Key Concepts and Formulas

• Set Theory: Understanding of sets, universal set, subsets, and operations like union ($\cup$) and intersection ($\cap$).
• Principle of Inclusion-Exclusion (PIE): For three sets A, B, and C, $n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(C \cap A)] + n(A \cap B \cap C)$. This formula helps calculate the number of elements in the union of sets.
• Floor Function: To find the number of multiples of an integer $k$ up to $N$, we use $\lfloor N/k \rfloor$.

Step-by-Step Solution

Let $U$ be the universal set of all students, so $n(U) = 140$. Let $M$ be the set of students who opted for Mathematics. Let $P$ be the set of students who opted for Physics. Let $C$ be the set of students who opted for Chemistry.

We want to find the number of students who did not opt for any of the three courses, which is $n(U) - n(M \cup P \cup C)$. We will use the Principle of Inclusion-Exclusion to find $n(M \cup P \cup C)$.

Step 1: Calculate the number of students in each individual course.

• Mathematics (M): Students whose number is even (divisible by 2).

  • The numbers are 2, 4, 6, ..., 140.
  • $n(M) = \left\lfloor \frac{140}{2} \right\rfloor = 70$.

• Physics (P): Students whose number is divisible by 3.

  • The numbers are 3, 6, 9, ..., 138.
  • $n(P) = \left\lfloor \frac{140}{3} \right\rfloor = \lfloor 46.66... \rfloor = 46$.

• Chemistry (C): Students whose number is divisible by 5.

  • The numbers are 5, 10, 15, ..., 140.
  • $n(C) = \left\lfloor \frac{140}{5} \right\rfloor = 28$.

Step 2: Calculate the number of students in the pairwise intersections of courses.

• Mathematics and Physics ($M \cap P$): Students whose number is divisible by both 2 and 3. This means the number is divisible by LCM(2, 3) = 6.

  • $n(M \cap P) = \left\lfloor \frac{140}{6} \right\rfloor = \lfloor 23.33... \rfloor = 23$.

• Physics and Chemistry ($P \cap C$): Students whose number is divisible by both 3 and 5. This means the number is divisible by LCM(3, 5) = 15.

  • $n(P \cap C) = \left\lfloor \frac{140}{15} \right\rfloor = \lfloor 9.33... \rfloor = 9$.

• Chemistry and Mathematics ($C \cap M$): Students whose number is divisible by both 5 and 2. This means the number is divisible by LCM(5, 2) = 10.

  • $n(C \cap M) = \left\lfloor \frac{140}{10} \right\rfloor = 14$.

Step 3: Calculate the number of students in the intersection of all three courses.

• Mathematics, Physics, and Chemistry ($M \cap P \cap C$): Students whose number is divisible by 2, 3, and 5. This means the number is divisible by LCM(2, 3, 5) = 30.
  • $n(M \cap P \cap C) = \left\lfloor \frac{140}{30} \right\rfloor = \lfloor 4.66... \rfloor = 4$.

Step 4: Apply the Principle of Inclusion-Exclusion to find the number of students who opted for at least one course.

Using the PIE formula: $n(M \cup P \cup C) = n(M) + n(P) + n(C) - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + n(M \cap P \cap C)$ $n(M \cup P \cup C) = (70 + 46 + 28) - (23 + 9 + 14) + 4$ $n(M \cup P \cup C) = 144 - 46 + 4$ $n(M \cup P \cup C) = 98 + 4$ $n(M \cup P \cup C) = 102$ So, 102 students opted for at least one of the three courses.

Step 5: Calculate the number of students who did not opt for any course.

The number of students who did not opt for any course is the total number of students minus the number of students who opted for at least one course. $n(\text{none}) = n(U) - n(M \cup P \cup C)$ $n(\text{none}) = 140 - 102$ $n(\text{none}) = 38$

Common Mistakes & Tips

• LCM is Crucial: When finding the intersection of two conditions (e.g., divisible by 2 AND 3), always use the Least Common Multiple (LCM) of the divisors. For coprime numbers, the LCM is their product.
• Floor Function Accuracy: Ensure you use the floor function ($\lfloor x \rfloor$) to count multiples. Do not round to the nearest integer or round up.
• PIE Structure: Carefully apply the PIE formula. Remember to subtract pairwise intersections and add the triple intersection. A common error is forgetting the signs or miscalculating the terms.

Summary

We used the Principle of Inclusion-Exclusion to determine the number of students who opted for at least one of the three courses (Mathematics, Physics, Chemistry) based on the divisibility of their student numbers. By calculating the cardinalities of individual sets, pairwise intersections, and the triple intersection, we found that 102 students opted for at least one course. Subtracting this from the total number of students (140) gives us the number of students who did not opt for any course, which is 38.

The final answer is $\boxed{38}$.