$153 = 3^2 \cdot 17$ so lets assume there are $s_3$ $3$-Sylow-Subgroups and $s_{17}$ $17$-Sylow-Subgroups. We know that $s_3 \mid 153$ so $s_3 \in \{1,3,9,17,51,153\}$ and $s_{17} \in \{1,17,51\}$. Due to the rule that the number of $p$-Sylow-Subgroups $s$ must satisfy
$ 1 + pk = s$
for some $k \in \mathbb{N}_0$. That leaves $1 = s_3 = s_{17}$. Is that correct? Now those two Subgroups $H_3, H_{17}$ are cyclic and abelain. How do I draw a conclusion about all possible groups of order 153 with this information?