Home > Just Math
|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
re: How the numbers get there- I EXPLAIN
| Comment 10 of 11 |
(In reply to How the numbers get there
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
Please log in:
About This Site
New Comments (6)
Top Rated Problems
This month's top
Most Commented On