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# math question?

If J={2,4,6,…,2018,2020}, E={3,6,9,…,2016,2019}, and F={5,10,15,…,2015,2020}, how many elements are there in the set J∩(E∪F)?

### 2 Answers

- PuzzlingLv 74 weeks agoFavorite Answer
E is the set of multiples of 3 between 1 and 2020.

F is the set of multiples of 5 between 1 and 2020.

J is the set of multiples of 2 (even numbers) in that range.

So basically you are looking for the multiples of 6 or 10 between 1 and 2020.

1/6 of the numbers are even multiples of 3 (aka multiples of 6).

⌊2020/6⌋ = 336

1/10 of the numbers are even multiples of 5 (aka multiples of 10).

⌊2020/10⌋ = 202

But we can't blindly just add these two numbers because we would be double-counting multiples of both. We need exclude even multiples of 15 (aka multiples of 30) which would have been counted twice.

1/30 of the numbers are multiples of 30.

⌊2020/30⌋ = 67

So by the inclusion-exclusion principle, we would add the multiples of 6 and 10, then subtract the multiples of 30.

336 + 202 - 67

= 538 - 67

= 471

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- davidLv 74 weeks ago
J is even numbers

E is multiples of 3 ... all are not even

but a sumset of E is mutipes of 6 which are even === F is immaterial

6, 12, 18 ... 2016

which is arithmetic .. d= 6 ... a1 = 6

an = 2016 = 6 + 6(n-1) <<< solve for n

n = 336 <<< answer

- atsuoLv 64 weeks agoReport
For example , 10 is contained in J∩(E∪F) but 10 is not a multiple of 6 .

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