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Math Help - word problem

  1. #1
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    word problem

    44 trees are planted in a circle. there are 44 sparrows sitting on these trees (one bird on each tree). Once in a while two sparrows fly to other trees as follows: one of them flies to the next tree clockwise. the other one flies to the next tree counterclockwise. can they all gather onthe same tree at some point?
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  2. #2
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    Hello spred

    Welcome to Math Help Forum!
    Quote Originally Posted by spred View Post
    44 trees are planted in a circle. there are 44 sparrows sitting on these trees (one bird on each tree). Once in a while two sparrows fly to other trees as follows: one of them flies to the next tree clockwise. the other one flies to the next tree counterclockwise. can they all gather onthe same tree at some point?
    No.

    I think it's easier to imagine the reverse problem: if all 44 sparrows are in one tree, can they move to occupy all 44 trees? Suppose the trees are numbered 1 -44, and they all start in tree #1. In pairs, then, they could move outwards until trees 2 - 22 on one side and 24 - 44 on the other were occupied, leaving two birds in the first tree.

    The other 42 birds now play no further part, since they are in their correct positions, and the problem reduces to just the two remaining birds. They can now only move in such a way as to maintain symmetrical positions either side (or on) the line joining tree #1 to tree #23. Hence it is impossible for one to remain on (or return to) tree #1 while the other moves to occupy tree #23.

    Clearly, we can reverse this, and, starting with one bird in each tree move the birds in pairs to occupy tree #1 until all the birds are there except the one in tree #23. It is then clearly impossible to navigate this bird into tree #1 without disturbing at least one of the 43 birds already there.

    Grandad
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  3. #3
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    Quote Originally Posted by spred View Post
    44 trees are planted in a circle. there are 44 sparrows sitting on these trees (one bird on each tree). Once in a while two sparrows fly to other trees as follows: one of them flies to the next tree clockwise. the other one flies to the next tree counterclockwise. can they all gather onthe same tree at some point?
    I'm not sure I follow Grandad's solution. Grandad, are you assuming the birds retain a symmetrical configuration? I don't see why they have to. Maybe I just don't understand, which is likely.

    Anyway, here is another approach. Number the trees from 0 to 43, counterclockwise. Let's suppose the birds all end up in tree number 0. If not, just renumber the trees.

    Count a move of one tree counterclockwise as a move of distance 1 and a move of one tree clockwise as a move of distance -1. The bird in tree 0 must move a total distance of 0, modulo 44, to end up in tree 0. The bird in tree 1 must move a total distance of -1 = 43, modulo 44, to end up in tree 0. And in general the bird in tree n must move a total distance of -n = 44-n, modulo 44, to end up in tree 0.

    So the total distance moved by all the birds must be 0 + 1 + 2 + ... + 43 = 946 = 22 (modulo 44). But since the moves come in pairs of +1 and -1, the total distance moved (again, this is all modulo 44) must be 0. So it can't be done.
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  4. #4
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    Hello everyone
    Quote Originally Posted by awkward View Post
    I'm not sure I follow Grandad's solution. Grandad, are you assuming the birds retain a symmetrical configuration? I don't see why they have to.
    I'm not saying that they have to, but if they do then (using your numbering from 0 to 43) if bird #22 stays still, and, in pairs equidistant from #22, they move towards #0 - and therefore retain the symmetry about the line joining #0 to #22 - you can get 43 of the birds onto tree #0. Bird #22, meanwhile, still hasn't moved and is 22 steps away from tree #0, and can't move without disturbing the birds that are 'home'.

    Clearly this confirms the answer you got using your mod 44 method (which I prefer to mine, as being rather more mathematical!).

    Grandad
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