Friday, September 27, 2013

Thinking Caps...


Couple puzzles for today (one easy, I think, one less-so):

1) Since I brought up Raymond Smullyan in the last post, the first comes from him, adapted from the "Mathematical People: Profiles and Interviews" book I've also  mentioned earlier. This seems surprisingly easy for a Smullyan puzzle, but perhaps it's one of those tricky ones that some people see through right away and others have more difficulty with:

100 politicians attended a certain convention. Each politician was either crooked or honest. The following two facts are known to be true:

a. At least one of the politicians was honest.
b. Given any two of the politicians, at least one of the two was crooked.

From these two facts can it be determined how many of the politicians total were honest and how many were crooked?


2) The 2nd problem I've adapted from a recent Matt Parker tweet:

It's possible to arrange all the numbers from 1 to 17 so that every adjacent pair adds up to a square number (...every number except the two end numbers will be part of 2 adjacent pairs, and each pair sums to a square no.).
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1st answer:  99 politicians must be crooked and 1 honest (to insure the two conditions are met).

2nd problem:  there may be more than one answer, but the first one I arrived at runs like this:
1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16, 8, 17
...see first comment below for a correct solution!



2 comments:

Neil Ostrove said...

That doesn't work; 8+16=24 which is not a square. 17,8,1,15,10,6,3,13,12,4,5,11,14,2,7,9,16 seems to.

"Shecky Riemann" said...

Thanks!! you're right (duhh on my part); will make correction above.