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Romanian Mathematical Olympiad - District Round

Algebra Difficulty 5.8 AIME, harder Prove it Romania

Find all the pairs $(x, y)$ of real numbers fulfilling
$3 \cdot \left\{ \frac{3x+2}{3} \right\} + 4 \cdot \left\lfloor \frac{4y+3}{4} \right\rfloor = 4 \cdot \left\{ \frac{4y+3}{4} \right\} + 3 \cdot \left\lfloor \frac{3x+2}{3} \right\rfloor = 18.$

Solution

Since $0 \le \{a\} < 1$ and $\lfloor b \rfloor$ is an integer, the equality $3\{a\} + 4\lfloor b \rfloor = 18$ is possible only when $\{a\} = \frac{2}{3}$ and $\lfloor b \rfloor = 4$ .
Since $0 \le \{c\} < 1$ and $\lfloor d \rfloor$ is an integer, the equality $4\{c\} + 3\lfloor d \rfloor = 18$ is possible only when $\{c\} = 0$ , $\lfloor d \rfloor = 6$ (I) or $\{c\} = \frac{3}{4}$ , $\lfloor d \rfloor = 5$ (II).

Case (I) yields $\frac{3x+2}{3} = 6 + \frac{2}{3}$ and $\frac{4y+3}{4} = 4$ , hence $x = 6$ , $y = \frac{13}{4}$ .
Case (II) yields $\frac{3x+2}{3} = 5 + \frac{2}{3}$ and $\frac{4y+3}{4} = 4 + \frac{3}{4}$ , so $x = 5$ , $y = 4$ .

The required pairs are $(6, \frac{13}{4})$ and $(5, 4)$ .

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