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

For every positive odd prime $p$ except 5, there is a nine-number $999...9$ divisible by $p$.

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

 $\displaystyle 9,\,99,\,999,\,\ldots,\,\underbrace{99...9}_{p\;\mathrm{nines}}.$ (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

 $\displaystyle 1,\,2,\,3,\,\ldots,\,p\!-\!1.$ (2)

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

 $b\!-\!a\;=\;999...9000...0,$

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  $b\!-\!a=999...9\!\cdot\!1000...0$  and 2 and 5 are the only prime factors of the latter factor (http://planetmath.org/Product), $p$ must divide the former factor $999...9$ (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 $111...1_{\mathrm{two}}$ divisible by $p$.

Title divisibility of nine-numbers DivisibilityOfNinenumbers 2013-03-22 19:04:43 2013-03-22 19:04:43 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 11A63 msc 11A05