Find all the digits $a,b$ in such a way that $n=62ab427$ is multiple of $99$

Question

Find all the digits $(a,b)$ in such a way that $n=62ab427$ is a multiple of $99$

My try

If its a multiple of $99$ it should be divisible by 9 and 11, so

Divisible by 9 : $6+2+a+b+4+2+7 = a+b+21= 9u$ with $u$ $\in$ $\Bbb N^+$

but $21$ $\le$ $a+b+21$ $\le$ 39 because $a+b$ $\le$ $18$

Divisible by 11: $6-2+a-b+4-2+7$ = $a-b+13 = 0$ or $a-b+13=11v$ with $v \in \Bbb N^+$

I can't see how to continue this problem and i don't know if my approaches are good or not.

Answer 1

I think your try is good. I'm not sure where your Problems come from.

1. You need $a+b+21$ to be divisible by $9$, so $a+b = 6$ or $a+b=15$

2. You need $a-b+13$ to be divisible by $11$, so $a-b = -2$ or $a-b = 9$

There is only one combination (out of 4) that leads to a result:

$a+b=6$, $a-b=-2$, then $a=2$ and $b=4$

Answer 2

$\textbf{Hint:}$

$a+b+21=9u\Longrightarrow a+b=6$ or $a+b=15$

$a-b+13=11v\Longrightarrow a-b=9$ or $a-b=-2$