ISI2014-DCG-15

Question

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

• A. $A_3$
• B. $A_4$
• C. $A_5$
• D. $A_6$

Comments

Is the answer ?

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A5 is the answer?

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How??

Answer 1

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

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

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

So the correct answer is C.