In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is:

We are given that the students are numbered from $$1$$ to $$140$$, so the total number of students is

$$|U| = 140.$$

Define three sets:

$$M = \{\text{students whose numbers are divisible by }2\},$$

$$P = \{\text{students whose numbers are divisible by }3\},$$

$$C = \{\text{students whose numbers are divisible by }5\}.$$

Our task is to find the number of students who do not belong to any of these three sets, i.e. the number of students in the complement set $$U - (M\cup P\cup C).$$

First we count each set individually.

• A number is divisible by $$2$$ if it is even. The largest even number ≤ $$140$$ is $$140$$ itself, and half of the numbers from $$1$$ to $$140$$ are even, so

$$|M| = \left\lfloor\frac{140}{2}\right\rfloor = 70.$$

• A number is divisible by $$3$$ if it is a multiple of $$3$$. The greatest multiple of $$3$$ in our range is $$138$$, so

$$|P| = \left\lfloor\frac{140}{3}\right\rfloor = 46.$$

• A number is divisible by $$5$$ if it ends in $$0$$ or $$5$$. The greatest multiple of $$5$$ ≤ $$140$$ is $$140$$, hence

$$|C| = \left\lfloor\frac{140}{5}\right\rfloor = 28.$$

Next we count the pairwise intersections, that is, the students who satisfy two divisibility conditions at once.

• A number is divisible by both $$2$$ and $$3$$ exactly when it is divisible by their least common multiple $$\text{lcm}(2,3)=6$$. Hence

$$|M\cap P| = \left\lfloor\frac{140}{6}\right\rfloor = 23.$$

• A number is divisible by both $$2$$ and $$5$$ when it is a multiple of $$\text{lcm}(2,5)=10$$, so

$$|M\cap C| = \left\lfloor\frac{140}{10}\right\rfloor = 14.$$

• A number is divisible by both $$3$$ and $$5$$ when it is a multiple of $$\text{lcm}(3,5)=15$$, therefore

$$|P\cap C| = \left\lfloor\frac{140}{15}\right\rfloor = 9.$$

Finally, we count the triple intersection, i.e. numbers divisible by $$2,3$$ and $$5$$ simultaneously. These are multiples of $$\text{lcm}(2,3,5)=30$$, so

$$|M\cap P\cap C| = \left\lfloor\frac{140}{30}\right\rfloor = 4.$$

Now we apply the principle of inclusion-exclusion for three sets. The formula is

$$|M\cup P\cup C| = |M| + |P| + |C| - |M\cap P| - |M\cap C| - |P\cap C| + |M\cap P\cap C|.$$

Substituting every value we have just calculated,

$$\begin{aligned} |M\cup P\cup C| &= 70 + 46 + 28 \\ &\quad - (23 + 14 + 9) \\ &\quad + 4 \\ &= 144 - 46 + 4 \\ &= 98 + 4 \\ &= 102. \end{aligned}$$

Thus $$102$$ students chose at least one of the three courses. The students who chose none are the complement of this union inside the universal set of $$140$$ students. Therefore,

$$\text{Number of students choosing no course} = |U| - |M\cup P\cup C| = 140 - 102 = 38.$$

Hence, the correct answer is Option C.