Problem on Function

$f(2^{n}) = f(2^{n - 1})\cdot f(2) = f(2^{n - 2})\cdot f(2)\cdot f(2) = .......= (f(2))^{n} = 2^{n}$

$\Rightarrow f(2^{a}) = 2^{a} , f(2^{a+1}) = 2^{a+1}$

Now , there are $2^{a} - 1$ integers between $2^a$ and $2^{a+1}$ , so the function can be applied on $2^a - 1$ values and it corresponds to exactly $2^a - 1$ values and since the function is increasing , therefore for all values of $x$ , $f(x) = x$ $\forall x \in (2^a , 2^a + 1 , \cdots , 2^{a+1})$

Therefore $\forall x \in \mathbb N , f(x) = x$

And hence $f(1983) = 1983$