Question

Find the total number of natural numbers which are multiples of $7$ or $3$ or both and lie between $200$ and $500$.

Answer 1

First numbers divisible by $7$ between $200$ and $500$ is $203$ and last number is $497$,

$\therefore$ Number of terms divisible by $7$ between $200$ and $500$ is

$497=203+(n_1-1)7$ $\Rightarrow 497=203+7n_1-7$ $\Rightarrow 7n_1=301\Rightarrow n_1=43$

Similarly, first numbers divisible by $3$ between $200$ and $500$ is $201$ and last number is $498$

$\therefore$ number of terms divisible by $3$ between $200$ and $500$ is

$498=201+(n_2-1)3$ $\Rightarrow 498=201+3n_2-3\Rightarrow n_2=100$

Now first number divisible by $7$ and $3$ i e $21$ between $200$ and $500$ is $210$ and last number is $483$

$\therefore$ number of terms divisible by $21$ between $200$ and $500$ is

$483=210+(n_3-1)21$ $\Rightarrow 483=210+21n_3-21\Rightarrow n_3=14$

Therefore, the total number of natural numbers which are multiples of $7$ or $3$ or both and lie between $200$ and $500$ is

$=n_1+n_2-n_3$ $=43+100-14=129$