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Old 04-27-2007   #1
Sillysidley

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Default Inclusion Exclusion

Let U=(1,2,3..,1000) and let A2,A3, and A5 be the numbers that are divisible by 2,3, and 5 respectively. Find:
a. The intersection of all 3 sets (I solved this already)
b. The intersection of A2,A3, and the compliment of A5
c. The intersection of A2, the compliment ofA3, and A5.
d. The intersection of the compliment of A2,A3, and A5
e. The intersection of A2, the compliment of A3, and the compliment of A5
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Last edited by Sillysidley; 04-27-2007 at 06:30 PM..
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Old 04-28-2007   #2
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a) To solve this, we need to find all numbers x where 2a=x, 3b=x and 5c=x.
These are all primes, so we could try multiplying.
2*3*5=30
4*3*5=60
6*3*5=90
8*3*5=120
We see a pattern here. Since it must end in 0 or 5, it must be even, and divisible by 3, this is easy. The intersection is I=(30,60,90,120,150...) all the way up to 1,000.

Is it right, somedude? I'll do the rest later...
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Old 04-28-2007   #3
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BTW, it's spelled "complement", not "compliment".
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Old 04-28-2007   #4
Sillysidley

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Oh well.
All you really need to do is draw a Venn Diagram Don't worry, I did it using the Inclusion Exclusion Property.
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Old 04-28-2007   #5
Sillysidley

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Quote:
Originally Posted by Archive View Post
a) To solve this, we need to find all numbers x where 2a=x, 3b=x and 5c=x.
These are all primes, so we could try multiplying.
2*3*5=30
4*3*5=60
6*3*5=90
8*3*5=120
We see a pattern here. Since it must end in 0 or 5, it must be even, and divisible by 3, this is easy. The intersection is I=(30,60,90,120,150...) all the way up to 1,000.

Is it right, somedude? I'll do the rest later...
You just divide 1000 by 30 and round down.
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Old 04-29-2007   #6
Sillysidley

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Ok I figured them all out, using a Venn Diagram
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Old 04-29-2007   #7
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I hate Venn Diagrams. I just solved the last one NOT using a venn diagram.
I wonder if you can make venn diagrams in ...
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Old 05-01-2007   #8
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5 times 3 times 2 us 30 count by 30 to 1000
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Old 05-01-2007   #9
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That's been confirmed already. Man, Reds, you wouldn't survive a minute on AoPS.
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Old 05-01-2007   #10
Sillysidley

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For the last time, instead of counting from 30 to 1000, just divide 1000 by 30!
Oh, everyone would yell at him for confirming an already said answer,
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