{Number theory} Integer solutions

AI Thread Summary
The discussion revolves around finding the integer solutions for the equation x_1 + x_2 + ... + x_251 = 708, which is identified as a "stars and bars" problem. The initial assumption was that the number of solutions for another equation y_1 + y_2 + ... + y_n = 708, where n ≠ 251, would be the same, leading to a proposed solution of n = 458 after corrections. There is some confusion regarding the notation, particularly the interpretation of "a certain number of solutions" versus the number of variables. The concept of "stars and bars" is clarified as a method for distributing identical objects into distinct groups, which is relevant to solving the problem. The conversation highlights the importance of understanding the notation and the conditions for positive integer solutions.
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Homework Statement


##x_1+x_2 \cdots x_{251}=708## has a certain # of solutions in positive integers ##x_1 \cdots x_{251}##
Now the equation ##y_1+y_2 \cdots y_{n}=708## also has the same number of positive integer solutions ##y_1, \cdots y_n## Where ##n \neq251## What is ##n##

Homework Equations


I think this is a stars and bars problem but I'm not super familiar with it still

The Attempt at a Solution


So looking at the stars and bars page it seems that ##{m \choose k}={m \choose m-k}## so then would ##n## in this case just be ##457##? In this case ##m=707## and ##k=250##

Edit figured it out it is ##458## I just forgot to add 1 back to the original ##n##
 
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I don't think I understand the notation here. You say this problem "has a certain # of solutions in positive integers x 1 ⋯x251" I would take that to mean that it has 251 solutions. You then say "y1+y2 \cdots yn=708 also has the same number of positive integer solutions y1 ,⋯yn y_1, \cdots y_n Where n≠251".

If the first equation has 251 solutions and the next has "the same number" how is it not 251?

And I have no idea what a "stars and bars problem" and a "stars and bars page" are!
 
HallsofIvy said:
And I have no idea what a "stars and bars problem" and a "stars and bars page" are!
I believe it refers to the problem of how many ways of placing r identical objects into n distinct buckets. Maybe the 'bars' represent the divisions between the buckets.
The posted solution, after correction, looks right.
 
HallsofIvy said:
I don't think I understand the notation here. You say this problem "has a certain # of solutions in positive integers x 1 ⋯x251" I would take that to mean that it has 251 solutions.
No, the OP simply means that it "has j solutions", where j is unknown.

HallsofIvy said:
And I have no idea what a "stars and bars problem" and a "stars and bars page" are!
This refers to an interesting class of problems with an elegant path to solution; there are pages that explain this further on both Mathworld and Wikipedia which can be found using a well-known search engine.

Is the question asking for the number of unique solutions, or are permutations of ## x_i ## permitted?
 
... and have you dealt properly with solutions ending with a bar (that will not sum to 708) and solutions starting with one or more stars that will not satisfy that ## x_i ## are positive?

Well done for spotting the stars and bars analogue though, it is not immediately obvious.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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