Difference between revisions of "2023 IOQM/Problem 3"
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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 01:50, 4 May 2024
Problem
Let α and β be positive integers such that ![\[\frac{16}{37}<\frac{\alpha}{\beta}<\frac{7}{16}\]](http://latex.artofproblemsolving.com/c/5/a/c5a37c4a86d6246fea22dacdffada434f87c22a5.png)
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
