Question:
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is_______.
Solution:
$\mathrm{A}=4$ - digit numbers divisible by 3
$A=1002,1005, \ldots, 9999$
$9999=1002+(n-1) 3$
$\Rightarrow(\mathrm{n}-1) 3=8997 \Rightarrow \mathrm{n}=3000$
$\mathrm{B}=4$-digit numbers divisible by 7
$\mathrm{B}=1001,1008, \ldots, 9996$
$\Rightarrow 9996=1001+(\mathrm{n}-1) 7$
$\Rightarrow \mathrm{n}=1286$
$A \cap B=1008,1029, \ldots, 9996$
$9996=1008+(n-1) 21$
$\Rightarrow \mathrm{n}=429$
So, no divisible by either 3 or 7
$=3000+1286-429=3857$
total 4 -digits numbers $=9000$
required numbers $=9000-3857=5143$