SWEEEEEEEEET Let's show a more general result, i.e. let $n \geq 3$ be an odd positive integer and consider $A = \{1,2,\dots , 2n\}$ and $M = \{n, 2n+1, 3n+2\}$. We show that there are exactly $2^{(n+1)/2}$ $M$-partitions. Define $X = \{1,2,\dots , n-1\}$ $Y = \{n, n+1\}$ $Z = \{n+2, n+3, \dots , 2n\}$ Now, note that (this is the crucial observation?) for any element $x \in X$, there are exactly two other elements $y, y' \in A$ such that $x+y$ and $x+y'$ are in $M$. Furthermore, one of $y$ and $y'$ is in $X$ and the other is in $Z$. (Note that clearly we have $x \neq y, y'$ as the three elements in $M$ are odd.) Similarly, for any element $z \in Z$, the same holds. Consider the $4$-tuples $(k, n-k, n+k+1, 2n-k+1)$ for $1\leq k \leq \frac{n-1}{2}$ and note that the sum of any two adjacent terms (and the first and last) is in $M$. Moreover, the union of all such $4$-tuples for $k = 1,2,\dots , \frac{n-1}{2}$ will be the union of $X$ and $Z$. Note that these adjacent sums are the ONLY possible sums of elements in $A\Y$ which in $M$, by our previous observations. Thus, we either have $k, n+k+1 \in A_1$ and $n-k, 2n-k+1 \in A_2$ or $n-k, 2n-k+1 \in A_1$ and $k, n+k+1 \in A_2$ for $k = 1,2,\dots , \frac{n-1}{2}$ Now, we may account for $Y$ by noting that either $n \in A_1$ or $n+1 \in A_1$ and only one element in $Y$ can be in each $A_i$. Thus, we have a total of $2^{(n-1)/2} \cdot 2 = 2^{(n+1)/2}$ possible $M$-partitions!

well done!!!! have u solved it urself. the proof is elementry one but to check its validity u need to think a lot. but how come u created X,Y and Z? how much time did it take to solve this problem? regards amir siddique

Thanks! Yeah, I admit I left out some details, but I guess I started out trying to characterize all pairs of elements which sum to an element in $M$. So, just looking at a random element of $A$ (let's say $1$ for now)... if $1\in A_1$ we have $1000, 2002 \in A_2$, so $1002, 1003 \in A_1$, etc... until it "circles" around and repeats again. This happens with all elements except those in $Y$, so it seems natural to create $X$ and $Z$ to characterize this. Now, taking another element besides $1$, you can basically use the same idea. Also, I suppose generalizing it actually makes the problem easier to think about... (I don't really remember how long I spent on it--I think I worked on it before without much progress and gave up, but returned to it recently and got it in an hour or so.)