Из пункта А в пункт В, расстояние между которыми равно 208 км, вышел катер. Дойдя до пункта В, он вернулся в пункт отправления, затратив на обратный путь на 5 часов меньше. Найдите собственную скорость катера, если скорость течения реки равна 5 км/ч. Ответ дайте в км/ч.

Let $$v$$ be the speed of the boat in still water. The speed downstream is $$v+5$$ km/h, and the speed upstream is $$v-5$$ km/h. The time taken to travel downstream is $$t_{down} = \frac{208}{v+5}$$ hours. The time taken to travel upstream is $$t_{up} = \frac{208}{v-5}$$ hours. We are given that $$t_{up} = t_{down} - 5$$. So, $$\frac{208}{v-5} = \frac{208}{v+5} - 5$$. Multiplying by $$(v-5)(v+5)$$, we get $$208(v+5) = 208(v-5) - 5(v-5)(v+5)$$. This simplifies to $$208v + 1040 = 208v - 1040 - 5(v^2 - 25)$$. Thus, $$2080 = -5v^2 + 125$$, which means $$5v^2 = 2080 - 125 = 1955$$. Therefore, $$v^2 = \frac{1955}{5} = 391$$. So, $$v = \sqrt{391} \approx 19.77$$ km/h. However, the problem implies a simpler solution, likely with integer speeds. Let's re-examine the equation: $$\frac{208}{v-5} - \frac{208}{v+5} = 5$$. $$\frac{208(v+5) - 208(v-5)}{(v-5)(v+5)} = 5$$. $$\frac{208v + 1040 - 208v + 1040}{v^2 - 25} = 5$$. $$\frac{2080}{v^2 - 25} = 5$$. $$2080 = 5(v^2 - 25)$$. $$416 = v^2 - 25$$. $$v^2 = 416 + 25 = 441$$. $$v = \sqrt{441} = 21$$ km/h. The speed of the boat in still water is 21 km/h.