<s>[INST] $A289B$ is divisible by $90$, then what is $A+B$? It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?

My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
since it is divisible by $90$, the sum of the remainders of $A \times 1000$ and $2890$ being divided by 90, must be divisible by 90.
We have  $mod(2890, 90) = 10$.
Now we may find $A$ such that $mod(A \times 10000, 90) = 80$.
$$mod(10000, 90) = 10 \implies 10000 = (90 \times 111 + 10) $$
so $A$ must be $8$, since
$$ 8 \times 10000 + 2890 = 8 \times (90 \times 111 + 10) + (90 \times 32 + 10) = 90 \times (888+32) + 90 $$
is divided by 90.
$$A+B = 8$$

Is this the best approach..? Thanks. 
 [/INST] You are going to a lot of effort to avoid noting that if $A2890=A289*10$ is divisible by $90$ and $\gcd(9,10) = 1$ then $A289$ is divisible by $9$.
And $A289 = A28*10 + 9$ being divisible by $9$ means $A28*10 + 9 \equiv A28 \equiv A27 + 1 \equiv 100A + 27 + 1 \equiv A + 1 \mod 9$.  And $A289\equiv 0 \mod 9$ so $A+1 \equiv 0 \mod 9$ so $A = 8$.
So $A+B = 8+0= 8$.
That's assuming you are not aware that if a number is divisible by $9$ then the sum of it's digits is divisible by $9$.  Which can be proven the exact same way.
$n = \sum_{k=0}n a_k 10^k = \sum_{k=0} a_k (10^k - 1) + \sum_{k=0}a_k$ and as $(10^k - 1) = 9*(\underbrace{1111....111}_{k}) \equiv 0 \mod 9$, we may conclude that  $n \mod 9 \equiv \sum_{k=0}a_k \mod 9$.
</s>