What is the smallest possible value of y for given x and y?

[anemone]

Let $x,\,y$ be positive integers such that $\dfrac{7}{10}<\dfrac{x}{y}<\dfrac{11}{15}$. Find the smallest possible value of $y$.

 

[alexmahone]

Let $x,\,y$ be positive integers such that $\dfrac{7}{10}<\dfrac{x}{y}<\dfrac{11}{15}$. Find the smallest possible value of $y$.

Spoiler

$\dfrac{7}{10}<\dfrac{x}{y}<\dfrac{11}{15}$

Multiplying throughout by $30y$, we get

$21y<30x<22y$

So we have to find a multiple of 30 lying between multiples of 21 and 22.

Multiples of 30 are 30, 60, 90, 120, 150, 180, 210, ...

Writing down multiples of 21 and 22,

21 22 y=1
42 44 y=2
63 66 y=3
84 88 y=4
105 110 y=5
126 132 y=6
147 154 y=7

Aha...150 lies between 147 and 154 for y=7.
So, the smallest possible value of y is 7.

(Sorry that my solution is not elegant.)

 

[anemone]

Thanks for participating, Alexmahone! Your answer is correct!

...

(Sorry that my proof is not elegant.)

I think as long as a solution led to the correct solution, it can be deemed as an elegant solution, no?

 

[alexmahone]

Thanks for participating, Alexmahone! Your answer is correct!
I think as long as a solution led to the correct solution, it can be deemed as an elegant solution, no?

You're welcome!

But my solution is just brute-force. I hope you or someone else has a better solution.

 

[anemone]

Solution from other:

Spoiler

Suppose $y \le 6$. Then $y$ is a divisor of 60 and hence $\dfrac{x}{y}=\dfrac{a}{60}$, where $a$ is a composite number.

But notice that $\dfrac{7}{10}=\dfrac{42}{60}$ and $\dfrac{11}{15}=\dfrac{44}{60}$, this tells us $a=43$, which leads to a contradiction.

Hence, the smallest possible value of $y$ is 7.

 

[Albert1]

Let $x,\,y$ be positive integers such that $\dfrac{7}{10}<\dfrac{x}{y}<\dfrac{11}{15}$. Find the smallest possible value of $y$.

Spoiler

$\dfrac {21y}{30y}<\dfrac {30x}{30y} <\dfrac {22y}{30y}$
$21y<30x ----(1)$
$30x<22y----(2)$
now we must find the smallest values of $x,y$ satisfying both (1) and (2)
and we get :$x=5,y=7$