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Let $\mathbb{N}=\{1,2,3, \dots\}$ be the set of natural numbers. For each $n \in \mathbb{N}$, define $A_n=\{(n+1)k, \: k \in \mathbb{N} \}$. Then $A_1 \cap A_2$ equals

  1. $A_3$
  2. $A_4$
  3. $A_5$
  4. $A_6$
in Set Theory & Algebra by Veteran (431k points)
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Is the answer ?
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A5 is the answer?
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How??

1 Answer

+3 votes

$A_1=\{2k: k\in \mathbb{N}\}=\{2,4,6,8,10,12,14,16,18,...,2k,...\}$

$A_2=\{3k: k\in \mathbb{N}\}=\{3,6,9,12,15,18,21,24,27,...,3k,...\}$

$\begin{align} \therefore A_1 \cap A_2&=\{6,12,18,...,6k,...\}\\&=\{6k:k \in \mathbb{N}\}\\&=\{(5+1)k:k \in \mathbb{N}\}\\&=A_5\end{align}$

So the correct answer is C.

 

by Active (3.5k points)

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