If {N_a} = {an : n in N},then the set {N_6} c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that ${N_a} = \{an : n \in N\}$.This represents the set of all multiples of $a$.Therefore,${N_6} = \{6, 12, 18, 24, 30, 36, 42, 48, \dots\}$

Vedclass · Accepted Answer

${N_{24}}$

Vedclass · Answer

(C) Given that ${N_a} = \{an : n \in N\}$.This represents the set of all multiples of $a$.Therefore,${N_6} = \{6, 12, 18, 24, 30, 36, 42, 48, \dots\}$ and ${N_8} = \{8, 16, 24, 32, 40, 48, \dots\}$.The intersection ${N_6} \cap {N_8}$ consists of all numbers that are multiples of both $6$ and $8$.The common multiples of $6$ and $8$ are the multiples of their least common multiple,$LCM(6, 8)$.$LCM(6, 8) = 24$.Thus,${N_6} \cap {N_8} = \{24, 48, 72, \dots\} = {N_{24}}$.

If ${N_a} = \{an : n \in N\}$,then the set ${N_6} \cap {N_8} = $

Similar Questions

Consider the experiment of rolling a die. Let $A$ be the event 'getting a prime number' and $B$ be the event 'getting an odd number'. Write the set representing the event '$A$ but not $B$'.

If $A, B$ and $C$ are non-empty sets,then $(A - B) \cup (B - A)$ equals:

Consider the sets $X$ and $Y$ where $X = \{ \text{Ram, Geeta, Akbar} \}$ and $Y = \{ \text{Geeta, David, Ashok} \}$. Find $X \cap Y$.

If $X = \{a, b, c, d\}$ and $Y = \{f, b, d, g\},$ find $X \cap Y$.

Let $A = \{1, 2, 3, 4\}$ and $B = \{2, 3, 4, 5, 6\}$. Then $A \cap B$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module