The number of 4-digit numbers

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 $=9000-3857=5143$