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04272007  #1 
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Inclusion Exclusion
Let U=(1,2,3..,1000) and let A_{2},A_{3}, and A_{5} 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 A_{2},A_{3}, and the compliment of A_{5} c. The intersection of A_{2}, the compliment ofA_{3}, and A_{5}. d. The intersection of the compliment of A_{2},A_{3}, and A_{5} e. The intersection of A_{2}, the compliment of A_{3}, and the compliment of A_{5}
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. Last edited by Sillysidley; 04272007 at 06:30 PM.. 
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04282007  #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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04282007  #3 
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BTW, it's spelled "complement", not "compliment".

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04282007  #4 
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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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04282007  #5  
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Quote:
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04292007  #6 
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Ok I figured them all out, using a Venn Diagram
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04292007  #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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05012007  #8 
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5 times 3 times 2 us 30 count by 30 to 1000
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05012007  #9 
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That's been confirmed already. Man, Reds, you wouldn't survive a minute on AoPS.

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05012007  #10 
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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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