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Topic: 8-digit squares (Read 998 times) |
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fatball
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8-digit squares
« on: Jan 22nd, 2006, 8:49pm » |
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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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Eigenray
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Re: 8-digit squares
« Reply #1 on: Jan 22nd, 2006, 10:01pm » |
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| hidden: | > chrem([[0,1],[1,0]],[2^8,5^8]);
[87109376, 12890625] |
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| « Last Edit: Jan 22nd, 2006, 10:05pm by Eigenray » |
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fatball
Senior Riddler
Can anyone help me think outside the box please?
Gender:
Posts: 315
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Re: 8-digit squares
« Reply #2 on: Jan 23rd, 2006, 10:45am » |
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Very neatly answered. Numbers with this property are called Automorphic Numbers and a brief discussion can be found here or there.
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| « Last Edit: Jan 23rd, 2006, 10:47am by fatball » |
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Eigenray
wu::riddles Moderator
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Re: 8-digit squares
« Reply #3 on: Jan 24th, 2006, 6:05am » |
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In fact, [xn,yn] := chrem([[0,1],[1,0]],[2n,5n]) 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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