2014 AMC 8 Problem 21 Divisibility (Number Theory)

Understand the 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 $C$? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$
Source of the problem

2014 AMC 8 Problem 21

Topic

Divisibility Number Theory

Difficulty Level
Easy
Suggested Book
Mathematical Circles

Start with hints

Do you really need a hint? Try it first!

Recall  the rule for divisibility by 3 . Then proceed to apply .
https://en.wikipedia.org/wiki/Divisibility_rule#Divisibility_by_3_or_9  \( \\ \) We came to know that any number N (in base 10) is divisible by 3 iff the sum of all decimal digits of the number is divisible by 3 .Alternatively any number N (in base 10) is congruent to 0 mod 3 iff the sum of all decimal digits of the number is congruent to 0 mod 3 .
So , $$ 74A52B1 \ and \ \ 326AB4C  \ \ are \ multiples \ of \  3 \\ \Rightarrow 7+4+A+5+2+B+1 \equiv 0 (mod \ 3)  \ and  \ 3+2+6+A+B+4+C \equiv 0 (mod \ 3)  \\ \Rightarrow A+B \equiv 2 (mod \ 3) \ and A+ B+C \equiv 0 (mod \ 3) \\ $$
$$  A+B \equiv 2 (mod  \ 3) \ and  \ \ A+ B+C \equiv 0 (mod 3) \\ \Rightarrow C \equiv 1 (mod \ 3) $$ So , C can be 1 , 4  or 7 . So option (A) is right .

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