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 999⁢…⁢9 divisible by p.

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

9, 99, 999,…,99⁢…⁢9⏟p⁢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

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= 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⋅1000⁢…⁢0  and 2 and 5 are the only prime factorsMathworldPlanetmath 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⁢…⁢1two 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