2014 AMC 8 Problems/Problem 21

Revision as of 16:47, 10 October 2015 by Hubzy (talk | contribs) (Solution)

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 $\textbf{(A) }1$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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