What is the time by now?

From $\dfrac{\alpha}{\beta} < \dfrac {17}{77}$ . We note that $\alpha = \lfloor \frac {17}{77} \beta \rfloor$ will satisfy the upper limit condition, when $\beta < 77$ . Our task is the find the $\beta < 77$ such that $\dfrac {\lfloor \frac {17}{77} \beta \rfloor}{\beta} > \dfrac{43}{197}$ that is satisfying the lower limit condition, $\Rightarrow \lfloor \frac {17}{77} \beta \rfloor > \frac{43}{197} \beta$ .

Let us now check the minimum $\beta$ , when $\lfloor \frac {17}{77} \beta \rfloor = 1,2,3,...$ and check if the condition $\lfloor \frac {17}{77} \beta \rfloor > \frac{43}{197} \beta$ is met.

\(\begin{array} {} \lfloor \frac {17}{77} \beta \rfloor = 1 & \Rightarrow \beta = 5 & \Rightarrow \frac{43}{197} \beta = 1.091 \color{red}{>} 1 \\ \lfloor \frac {17}{77} \beta \rfloor = 2 & \Rightarrow \beta = 10 & \Rightarrow \frac{43}{197} \beta = 2.182 \color{red}{>} 2 \\ \lfloor \frac {17}{77} \beta \rfloor = 3 & \Rightarrow \beta = 14 & \Rightarrow \frac{43}{197} \beta = 3.055 \color{red}{>} 3 \\ \lfloor \frac {17}{77} \beta \rfloor = 4 & \Rightarrow \beta = 19 & \Rightarrow \frac{43}{197} \beta = 4.147 \color{red}{>} 4 \\ \lfloor \frac {17}{77} \beta \rfloor = 5 & \Rightarrow \beta = 23 & \Rightarrow \frac{43}{197} \beta = 5.020 \color{red}{>} 5 \\ \lfloor \frac {17}{77} \beta \rfloor = 6 & \Rightarrow \beta = 28 & \Rightarrow \frac{43}{197} \beta = 6.111 \color{red}{>} 6 \\ \lfloor \frac {17}{77} \beta \rfloor = 7 & \Rightarrow \beta = 32 & \Rightarrow \frac{43}{197} \beta = 6.984 \color{blue}{<} 7 \end{array} \)

Therefore the minimum possible value of $\beta = \boxed{32}$ .

The left continued fraction is $[0,4,1,1,18/7 ]$ . The right continued fraction is $[0,4,1,1,8]$ . So the fractions in between it are of the form $[0,4,1,1,x]$ for some $x$ between $18/7$ and $8$ . That is, we have $\frac{\alpha}{\beta} = \frac{2x+1}{9x+5}$ for some rational $x$ between $18/7$ and $8$ . (I guess I could have gotten to that last sentence without continued fractions, now that I look at it.)

Let $x = p/q$ with $p$ and $q$ positive coprime integers. Then $\beta = 9p + 5q$ (exercise: $2p+q$ and $9p+5q$ are relatively prime!), and this is what we want to minimize. It's clear that $p =3$ , $q = 1$ , $\beta = 32$ is best possible, since for all other $p/q > 18/7$ , $p$ has to be at least $4$ and that's already too big.

Solve Diophantine equations, 197x-43y>0, 17y-77x>0, using wolfram alpha. You get ( x/y) = (7/32), (9/41), (11/50), (13/59), (15/68), (19/87) and (31/142) as possible pairs. Thus, choosing the lowest pair (7/32) gives answer = 32.

If a, b, c and d are positive and $\frac{a}{b} < \frac{c}{d}$ , then $\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}$ .

This means that:

$\frac{43}{197} = \frac{43y}{197y} < \frac{43y + 17x}{197y + 77x} < \frac{17x}{77x} = \frac{17}{77}$ , where x and y are positive integers.

$\frac{\alpha}{\beta} = \frac{43y + 17x}{197y + 77x}$ WLOG we can assume that $\gcd(x, y) = 1$ .

$\gcd((43y + 17x), (197y + 77x)) = k$

$\gcd(k, x) = 1$ or $\gcd(k, y) = 1$ or both. If this wasn't true, then $k$ would have to divide both $17(x + y)$ and $77(x + y)$ , which means dividing both 77 and 17.

$k | 43(197y + 77x) - 197(43y + 17x)$ and $k | 17(197y + 77x) - 77(43y + 17x)$ , which means $k | 38$ .

If we assume $k = 38$ , we must find the smallest possible positive integer pair $(x, y)$ such that $38 | (17x + 43y)$ and $38 | (77x + 197y)$ , which is $(3, 5)$ .

If we assume $k = 19$ , the smallest possible integer pair that satisfies $19 | (17x + 43y)$ and $19 | (77x + 197y)$ is $(2, 5)$ .

Correct solution (the one which produces smaller $\beta$ ) is $(x, y) = (3, 5)$

$a=\frac{43}{197}=.218274,....~~b=\frac{17}{77}=.220779,......~~~c=\dfrac{2}{a+b}=4.56\\If~\dfrac{\alpha_o}{\beta}<a,~and~ \dfrac{\alpha_o}{\beta+1}>b~~Then~~\alpha>\alpha_o.~~Min~\beta,~\alpha~also~ is~ min.~\\ Assume~ \alpha_o, ~try~ \beta=\alpha_o*c. ~~~~~~~~Start ~with~\alpha_o=10~(Since~17~is~in~b)\\ \alpha_o=10,~~~\beta=10*c =45.6 ~say~45,~\dfrac{\alpha_o}{\beta}=.2222~~ >~b.\\ \alpha_o=8,~~~\beta=8*c =36.5 ~say~37,~\dfrac{\alpha_o}{\beta}=.216~~ <~a. \\\alpha_o=7,~~~\beta=7*c =31.9 ~say~\color{#D61F06}{32},~\dfrac{\alpha_o}{\beta}=.218756~\color{#3D99F6}{between~a~and~b.} \\~very ~near~a~and~less~than~b.~so~try~\beta=33.\\ \dfrac{\alpha_o}{\beta}=.212121<a. ~32~a~good ~candidate. See~if~\alpha_o=6 ~is~~ possible.\\\alpha_o=6,~~~\beta=6*c =27.4 ~say~28,~\dfrac{\alpha_o}{\beta}=.2214>b.\\so~\beta=29~give~\dfrac{\alpha_o}{\beta}=.207<a.~~No~value~between~a~and~b.\\So~7~and ~32~is~ minimum.\\ For~ an~\alpha~\text{if there is no value between a and b , go to higher value.}\\If~the~first~value~was ~\dfrac{10}{45}~take ~now~\dfrac{K*10}{K*45}~\\Where K=1,2,3,4,....$