An algebra problem by Aaron Jerry Ninan

If $f(1)=1$ then $f(f(1))=1$ but $f(f(1))=3$ , so $f(1)$ isn't equal to $1$ If $f(1)>2$ then $f(2)>3$ , so $f(f(1))>3$ , but $f(f(1))=3$ , so $f(1)$ isn't greater than $2$

Thus $f(1)=2$ , and $f(2)=3$ . Similarly, it follows that $f(3)=f(f(2))=6$ , and $f(6)=f(f(3))=9$ , so $9 > f(5) > f(4) > 6$ , so $f(5)=8$ and $f(4)=7.$

Now we have found the following:

$f(1)=2,f(2)=3,f(3)=6,f(4)=7,f(5)=8,f(6)=9$

Through induction, it is simple to prove that $f(3^n)=2 \times 3^n$ and that $f(2 \times 3^n)=3^{n+1}$ .

Since $f(n)$ is a strictly increasing function $f(n+1) \ge f(n)+1$ . This further implies that $f(n+m) \ge f(n)+m$ .

We thus have that $f(3^n+k)$ (where $3^n \ge k \ge 0$ ) $2 \times 3^n+k=f(2 \times 3^n) -(3^n-k) \ge f(3^n+k) \ge f(3^n)+k= 2 \times 3^n+k$

Thus $f(3^n+k)=2 \times 3^n +k$ where $3^n \ge k \ge 0$ . It also follows that $f(2 \times 3^n+k)=3^{n+1}+3k$ where $3^n \ge k \ge 0$

Since $f(2001)=f(2 \times 729+543)=f(2 \times 3^6+543)=3^7+3 \times 543=3816$