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October 22nd, 2007, 03:46 PM  #1 
Newbie Joined: Oct 2007 From: Belgrade, Serbia Posts: 2 Thanks: 0  Ndigit numbers and their squares
I noticed that, for a given N, there are always one or two Ndigit numbers whose digits are equal to last N digits of square of that number. For example: N=1: 1, 5, 6 (1^2=1, 5^2=25, 6^2=36) N=2: 25, 76 (25^2=625, 76^2=5776) N=3: 625, 376 (625^2=390625, 376^2=141376) N=4: 9376 (9376^2=87909376) N=5: 90625 (90625^2=8212890625) N=6: 890625, 109376 (890625^2=793212890625, 109376^2=11963109376) N=7: 2890625, 7109376 (2890625^2=8355712890625, 7109376^2=50543227109376) N=8: 12890625, 87109376 (12890625^2=166168212890625, 87109376^2=7588043387109376) and so on. In these examples, I also noticed that, if for a given N (greater than 1) there are two numbers with this feature, than their sum is 10^N+1: 25 + 76 = 10^2 + 1 ... 890625 + 109376 = 10^6 + 1 ... 12890625 + 87109376 = 10^8 + 1 ... I've searched on the net so I found that this IS a rule, but nowhere I could find its proof. So if anybody knows how to prove that statement, or to give a link to the proof, I would really appreciate. 
October 22nd, 2007, 05:08 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
This is related to the 10adic numbers. Do a Google search for "padic numbers" and see if there's an introduction you can read. In the 10adics, you can extend this all the way, making numbers that extend forever to the left. 

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