Puzzle #122: Sparrows on Trees

This is a very cute puzzle borrowed from the book Mathematical Circles (Russian Experience) by Dmitri Fomin, Sergey Genkin and Ilia Itenberg. It is a wonderful book for middle graders.

There are six sparrows sitting on six trees, one sparrow on each tree. The tress stand in a row, with 10 meters between any two neighbouring trees.If a sparrow flies from one tree to another, then at the same time some other sparrow flies from some other tree to another the same distance away, but in the opposite direction. Is it possible for all the sparrows to gather on one tree?  What if there are seven tree and seven sparrows?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy flying on the trees!

This entry was posted in Puzzles and tagged , . Bookmark the permalink.

2 Responses to Puzzle #122: Sparrows on Trees

  1. Not possible in 6.
    Solution:
    Let us treat distance between trees as 1 unit. Let us define a metric X = Sum of the distance of all the birds from the first tree. The initial value of X is 0+1+2+3+4+5 = 15 units. Whenever the birds fly, as it can be proven, that X remains invariant. For all the birds to be on the same tree, X either becomes 0 or a multiple of 6. Hence its not possible.

    Check that this problem does not arise when there are 7 birds. X is then 21 units and then all the birds potentially could be on the fourth tree (Tree with 3 units distance from left most tree). This does not prove anything though. We need to give a construction on how it would happen to prove that a solution exists.
    1) Birds on 0,1,2,3,4,5,6
    2) Birds on 1,2,3,3,3,4,5 (Bird from first tree moves 3 units right and last tree moves 3 units left)
    3) Birds on 2,3,3,3,3,3,4 (Bird from second tree moves 2 units right and second last tree moves 2 units left)
    4) Birds on 3,3,3,3,3,3,3 (Other two birds come to the fourth tree as well)
    Hence proved.

  2. Prakhar says:

    With the 6 sparrows, gathering the sparrows on one tree is equivalent to multiplying an odd integer by 10; however as described, a given move leads to a net displacement of 0. Therefore it is not possible to have all sparrows locate themselves into a single tree. With 7 sparrows, we see a net displacement of 10 times an even integer. So it is possible for the sparrows to gather on one tree.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s