QJEEM21Ma03f3: JEE Main 2021 Mathematics - Class 11 JEEM21Ma03f3 Solution: The number of 4-digit numbers which are neither multiple of 7…

A = 4-digit numbers divisible by 3

A = 1002, 1005, ....., 9999.

9999 = 1002 + (n $$-$$ 1)3

$$\Rightarrow$$ (n $$-$$ 1)3 = 8997 $$\Rightarrow$$ n = 3000

B = 4-digit numbers divisible by 7

B = 1001, 1008, ......., 9996

$$\Rightarrow$$ 9996 = 1001 + (n $$-$$ 1)7

$$\Rightarrow$$ n = 1286

A $$\cap$$ B = 1008, 1029, ....., 9996

9996 = 1008 + (n $$-$$ 1)21

$$\Rightarrow$$ n = 429

So, no divisible by either 3 or 7

= 3000 + 1286 $$-$$ 429 = 3857

total 4-digits numbers = 9000

required numbers = 9000 $$-$$ 3857 = 5143