Let

• $M$ = set of students whose numbers are divisible by $2$ (Mathematics)
• $P$ = set of students whose numbers are divisible by $3$ (Physics)
• $C$ = set of students whose numbers are divisible by $5$ (Chemistry)

Total students $= 140$.

We need the number of students who did not opt for any course, i.e. those not in $M \cup P \cup C.$

So first find $|M \cup P \cup C|$ using inclusion-exclusion.

Count each set: $|M| = \left\lfloor \frac{140}{2} \right\rfloor = 70$ $|P| = \left\lfloor \frac{140}{3} \right\rfloor = 46$ $|C| = \left\lfloor \frac{140}{5} \right\rfloor = 28$

Count pairwise intersections:

• Divisible by both $2$ and $3$ means divisible by $\operatorname{lcm}(2,3)=6$: $|M \cap P| = \left\lfloor \frac{140}{6} \right\rfloor = 23$
• Divisible by both $2$ and $5$ means divisible by $10$: $|M \cap C| = \left\lfloor \frac{140}{10} \right\rfloor = 14$
• Divisible by both $3$ and $5$ means divisible by $15$: $|P \cap C| = \left\lfloor \frac{140}{15} \right\rfloor = 9$

Count the triple intersection:

• Divisible by $2,3,5$ means divisible by $\operatorname{lcm}(2,3,5)=30$: $|M \cap P \cap C| = \left\lfloor \frac{140}{30} \right\rfloor = 4$

Apply inclusion-exclusion: $|M \cup P \cup C| = |M|+|P|+|C|-|M\cap P|-|M\cap C|-|P\cap C|+|M\cap P\cap C|$

$= 70+46+28-23-14-9+4$ $= 102$

Therefore, number of students who did not opt for any of the three courses: $140-102=38$

Hence the correct option is: $\boxed{\text{D: }38}$