2023 IOQM/Problem 3

Problem

Let α and β be positive integers such that $\frac{16}{37}<\frac{\alpha}{\beta}<\frac{7}{16}$ Find the smallest possible value of β .

Solutions

First reciprocal $\frac{\alpha}{\beta}$ to get a smaller number, $\frac{16}{7}<\frac{\beta}{\alpha}<\frac{37}{16}$ . This gives , $\frac{16\alpha}{7}<\beta<\frac{37\alpha}{15}$ . Now $\frac{16\alpha}{7}\approx 2.28\alpha$ and $\frac{37\alpha}{16} \approx 2.31\alpha$ . So we have $2.28\alpha<\beta<2.31\alpha$ . To make $\beta$ minimum we can put $\alpha=10$ to get $22.8<\beta<23.1$ . This gives $\boxed{\beta=23}$

~Lakshya Pamecha and A. Mahajan Sir's

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2023 IOQM/Problem 3

Problem

Solutions