MCQ
If ${N_a} = \{ an:n \in N\} ,$ then ${N_3} \cap {N_4} = $
- A${N_7}$
- ✓${N_{12}}$
- C${N_3}$
- D${N_4}$
$= \{12, 24, 36......\} =$ ${N_{12}}$.
Trick : ${N_3} \cap {N_4} = {N_{12}}$
[ $\because $ $3, 4$ are relatively prime numbers].
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