A problem about elements in a set

mathlover28

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I just come across difficulties when handling this problem.

I know a must be odd and b must be even. But how would I count how many elements in the set A and B, other than listing out all the elements?

Appreciate your help. Thanks.
 

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I just come across difficulties when handling this problem.
I know a must be odd and b must be even. But how would I count how many elements in the set A and B, other than listing out all the elements?
In the textbook from which this question comes does N\mathbb{N} contain 0 ?0~? or not?
 
In my course, natural number does not contain 0.
That makes a difference. It means that set A\mathcal{A} contains all odd positive integers 3,5,7.40393,5,7.\cdots 4039.
How many are in set A ?\mathcal{A}~?
If you follow the posting guidelines then you must post your work.
 
That makes a difference. It means that set A\mathcal{A} contains all odd positive integers 3,5,7.40393,5,7.\cdots 4039.
How many are in set A ?\mathcal{A}~?
If you follow the posting guidelines then you must post your work.
There are 2019 elements in set A.

This is what I literally wrote down for this problem:

Note that A contains odd integers {3,5,7,9,...,4039}. For the set A and B, the elements must also be odd.
i.e. Both a and b+1 are odd. Hence b is even and b >= 2.

Then I am struggling do I really need to count all the elements or not.

Any more hints? Thanks.
 
There are 2019 elements in set A. This is what I literally wrote down for this problem:
Note that A contains odd integers {3,5,7,9,...,4039}. For the set A and B, the elements must also be odd.
i.e. Both a and b+1 are odd. Hence b is even and b >= 2.
Then I am struggling do I really need to count all the elements or not.
153=337515^3=3375 BUT 163=409616^3=4096 this limits the size of numbers in (\\mathcal{A}\cap\mathcal{B}\)
Since B={k(2j+1)3: {k,j}N\mathcal{B}=\{k(2j+1)^3:~\{k,j\}\subset\mathbb{N}, we can see for example k=1 & j15k=1~\&~j\le 15 would give numbers in both sets.
k=3 & j11k=3~\&~j\le 11 would give numbers in both sets BUT not k=3 & j=13k=3~\&~j= 13 would give numbers in both sets.
Can you finish?
 
153=337515^3=3375 BUT 163=409616^3=4096 this limits the size of numbers in AB\mathcal{A}\cap\mathcal{B}
Since B={k(2j+1)3: {k,j}N\mathcal{B}=\{k(2j+1)^3:~\{k,j\}\subset\mathbb{N}, we can see for example k=1 & 2j+115k=1~\&~2j+1\le 15 would give numbers in both sets.
k=3 & 2j+111k=3~\&~2j+1\le 11 would give numbers in both sets BUT not k=3 & 2j+1=13k=3~\&~2j+1= 13 would give numbers in both sets.
Can you finish?
EDITED
 
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