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}\leq\frac{\alpha}{\beta}\leq\frac{7}{16}</cmath>
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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==
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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 02:50, 4 May 2024

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