The number of rational terms in the binomial expansion of (√[3]2+√[4]5)^100 is

The general term is $T_{r+1}=\binom{100}{r}\,2^{(100-r)/3}\,5^{r/4}$. It is rational only if both exponents are integers: $4\mid r$ and $3\mid(100-r)$. The second condition means $r\equiv100\equiv1\pmod 3$. Combining $r\equiv0\pmod4$ and $r\equiv1\pmod3$ gives $r\equiv4\pmod{12}$. With $0\le r\le100$ the admissible values are $r=4,16,28,\dots,100$, an AP with common difference $12$: $100=4+12(k-1)\Rightarrow k=9$. So there are $9$ rational terms. Counting only multiples of $4$ would give $26$, and only the second condition gives $34$, both of which are wrong.