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Homedivisibility of nine-numbers

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# 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 $999...9$ 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 $b$ ($a<b$), having the same remainder. The difference $b\!-\!a$ then has the decadic representation 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, $p$ 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 positional digital systems, one can write propositions analogous to the above one concerning the decadic system, for example in the dyadic (a.k.a. binary) digital system:

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

## Mathematics Subject Classification

11A63*no label found*11A05

*no label found*

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