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Two happy ends (Posted on 2011-03-07) Difficulty: 3 of 5
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    
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Hints/Tips re: How the numbers get there- I EXPLAIN | Comment 10 of 11 |
(In reply to How the numbers get there by Charlie)

The list is redundant once you perceive that the transformation of the 3-digit number 100a+10b+c into a^2+b^2+c^2 produces always a lesser number, so in few steps a 2-digit number must be  reached, - from this moment on  the reduction does not necessarily hold and the process should be examined.

Proof: Comparing  100a+10b+c and a^2+b^2+c^2 we get:

100a-a^2>99         ....FROM 99 TO 819

10b-b^2>=0            ''''0 TO 25

c-c^2>=-72               '''''-72 TO 0

so the number goes down by at least 27(109==>82   ) and at most by 844 (950==>106 and 951==>107 & later below 100..).

999==>243  IS NOT AN EXTREMUM CASE.

Edited on March 9, 2011, 9:59 pm
  Posted by Ady TZIDON on 2011-03-09 19:43:15

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