# Thread: 26th digit

1. can we use the Divisibility rule for 13?
http://en.wikipedia.org/wiki/Divisibility_rule
tnx

2. Originally Posted by malaygoel
I would be happy if you post here the proof of the statement in the quoted work.But the question still remains:Is there any criteria by which we can say that " $\frac{10^k - 1}{9}$ is prime or composite depends on the value of k"?
Similarly, what can you say about the numbers $10^k + 1(k\geq4)$.
Keep Smiling
Malay
Assume,
$\frac{10^k-1}{9}$ is a prime and $k$ is not.
Then,
$\frac{(10^a)^b-1}{9}$ where $a,b$ are proper non-trivial factors of $k$.
Using the identity,
$x^n-y^n=(x-y)(x^{n-1}y+...+xy^{n-1})$
Thus,
$\frac{(10^a-1)((10^a)^{b-1}+(10^a)^{b-2}+...+1)}{9}$
Using, this identity again,
$\frac{(10-1)(10^{a-1}+...+1)((10^a)^{b-1}+...+1)}{9}$
You are left with,
$(10^{a-1}+...+1)((10^a)^{b-1}+...+1)$
Since, $a,b\not = 1$ These two factors are not trival. Thus,
$\frac{10^k-1}{9}$ cannot be prime a contradiction.

3. Originally Posted by ThePerfectHacker
Assume,
$\frac{10^k-1}{9}$ is a prime and $k$ is not.
Then,
$\frac{(10^a)^b-1}{9}$ where $a,b$ are proper non-trivial factors of $k$.
Using the identity,
$x^n-y^n=(x-y)(x^{n-1}y+...+xy^{n-1})$
Thus,
$\frac{(10^a-1)((10^a)^{b-1}+(10^a)^{b-2}+...+1)}{9}$
Using, this identity again,
$\frac{(10-1)(10^{a-1}+...+1)((10^a)^{b-1}+...+1)}{9}$
You are left with,
$(10^{a-1}+...+1)((10^a)^{b-1}+...+1)$
Since, $a,b\not = 1$ These two factors are not trival. Thus,
$\frac{10^k-1}{9}$ cannot be prime a contradiction.
Great!
You proved if k is composite
$\frac{10^k-1}{9}$ is composite.
What if k is prime?

KeepSmiling
Malay

4. Originally Posted by malaygoel
What if k is prime?
Unknown.
Sometimes prime and sometimes is not.
If it were, then we would have the very first case of a prime producing phunction!

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