Algebra and Divisibility

Divisibility test of 11 is that if you sum every second digit and then subtract all other digits and the answer is 0 or 11.

So

$8-k+6-k+k-5=11\quad OR\quad 0\\ 9-k=11\quad OR\quad 0$

Now we need a even answer so

$9-k=0\\ k=9$

Can Also be done without using Divisibility test

The number

$8k6kk5 \% 11 = 0$

$8k6kk5$ can be written as $806005$ + $k0kk0$

$806005 \% 11 = 2$

$k0kk0$ can be written as $k(10000 + 100 + 10) = k(10110)$

$10110 \% 11 = 1$

So , $(2 + k) \% 11 = 0$

$k = 9$

The number can be divided by 11 if the sum of the digit in the odd and even is same, 8k6kk5 so, 8 (first digit) k (second digit) 6 (third digit) k (fourth digit) k (fifth digit) 5 (sixth digit)

so, 8+6+k=k+k+5 k=9