Let mathrm{A}={mathrm{n} ∈[100,700] cap mathrm{N}: mathrm{n} is neither a multiple of 3 nor a mult

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1.Set Theory

medium

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

$300$

$280$

$310$

$290$

(JEE MAIN-2024)

Solution

$ \mathrm{n}(3) \Rightarrow \text { 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$ multiple of $4$

$ 100,104,108, \ldots ., 700 $

$ T_n=700=100+(n-1)(4) $

$ n=151 $

$ n(4)=151 $

$ \mathrm{n}(3 \cap 4) \Rightarrow \text { multiple of } 3 \& 4 \text { both } $

$ 108,120,132, \ldots ., 696 $

$ T_n=696=108+(n-1)(12) $

$ \mathrm{n}=50 $

$ \mathrm{n}(3 \cap 4)=50 $

$ \mathrm{n}(3 \cup 4)=\mathrm{n}(3)+\mathrm{n}(4)-\mathrm{n}(3 \cap 4) $

$ \quad=200+151-50 $

$ =301$

$\mathrm{n}(\overline{3 \cup 4})=$ Total $-\mathrm{n}(3 \cup 4)=$ neither a multiple of $3$ nor a multiple of $4$

$=601-301=300$

Standard 11

Mathematics

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