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 1

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 $\boxed{\textbf{(A) }1}$ .

Solution 2

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 5 or 8. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add 5 to 15, to get 20. We then see that C = 1 or 7, 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 in order to be a multiple of three. We take the common numbers we got from both these equations, which are 1 and 7. However, in the answer choices, there is no 7, but there is a 1, so $\boxed{\textbf{(A) }1}$ is your answer. :)

~ UnicornFrappacino

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Another solution is using the divisbility rule of 3. Just add the digits of each number together.

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2014 AMC 8 Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

See Also