divisibility of nine-numbers

We know that 9 is divisible by the prime number 3 and that 99 by another prime number 11.  If we study the divisibility other “nine-numbers” by primes, we can see that 999 is divisible by a greater prime number 37 and 9999 by 101 which also is a prime, and so on.  Such observations may be generalised to the following

Proposition.  For every positive odd prime $p$ except 5, there is a nine-number $999\mathrm{\dots }9$ divisible by $p$.

Proof.  Let $p$ be a positive odd prime $\ne 5$.  Let’s form the set of the integers

$9,\mathrm{\hspace{0.17em}99},\mathrm{\hspace{0.17em}999},\mathrm{\dots },\underset{p\mathrm{nines}}{\underbrace{99\mathrm{\dots }9}}.$(1)

We make the antithesis that no one of these numbers is divisible by $p$.  Therefore, their least nonnegative remainders modulo $p$ are some of the $p-1$ numbers

$1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots },p-1.$(2)

Thus there are at least two of the numbers (1), say $a$ and $b$ (), having the same remainder.  The difference$b-a$ then has the decadic of the form

$$b-a=\mathrm{\hspace{0.33em}999}\mathrm{\dots }9000\mathrm{\dots }0,$$

which comprises at least one 9 and one 0.  Because of the equal remainders of $a$ and $b$, the difference is divisible by $p$.  Since  $b-a=999\mathrm{\dots }9\cdot 1000\mathrm{\dots }0$  and 2 and 5 are the only prime factors of the latter factor (http://planetmath.org/Product), $p$ must divide the former factor $999\mathrm{\dots }9$ (cf. divisibility by prime).  But this is one of the numbers (1), whence our antithesis is wrong.  Consequently, at least one of (1) is divisible by $p$.

In other http://planetmath.org/node/3313positional digital systems, one can write propositions analogous to the above one concerning the decadic system, for example in the dyadic (a.k.a. digital system:

Proposition.  For every odd prime $p$, there is a number $111\mathrm{\dots }{1}_{\mathrm{two}}$ divisible by $p$.