Question

Consider three sets $A,B,C$ such that set $A$ contains all three digit numbers that are multiples of $4,$ set $B$ contains all three digit even numbers that are multiples of $3$ and set $C$ contains all three digit numbers that are multiples of $5.$ Then the number of elements are present in $n(A\cup B\cup C)$ is

Answer 1

$A=\{100,104,108,...,996\}$
$=\{4\times 25,4\times 26,...,4\times 249\}\to 225\text{elements}$
$B=\{102,108,114,...,996\}$
$=\{6\times 17,6\times 18,...,6\times 166\}\to 150\text{elements}$
$C=\{100,105,110,...,995\}$
$=\{5\times 20,5\times 21,...,5\times 199\}\to 180\text{elements}$

$A\cap B=\{108,120,132,...,996\}$
$\Rightarrow$ All $3$ digit multiples of $12$
$\Rightarrow 75$ elements.
$B\cap C=\{120,150,180,...,990\}$
$\Rightarrow$ All $3$ digit multiples of $30$
$\Rightarrow 30$ elements.
$C\cap A=\{100,120,140,160,...,980\}$
$\Rightarrow$ All $3$ digit multiples of $20$
$\Rightarrow 45$ elements.

$A\cap B\cap C=\{120,180,...,960\}$
$\Rightarrow$ All $3$ digit multiples of $60$
$\Rightarrow 15$ elements.
$n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)–n(A\cap B)-n(B\cap C)–n(C\cap A)+n(A\cap B\cap C)$
$=225+180+150–75–30–45+15=420$