If {N_a} = {an : n in N},then {N_5} cap {N_7} | vedclass

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(C) Given that ${N_a} = \{an : n \in N\}$,which represents the set of all multiples of $a$.Thus,${N_5} = \{5, 10, 15, 20, 25, 30, 35, 40, \dots\}$ and

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${N_{35}}$

Vedclass · Answer

(C) Given that ${N_a} = \{an : n \in N\}$,which represents the set of all multiples of $a$.Thus,${N_5} = \{5, 10, 15, 20, 25, 30, 35, 40, \dots\}$ and ${N_7} = \{7, 14, 21, 28, 35, 42, \dots\}$.The intersection ${N_5} \cap {N_7}$ consists of numbers that are multiples of both $5$ and $7$.Since $5$ and $7$ are relatively prime,their least common multiple $(LCM)$ is $5 	imes 7 = 35$.Therefore,the common multiples are multiples of $35$,which is denoted as ${N_{35}}$.

If ${N_a} = \{an : n \in N\}$,then ${N_5} \cap {N_7} = $

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