4. Three boys participated in a "Russian Lotto" draw. Misha won 943 rubles more than Kolya, Vitya - 127 rubles more than Misha, and Misha and Kolya together - 479 rubles more than Vitya. How much money did each win?
4. First method of solving. Let's make a brief record of the problem's condition in the form of a drawing:

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-092.jpg?height=403&width=711&top_left_y=1312&top_left_x=593)

From the drawing, it is clear that

1) $127+479=606$ (rubles) - Kolya won;
2) $606+943=1549$ (rubles) - Misha won;
3) $1549+127=1676$ (rubles) - Vitya won.

Second method of solving (by Alena Burmistrova, student of School No. 24).

1) $943+127=1070$ (rubles) - Vitya won more than Kolya;
2) $1070+479=1549$ (rubles) - Misha won;
3) $1549-943=606$ (rubles) - Kolya won;
4) $1549+127=1676$ (rubles) - Vitya won.

Third method of solving (by Ilya Mikhailov, student of School No. 12).

According to the problem's condition, we can form the equation:

$$
\mathrm{B}=\mathrm{K}+943+127 \text{ or } \mathrm{B}=\mathrm{K}+1070.
$$

Additionally, it is known that $\mathrm{M}+\mathrm{K}=\mathrm{B}+479$.

Using the first equation, we get:

$$
\mathrm{M}+\mathrm{K}=\mathrm{K}+(1070+479)
$$

Since the sums in the left and right parts are equal and their first addends are equal, the second addends will also be equal: $\mathrm{M}=1070+479$ or $\mathrm{M}=1549$.

Answer: Kolya won 606 rubles, Misha won 1549 rubles, Vitya won 1676 rubles.