# Too early in the morning to have SODA?

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:

$$COCA + COLA = SODA$$

• deleted image. feel free to rollback if needed :) – Omega Krypton Jun 17 at 8:40
• Am I the only one who thought "Too early in the morning" was a hint? – Mr Lister Jun 18 at 12:10
• I don’t know..when I got up little early yesterday, title popped in my head – Uvc Jun 18 at 12:13

$$2C+1=S,C+L=D+10$$, $$A=0,O=9$$. (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 $$3930+3980=7910$$.

• hi, nice try! +1 – Omega Krypton Jun 17 at 8:55
• Got it!!....deceptively unique – Uvc Jun 17 at 8:56
• Yeah deceptively. :) – r_64 Jun 17 at 8:57

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

• Easiest deductions for me =) – Montolide Jun 17 at 18:41

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

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

• On the right track..I think it is unique – Uvc Jun 17 at 8:51
• Keep going..eventually you will get there – Uvc Jun 17 at 8:55