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seen Dec 5 '12 at 22:52

Dec
5
comment Another Ordered Quadruples Question
Also for the part where you get $2^4*C(49,3)=294784$ , when I calculate $C(49,3)$ I do get 18424 but when I do $2^4*C(49,3)$ I get 147392 am I making a mistake?
Dec
5
comment Another Ordered Quadruples Question
Do you say positive integers and not just integers because the value of the x's will definitely be positive since we are looking for the absolute value? More to the point do we not have to count the possibilities for if the value of x is negative?
Dec
5
asked Another Ordered Quadruples Question
Dec
5
comment Ordered Quadruples Question
I reread all of this when I woke up this morning and now I believe that I do understand the question and the answer, does the fact that it is asking about the set of natural numbers raised to the fourth not matter?
Dec
5
awarded  Commentator
Dec
5
comment Ordered Quadruples Question
Ok I think I understand thank you very much, just to be clear x1+x2+x3+x4<=50 is the same as x1+x2+x3+x4+x5=50?
Dec
5
comment Ordered Quadruples Question
Thank you that was very helpful. How does the answer change when you consider <= 50 instead of = 50 and is the answer the same whether you have 4 or 5 x's?
Dec
5
comment Ordered Quadruples Question
I'm sorry I just realized I made a mistake in the problem I didn't mean quadrupled I meant raised to the 4th power.
Dec
5
comment Ordered Quadruples Question
The number of solutions.
Dec
5
comment Ordered Quadruples Question
Yes, please more help I am still baffled.
Dec
5
comment If p is a prime number greater than 2 and k is a natural number so that k<p, how can I prove that…
How do I go about the congruence part of the question?
Dec
5
asked Ordered Quadruples Question
Dec
5
awarded  Student
Dec
5
comment If p is a prime number greater than 2 and k is a natural number so that k<p, how can I prove that…
I wasn't sure if you were asking if I was of the belief that 0 was a natural number or not but now I understand your question, no we say that 0 is not a natural number.
Dec
5
comment If p is a prime number greater than 2 and k is a natural number so that k<p, how can I prove that…
0 is not ∈ N correct?
Dec
5
comment If p is a prime number greater than 2 and k is a natural number so that k<p, how can I prove that…
There can not be a factor for the denominator since p is only divisible by 1 and itself.
Dec
5
comment If p is a prime number greater than 2 and k is a natural number so that k<p, how can I prove that…
I do I'm having trouble with the congruency part of it, I haven't done congruences in a while and was shaky at best when I did do them.
Dec
5
asked If p is a prime number greater than 2 and k is a natural number so that k<p, how can I prove that…