If Na=an: n∈N, then N3∩N4=

N3∩N4=3,6,9,12,15.....∩4,8,12,16,20....=12, 24, 36....=N12.[∵3, 4 are relatively prime numbers].

If Na=an: n∈N, then N3∩N4=

N3∩N4=3,6,9,12,15.....∩4,8,12,16,20....=12, 24, 36....=N12.[∵3, 4 are relatively prime numbers].

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If $N_{a} = \{an : n \in N\}, then N_{3} \cap N_{4} =$

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$N_{7}$

$N_{12}$

$N_{3}$

$N_{4}$

answer is B.

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Detailed Solution

$N_{3} \cap N_{4} = \{3, 6, 9, 12, 15 . . . . .\} \cap \{4, 8, 12, 16, 20 . . . .\}$

$= \{12, 24, 36 . . . .\} = N_{12} .$

$[∵ 3, 4 are relatively prime numbers] .$

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If $N_{a} = \{an : n \in N\}, then N_{3} \cap N_{4} =$