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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

Solve for x ;

10^n+10^n-1+10^n-2+...+10^1+1 congruent to 0 (mod 59)

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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

I'm sorry that was (solve for n)

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The LHS is the sum of (n+1) terms of a geometric sequence with common ratio 10. What happens when you simplify it?

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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 166
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Since 59 and 9 = 10 − 1 are coprime, any solution to

is also a solution to

i.e. to

and vice versa. By Fermat’s little theorem, as 59 is prime,

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

then *n*+1 will be a multiple of 58 and the general solution will be *n* = 58*k* − 1, *k* = 1, 2, 3, ….

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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

Thanks but what was fermat's little theorem and how you conclude n+1=58 by that?

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**Alg Num Theory****Member**- Registered: 2017-11-24
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So there you have it:

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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

I still can't understand how you've found n+1=58 in step 3 and 4

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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

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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**Alg Num Theory****Member**- Registered: 2017-11-24
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Let me put this in that language of group theory. We have

because 58 is the order of *G*, the multiplicative group of nonzero integers modulo 59 (a prime). If

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*).

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**Abbas0000****Member**- Registered: 2017-03-18
- Posts: 29

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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