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Solve for x ;
10^n+10^n-1+10^n-2+...+10^1+1 congruent to 0 (mod 59)
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I'm sorry that was (solve for n)
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Since 59 and 9 = 10 − 1 are coprime, any solution to
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is also a solution to
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i.e. to
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and vice versa. By Fermat’s little theorem, as 59 is prime,
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Therefore the smallest positive value of n+1 must divide 58. Clearly 10 ≢ 1 (mod 59) and 10² = 100 ≢ 1 (mod 59). If you can show that
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then n+1 will be a multiple of 58 and the general solution will be n = 58k − 1, k = 1, 2, 3, ….
Me, or the ugly man, whatever (3,3,6)
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Thanks but what was fermat's little theorem and how you conclude n+1=58 by that?
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So there you have it:
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Me, or the ugly man, whatever (3,3,6)
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I still can't understand how you've found n+1=58 in step 3 and 4
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If you have multiplied 10^29 by 10^29 than ok. But how we should know to find 10^29 and than result its congruent to -1 in mod 59 and THEN multiply it by it-self to find out this answer?
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Let me put this in that language of group theory. We have
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because 58 is the order of G, the multiplicative group of nonzero integers modulo 59 (a prime). If
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then m must be a multiple of r, the order of 10 in the group G. Now r must be a divisor of the group order |G| = 58, i.e. its possible values are 1, 2, 29, 58. In my posts above, I eliminated 1, 2, 29. Therefore r = 58 (i.e. 10 is generator of the cyclic group G).
Me, or the ugly man, whatever (3,3,6)
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Thanks again but another question: how can we find the group order because in the answer above,you mentioned it's 58 but how did you measured it before finding the answer?((sorry for asking a lot ))
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