Math Is Fun Forum

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?

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 )

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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?

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).

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

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Numbers are the essence of the Universe

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.

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mathbalaji

Member

Registered: 2007-05-29

Posts: 0

Re: Help me with thi...

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

sk

Member

Registered: 2006-08-04

Posts: 19

Re: Help me with thi...

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

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: 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?

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".

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"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..."