Question

How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?​

Answer 1

We are asked to find the number of numbers that are divisible by none of 2, 5 and 7. Instead, let us find the number of numbers that are divisible by atleast one of 2, 5 and 7.

Recall that n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C).

Therefore, Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) - Number of Multiples of (2 and 5) - Number of Multiples of (5 and 7) - Number of Multiples of (7 and 2) + Number of Multiples of (2 and 5 and 7)

Number of Multiples of (2 or 5 or 7) = Number of Multiples of (2) + Number of Multiples of (5) + Number of Multiples of (7) - Number of Multiples of (10) - Number of Multiples of (35) - Number of Multiples of (14) + Number of Multiples of (70)

Number of Multiples of (2 or 5 or 7) = 60 + 24 + 17 - 12 - 3 - 8 + 1 = 79

Since we know the number of integers that can be divided by atleast one of 2, 5 and 7 is 79. We can arrive at the number of integers that can be divided by none of 2, 5 and 7, by subtracting from the whole of 120.

No of integers of 1, 2, … , 120, that are divisible by none of 2, 5 and 7 = 120 - 79 = 41

Answer

41

Answer 2

Answer:

please don't post your answer in my question