Question

Take any three digit number and write it two times to make a $6-digit$ number. Verify whether it is divisible by both $7$ and $11$

Answer 1

$\left(i\right)$ The Divisibility Rule of $11$ is $\Rightarrow$ The alternative sum of the digits of Number is divisible by $11$
$\left(ii\right)$ for divisible by $7\Rightarrow$ Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7, so is the original number.
Let the 3 digit number be $abc$ and writing it two times we get $=abcabc$
Now, $abc\times 1001=abcabc$

hence, $1001$ is a factor of $abcabc$
and $1001$ is divisible by both $7$ and $11$ (as $100-2=98$ applying $\left(i\right)$ again $9-(8\ast 2)=-7$ which is divisible by $7$
and Applying $\left(ii\right)1-0+0-1=0$ which is divisible by $11$)
Therefore $1001$ is divisible by $11$ and $7$
Hence we can see that whatever be the value of $abc$, $abcabc$ is always be divisible by $7$ and $11$