2025/005) Consider the formation of integer, $x = 1555555\cdots 526$ . How to show that whatever amount of digit 5 is placed between $1 \& 2 , 7 ∣ x$

We have

$x = 1555555\cdots 526$ number of 5's >=0$

so $x + 29 = 1555555\cdots 5$

Let it be a n digit number 

so $x + 29 = 10^{n-1} +  5 *111\cdots 1$                                  n-1 1's

or  $x + 29 = 10^{n-1} + \frac{5}{9}(10^{n-1} -1 )$

as 7 is not a factor of 9 

So we have $9(x+29) = 9 * 10^{n-1} + 5 * 10^{n-1} -5$

or $9x + 9 * 29 -5 = 14 * 10^{n-1}$

so $9x + 266$ is divisible by 7

as $266$ is divisible by $7$ so $9x$ is divisible by $7$ and $7$ is coprime to $9$ so $x$ is divisible by $7$