2. A collector has $N$ precious stones. If he takes away the three heaviest stones then the total weight of the stones decreases by $35 \%$. From the remaining stones if he takes away the three lightest stones the total weight further decreases by $\frac{5}{13}$. Find $N$.

A collector has $N$ precious stones. If he takes away the three heaviest stones, then the total weight of the stones decreases by $35 \%$. From the remaining stones, if he takes away the three lightest stones, the total weight further decreases by $\frac{5}{13}$. Find $N$.
2. 10
2. Let the total weight of the stones be 100. Then the weight of the three heaviest stones is 35 and that of the three lightest stones is $(100-35) \times \frac{5}{13}=25$. The remaining $N-6$ stones, of total weight $100-35-25=40$, have an average weight between $\frac{25}{3}$ and $\frac{35}{3}$. Since $40 \div \frac{35}{3}>3$ and $40 \div \frac{25}{3}<5$, we must have $N-6=4$, from which the answer $N=10$ follows.