Key Concepts and Formulas

• Union of Sets: The smallest subset of a universal set $X$ that contains both sets $A$ and $B$ is their union, denoted by $A \cup B$.
• Principle of Inclusion-Exclusion for Two Sets: For any two finite sets $A$ and $B$, the number of elements in their union is given by: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ This formula accounts for elements that are common to both sets ($A \cap B$) to avoid double-counting.
• Counting Multiples: The number of multiples of an integer $k$ in the range from 1 to $N$ (inclusive) is given by $\lfloor N/k \rfloor$.

Step-by-Step Solution

Step 1: Understand the Universal Set $X$ The problem defines the universal set $X$ as all natural numbers from 1 to 50. $X = \{n \in \mathbb{N} : 1 \le n \le 50\} = \{1, 2, 3, \dots, 50\}$ The total number of elements in $X$ is $n(X) = 50$.

Step 2: Determine Set $A$ and its Cardinality $n(A)$ Set $A$ contains all elements in $X$ that are multiples of 2. To find the number of such elements, we use the formula for counting multiples. $A = \{n \in X : n \text{ is a multiple of 2}\}$ The number of multiples of 2 between 1 and 50 is: $n(A) = \left\lfloor \frac{50}{2} \right\rfloor = 25$ So, there are 25 elements in set $A$.

Step 3: Determine Set $B$ and its Cardinality $n(B)$ Set $B$ contains all elements in $X$ that are multiples of 7. To find the number of such elements, we use the formula for counting multiples. $B = \{n \in X : n \text{ is a multiple of 7}\}$ The number of multiples of 7 between 1 and 50 is: $n(B) = \left\lfloor \frac{50}{7} \right\rfloor = \lfloor 7.142... \rfloor = 7$ So, there are 7 elements in set $B$.

Step 4: Determine the Intersection Set $A \cap B$ and its Cardinality $n(A \cap B)$ The intersection $A \cap B$ consists of elements that are common to both $A$ and $B$. This means these elements must be multiples of both 2 and 7. A number that is a multiple of both 2 and 7 must be a multiple of their Least Common Multiple (LCM). The LCM of 2 and 7 (since they are coprime) is: $\text{LCM}(2, 7) = 2 \times 7 = 14$ So, the elements in $A \cap B$ are the multiples of 14 in $X$. $A \cap B = \{n \in X : n \text{ is a multiple of 14}\}$ The number of multiples of 14 between 1 and 50 is: $n(A \cap B) = \left\lfloor \frac{50}{14} \right\rfloor = \lfloor 3.571... \rfloor = 3$ So, there are 3 elements in the intersection of $A$ and $B$.

Step 5: Apply the Principle of Inclusion-Exclusion The problem asks for the number of elements in the smallest subset of $X$ containing both $A$ and $B$, which is $n(A \cup B)$. Using the Principle of Inclusion-Exclusion: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ Substitute the values calculated in the previous steps: $n(A \cup B) = 25 + 7 - 3$ $n(A \cup B) = 32 - 3$ $n(A \cup B) = 29$ Therefore, the number of elements in the smallest subset of $X$ containing both $A$ and $B$ is 29.

Common Mistakes & Tips

• Misinterpreting "Smallest Subset": Always remember that the smallest subset containing two sets is their union ($A \cup B$).
• Double-Counting: A common error is to simply add $n(A)$ and $n(B)$ without subtracting the intersection. The Principle of Inclusion-Exclusion is essential to avoid this.
• LCM for Intersection: When sets are defined by multiples of different numbers, their intersection consists of multiples of the LCM of those numbers.

Summary We are asked to find the cardinality of the union of set $A$ (multiples of 2 in $X$) and set $B$ (multiples of 7 in $X$), where $X = \{1, 2, \dots, 50\}$. We calculated $n(A) = 25$, $n(B) = 7$, and $n(A \cap B)$ by finding the multiples of LCM(2,7)=14, which is $n(A \cap B) = 3$. Applying the Principle of Inclusion-Exclusion, $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 25 + 7 - 3 = 29$.

The final answer is $\boxed{29}$.