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October 22nd, 2007, 03:46 PM   #1
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N-digit numbers and their squares

I noticed that, for a given N, there are always one or two N-digit 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.
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October 22nd, 2007, 05:08 PM   #2
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This is related to the 10-adic numbers. Do a Google search for "p-adic numbers" and see if there's an introduction you can read.

In the 10-adics, you can extend this all the way, making numbers that extend forever to the left.
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