[ [ $\left[\begin{array}{l}\text { Divisibility of numbers. General properties } \\ \text { p }\end{array}\right]$

Find such an $n$ that the number $10^{n}-1$ is divisible by a) 7 ; b) 13 ; c) 91 ; d) 819.
$10^{6}-1$ is divisible by $10^{3}+1=1001=7 \cdot 11 \cdot 13$, and therefore, by 7, and 13, and $91=7 \cdot 13$. In addition, $10^{6}-1=$ 999999 is divisible by 9, and therefore, by

$819=9 \cdot 91$.

## Answer

$n=6$.