Number cards game

After the game, the total points of $A,B,C$ are $39$ points. So the number of draws and the sum of the numbers on the cards are the divisors of $39$ .

Moreover, the divisors of $39$ are $1,3,13,39$ . The smallest possible sum of the numbers on the cards is $1+2+3=6$ and base on the information in the problem, the smallest possible number of draws is $2$ .

Base on the information above, we can conclude that the number of draws is $3$ and the sum of the numbers on the cards is $13$ .

Now, we call $I,II,III$ are the cards with number $a,b,c$ . Because $A$ received $20 points$ after $3$ draws, so the smallest possible value on the card with highest score (card $III$ ) is $7$ because $7\times 3=21>20$ $*$ .

$B$ received card $III$ in the last draw, so the maximum possible value on card $III$ is $8$ (in the case the card with lowest score (card $I$ ) had the value of 1 and $B$ received this card twice)

The maximum possible value on card $I$ is $2$ . Because if the maximum possible value on this card is $3$ , so the maximum possible value on card $III$ is: $13-3-4=6$ (does not satisfy $*$ )

Now, from the information above, there are 4 possible cases of the numbers on the cards:

$8;4;1$

$8;3;2$

$7;5;1$

$7;4;2$

From these 4 cases, the first case (from the left) is the only case which can give us $20 points$ after $3 draws$

Now, here are the points of the players after $3 draws$ (You can easily prove this!)

$A$ : $8+8+4=20$

$B$ : $1+1+8=10$

$C$ : $4+4+1=9$

In the last draw, $B$ received $8 points$ , so $A$ received $4 points$ and $C$ received $1 point$ .

So in the first draw, $C$ received $4 points$

Finally, the result of the product we have to find is $4\times 3=12$ ( $4$ is the value of this card and $3$ is the number of the person)