Let $f$ be a function on the positive integers, i.e., $f\mathit{ : }\mathbb{\text{ℕ}}\text{→}\mathbb{\text{ℤ}}$with the following properties:

(i) $f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{=}\mathit{2}$

(ii) $f\mathit{(}m\mathit{\times }n\mathit{)=}f\mathit{(}m\mathit{\left)}f\mathit{\right(}n\mathit{)}$ for all positive integers $m$and $n,$

(iii) $f\mathit{\left(}m\mathit{\right)}\mathit{ }\mathit{>}f\mathit{\text{(n) for m>n.}}$

Find $f\mathit{\left(}\mathit{1998}\mathit{\right)}$

$$\begin{gathered} \mathit{2}\mathit{=}f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{=}f\mathit{(}\mathit{1}\mathit{\times }\mathit{2}\mathit{)}\mathit{=}f\mathit{\left(}\mathit{1}\mathit{\right)}\mathit{\times }f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{=}f\mathit{\left(}\mathit{1}\mathit{\right)}\mathit{\times }\mathit{2} \\ \mathit{\therefore }\mathit{ }f\mathit{\left(}\mathit{1}\mathit{\right)}\mathit{=}\frac{\mathit{2}}{\mathit{2}}\mathit{=}\mathit{1} \end{gathered}$$

Now ,$\mathit{ }f\mathit{\left(}\mathit{4}\mathit{\right)}\mathit{>}\mathit{ }f\mathit{\left(}\mathit{3}\mathit{\right)}\mathit{>}f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{=}\mathit{2}$

and $\mathit{ }f\mathit{\left(}\mathit{4}\mathit{\right)}\mathit{=}f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{\times }f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{=}\mathit{4}$

and so, $4>f\left(3\right)>2$ and $f\left(3\right)$ is an integer, hence $f\left(3\right)=3$

and $\mathit{ }f\mathit{\left(}\mathit{6}\mathit{\right)}\mathit{>}f\mathit{\left(}\mathit{5}\mathit{\right)}\mathit{>}f\mathit{\left(}\mathit{4}\mathit{\right)}$

$$\begin{gathered} \mathit{\Rightarrow }f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{>}f\mathit{\left(}\mathit{3}\mathit{\right)}\mathit{>}f\mathit{\left(}\mathit{5}\mathit{\right)}\mathit{>}\mathit{4} \\ \mathit{\Rightarrow }\mathit{2}\mathit{\times }\mathit{3}\mathit{>}f\mathit{\left(}\mathit{5}\mathit{\right)}\mathit{ }\mathit{>}\mathit{4} \\ \mathit{\Rightarrow }f\mathit{\left(}\mathit{5}\mathit{\right)}\mathit{=}\mathit{5} \end{gathered}$$

So, we guess that $f\left(n\right)=n$. Let us prove it.

We will use mathematical induction for proving.

$f\left(n\right)=n$ is true for $n=1,2$

Let us assume that the result is true for all $m<n$, and then we shall prove it for $n$, where $n>2$.

If $n$is even, then let $n=2m$

$f\mathit{\left(}n\mathit{\right)}\mathit{=}f\mathit{\left(}\mathit{2}m\mathit{\right)}\mathit{=}f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{\times }f\mathit{\left(}m\mathit{\right)}\mathit{=}\mathit{2}\mathit{\times }m\mathit{=}\mathit{2}m\mathit{=}n\mathit{.}$

If $n$ is odd and $n=2m+1$, then $n>2m$

$$\begin{gathered} \mathit{2}m\mathit{<}\mathit{2}m\mathit{+}\mathit{1}\mathit{<}\mathit{2}m\mathit{+}\mathit{2} \\ \mathit{\Rightarrow }f\mathit{\left(}\mathit{2}m\mathit{\right)}\mathit{<}f\mathit{(}\mathit{2}m\mathit{+}\mathit{1}\mathit{)}\mathit{<}f\mathit{(}\mathit{2}m\mathit{+}\mathit{2}\mathit{)} \\ \mathit{\Rightarrow }f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{.}f\mathit{\left(}m\mathit{\right)}\mathit{<}f\mathit{(}\mathit{2}m\mathit{+}\mathit{1}\mathit{)}\mathit{<}f\mathit{\left(}\mathit{2}\mathit{\right)}\mathit{.}f\mathit{(}m\mathit{+}\mathit{1}\mathit{)} \\ \mathit{\Rightarrow }\mathit{2}m\mathit{<}f\mathit{(}\mathit{2}m\mathit{+}\mathit{1}\mathit{)}\mathit{<}\mathit{2}m\mathit{+}\mathit{2} \end{gathered}$$

There is exactly one integer $2m+1$ between $2m$ and $2m+2$ and hence,

$\mathit{\text{f(n)=f(2m+1)=(2m+1)=n}}$

Thus $f\left(n\right)=n$ for all $n\in \mathrm{\mathbb{N}}$

Hence, $f\left(99\right)=99$