All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Two happy ends (Posted on 2011-03-07)
Consider a series of numbers, defined as follows:
Starting with any natural number, each member is a sum of the squares of the previous member`s digits.

Prove : The series always reaches either a stuck-on-one sequence: 1,1,1… or a closed loop of the following 8 numbers: 145,42,20,4,16,37,58,89, ...

Ex1: 12345,55,50,25,29,85,89,145….. etc
Ex2: 66,72,53,34,25,29,85,89,145…
Ex3: 91,10,1,1,1…..

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): solution NOT enough | Comment 4 of 11 |

It seems he did, along with Jer's comment. Jer notes every number with more than 3 digits is mapped to a number with strictly fewer digits. Thus clearly in a finite number of applications, one will reach a number with 3 or fewer digits.

Then he simply checks those thousand numbers (excluding zero) and shows every chain that continues on from one of them will either end in the cycle with 4 or with 1. That is what was shown in the computer output.

In other words, every chain that begins with a natural number must include a number with 3 or fewer digits, and considering the chain that starts there shows it must eventually reach one of the two desired cycles, indicating that the whole chain must then reach one of the two desired cycles.

 Posted by Gamer on 2011-03-08 05:17:54

 Search: Search body:
Forums (0)