Question 1148455: (a) How many integers from 1 through 201 are multiples of 6 or multiples of 11?
(b) How many integers from 1 through 201 are neither multiples of 6 nor multiples of 11?
(c) How many integers from 1 through 201 are multiples of both 6 and 11?
(d) How many integers from 1 through 201 are multiples of either 6 or of 11 but not multiple of both?
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 201/6 is just under 34, so there are 33 numbers that are multiples of 6
201/11 is just over 18, so there are 18 numbers that are multiples of 11
multiples of both at the same time are at 66 132 and 198, 3 numbers
Either or but not both include the sum of 34 and 18 minus 6 where it is both (2 numbers aren't counted for each of 66 132 198) or 45.
Answer by ikleyn(52851)  (Show Source):
You can put this solution on YOUR website! .
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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