A=4− digit numbers divisible by 3 A=1002,1005,…,9999 9999=1002+(n−1)3 ⇒(n−1)3=8997⇒n=3000 B=4− digit numbers divisible by 7 B=1001,1008,…,9996 ⇒9996=1001+(n−1)7 ⇒n=1286 A∩B=1008,1029,…,9996 9996=1008+(n−1)21 ⇒n=429
So, no divisible by either 3 or 7 =3000+1286−429=3857
total 4-digits numbers =9000
required numbers =9000−3857=5143