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
Proof. Let be a positive odd prime . Let’s form the set of the integers
We make the antithesis that no one of these numbers is divisible by . Therefore, their least nonnegative remainders modulo are some of the numbers
which comprises at least one 9 and one 0. Because of the equal remainders of and , the difference is divisible by . Since and 2 and 5 are the only prime factors of the latter factor (http://planetmath.org/Product), must divide the former factor (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 .
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 , there is a number divisible by .
|Title||divisibility of nine-numbers|
|Date of creation||2013-03-22 19:04:43|
|Last modified on||2013-03-22 19:04:43|
|Last modified by||pahio (2872)|