Maths Olympiad Prep

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Number theory Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let N=5AB37C2N=\overline{5 A B 37 C 2}, where A,B,CA, B, C are digits between 00 and 99, inclusive, and NN is a 77-digit positive integer. If NN is divisible by 792792, determine all possible ordered triples (A,B,C)(A, B, C).

Solution

Solution:

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

First, note that 792=23×32×11792 = 2^{3} \times 3^{2} \times 11. So we get that
8N87C2810C+6C=1,5,99N95+A+B+3+7+C+2A+B+C=1,10,1911N115A+B3+7C+2A+BC=11,0 \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 2B2B, which must be even, we get that B=4,5B = 4, 5.

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

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