Help 10 year old maths question.... !
This has got me baffled... But is it Sunday morning. :D
2 families of frogs (three in each family), need to swop sides of the pond. There are 7 Lily pads on which to jump over each other on. They can only jump onto empty lilies, only jump over one frog at a time and can't jump backwards. WTF. :confused: Can anyone help me out so I can tell him 1, how to do it, and 2, what is the smallest amount of jumps required to do this. It also has to be repeated for 4, 5 and 6 frog per family. Thanks in advance. :D |
swim ???
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Google seems to think...
one family moves at a time each frog jumps 5 times over the lily pads following one after the other.making 15 jumps total for family of 3 4 frogs = 6 jumps per frog 5 frogs = 7 jumps per frog 6 frogs = 9 jumps per frog there is a sequel pattern as the frogs increase by 1 the jumps increase by 1 and the lily pads increase by 3 |
I've worked it out but it's far too complicated to write down :freak3:
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:lol1:
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What sort of maths question is THAT?!
Teachers always used to tell me that what I learnt in school would help me later in life... yea, i'm sure this one is guna serve your young'n well in the future! |
10 years old!!!!!!
haven't got a clue, but i agree; how is that going to help later in life?? springwatch?!!??! |
this is easy. Recursive descent analysis. Work out the answer for one frog on each side. Apply the various rules you discover to make it work for two on each side, establish the pattern and simplify the formula. Now you can do it for a million frogs.
(p.s. - One frog doesn't tell you that much if you don't make him jump empty lillies when available) |
And people complain that exams/school is easier than it was in their day ;)
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Yea but we could never find the answer to our homework by simply searching on fecking YouTube :thumb: :D
Slide rule and logbook in your day Trout? :lol: |
No - I got the butler to do my homework ;) :p
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:D
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