Maths Olympiad Prep

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, 2021

Number theory Difficulty 4.8 AIME Find the answer United States

Problem:

For each positive integer 1m101 \leq m \leq 10, Krit chooses an integer 0am<m0 \leq a_{m} < m uniformly at random. Let pp be the probability that there exists an integer nn for which nam(modm)n \equiv a_{m} \pmod{m} for all mm. If pp can be written as ab\frac{a}{b} for relatively prime positive integers aa and bb, compute 100a+b100a + b.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Tuples of valid ama_{m} correspond with residues mod lcm(1,2,,10)\operatorname{lcm}(1,2, \ldots, 10), so the answer is
lcm(1,2,,10)10!=2332572834527=11440 \frac{\operatorname{lcm}(1,2, \ldots, 10)}{10!} = \frac{2^{3} \cdot 3^{2} \cdot 5 \cdot 7}{2^{8} \cdot 3^{4} \cdot 5^{2} \cdot 7} = \frac{1}{1440}

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