The title is the problem: how many rational terms are there in $(\sqrt{2}+\sqrt[3]{3})^{100}$?
I am very close to the answer and intuitively I know the answer, but I've been struggling for an hour already to get past a certain step and I'm starting to think I cannot go further and have to start guessing. Here's what I did.
The $n$th term in the binomial expansion is $T_{n+1} =C_{100}^n\cdot(\sqrt{2})^n\cdot(\sqrt[3]{3})^{100-n}$, where $C_{a}^b = \frac{a!}{b!(a-b)!}$. Obviously, for $T_{n+1}$ to be a rational term, $n$ must be even and $100-n$ must be divisible by $3$ $\Rightarrow$
$$\begin{cases} n\equiv0\ (\text{mod } 2) \\ (100-n)\equiv0 \text{ (mod } 3)\end{cases} \iff \\[5ex] \iff \begin{cases}n=2k,\ k \in \mathbb{N} \\ n - 1 = 3j, \text{(subtracted 99(}\equiv 0 \text{ (mod 3)}) \text{ and multiplied by -1)} \ j\in\mathbb{N}\end{cases} \iff \begin{cases}n=2k \\ n =3j+1\end{cases}$$
At this point, I know I can simply count all even numbers that also comply with the second equation in the system, but is there no way to reach a formula such as $n=a\cdot z + b$, where $z \in\mathbb{Z}$, so that I'm able to pick positive integers until plugging them into this formula no longer returns integers $\leq 100$? To check for each even number those two conditions works with small $n$'s, so for anything even a bit greater counting won't work anymore.