Prove that p divides infinitely many numbers...

**NikoBellic** · February 28th 2010, 11:11 AM

Let p>5 be a prime. Prove that p divides infinitely many numbers of the (base ten) form:

1, 11, 111, 1111, 11111, 111111, ...

**chiph588@** · February 28th 2010, 04:55 PM

Let $\displaystyle a=\phi(p)=p-1$ .

Define $\displaystyle x=\sum_{i=0}^{a-1} 10^i = \underbrace{11\cdots 1}_a$ .

Note $\displaystyle 9x=9\sum_{i=0}^{a-1} 10^i = 10^a-1 \equiv 0 \mod{p}$ , since $\displaystyle (10,p)=1$ i.e. $\displaystyle p \neq 2,5$ .
Thus $\displaystyle p\neq 3 \implies x\equiv 0 \mod{p}$ .

Now let $\displaystyle x_i = x\cdot 10^{i\cdot a}$
and define $\displaystyle S=\left\{\sum_{i=0}^{n} x_i | n\in\mathbb{N}\right\}$ .

Observe if $\displaystyle b=\sum_{i=0}^{n} x_i \in S$ , then $\displaystyle b=\underbrace{11\cdots1}_{(n+1)\cdot a}$ .

But $\displaystyle b=\sum_{i=0}^{n} x\cdot 10^{i\cdot a} = x\sum_{i=0}^{n}10^{i\cdot a} \equiv 0 \mod{p}$ since $\displaystyle x\equiv 0 \mod{p}$ .
Hence $\displaystyle b\equiv 0 \mod{p} \; \forall \; b\in S$ .

Since $\displaystyle S$ is not a finite set, we are done.

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