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The solution to part (a) given by @boreal is WRONG.
I came to bring the correct solution.
The number of integers from 1 to 201 (inclusive) that are multiple of 6 is 33
(notice that = 33.5, approximately).
The number of integers from 1 to 201 (inclusive) that are multiple of 11 is 18
(notice that = 18.3, approximately).
The number of integers from 1 to 201 (inclusive) that are multiple of 6 and of 11 simultaneously is 3: 66, 132 and 198).
From it, the number of integers from 1 to 201 (inclusive) that are multiple of 6 or 11 is 33 + 18 - 3 = 48. ANSWER
We must first add 33 and 18 and then subtract the "3" from this sum (!)
It is because when we add 33 and 18, we count TWICE the integers in the intersection of these subsets (!)
Solved.
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See the lesson
- Counting elements in sub-sets of a given finite set
in this site, where you can find other similar solved problems.
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