[ $\left[\begin{array}{ll}\text { Motion Problems } & \text { Processes and Operations }\end{array}\right]$ [ Divisibility of Numbers. General Properties ]

Two frogs, Kva and Kvi, are participating in a "race" - 20 meters forward in a straight line and back. Kva covers 6 dm in one jump, while Kvi only 4, but Kvi makes three jumps in the time it takes her opponent to make two. Given these conditions, what is the possible outcome of the competition?
Think about how many jumps the second frog will have to make.

## Solution

The distance of 12 dm is covered simultaneously by both frogs. However, here they need to cover the distance of 20 m twice. 20 m = 200 dm is divisible by 4, so the second frog will jump exactly 40 m. But 200 is not divisible by 6, which means the first frog will have to cover a greater distance (specifically, 204 dm each way). Therefore, the second frog will win.