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

Question

Accepted Answer

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$n \in[100,700]$

$n(A)=$ Total $-$ (multiple of $3$ + multiple of 4) + (multiple of 12)

Total $=601$

Multiple of $3=102,105, \ldots, 699$

$$\begin{aligned} & n=699=102+(n-1) 3 \\ & \Rightarrow n=200 \end{aligned}$$

Multiple of $4=100,104 \ldots ., 700$

$$\begin{aligned} & n=700=100+(n-1) 4 \\ & \frac{600}{4}+1=n \\ & \Rightarrow n=151 \end{aligned}$$

Multiple of $12=108,120 \ldots .696$

$$\begin{aligned} & n=696=108+(n-1) 12 \\ & n=50 \\ & \therefore n(A)=601-(200+151)+50 \\ & =300 \end{aligned}$$

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