# Thread: Prove that p divides infinitely many numbers...

1. ## Prove that p divides infinitely many numbers...

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

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

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

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

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

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

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

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

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