 348$ Writing problem as $1A \times 2B = 34C \implies 10(2A + B) +AB= 140 +C$ Observe that $140\leq 10(2A+B) +AB < 150\implies 14 \leq 2A+B+ \dfrac{AB}{10} < 15$ As the largest possible value of $A=7$ giving us $0\leq B < 1\implies B=0$ and since $10=2\times 5$ and if $B=5$ then $A$ isnot integer (from above inequality ). Therefore $A=5$ and then $4 \leq B+\frac{B}{2} < 5\implies 8 \leq 3B \leq 10$ . Giving only possible $B=3$ . For $B=2$ does not hold good for $60\leq 11A<65$ meaning $A=2$ and $10\leq B +\frac{B}{5} <11\implies 50\leq 6B<55$ giving us next possible value $B=9$ .

Thus, $17\times 20= 340 \ , 15\times 23 = 345, \ 12 \times 29= 348$ and the answer is yes

$15\times 23 = 345$

15 23=345 and 12 29=348