2014 AMC 8 Problems/Problem 21

Problem

The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

Solution

The sum of a number's digits $\mod{3}$ is congruent to the number $\pmod{3}$ . $74A52B1 \mod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \mod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\equiv 0 \pmod{3}$ , and therefore $C\equiv 1\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\textbf{(A) }1$ .

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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2014 AMC 8 Problems/Problem 21

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