parmenides51 wrote: Show that if a $6n-$digit number is divisible by $7$, then the number that results from moving the ones digit to the beginning of the number is also a multiple of $7$. can someone pls give an example of 6n digit number? is it like 123456 is a 6n digit number

Let the unit digit of the $6n$ digit number be $a_0$, and let the number itself be $7k$ for some integer $k$. Now the new number is the same as $a_0 \cdot 10^{6n-1} +\frac{7k -a_0}{10}$. We just need to prove that this new expression is $\equiv 0 (mod ~7)$. Now $a_0 \cdot 10^{6n-1} \equiv a_0 \cdot 3^{6n-1} \equiv a_0 \cdot \frac{3^{6n}}{3} \equiv \frac{a_0}{3} $ (Since $729 \equiv 1$).$\frac{7k -a_0}{10} \equiv \frac{-a_0}{3}$. So the whole expression is $\equiv \frac{a_0 -a_0}{3} \equiv 0$, and we're done. @above I think it just means that the number of digits in the number is a multiple of $6$.