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 Erase the right 90 (Posted on 2016-01-28)
101 distinct real numbers are written in any order

Prove that it is possible to erase 90 of them leaving a sequence of 11 numbers that are in either a strictly increasing or strictly decreasing order.

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 re: proof on the internet (spoiler) | Comment 3 of 5 |
(In reply to proof on the internet (spoiler) by Charlie)

Actually, I find the puzzle itself difficult to follow.

On my bookshelf I have the ten-volume set of K.Sengupta's collected problems. As is well known, by some whimsy of the author, each volume has a real number 1.x, 2.x, 3.x, etc. We shall call these 1,2,3.. for short.

Assume I order them: 1,10,2,9,3,8,4,7,5,6. I see no continuous subsequence of four here which is either strictly increasing or decreasing.

But if the subsequence can be non-continuous, e.g. we can rub out 10, and select 1,2,3,4 as our four-part sequence, then a more challenging problem (by far) is to arrange volumes 1 to 9 in some order such that there is no non-continuous sequence of four that is either strictly increasing or strictly decreasing, given that either is acceptable.

Edited on January 29, 2016, 7:47 am
 Posted by broll on 2016-01-29 07:37:57

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