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Title: 8-digit squares Post by fatball on Jan 22nd, 2006, 8:49pm Find all 8-digit natural numbers n such that n2 ends in the same 8 digits as n. Numbers are written in standard decimal notation, with no leading zeroes. |
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Title: Re: 8-digit squares Post by Eigenray on Jan 22nd, 2006, 10:01pm [hideb]> chrem([0,1],[1,0],[2^8,5^8]); [87109376, 12890625][/hideb] |
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Title: Re: 8-digit squares Post by fatball on Jan 23rd, 2006, 10:45am [hide]Very neatly answered.[/hide] Numbers with this property are called Automorphic Numbers and a brief discussion can be found here (http://en.wikipedia.org/wiki/Automorphic_number) or there (http://mathworld.wolfram.com/AutomorphicNumber.html). |
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Title: Re: 8-digit squares Post by Eigenray on Jan 24th, 2006, 6:05am In fact, [xn,yn] := [hide]chrem([0,1],[1,0],[2n,5n])[/hide] is a pair of orthogonal idempotents of the ring R=(Z/10n), the only pair other than [0, 1]. This gives the decomposition R = Rxn [oplus] Ryn, giving an explicit inverse to the ring isomorphism (Z/10n) ~= (Z/5n) x (Z/2n), (r mod 10n) -> (r mod 5n, r mod 2n) (axn + byn mod 10n) <- (a mod 5n, b mod 2n). Moreover, we have xn+1 = xn mod 10n, so that, viewing xn as an integer between 0 and 10n, x = lim xn = x1 + (x2-x1) + (x3-x2) + ... = ...109376 exists as a 10-adic integer. Defining also y = lim yn = ...890625, this gives a non-trivial pair of zero-divisors in the 10-adics Z10 ~= Z2 x Z5. |
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