INMO 2005

First of all, I am reversing the guven equation , which forms
$\dfrac{197}{43} > \dfrac {b}{a} > \dfrac{77}{17}$

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

$4 + \dfrac {25}{43} > \dfrac{b}{a} > 4 + \dfrac{9}{17}$

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

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

Now, we can say that $4 + \dfrac {25}{43} > 4 + \dfrac{x}{a} > 4 + \dfrac{9}{17}$

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

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

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

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

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 $a$ i.e. $7$ .
Then $b$ would be $(4a + x) = [(4×7) + 4] = [28+4] = \boxed {32}$