QJEEM24Maa70e: JEE Main 2024 Mathematics - Class 11 JEEM24Maa70e Solution: Let is neither a multiple of 3 nor a multiple of 4 . Then the…

$n \in[100,700]$

$n(A)=$ 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 $12=108,120 \ldots .696$

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