Let $A=\{n \in[100,700] \cap \mathbb{N}: n$ is neither a multiple of 3 nor a multiple of 4$\}$. Then the number of elements in $A$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{n}(3) \Rightarrow 	ext { multiple of } 3 \ & 102,105,108, \ldots . ., 699 \ & \mathrm{~T}_{\mathrm{n}}=699=102+(\mathrm{n}-1)(3) \ & \mathrm{n}=200 \ & \mathrm{n}(3)=200 \ & \because \mathrm{n}(4) \Rightarrow 	ext { multiple of } 4\end{aligned}$ $\begin{aligned} & 100,104,108, \ldots ., 700 \ & T_n=700=100+(n-1)(4) \ & n=151 \ & n(4)=151 \ & n(3 \cap 4) \Rightarrow 	ext { multiple of } 3 \& 4 	ext { both } \ & 108,120,132, \ldots ., 696 \ & T_n=696=108+(n-1)(12) \ & n=50 \ & n(3 \cap 4)=50 \ & n(3 \cup 4)=n(3)+n(4)-n(3 \cap 4) \ & \quad=200+151-50 \ & \quad=301 \ & n(\overline{3 \cup 4})=	ext { Total }-n(3 \cup 4)=	ext { neither a multiple } \ & 	ext { of } 3 	ext { nor a multiple of } 4 \ & =601-301=300\end{aligned}$

Let $A=\{n \in[100,700] \cap \mathbb{N}: n$ is neither a multiple of 3 nor a multiple of 4$\}$. Then the…

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