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If $p$ and $q$ are positive integers such that $\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$ then the smallest possible value of $q$ is:

$(A)\quad 60;\quad (B)\quad 30;\quad (C)\quad 25;\quad (D)\quad 7$.

What is the correct way to solve this kind of problems? I have tried to simplify the given inequality:

$$0.70<\frac{p}{q}< \approx0.73 \quad\text{ or }\quad \frac{21}{30}<\frac{p}{q}<\frac{22}{30}$$

Thank you in advance!

Community
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keylin.rion
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    Don't use decimals. Your fraction representation is good. Does $q = 30$ work based on your work (meaning can you squeeze a $p$ between $21$ and $22$)? If not, what's an obvious next choice for a common denominator to work with? Can you find a $p$ that works for that value of $q$? – Cameron L. Williams Sep 11 '19 at 14:05
  • Not difficult enough it seems. $q=30$ does not work. The next choice for a common denominator is $60$. $\dfrac{42}{60}<\dfrac{p}{60}<\dfrac{44}{60}$. Thank you! I appreciate it. – keylin.rion Sep 11 '19 at 14:12
  • @keylin.rion Happy to help. Great work. – Cameron L. Williams Sep 11 '19 at 14:13
  • Actually, $q=60$ is not the smallest value as I wrote above. – richrow Sep 11 '19 at 14:14
  • Whooooops. I didn't see that the prompt says smallest. I should probably learn to read more gooder. – Cameron L. Williams Sep 11 '19 at 14:19
  • @richrow, May I ask you how did you come up with the idea that we can take $p=7+11=18$ and $q=10+15=25$? – keylin.rion Sep 11 '19 at 14:35
  • You can prove the general statement: for positive reals $a, b, c, d$ if $\frac{a}{b}<\frac{c}{d}$ then $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$. Such transformations are also used in the theory of Farey sequence. – richrow Sep 11 '19 at 17:43

3 Answers3

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If $7$ worked we would have

\begin{align} &\frac{21}{30}<\frac{p}{7}< \frac{22}{30}\\ \iff&\frac{21*7}{30*7}<\frac{30p}{30*7}< \frac{22*7}{30*7} \end{align} So $7$ is a solution if and only if there is a multiple of $30$ between $21*7$ and $22*7$. Since $21*7=147$ and $22*7=154$ we have a multiple of $30,$ namely $150$. So $p=\frac{5}{7}$ is between those numbers.

To prove that this is the smallest just notice that every number smaller than $7$ other than $4$ is a divisor of $30$ so their fractions can be written as $\frac{p}{30}.$ And obviously it is impossible to have

\begin{align*} \frac{21}{30}<\frac{p}{30}<\frac{22}{30} \end{align*}

As for $4$ if $4$ was possible we would have

\begin{align} &\frac{21}{30}<\frac{p}{4}<\frac{22}{30}\\ \iff & \frac{21*2}{30*2}<\frac{p*15}{4*15}<\frac{22*2}{30*2} \end{align} and there are no multiples of $15$ between $42$ and $44$.

pm2595
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    Since 7 works and it's the smallest option given, it has to be right. You can often make use of the properties of the options in multiple choice question. – Joonas Ilmavirta Sep 11 '19 at 14:35
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    You are right, the second part was unnecessary looking at the multiple choices options. My first answer, before the edit, only proved that 7 was possible and since 7 was the smalles option it had to be it. Then, I started thinking why would the numbers smaller than 7 not work, and decided to edit and add that explanation as well, more as a curiosity. But if the only goal was to the answer the question asked as it is, I agree that as soon as we prove that 7 is a solution we should stop. – pm2595 Sep 11 '19 at 14:45
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The way to find the smallest denominator "from scratch" is with continued fractions.

Begin by rendering the proposed bounds thusly:

$\dfrac{7}{10}=\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{3}}}}$

$\dfrac{11}{15}=\dfrac{1 }{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{1+\dfrac{1}{3}}}}}$

The upper layers of the continued fractions are identical but they eventually become different when we get down to the layers in blue. We may now replace those entries with the smallest whole number lying between, thus

$1+\dfrac{1}{3}<2<3$

So the smallest denominator fraction meeting the betweenness criterion will be

$\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\color{blue}{2}}}}=\dfrac{5}{\color{blue}{7}}$

Oscar Lanzi
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  • Thank you for the response. I appreciate it. I do not understand how you rendered the proposed bounds. – keylin.rion Sep 11 '19 at 19:32
  • Read up.. – Oscar Lanzi Sep 11 '19 at 20:43
  • Thanks! I got how to make a continued fraction. To be honest, after rendering the proposed bounds, I don't understand how your solution goes. – keylin.rion Sep 11 '19 at 20:44
  • The value of a continued fraction is a monotonic function of any part that lies below a given layer. Here the stuff below the third layer is $1+1/3$ in one case and $3$ in the other, putting $2$ there with the function being monotonic is sure to give an overall value that's between the given bounds. – Oscar Lanzi Sep 11 '19 at 21:21
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We can take the common denominator: $$ \frac{21q}{30q}<\frac{30p}{30q}<\frac{22q}{30q},\qquad\text{or}\qquad 21q<30p<22q. $$

Then you go from the smallest $q$: $$ 21\cdot7=147<30p<22\cdot7=154, \qquad \ldots 120,150,180\ldots\\ $$ We see that $30p=150$ lies between $147$ and $154$.

Vasily Mitch
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