JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 17)
The number of 4-digit numbers which are neither multiple of 7 nor multiple of 3 is ____________.
Answer
5143
Explanation
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
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
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