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 $2011$?

Eric Naslund
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Lazar Ljubenović
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2 Answers2


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.

Eric Naslund
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  • Thank you. Are there any general directions how I should think when solving problems like these? I couldn't even classify this in "number theory", as someone added a tag, I though it was more combinatorics. Some links? – Lazar Ljubenović Feb 12 '12 at 15:44
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    @LazarLjubenović: One thing I suggest, is just go and try easy cases. Try things and see what happens! Can you construct 2011? What happens if $a=1$? What happens if $k$ is $1$, $2$, $3$, $4$, $5$, or $6$. My solution comes purely from thinking about these simple cases. Often, even for difficult problems, you'd be surprised how often, thinking about the small cases, or just carrying through a calculation can lead you to the proof, and help understand what is going on. – Eric Naslund Feb 12 '12 at 15:47

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.

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