I wandered if anyone could give me as much as possible info on the task I need to investigate at:

http://maths.fallibroome.cheshire.sch.uk/leapfrog.swf

I have worked out that frog 3 blue frogs and 3 green frogs it will take me 15 moves.

I have worked out that frog 3 blue frogs and 4 green frogs it will take me 19 moves.

I have worked out that frog 4 blue frogs and 4 green frogs it will take me 24 moves.

I have worked out that frog 4 blue frogs and 5 green frogs it will take me 29 moves.

I have worked out that frog 5 blue frogs and 5 green frogs it will take me 35 moves.

If n represents the number of frogs in a team then:

For:

n = 1 we have 3 = n * 3 moves

n= 2 we have 8 = n * 4 moves

n = 3 we have 15 = n * 5 moves

n = 4 we have 24 = n * 6 moves

n = 5 we have 35 = n * 7 moves

n = 6 we have 48 = n * 8 moves

Or more generally if there are n frogs on each side, then the minimum number of moves will be n*(n+2)

This formula does not work if the number of frogs in each team is uneven

If n represents the number of frogs in a team and m the number of frogs in the other then:

For:

n = 1 and m = 6 we have 13 = 1 * 6 + 1 + 6 moves

n = 4 and m = 3 we have 19 = 4 * 3 + 4 + 3 moves

n = 2 and m = 1 we have 5 = 2 * 1 + 2 + 1 moves

n = 5 and m = 6 we have 41 = 5 * 6 + 5 + 6 moves

n = 5 and m = 5 we have 35 = 5 * 5 + 5 + 5 moves

So if there are n frogs on one side and m on the other, then the minimum number of moves will be n*m+n+m

Other patterns notices but I can't explain why (please help!)

- For each frog extra that one team has, there will be a repetition of the jumps and slides. The string of jumps and slides is also symmetric.

- If there are the same number of frogs in each team, then each frog will have to move n+1 times.