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

Question

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

Professor PC Thomas Classes · Answer

(b): Multiples of 3 are $102,105, \ldots, 699$ $$ \begin{aligned} & a_{n}=699=102+(n-1) \cdot 3 \ \Rightarrow \quad & 597=3(n-1) \Rightarrow n=200 \end{aligned} $$ Now, multiples of 4 are $100,104, \ldots, 700$ $$ \begin{aligned} & a_{m}=700=100+(m-1) \cdot 4 \ \Rightarrow \quad & m=151 \end{aligned} $$ Multiples of 3 and 4 are 108, 120, ..., 696 $$ \begin{aligned} & a_{p}=696=108+(p-1) \cdot 12 \ \Rightarrow \quad & p=50 \ & n(3 \cup 4)=n(3)+n(4)-n(3 \cap 4) \ = & n+m-p=200+151-50=301 \end{aligned} $$ $	herefore \quad$ The number of elements in $A$ $$ =601-301=300 $$