 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Simultaneous Settlement II (Posted on 2013-09-26) Determine the total number of nonnegative integer solutions to this system of simultaneous equations:

(A) P*Q*R*S*T = 7X*11Y*13Z*79W, and:
(B) X+Y+Z+W = 44

Prove that there are no others.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) re: solution Comment 3 of 3 | (In reply to solution by Charlie)

your simplification gave me inspiration to revisit my approach from a combinatorial perspective.  This is a good example of depending too much on a computer for me because I didn't catch that my phones CAS failed to factorize the x^2+5x+6 in my f5(x) function and thus I did not realize that f5(x)=C(x+4,4).

Using this simplified form I was able to evaluate the nested summation for general n and got that when
x+y+z+w=n the number of solutions is C(n+19,19)
and indeed
C(44+19,19)=6,131,164,307,078,475 as has been shown.

Now I believe this can be explained rather simply when we look at this from a combinatorical perspective.  If you look at what we are really doing, in the end we are simply taking 44, representing it as a sum of 4 nonnegative integers, and taking each of those nonnegative integers and representing them as a sum of 5 nonnegative integers.  Well this is the same as representing 44 as the sum of 4*5=20 nonnegative integers and then simply adding them in groups of 5 to get x,y,z,w.  Thus for the total number of solutions we get C(n+19,19)

This can then be expanded to the more general problem:
x1*x2*...*xa = p1^y1*p2^y2*...*pb^yb
y1+y2+...+yb=n
where we are solving for a x's and b y's as nonegative integers, the b p's are given primes and n is a given nonnegative integer.
Then using the argument above we get the number of solutions to be:
C(n+ab-1,ab-1)

 Posted by Daniel on 2013-09-26 15:14:27 Please log in:

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