$A289B$ is divisible by $90$, then what is $A+B$?

Question

It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?

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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.

$$A+B = 8$$

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Is this the best approach..? Thanks.

Answer 1

If an integer is divisible $9$ then the sum of the digits is also divisible by $9$. So

$$A+2+8+9 +B \equiv A+B + 1 \equiv 0 \pmod{9}$$ shows $A+B \equiv_9 8$. Then since $B=0$ like you noted, $A+B=8$.

Answer 2

You can simply take the sum of digits $A+2+8+9+0 \equiv A + 1 \equiv 0 \pmod 9$, so as to know that $A \equiv 8 \pmod 9$ is the desired solution because $10^k \equiv 1 \pmod 9$.

Answer 3

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 $n \mod 9 \equiv \sum_{k=0}a_k \mod 9$.