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#1 2007-05-29 01:06:48

sk
Member
Registered: 2006-08-04
Posts: 19

Help me with thi...

Let X be the super set..
and A be the subset of X..   the complement of A is not in X.. Is there any possibility to have  sets like this?

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#2 2007-05-29 01:38:03

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Help me with thi...

by defintion alone, thats impossible:

(*please say im not wrong someone tongue )


The Beginning Of All Things To End.
The End Of All Things To Come.

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#3 2007-05-29 18:58:40

sk
Member
Registered: 2006-08-04
Posts: 19

Re: Help me with thi...

Yes.. thanks for the reply and i have one more doubt..

Let X be the super set..  and M be the collection of subsets of X .. then A belongs to M.. but the complement of A does not belongs to X..     Can any one give me an example for this type?

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#4 2007-05-29 19:01:14

sk
Member
Registered: 2006-08-04
Posts: 19

Re: Help me with thi...

Help me with this problem....

A, B, C, and D represent different digits. AACA, ADDD, BCDB, and BDAC represent four different four-digit prime numbers. Determine the sum of all four four-digit prime numbers (AACA + ADDD + BCDB + BDAC).

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#5 2007-05-29 19:24:07

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: Help me with thi...

The only clue I have is
Since A,B,C,D are distinct , and primes can't be divides by 2, or 5.
A,B,C,D={1,3,7,9},A+B+C+D=20
S=(2A+2B)1000+(A+2D+C)100+(A+2D+C)10+(A+B+C+D)=2000(A+B)+(A+2D+C)110+20


Numbers are the essence of the Universe

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#6 2007-05-29 21:57:23

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Help me with thi...

Stanley has already deduced that primes have to end in 1, 3, 7 or 9 and that as your four numbers end in all the letters between them, then the letters must be some arrangement of 1, 3 7 and 9.

I went and looked at a big list of primes to see which of the sixteen numbers in the form ADDD were prime. It turns out that there are three: 1777, 1999 and 7333. This also means that A is 1 or 7.

Then I looked for numbers in the form AACA which started with 1 or 7.
There are only two: 1171 and 7717. This is useful because it means that 1 and 7 go with A and C, although we don't know which way around yet.
Combining this knowledge with the ADDD information gives us two possible combinations:

#1
A1, B3, C7, D9

#2
A7, B9, C1, D3

This means that one of the following sets must contain all primes:

#1: {1171, 1999, 3793, 3917}
#2: {7717, 7333, 9139, 9371}

We already know that the first two numbers in each set are prime.
Checking the prime list, 3793 and 3917 are both there and neither of 9139 or 9371 are.
Therefore, #1 is the correct combination.

Your final answer is 1171 + 1999 + 3793 + 3917 = 10880.


Why did the vector cross the road?
It wanted to be normal.

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#7 2007-05-29 22:46:28

mathbalaji
Member
Registered: 2007-05-29
Posts: 0

Re: Help me with thi...

mathsyperson! Excellent! That was really a great answer and great explanation!

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#8 2007-05-29 23:19:47

sk
Member
Registered: 2006-08-04
Posts: 19

Re: Help me with thi...

mathsyperson! Excellent! That was really a great answer and fine explanation..

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#9 2007-05-30 02:17:37

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Help me with thi...

sk wrote:

Let X be the super set..
and A be the subset of X..   the complement of A is not in X.. Is there any possibility to have  sets like this?

Are you taking the complement with respect to X?  Then let:

X = {x, {x}}

Let A = {{x}}

Then A complement = {x}.  But {x} is in X.

Edit: But I guess it depends what you mean by "in".  Typically, you mean "is an element of".  You appear to be using "in" as "a subset of".


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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