194580

The number of red frogs multiply by the number of blue frogs equals the number of jumps and if you add the number of the two different coloured frogs it will equal the number of slides so the formula would be rb+r+b to get the number of moves if r represents the number of red frogs and b represents the number of blue frogs. At first I used 2 red frogs and 2 blue frogs and the number of moves were 8 and 3 frogs of each colour and got 15. I notice that 2 red frogs multiplied by 2 blue frogs then doubled would get 8. However this did not work for 3 red and 3 blue frogs so I tried to but it in another way. I found that another way was 2(2+2) and this would also give the answer 8 so I tried the method of 3 red and 3 blue frogs and it got me the answer 15. After many tries I came to a conclusion that with the same number of frogs on each side the method{2(2+2) = a(a+a)} works so then I tried it on an occasion when the number of frogs aren't the same. When I tried 2 red frogs and 3 blue frogs I found out that either way it would be one less or one too more. From this I figure out that this equation a(a+a) is the same as a*a+a+a and if the numbers are different then it would be a*b+a+b. The main methods used was first looking for a pattern that I wrote it in algebra and then expanded it to get the solution.