Competition problem — Algebra — Olympiad Maths Prep

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Algebra Difficulty 4.3 AIME Find the answer Ukraine

Using the numbers $1, 2, \ldots, 20$ (each number once) as denominators and numerators, construct $10$ fractions with integer sum.

Solution

$\frac{17}{3} + \frac{13}{2} + \frac{11}{6} + \frac{19}{1} + \frac{14}{7} + \frac{18}{9} + \frac{20}{10} + \frac{16}{8} + \frac{15}{5} + \frac{12}{4} = 14 + 19 + 2 + 2 + 2 + 2 + 3 + 3 = 47.$

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