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Title: Repunit Quadratics Post by Sir Col on Nov 9th, 2006, 3:15pm A repunit is defined as a positive integer consisting entirely of the digit 1. Let R be any repunit. (a) Prove that 9R2+2R is always a repunit. (b) Find the value of the coefficients of the "next" quadratic, aR2+bR+c, which always produces a repunit. (c) Show that infinitely many such quadratics exist and generalise the sequence of coefficients (a,b,c). |
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Title: Re: Repunit Quadratics Post by Sameer on Nov 9th, 2006, 4:17pm 1. [hide] I have a weird method. Let Rn=1111.....1 n digits = R (for calculation down) Rn-1=111111....1 n-1 digits = R' (for calculation down) Then R = 10R' + 1 R2 = 100R'2 + 20R' + 1 9R2 = 900R'2 + 180R' + 2 9R2 + 2R = 900R'2 + 200R' + 11 as 2R = 20R' + 2 As seen on RHS, first and second term's have lower two digits 00 in addition with 11 giving the RHS to have 11 in least significant digits. RHS = 100(9R'2+2R')+11 Thus if we prove that the inside of bracket is repunit we can prove the whole thing is repunit. However this is same as proving the original equation, so by PMI (assuming k and proving for k+1) we can say the original equation is repunit... [/hide] |
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Title: Re: Repunit Quadratics Post by Icarus on Nov 9th, 2006, 6:31pm Note that all repunits are of the form (10n - 1)/9 The condition that a quadratic of repunits gives repunits is, for all n, there is a k such that a(102n - 2*10n + 1)/9 + b(10n - 1) + 9c = 10k - 1 Rewriting the expression multiplying a gives a(102n - 1)/9 - 2a(10n - 1)/9 + 4a/9. The only term then in the entire equation that is not obviously integral is 4a/9. Hence 9 divides a: say a = 9t, and we have the equation: t102n - 2t10n + t + b10n - b + 9c = 10k - 1 Since k has to be >=n, we have t - b + 9c = -1 mod 10n. By choosing n large enough, 10n is much larger than all the values involved, so it must be that t - b + 9c = -1. Applying this identity to the equation before gives t102n - 2t10n - b10n = 10k. If n is taken much larger than 2t and b, this can only be if 2t - b = 0. Hence t = 9c + 1, b = 18c + 2, a = 81c + 9. c = 0 gives the "first" solution of 9R2 + 2R. The next solution is 90R2 + 20R + 1. Lest anyone think a different pattern is forming, the third is 171R2 + 38R + 2. |
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Title: Re: Repunit Quadratics Post by Eigenray on Nov 9th, 2006, 8:27pm on 11/09/06 at 18:31:12, Icarus wrote:
That 4a/9 term shouldn't be there, so you can't immediately conclude 9|a. However the next step works anyway, in that a - 9b + 81c = -9 modulo arbitrarily high powers of 10, hence equality holds, and proceed as before. Quote:
But we still need to have t*102n = 10k! Hence t = 9c+1 is a power of 10, i.e., c is itself a repunit. My solution used the higher order terms instead: a(10n-1)2 + 9b(10n-1) + 81c = 9(10k-1), and dividing through by 102n shows 9 10k-2n -> a as n,k->infinity. That is, k-2n -> log10(a/9). But the only way a sequence of integers can converge is if they are eventually constant, and the limit is an integer. Hence for sufficiently large n, k-2n=r, and a = 9*10r. Subtracting the term a*102n = 9*102n+r = 9*10k, the equation becomes -2a 10n + a + 9b 10n - 9b + 81c = -1. Since this holds for all n (sufficiently large, at least), we must have -2a + 9b = 0, and a - 9b + 81c = -1. Thus a = 9*10r b = 2/9 a = 2*10r, c = (9b - a - 1)/81 = (10r-1)/9. Conversely, we have 9*10rR2 + 2*10rR + (10r-1)/9 = 10r(9R2 + 2R) + (10r-1)/9 is a repunit by (a). |
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Title: Re: Repunit Quadratics Post by Icarus on Nov 10th, 2006, 3:48pm Sign errors and ignored coefficients. :-[ I've got to start checking what I write more thoroughly! |
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