How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?
Explanation:
There are total 120 numbers given.
Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)
We know, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).
n(x) = number of multiples of x.
∴ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(2 ∩ 5) - n(5 ∩ 7) - n(7 ∩ 2) + n(2 ∩ 5 ∩ 7)
⇒ n(2 ∪ 5 ∪ 7) = n(2) + n(5) + n(7) - n(10) - n(35) - n(14) + n(70)
⇒ n(2 ∪ 5 ∪ 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79.
Now, Integers not divisible by 2, 5 and 7 = 120 – (numbers that are divisible by at least one of 2, 5 and 7)
= 120 – 79 = 41
∴ Out of 120 numbers given, 41 numbers are not divisible by any of 2, 5 and 7.
Hence, option (b).
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