Q112 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
$\text { If } a \mathcal{N}=\{a x: x \in \mathcal{N}\} \text {, then } 3 \mathcal{N} \cap 7 \mathcal{N} \text { is }$
- A$10 \mathcal{N}$
- B$4 \mathcal{N}$
- Correct answerC$21 \mathcal{N}$
- D$3 \mathcal{N}$
Solution
The set $a\mathcal{N}$ is defined as the set of all multiples of $a$ where $a \in \mathbb{N}$ and $\mathcal{N} = \{1, 2, 3, \dots\}$.
The set $3\mathcal{N}$ consists of all natural numbers that are multiples of $3$, i.e., $3\mathcal{N} = \{3, 6, 9, 12, 15, 18, 21, 24, \dots\}$.
The set $7\mathcal{N}$ consists of all natural numbers that are multiples of $7$, i.e., $7\mathcal{N} = \{7, 14, 21, 28, 35, \dots\}$.
The intersection $3\mathcal{N} \cap 7\mathcal{N}$ consists of all natural numbers that are multiples of both $3$ and $7$.
A number is a multiple of both $3$ and $7$ if and only if it is a multiple of the least common multiple of $3$ and $7$. Since $3$ and $7$ are coprime, $\text{lcm}(3, 7) = 3 \times 7 = 21$.
Therefore, $3\mathcal{N} \cap 7\mathcal{N} = 21\mathcal{N}$.
Answer: $21 \mathcal{N}$
The set $3\mathcal{N}$ consists of all natural numbers that are multiples of $3$, i.e., $3\mathcal{N} = \{3, 6, 9, 12, 15, 18, 21, 24, \dots\}$.
The set $7\mathcal{N}$ consists of all natural numbers that are multiples of $7$, i.e., $7\mathcal{N} = \{7, 14, 21, 28, 35, \dots\}$.
The intersection $3\mathcal{N} \cap 7\mathcal{N}$ consists of all natural numbers that are multiples of both $3$ and $7$.
A number is a multiple of both $3$ and $7$ if and only if it is a multiple of the least common multiple of $3$ and $7$. Since $3$ and $7$ are coprime, $\text{lcm}(3, 7) = 3 \times 7 = 21$.
Therefore, $3\mathcal{N} \cap 7\mathcal{N} = 21\mathcal{N}$.
Answer: $21 \mathcal{N}$
Hence, the correct answer is option: 'C'.