Problem: For $k \geq 3$, we define an ordered $k$-tuple of real numbers $\left(x_1, x_2, \ldots, x_k\right)$ to be special if, for every $i$ such that $1 \leq i \leq k$, the product $x_1 \cdot x_2 \cdots x_k=x_i^2$. Compute the smallest value of $k$ such that there are at least 2009 distinct special $k$-tuples. Solution: Immediately, a red flag should be going off; we're being asked to find a finite number $k$ so that the number of $k$-tuples over the *real* numbers satisfying a property is at least $2009$. If the problem is written well, that signals at us that there is some massive level of control that the condition puts on our tuples. After all, there sure are lots of real numbers, let alone the cartesian product $\mathbb{R}^k$... We might first think about the control of $0$ in a large product. Notice that if even one of the $x_{i}s is zero, then $x_{j} = 0$ for all $j \in \{ 1,2,\dots, k \}$ (otherwise we would have a non-zero number is equal to a product that has a zero in it, which is zero, which is impossible). So, we've dealt with the case when the special $k$-tuple has a zero in it. Otherwise, $x_{1},x_{2}, \dots, x_{k} \neq 0$. Assume that we have a special, $k$-tuple $\left(x_1, x_2, \ldots, x_k\right)$, none of whose entries are $0$. Then, we know that $x_1 \cdot x_2 \cdots x_k=x_1^2$, $x_1 \cdot x_2 \cdots x_k=x_2^2$, $x_1 \cdot x_2 \cdots x_k=x_3^2$, and so on and $x_1 \cdot x_2 \cdots x_k=x_k^2$. Multiplying all of these together gives us $(x_1 \cdot x_2 \cdots x_k)^k = (x_1 \cdot x_2 \cdots x_k) \iff (x_1 \cdot x_2 \cdots x_k)^{k-1} = 1 \implies x_1 \cdot x_2 \cdots x_k=1$. Therefore, $x_{1}^{2}=1, x_{2}^{2}=1, \dots, x_{k}^{2}=1 \iff (x_{1},x_{2}, \dots, x_{k}) = (\pm 1, \pm 1, \dots, \pm 1)$. Now, we must be careful because we must ensure the product is still positive, meaning we need to somehow count the number of ways to have there be an even number of negative ones appear in the special tuple. A greedy approach works here; if we simply select each value as we go (for $x_{1}$ and then $x_{2}$ and so on), we can stop before we get to $x_{k}$ and observe our current product, which is either positive or negative. Then, the positivity or negativity determines the value of $x_{k}$ for us. So, there are $2^{k-1}+1$ total special $k$-tuples, and $2^{k-1}+1 > 2009$ when $k = 12$.