Let A=n [100,700] N:n is neither a multiple of 3 nor a multiple of 4.

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

Accepted Answer

$$n \in[100,700]$$

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

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

Solutions · 1