3. The brothers found a treasure of gold and silver. They divided it so that each received 100 kg. The elder brother received $1 / 5$ of all the gold and $1 / 7$ of all the silver, while the younger brother received $1 / 7$ of all the gold. What share of the silver did the younger brother receive?
# Solution.

1) Let the mass of gold in the treasure be $z$ kg, and the mass of silver be $s$ kg. The eldest brother received $z/5 + s/7$ kg; this is less than $(z+s)/5$, but more than $(z+s)/7$. From this, it follows that the number of brothers is more than five but less than seven, meaning there are six of them.
2) Now we get the system of equations: $\mathrm{z} / 5 + \mathrm{s} / 7 = 100, \mathrm{z} + \mathrm{s} = 600$.

Solving: $z / 5 + s / 7 = z / 6 + s / 6; (1 / 5 - 1 / 6) z = (1 / 6 - 1 / 7) s; z / 30 = s / 42; z = (5 / 7) s$.

Since $\mathrm{z} + \mathrm{s} = 600$, then $\mathrm{z} = 250, \mathrm{~s} = 350$.

3) The youngest brother received 250/7 kg of gold; thus, he received 100 - 250/7 = 450/7 kg of silver. The share of this silver from the total silver in the treasure is 450/7 : 350 = 9/49.

Answer: $9 / 49$.

Criteria.

If it is not proven (but simply assumed out of the blue) that there are exactly six brothers, the problem is not solved (no more than 3 points).

Errors. Arithmetic errors with the correct approach to the solution - minus 2 points.

Advancements.

It is stated that there are definitely six brothers - 1 point (2 points if with proof). The amount of gold and silver is found - an additional third point can be added.