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Find the smallest integer $n$ such that $\frac{11}{15}>\frac{m}{n}>\frac{7}{10}$ and $m$ is an integer.

There are multiple ways to solve this problem. We can get rid of the fractions by multiplying by $30n$ and do casework, we can convert to decimals and do casework, heck, we can graph and do casework. I know that this problem is really simple, but I am just looking for a way to solve this problem without casework.

TheBestMagician
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  • Farey sequence – Alexey Burdin Aug 13 '20 at 19:51
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    But that is not the smallest possible denominator. $\frac57$ works as well. Ah, I see your edit. How can we apply the Farey sequence to this problem? – TheBestMagician Aug 13 '20 at 19:55
  • You need to find minimal $n$ such that $\left\lfloor n\cdot \frac{11}{15}\right\rfloor=\left\lceil n\cdot \frac{7}{10}\right\rceil$. Knowing $n=7$ you'll have to check every $n<7$, yes, it's a casework, but Farey sequences don't work directly here as $\frac{11}{15}$ and $\frac{7}{10}$ are not neighbours: $11\cdot 10-15\cdot 7\ne 1$, so building Farey sequences to order $15$ will take more time than "casework". But it's a direct approach. In the end you'll get $\frac{7}{10}<\frac{5}{7}<\frac{8}{11}<\frac{11}{15}$ each of them are neighbours, so nothing with a less denominator in between. – Alexey Burdin Aug 13 '20 at 20:08

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It is clear that $m<n<2m$. So we can write $n=m+r$ with $0 < r <m$. Then we have $$\frac{7}{10} < \frac{m}{m+r}<\frac{11}{15}.$$ This is same as saying $$\frac{7r}{3} < m<\frac{11r}{4}.$$ Now we want the smallest $r$ (so that $n$ is smallest) for which $m$ is an integer. With $r=1$, the interval $\left(\frac{7}{3},\frac{11}{4}\right)$ will not contain any integer. With $r=2$ it does because $5 \in \left(\frac{14}{3},\frac{22}{4}\right)$, so $m=5$ and correspondingly $n=7$ is the smallest.

Anurag A
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