Competition problem — Number theory

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Number theory Difficulty 4.6 AIME Prove it Ukraine

There are $22$ cards, where the numbers $1, 2, \ldots, 22$ are written. Using these cards one formed $11$ fractions. What is the greatest possible number of integer numbers among the fractions?

Solution

The numbers $13$ , $17$ , $19$ may form an integer only if they stand at the numerator position and $1$ stands in the denominator position. Hence, at least one fraction cannot be integer. However, $10$ numbers may occur to be integer:
$\frac{22}{11}, \frac{14}{7}, \frac{15}{5}, \frac{21}{3}, \frac{20}{10}, \frac{18}{9}, \frac{16}{8}, \frac{12}{6}, \frac{19}{1}, \frac{4}{2}, \frac{13}{17}$

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