An idea of probability can be used here. Let $p, ~ q,~ r$ be the probability of having a number divisible by $2, ~3, ~5$ respectively.
If we notice from the set of natural numbers $\mathbb{N}=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,\cdots,\infty\}$, we observe that any one of every $2$ consecutive numbers is divisible by $2$. Again, any one of every $3$ consecutive numbers is divisible by $3$. The same goes for $5$. Therefore $$p = \frac{1}{2}\\q = \frac{1}{3}\\r = \frac{1}{5}$$
Now the complement of these probabilities are $p', ~ q', ~ r'$ meaning that the probabilities of having a number NOT divisible by $2,~3,~5$ respectively. Therefore, $$p' =1- \frac{1}{2}=\frac{1}{2}\\q' =1- \frac{1}{3}=\frac{2}{3}\\r' =1- \frac{1}{5}=\frac{4}{5}$$
So the probability of having a number NOT divisible by 2 or 3 or 5 is $\mathrm{P}= p'\times q'\times r'$. [It can be thought of De Morgan's law, $(p \vee q \vee r)'=p' \wedge q' \wedge r'$ ]
$$ \therefore \mathrm{P}=p' \times q' \times r' =\frac{1}{2} \times \frac{2}{3} \times \frac{4}{5}= \frac{4}{15}$$
The number of positive integers out of the first 720 positive integers NOT divisible by 2 or 3 or 5 is $ \left[ 720 \mathrm{P} \right]=\left[ 720\times \frac{4}{15} \right]=192$.
So the answer is $192$.