Number of positive integer divisors Let $n$ be an integer more than $9$. Find the number of positive divisors of $n$ lets call it $d(n)$ and take the difference from $n$. So the new number is $n-d(n)$. If we continue this process, we always end up with $2$. I don't know anybody worked this question before but somehow this question reminded me Collatz Conjecture. Anybody can give me an idea how to approach this problem ? Many thanks..
 A: If $n < (k+1)^2$, there's at most $2k$ divisors here.  
That's because if $x$ is a divisor of $n$ and greater than $\sqrt{n}$, the number $\frac{n}{x}$ must be an integer less than $\sqrt{n}$.  
So there are only about $\sqrt{n}$ candidates for $x$, and it implies that the number of divisors is at most about $2\sqrt{n}$. More technically, $d(n)$ is at most $2k$.  
The smallest $k$ here is exactly $\lfloor \sqrt{n} \rfloor$. So the upper bound of number of divisors is $2 \times \lfloor \sqrt{n} \rfloor$.  
Now, let's check about when will $n - d(n) \geq 9$ hold.
Here, $n - d(n) \geq n - 2 \times \lfloor \sqrt{n} \rfloor \geq n - 2\sqrt{n} \geq 9$. The range of $n - 2 \sqrt{n} \geq 9$ can be solved by translating to quadratic equation.
Let $N = \sqrt{n}$. The inequality will be $N^2 - 2N \geq 9$, and it will be always true when $N \geq 1+\sqrt{10}$.  
Now, we can say that $n - d(n) \geq 9$ when $n \geq 18 \geq (1+\sqrt{10})^2$.  
The final winning run is to check validity for $n=9, 10, 11, 12, 13, 14, 15, 16, 17$ and we can do inductive proof for $n \geq 18$ because we proved that the next number will always be at least $9$.
A: Hints:


*

*You can check it is true for $n=6,9,10,11,12$ 

*For $n \gt 12$ you could show $9 \le n-d(n) \lt n$

*Then use induction

