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Harvard-MIT November Tournament

Number theory Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $N=\overline{5 A B 37 C 2}$ , where $A, B, C$ are digits between $0$ and $9$ , inclusive, and $N$ is a $7$ -digit positive integer. If $N$ is divisible by $792$ , determine all possible ordered triples $(A, B, C)$ .

Solution

Solution:

Answer: $(0,5,5), (4,5,1), (6,4,9)$

First, note that $792 = 2^{3} \times 3^{2} \times 11$ . So we get that
$\begin{gathered} 8 \mid N \Rightarrow 8 \mid \overline{7 C 2} \Rightarrow 8 \mid 10C + 6 \Rightarrow C = 1, 5, 9 \\ 9 \mid N \Rightarrow 9 \mid 5 + A + B + 3 + 7 + C + 2 \Rightarrow A + B + C = 1, 10, 19 \\ 11 \mid N \Rightarrow 11 \mid 5 - A + B - 3 + 7 - C + 2 \Rightarrow -A + B - C = -11, 0 \end{gathered}$
Adding the last two equations, and noting that they sum to $2B$ , which must be even, we get that $B = 4, 5$ .

Checking values of $C$ we get possible triplets of $(0,5,5), (4,5,1)$ , and $(6,4,9)$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.