2. Winnie-the-Pooh stocked up on chocolate bars for the winter: $60\%$ of the total number were "Snickers", $30\%$ were "Mars", and $10\%$ were "Bounty". In the spring, it turned out that the number of "Bounty" bars eaten by Winnie-the-Pooh was $120\%$ of the number of "Mars" bars eaten and $30\%$ of the number of "Snickers" bars eaten. How many chocolate bars did Winnie-the-Pooh stock up for the winter if $2/3$ of all "Bounty" bars were left uneaten and no more than 150 "Snickers" bars were left uneaten.
# Solution.

Let there be $3 k$ "Bounty" chocolate bars in total. Then there were $9 k$ "Mars" bars and $18 k$ "Snickers" bars. Since $k$ "Bounty" bars were eaten, $\frac{k}{1.2}=\frac{5 k}{6}$ "Mars" bars were eaten. Therefore, $k$ is divisible by 6. "Snickers" bars eaten were $\frac{k}{0.3}=\frac{10 k}{3}$, and the remaining "Snickers" bars are $18 k-\frac{10 k}{3}=\frac{44 k}{3} \leq 150$. Thus, $k \leq \frac{450}{44}=10 \frac{5}{22}$. Since $k$ is divisible by $6, k=6$. The total number of chocolate bars was $18 k+9 k+3 k=30 k=180$.

Answer: 180.

Remarks for checking. Guessed answer without justification - 0 points.