Lusophon Mathematical Olympiad

Let $x_1, x_2, x_3,$ and $x_4$ be the amounts of money that the first, second, third, and fourth brothers have, respectively. According to the problem, we have the following equation describing their total amount of money:

$$x_1 + x_2 + x_3 + x_4 = 48$$

We are also given conditions on how these amounts are adjusted:

1. The first brother's money is increased by 3 Kwanzas: $x_1 + 3$.
2. The second brother's money is decreased by 3 Kwanzas: $x_2 - 3$.
3. The third brother's money is tripled: $3x_3$.
4. The fourth brother's money is reduced by a third: $\frac{x_4}{3}$.

These adjustments result in all brothers having the same amount of money. Let's denote this common amount by $c$. Therefore, we have the following equations:

$$x_1 + 3 = c$$
$$x_2 - 3 = c$$
$$3x_3 = c$$
$$\frac{x_4}{3} = c$$

Solving these equations for $c$ in terms of each brother's amount, we get:

$$x_1 = c - 3$$
$$x_2 = c + 3$$
$$x_3 = \frac{c}{3}$$
$$x_4 = 3c$$

Substituting these into the total money equation:

$$(c - 3) + (c + 3) + \frac{c}{3} + 3c = 48$$

Simplifying:

$$2c + \frac{c}{3} + 3c = 48$$

Combine terms:

$$5c + \frac{c}{3} = 48$$

Multiply the entire equation by 3 to eliminate the fraction:

$$15c + c = 144$$

$$16c = 144$$

Solving for $c$:

$$c = \frac{144}{16} = 9$$

Now, substitute $c = 9$ back into the expressions for $x_1, x_2, x_3,$ and $x_4$:

$$x_1 = 9 - 3 = 6$$
$$x_2 = 9 + 3 = 12$$
$$x_3 = \frac{9}{3} = 3$$
$$x_4 = 3 \times 9 = 27$$

Thus, the amounts of money each brother has are:

$$\boxed{6, 12, 3, 27}$$