Difference between revisions of "2023 IOQM/Problem 3"
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==Problem== | ==Problem== | ||
− | Let α and β be [[positive integers]] such that <cmath>\frac{16}{37} | + | Let α and β be [[positive integers]] such that <cmath>\frac{16}{37}<\frac{\alpha}{\beta}<\frac{7}{16}</cmath> |
Find the smallest possible value of β . | Find the smallest possible value of β . | ||
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+ | ==Solutions== | ||
+ | First reciprocal <math>\frac{\alpha}{\beta}</math> to get a smaller number, <math>\frac{16}{7}<\frac{\beta}{\alpha}<\frac{37}{16}</math>. This gives ,<math>\frac{16\alpha}{7}<\beta<\frac{37\alpha}{15}</math>. Now <math>\frac{16\alpha}{7}\approx 2.28\alpha</math> and <math>\frac{37\alpha}{16} \approx 2.31\alpha</math>. So we have <math>2.28\alpha<\beta<2.31\alpha</math>. To make <math>\beta</math> minimum we can put <math>\alpha=10</math> to get <math>22.8<\beta<23.1</math>. This gives <math>\boxed{\beta=23}</math> | ||
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+ | ~Lakshya Pamecha and A. Mahajan Sir's |
Latest revision as of 01:50, 4 May 2024
Problem
Let α and β be positive integers such that Find the smallest possible value of β .
Solutions
First reciprocal to get a smaller number, . This gives ,. Now and . So we have . To make minimum we can put to get . This gives
~Lakshya Pamecha and A. Mahajan Sir's