How many numbers, from 1 to 50 (both included) can you arrange in a row (one of each) so that each one, except the first and the last, is the sum or difference of its two neighbours?

Example: 3, 10, 7, 17, 24, 41.

10 = 3+7, 7 = 17-10, 17 = 24-7, 24 = 41-17.

(In reply to computer solution by Charlie)

41 / 25 / 16 / 9 / 7 / 2 / 5 / 3 / 8 / 11 / 19 / 30 / 49.

Starting from the middle (5) we have the "right leg" exactly +1, +1, +2, +3, +5, +8 than the corresponding number in "left leg".

These are the first Fibonacci numbers. Coincidence, or we could imply that if we extend the numbers to 80, the maximum sequence will be:

66 / 41 / 25 / 16 / 9 / 7 / 2 / 5 / 3 / 8 / 11 / 19 / 30 / 49 / 79?

 

 

Edited on November 20, 2008, 9:32 pm

Posted by pcbouhid on 2008-11-20 21:30:12