24. Use coins to stack a triangle, with $2^{10}$ coins on the bottom layer, and $2^{9}+1$ coins facing up. The principle for stacking upwards is: (1) each layer has one less coin than the adjacent lower layer; (2) if the two adjacent coins are both facing up or both facing down, the coin stacked between them faces up, and the coin stacked between coins with different faces faces down. Try to explain whether the top coin is facing up or down.
Solve for the $j$-th coin of the $i$-th layer with integer $a_{i, j}(1 \leqslant i \leqslant 2^{10}, i \geqslant j)$, where a coin with the head side up is +1, and a coin with the head side down is -1. According to the problem, list the array:
$$
\begin{array}{l}
a_{1.1} \\
\begin{array}{ll}
a_{2,1} & a_{2,2}
\end{array} \\
\begin{array}{llll}
a_{3,1} & a_{3,2} & a_{3,3} & \\
\ldots \ldots . & \ldots \ldots \ldots
\end{array} \ldots \ldots . \\
a_{2^{10}, 1} \quad a_{2^{10}, 2^{1} \cdots \cdots \cdots a_{2^{10}}, 2^{10}-1} \quad a_{2^{10}}^{10} 2^{10} \\
\end{array}
$$

According to the problem, we have
$$
a_{i-1, j-1}=a_{i, j-1} a_{i, j}\left(1 \leqslant i \leqslant 2^{10}, i \geqslant j\right) .
$$
(i) By mathematical induction, it is easy to prove
$$
a_{1,1}=a_{n, 1^{-1}}^{c_{0}^{0}} a_{n: 2}^{c_{1}^{1}} a_{n, 5}^{c^{2}} \cdots a_{n:-1}^{c-2} \cdots a_{n:-1}^{c-1}(2 \leqslant n \leqslant
$$
$2^{10}$ ).
Thus, $a_{1,1}=a_{2^{10}, 2}^{c_{2}^{0}-1} a_{2^{10}, 2}^{c_{2}^{1}-1} a_{2^{10}, 3}^{c_{2}^{2}-1} \cdots$
$$
a_{2^{10}, 2^{10}-1}^{c_{2}^{210}}, a_{2^{10}, 2^{10}}^{c_{2^{10}}^{20}-1}
$$
(ii) Next, we prove that $C_{2}^{10}-1\left(0 \leqslant m \leqslant 2^{10}-1\right)$ is odd.
$C_{2^{10}-1}=\frac{\left(2^{10}-1\right)\left(2^{10}-2\right) \cdots\left(2^{10}-m\right)}{1 \times 2 \times \cdots \times m}$
$=\frac{2^{10}-1}{1} \times \frac{2^{10}-2}{2} \times \cdots \times \frac{2^{10}-m}{m}$.
Let $r$ be the maximum value such that $2^{r} \mid i$. Clearly, for the fraction
$\frac{2^{10}-i}{i}$, the numerator and denominator can be reduced by $2^{2}$, and after reduction, both the numerator and denominator are odd, hence (10) shows that $C_{2}^{10}-1$ is odd.
(iii) According to the conclusion of (ii), and because $a_{2^{10},(}\left(1 \leqslant i \leqslant 2^{10}\right)$ is either +1 or -1, so
$$
a_{2^{10}, i}^{c_{2^{10}}^{1-1}}-a_{2^{10}, i}
$$

Thus,
$$
a_{1,1}=a_{2^{10}}, 1 a_{2^{10}}{ }^{1} \cdots a_{2^{10}} \cdot 2^{10} \text {. }
$$

In the bottom layer of $2^{10}$ coins, there are $2^{9}+1$ coins with the head side up, and $2^{9}-1$ coins with the head side down, so the right side of (6) has $2^{\circ}-1$ factors of -1, thus, $a_{1,1}=-1$.
Conclusion: The top coin is head side down.
(Wang Dazhu, Hubei Yuanjiang City Zhuan Dou Middle School, 431909)