We have to determine the missing digit $y$ in a $41$-digit number $N$, where the first $20$ digits are $8$'s and the last $20$ digits are $9$'s, such that $N$ is divisible by $7$.
i.e. $N= \overline{\underbrace{8\ldots8}_{20} \;y \; \underbrace{9\ldots9}_{20}}$ and must be divisible by $7$.
Kindly explain the steps to reach an answer.
Thanks!
the searched number is $$y=5$$ we use the rule that a numberr is divisible by $7$ if we group the digits of the given number in triplets starting from the right and adding and subtracting this numbers and so on thus we get $$100y+99-88=100y+11\equiv 2y+4\mod 7$$ with $$y=1,2,3,4,5,6,7,8,9$$ only $$y=5$$ is a solution since $$2\cdot 5+4\equiv 0 \mod 7$$
Hints : $10^n \equiv 3^n [7]$
$3^{6} \equiv 1 [7] $