INMO 2005 problem

First of all, I am reversing the given equation , which forms
$\dfrac{197}{43} > \dfrac {\beta}{\alpha} > \dfrac{77}{17}$

It can also be written as
$\dfrac {172+25}{43} > \dfrac {\beta}{\alpha} > \dfrac {68 + 9}{17}$

$4 + \dfrac {25}{43} > \dfrac{\beta}{\alpha} > 4 + \dfrac{9}{17}$

Now, we know that $\dfrac{197}{43} > 4$ and $\dfrac {77}{17} < 5$

So, we can say that the following equation is true.
$4 > \dfrac{\beta}{\alpha} > 5$

Now, we can say that $4 + \dfrac {25}{43} > 4 + \dfrac{x}{\alpha} > 4 + \dfrac{9}{17}$ (Here $x$ is positive because $\dfrac{\beta}{\alpha} > 4$ ).

Subtracting $4$ on all sides, we get
$\dfrac {25}{43} > \dfrac {x}{\alpha} > \dfrac {9}{17}$

Now, reciprocating the equation,
$\dfrac {43}{25} > \dfrac {\alpha}{x} > \dfrac {17}{9}$

Now, multiplying $x$ on all sides,
$\dfrac {43x}{25} > \alpha > \dfrac {17x}{9}$

Lets find the bounds of $\alpha$ for $x = 1,2,3,\cdots$ because we need the least value of $\beta$ .
We obtain the bounds as
$\left(1\dfrac {18}{25},1\dfrac {8}{9}\right),\left(3\dfrac {11}{25},1\dfrac {7}{9}\right),\left(5\dfrac {4}{9},4\dfrac{2}{3}\right)$

But, none of these pairs contain an integer between them,
So, lets proceed with $x = 4$ , the bounds obtained are
$6\dfrac {12}{25}$ for $\dfrac {43x}{25}$ , and $7\dfrac {5}{9}$ for $\dfrac {17x}{9}$

So, in this situation, take the smallest value for $\alpha$ i.e. $7$ .
Then $\beta$ would be $(4\alpha + x) = [(4×7) + 4] = [28+4] = 32$ .

So, $\log_2 x = \log_2 32 = \color{#69047E}{\boxed{5}}$ .

My solution is based on hit & trial & assumptions ,, so it would be very nice if someone posts a good solution to this , However here is how it goes;

Here we observe ,

$\dfrac {43}{197} \beta$ $<$ $\alpha$ $<$ $\dfrac {17}{77} \beta$ ,

Now,

$\dfrac {43}{197}$ $\sim$ $0.218$ &

$\dfrac {17}{77}$ $\sim$ $0.220$

Now we need the first number , $0.218 \beta$ to be just less than a postive integer while we need the second no. , $0.220 \beta$ to be just greater than the same positive integer. Plugging in $\beta = 32$ gives the least possible positive integral value for $\alpha$ such that $6.976$ $<$ $\alpha$ $<$ $7.040$ ,

which gives $\alpha$ $=$ $7$ ,

Hence required value of $\beta=32$ , & therefore our answer is :-

$\log_2 \beta$ $=$ $\log_2 32$ $=$ $\boxed {5}$ . $\textit{Thank you}$ .