Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**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)

Offline

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

I'm sorry that was (solve for n)

Offline

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

**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

Offline

**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 198
- Website

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, ….

Offline

**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?

Offline

**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 198
- Website

So there you have it:

Offline

**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

Offline

**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?

Offline

**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 198
- Website

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*).

Offline

**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 ))

Offline

Pages: **1**