A game of numbers: When can we have 2011? Two friends are playing a game. In every turn, after one of them says a number $k$, the other one has to say a number in form $a\cdot b$ where $a,b\in \mathbb{N}$ such that $a+b=k$ holds. The game continues in that way, from a just said number. 
From which numbers could the game have started if at one point one of the friends said number $2011$?
 A: You can start at $5$ with the sequence $5,6,9,20,96,1008,2012,2011$ and with any larger number since each step can reduce $k$ by $1$.
You cannot start with $1,2,3$ or $4$, since you cannot increase from these.
For such problems you can explore which possibilities lead to the final result.
A: Short answer: We can start at any number larger then $4$.  
Solution: Instead of $2011$, suppose we are given any integer $\alpha$, and we want to know when we can reach $\alpha$.  Suppose we start at $k$, and $k>4$.  Then choose $a=\lfloor \frac{k}{2}\rfloor$ and $b=\lceil\frac{k}{2}\rceil$.  (If $k$ is even, this is just $a=b=\frac{k}{2}$).  Then because $k>4$, we will have that $ab>k$, so the sequence grows and by repeating this it will eventually be larger then $\alpha$.  Now, we can decrease the sequence by $1$ at any point by choosing $a=1$, $b=k-1$, so once we have passed $\alpha$, we can just decrease by one at a time until we reach exactly $\alpha$.
For integers $k\leq 4$, notice that there is no choice of nonnegative $a$, $b$ such that $a+b=k$ and $a\cdot b>k$, so we are confined to this short interval.
