11. I $\pi$ cchi are animals that live in families of 1, 2, or 3 individuals. Originally, a family of $3 \pi$ cchi was imported to Italy. A family of $n \pi$ cchi reproduces by growing $2n-2$ new individuals, forming a total of $3n-2$, and dividing into new families (not necessarily two families with the same number of individuals divide in the same way). All families reproduce simultaneously at each brood. For example, after the first brood, there are definitely seven $\pi$ cchi, which could be divided into three families of two and one of one, or one of three and two of two, or seven of one, etc. How many possibilities are there for the total number of $\pi$ cchi after the seventh brood?
(A) 129
(B) 253
(C) $2^{7}$
(D) $\frac{2^{7}+3^{7}+1}{2}$
(E) $3^{7}$
11. The answer is (B). The number of $\pi$ chicks after 7 broods can take on any odd value between 7 and $2^{9}-1$, and no other. The answer to the question is therefore $\frac{2^{9}-1-7}{2}+1=2^{8}-3=253$, since this is the number of odd numbers between 7 and $2^{9}-1=511$.

Let's justify the previous statement through the following facts:

(a) a family consisting of 1 single $\pi$ chick will continue to consist of a single $\pi$ chick forever: obvious;

(b) a family consisting of $2 \pi$ chicks can give rise, after $n \geq 1$ broods, to no more than $2^{n+1}$ $\pi$ chicks;

(c) a family consisting of $3 \pi$ chicks can give rise, after $n \geq 1$ broods, to any number of $\pi$ chicks that is odd and between 7 and $2^{n+2}-1$.

We will prove statements (b) and (c) simultaneously by induction: assuming that (b) and (c) hold for some $n$, we will prove them for $n+1$. Note that the base case $n=1$ is obvious for both. Note also that the parity of the number of $\pi$ chicks does not change after a brood, so the parity condition stated in (c) is indeed necessary. Furthermore:

(b) in this case, after one brood we have $4 \pi$ chicks, which can live in four families of 1 (and in this case we will still have $4 \pi$ chicks in all subsequent generations), one family of 2 and two of 1, two families of 2, or one family of 3 and one of 1. In the various cases, statements (b) and (c) for $n$ broods show that - after further $n$ broods - we will have at most

$$
1+1+1+1, \quad 2^{n+1}+1+1, \quad 2^{n+1}+2^{n+1}, \quad\left(2^{n+2}-1\right)+1
$$

$\pi$ chicks. Since each of these numbers is less than or equal to $2^{n+2}$, this proves statement (b) for $n+1$ broods.

(c) in this case, after one brood the $7 \pi$ chicks can be distributed into families in the following ways:

$$
\begin{gathered}
3+3+1, \quad 3+2+2, \quad 3+2+1+1, \\
3+1+1+1+1, \quad 2+2+2+1, \quad 2+2+1+1+1, \\
2+1+1+1+1+1, \quad 1+1+1+1+1+1+1 .
\end{gathered}
$$

After further $n$ broods, using the inductive hypothesis as above, we obtain that in the various cases the total number of $\pi$ chicks is at most

$$
\begin{gathered}
2\left(2^{n+2}-1\right)+1, \quad\left(2^{n+2}-1\right)+2\left(2^{n+1}\right), \quad\left(2^{n+2}-1\right)+\left(2^{n+1}\right)+1+1, \\
\left(2^{n+2}-1\right)+1+1+1+1, \quad 3 \cdot\left(2^{n+1}\right)+1, \quad 2 \cdot\left(2^{n+1}\right)+1+1+1 \\
\left(2^{n+1}\right)+1+1+1+1+1, \quad 1+1+1+1+1+1+1 .
\end{gathered}
$$

It is easy to verify that each of these numbers is less than or equal to $2^{n+3}-1$. Finally, if after the first brood the $7 \pi$ chicks are organized into families as $3+3+1$, then after further $n$ broods the number of $\pi$ chicks will be of the form $d_{1}+d_{2}+1$, where $d_{1}, d_{2}$ can take on any odd value between 7 and $2^{n+2}-1$. The expression $d_{1}+d_{2}+1$ can then take on any odd value between $7+7+1=15$ and $2 \cdot 2^{n+2}-2+1=2^{n+3}-1$. It remains only to show that the number of $\pi$ chicks after $n+1 \geq 2$ broods can also be 7, 9, 11, or 13, but this is easy. Indeed, if at some point the $\pi$ chicks organize themselves into families of 1, from that point on their number will no longer grow, so it is sufficient to show that such numbers of $\pi$ chicks are realizable after at most two broods, for example as follows:

$$
\begin{gathered}
3 \rightarrow 7 ; \quad 3 \rightarrow 2+1+1+1+1+1 \rightarrow 4+1+1+1+1+1=9 \\
3 \rightarrow 2+2+1+1+1 \rightarrow 4+4+1+1+1=11 ; \quad 3 \rightarrow 2+2+2+1 \rightarrow 4+4+4+1=13
\end{gathered}
$$