Task A-2.8. (10 points)

Ivan, Stipe, and Tonći take turns throwing a die. Ivan goes first, then Stipe, and then Tonći, and after that, the order repeats. Each of them, when it is their turn, throws the die once until they get their first "six." After getting their first six, in each subsequent turn, Ivan throws the die four times, Stipe six times, and Tonći eight times.

Tonći was the last to get his first six, on his tenth throw, and the game ended then. If the die was thrown 47 times, determine who among them threw the die the most times.
## First Solution.

Each of them came up three times, i.e., ten rounds were played. If they had all rolled the dice only once, there would have been a total of 30 rolls.

The remaining $47-30=17$ dice rolls were made by Ivan and Stipe with additional rolls.

(2 points)

If Ivan rolled the dice three additional times in $x$ rounds, and Stipe rolled the dice five additional times in $y$ rounds, then $3 x+5 y=17$.

(3 points)

The numbers $x$ and $y$ must be natural numbers or 0, so the equation can be solved by substituting $y=0,1,2,3, \ldots$

For $y \geqslant 4$, the number $x$ is less than zero. The only solution is $x=4, y=1$.

(3 points)

Ivan rolled the dice four times with additional rolls, so he rolled the dice a total of $10+4 \cdot 3=22$ times.

Stipe rolled the dice with additional rolls only once, so he rolled the dice a total of $10+5=15$ times.

Ivan rolled the dice the most times.

(2 points)