Too early in the morning to have SODA?

Question

Each letter shown represent distinct digit...can vary from zero to nine.

$COCA$, $COLA$, $SODA$ are three concatenated numbers.

Figure these out from the following relation:

$COCA + COLA = SODA$

Answer 1

Based on Omega Krypton's answer,

$2C+1=S,C+L=D+10$, $A=0,O=9$. (Note that $O=9$ so $C+L$ carries.)
We also need that these digits $C,L,D,S$ are distinct between $1\sim 8$. ($0$ and $9$ are taken.)
If $C=1$ or $C=2$ then, since $D\ge 1$ we have $L\ge 9$ which is incorrect.
So $C=3$ and $S=7$. We have $L=8$ and $D=1$.
That is $3930+3980=7910$.

Answer 2

We have the following

 COCA
+COLA
-----
 SODA

First, from the ones column, we have $A+A \implies A$ which is only possible if $A=0$.

Next, notice something similar in the

hundreds place; $O+O \implies O$. Since $0$ is already taken and the only possibility without a carry over, we must have a carry over from the 10s, and $O=9$ is the only possibility. We will also have a carry over into the thousands.

Since we have a 4 digit number as the result, we know that

$0 \lt C \le 4$.

But:

-But $C=4 \implies S=9$ which is already taken by $O$.
-And $C=1 \implies L=9$ to achieve a carryover, which is taken by $O$.
-And $C=2 \implies L\in\{8,9\}$. But $L=9$ is taken, and $L=8 \implies D=0$ is also taken.

Thus,

$C=3$.

Also, we know

$S=7$ because the hundreds will carry over, and we also know that in order to carry over the 10s, we need $L\ge 7$. But $L=7$ and $L=9$ are taken leaving only $L=8$, and thus, $D=1$.

Thus, the solution is;

COCA+COLA=SODA, 3930+3980=7910

Answer 3

Since we know that

$A+A \equiv A \pmod {10}$

Therefore $A$

$=0$

Hundreds value must carry since $O \neq 0$

Therefore

$O+O+1 \equiv O \pmod {10}$

Therefore $O$

$=9$

We now get

$2C+1=S$
$C+L=D$

And since $S<9$

$0<C<4$

Then there are many possibilities... any relations I missed out?