Problem: A group of order $24$ must have either ? or ? Sylow $2$-subgroups.
I know from Cauchy's theorem that if $p$ is a prime and $G$ is a finite group such that p divides $G$, then $G$ has a subgroup of order $G$ and Sylow $3$ tells us that the number of Sylow $p$-subgroups is congruent to $1$ mod $p$ and divides the order of $G$, so this is what I've tried:
$|G|=24=(2^3)(3)$ and $s_p \equiv 1 \mod 2$ and $\frac{|G|}{|s_p|} = x \Rightarrow \frac{(2^3)(3)}{|s_p|} =x \Rightarrow$ $|s_p|$ divides $2^3$ or $|s_p|$ divides $3$. But only $|s_p|=3 \equiv 1 \mod 2$, so I must be doing something wrong.
$\bf{Additional}$: So maybe $1$ or $3$, Sylow $2$ subgroups?
Note that Sylow $2$-subgroups will have order 8. The divisors of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$. All of those are congruent to $0$ (mod $2$) except $1$ and $3$. Therefore, there must be either $1$ or $3$ Sylow $2$-subgroups of order $8$.