Factorising $99999\,00000\,99999\,00001$ I am reading an article in The Mathematical Gazette about factorising the 20 digit number
$$N=99999\,00000\,99999\,00001.$$
It is stated that $N=\dfrac{10^{25}+1}{10^5+1}$ (I understand why this is true), and consequently, if $p$ is a prime factor of $N$, then $p$ must be of the form $50k+1$. Why does this follow?

If $p$ is a prime factor of $N$, then it is a factor of $10^{25}+1$ and thus $$10^{25}\equiv -1 \implies 10^{50}\equiv 1\pmod p,$$
i.e. if $d$ is the order of $10$ modulo $p$, then $d$ divides $50$. But $d$ cannot divide $25$ (otherwise we would have $10^{25}\equiv 1\pmod p$), and $d$ cannot divide $10$ either (otherwise $10^{10}\equiv 1\pmod p$, but one can easily verify that $\gcd(N,10^{10}-1)=1$).
Therefore, $d=50$. And Little Fermat says that $d\mid(p-1)$.
