Find $a + b + c$, if $4a1b4c9$ is divisible by $99$.
Since it is divisible by $99$, it would be divisible by $11$ and $9$. Applying divisibility rules of $9$ and $11$:
$18 + (a + b + c)$, must be a multiple of $9$ (I);
$18 - (a + b + c)$, must be either $0$ or a multiple of $11$ (II).
From I, $a + b + c$ is clearly a multiple of $18$. Now, how to use this to single out either $0$ or a multiple of $11$ in equation II?
Say, if $a + b + c = 216$, then $18 - (a + b + c)$ would be a multiple of $11$. Again, if $a + b + c = 9$ or $18$ or $27$ or $45$ etc., then $18 - (a + b + c) = 0$ (since it won't be a multiple of $11$, from II), giving $a + b + c = 18$.
Please, give me hints to proceed further, or correct me if I am wrong anywhere. Thanks.
I followed this for divisibility rules - https://www.mathsisfun.com/divisibility-rules.html
Looks like you messed up your math:
Reading the question, it looks like you want the number $4a1b4c9$ to be divisible by $99$, or both $11$ and $9$.
Now, for it to be divisible by $9$, it follows that the sum of the digits are divisible by $9$:
$$18 + (a+b+c) \equiv 0 \mod 9$$
And for $11$, the difference between the sum of the alternating digits are either divisible by $0$ or $11$:
$$18 - (a+b+c) \equiv 0 \mod 11$$ $$18 - (a+b+c) = 0$$
If you notice from the first equation, $18$ if already divisible by $9$, so we only want $(a+b+c)$ to be divisible by $9$.
From the next $2$ equations we could let the quantity $18 - (a+b+c)$ be divisible by $0$, therefore letting the number $4a1b4c9$ be divisible by $11$. Thus, we get that:
$$a+ b+c = 18$$
And we are done.