Since both numbers are divisible by $3$ , the sum of their digits has to be divisible by three. $7+4+5+2+1=19$. To be a multiple of $3$ , $A+B$ has to be either $2$ or $5$ or $8 \ldots$ and so on. We add up the numerical digits in the second number; $3+2+6+4=15$. We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that $C=1,4$ or $7,10 \ldots$ and so on, otherwise the number will not be divisible by three. We then add 8 to 15 , to get 23 , which shows us that $C=1$ or $4$ or $7 \cdots$ and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1,4$, and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $(\mathbf{A}) 1$ is our answer. 
The $7$-digit numbers $\underline{74 A 52 B 1}$ and $\underline{326 A B 4 C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ? (2014 amc 8 Problem, Question \#21) 
a. $$1$$ 
b. $$2$$ 
c. $$3$$ 
d. $$5$$ 
e. $$8$$ 