divisibility of nine-numbers


We know that 9 is divisible by the prime numberMathworldPlanetmath 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 9999 divisible by p.

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

9, 99, 999,,999pnines. (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, 2, 3,,p-1. (2)

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

b-a= 99990000,

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=999910000  and 2 and 5 are the only prime factorsMathworldPlanetmath of the latter factor (http://planetmath.org/Product), p must divide the former factor 9999 (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 1111two divisible by p.

Title divisibility of nine-numbers
Canonical name DivisibilityOfNinenumbers
Date of creation 2013-03-22 19:04:43
Last modified on 2013-03-22 19:04:43
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 8
Author pahio (2872)
Entry type Theorem
Classification msc 11A63
Classification msc 11A05