5 digit number $a6a41$ divisible by 9 In the 5-digit number $a6a41$ each of the a's represent the same number. If the number is divisible by 9, what is the digit represented by $a$?
I first approached this by saying $$2a + 11$$ since it's divisible by 9. I got stuck because I didn't know what to equate it to. So I randomly said $2a + 11 = 27$ and then I got $ a = 8$ and indeed the number was divisble by 9. My question is, how do I know what to equate it to?
 A: Well if $(2a + 11)$ is a multiple of $9$, then
$$(2a + 11) = 0\ (mod \ 9)$$
so $$(2a + 2) = 0\ (mod\ 9)$$
and since $(2a + 2)$ is even, it follows that
$$(a + 1) = 0\ (mod\ 9)$$
A: Let's look a little more carefully at the divisibility rule you're applying.

A number is divisible by 9 if and only if the sum of its digits is divisible by 9

There is nothing overtly in here about equality, although there is an explanation with a type of equality I'll discuss in a second.
Bottom line is, no matter what digit you pick $a$ to be, as long as $2a+11$ is divisible by 9, that $a$ will work.
If you are learning modular arithmetic, then you really can find an equality. A number $n$ satisfies $n\equiv 0\pmod{9}$ iff 9 divides $n$.
So the rule you mentioned above can be rephrased as 

A number is equal to 0 mod 9 iff the sum of its digits is equal to 0 mod 9.

Solving $2a+11\equiv 0\pmod{9}$ for $a$ is not really much different than solving $2a+11=0$. You can say that $2a\equiv -11\pmod{9}$, and then you need to find an inverse for 2 (that us, a number to multiply it with that turns it into a 1. Module 9, it's easy to discover that $5$ does the trick. So, $a\equiv 5\cdot2a\equiv 5(-11)\equiv -55\pmod{9}$. To get $a$ to be a digit, we need to shift it by multiples of 9. As you can see, $-55\equiv -55+63\equiv 8\pmod{9}$.
A: In general, problems like this may have multiple solutions, so there may not be a unique $x$ in the equation $2a+11=x$. However, there are a few restrictions that apply: 


*

*$a$ must not be negative, so $x\geq 11$;

*$a< 10$, so $x<31$.


This leaves only $a=18,27$ to try. Moreover:


*

*$2a$ is even and 11 is odd, so $x$ must be odd as well,


so 27 is the only possibility.
