I'm trying to use the congruence relations from Sylow's Theorem to show that a group $G$ of order $|G|= 6435= 3^2 \cdot 5 \cdot 11 \cdot 13$ is not simple. To work with the least amount of Sylow subgroups, it's usually preferred to start working with the largest prime $p$ dividing the order of the group. So here's what I did.
It turns out that when I let $n_{13}$ denote the number of Sylow $13$-subgroups of $G$, I obtained the following: $$n_{13} \equiv 1(\text{mod }13), n_{13} | 3^2 \cdot 5 \cdot 11 \implies n_{13}=1,$$ since $3, 5, 3^2, 11, 3 \cdot 5, 3 \cdot 11, 3^2 \cdot 5, 5 \cdot 11, 3^2 \cdot 11, 3 \cdot 5 \cdot 11, 3^2 \cdot 5 \cdot 11 \not\equiv 1(\text{mod }13)$. So this means that $G$ has a unique Sylow $13$-subgroup, which is normal. Therefore, $G$ is not simple. On this particular link (https://crazyproject.wordpress.com/2010/07/14/no-simple-groups-of-order-2205-4125-5103-6545-or-6435-exist/), the fifth problem involves the same group $G$ and uses an alternative method dealing with factorials and subgroups of a particular index. It is sometimes much more time-consuming however. Was my method correct?
Unfortunately you are not correct because $3^2\times5\times11=495$ is actually $1\pmod{13}$.
However, we can fix things with a classic argument. Note that $n_{11}\in\{1,45\}$, $n_5\in\{1,11\}$, and $n_{13}\in\{1,495\}$. Then note the intersection of any two distinct subgroups of prime order is trivial (applying Lagrange's theorem and primality). Hence the number of elements in the group (if the group is simple) is at least $12\times495+10\times45+4\times11+1+1>6435$, which is a contradiction (the first $+1$ comes from the identity element and the second $+1$ comes from some element of order $3$ in the Sylow $3$-subgroup).