How do I show this result:
There are n 1's..
1111111..(n times)
Then if n = p * m where p and m are integers
=
11....1 (n 1's) = (11...1)*(100..0100...01......1), where the right factor in the right side is composed of 1, then p-1 0's, then 1, then p-1 0's, etc., up to the last 1 which is the m-th one.

Question

How do I show this result:
There are n 1's..
1111111..(n times)
Then if n = p * m where p and m are integers
 =>
11....1 (n 1's) = (11...1)*(100..0100...01......1), where the right factor in the right side is composed of 1, then p-1 0's, then 1, then p-1 0's, etc., up to the last 1 which is the m-th one.

abb0t · Answer

|dw:1359831565515:dw|

anonymous · Answer

@amistre64 , @Yahoo!

abb0t · Answer

A line over the top of the repeated number(s)

anonymous · Answer

I dont understand??

anonymous · Answer

Could you refer to the q again?

anonymous · Answer

@experimentX

experimentx · Answer

this is not always true. eg 111 = 37x3

experimentx · Answer

maybe 11111... = 99999../9 = (10^(n+1)-1)/9 ... 
If this n+1 is not prime, then (10-1)(10+1)(100+1) ... /9
something like that ... sorry, extremely busy due to exams!!

anonymous · Answer

1111111111111

anonymous · Answer

@Ineedanswersbadfail  ?? 
@experimentX It meant that 111, has 3 ones.
So 3=3x1.
Therefore it can only be expressed as 111*1

experimentx · Answer

That's a trivial case  when you the (number of digits +1) is prime ... I think we could use above idea.

Suppose (the number of digits +1)is not prime.
$$ 1111...111 \text{ (m times)}(m+1 = p q) \  = \frac{10^{m+1}-1}{9}=\frac{10^{pq}-1}{9} = \frac{10^p - 1}{9}(1 + 10^q + 100^q+ ... + 10^{(p-1)q}) \
= 111...111 \times 10010...01001$$