The number of terms with integral coefficients in the expansion of `(17^(1/3)+35^(1/2)x)^(600)` is (A) `100` (B) `50` (C) `150` (D) `101`

Explanation

The general term `T_{r+1}` in the given expansion is given by


`\ ^{600} C_{r}(17^{1 / 3})^{600-r}(35^{1 / 2} x)^{r}`


`=\ ^{600} C_{r} 17^{200-r / 3} times 35^{r / 2} x^{r}`


Now, `T_{r+1}` is an integer if `frac{r}{2}` and `frac{r}{3}` are integers for all `0 leq r leq 600`


...