Turtle Posted November 30, 2007 Author Report Share Posted November 30, 2007 Puts on straw hat, bibs, boots, grabs piece of straw to chew on. ... Good advice. What did Mark say about common sense again? Anyway, giddyup & here we go with chewin' straw in place. If you highlight my first post you will find my concurrance at bottom right justified. :rose: I just knew you had something up your sleeve, when you said:...an answer, if not a solution. but, even knowing you, it didn't cross my mind that's what you meant! :xmas_gift: That little double entendre was a quantum fluxuation...a little peek into the sense of humor of the Universe beating me to the punch line for a change. :prezsmall: Arguably getting off on the formatting is cut of the same cloth. Smoke 'em if ya got 'em I guess. :friday: I have a straw-in-the-mouth update on the matter of the student who is at the bottom of all this hay. I called my tutor friend to forward our information and as it turns this may all come to a satisfying conclusion. I live in Washington State where the new testing program (WASL) requires students to show and justify their math work. The student under the bales is a 11th grader recovering from severe head trauma, and so the tutoring. The deadline for the assignment is December 5th, and through our little international version of the 'telephone game' I hope to get the necessary & sufficient info down the pipe on schedule. Inasmuch as the local curriculums now include matrices for all students at this level, not only will this particular student get the juice by our sweat, but the tutors have what they need to service other students doing matrices. My tutor friend says "thanks" to you all. :cake: If anyone is inclined to look more deeply (and because I love to post peculiar lists ) here’s a (slightly oddly collated) list of all the valid row combinations: Oh yeah baby! I'm sure we can get someone to reformat it if need be. I have given it a few looks & it appears worthy of more. One of my first impulses was to take each equation and use it as initial conditions in a cellular automaton, and keeping all else equal, make a comparative analysis of the resulting patterns. :biggringift: That's a wrap.;) Quote Link to comment Share on other sites More sharing options...

Turtle Posted November 30, 2007 Author Report Share Posted November 30, 2007 Alrighty thens. :evil: In the above I presumed that the student is learning matrices and I really don't know it is so. :) Moreover, in my original post I note that I used 'other methods', or some such a matter of words, to describe how I concluded the list was flawed. Just so, and in line with the question of choosing equations, here attached is a scan of my work to those ends. The 2 dark dots mark the contradictory/inconsistent equalities. :cup: :phones: Y'all cain't bale no hay til' y'all cut sum grass. - Roger Thelonious George Quote Link to comment Share on other sites More sharing options...

Turtle Posted December 16, 2007 Author Report Share Posted December 16, 2007 Okally dokally Boys & Girls. Time to put those caps back on. :D I have spoken again with my tutor friend, and the telephone has played a role in our...erhrm... my error(s). :D :) The tutor called the student's teacher and we are closer now to the horse's mouth than ever. :) To whit, all that is actually given in the problem is the list of 10 weights, and they are correct as I gave them. However, the variable assignments I gave were my misunderstanding on the phone that they were given when in fact they were some of the tutor's initial attempts to analyze the problem. :doh: The teacher is not teaching matrices with this either. So, the games back on, and I admit I waited to post until I solved it to my satisfaction. So, with no blatant spoiler, here's the hints. Q, hit exactly on one part when he said The first useful thing to realize is that the weight of all 5 bales must be the sum of those 10 weights divided by 4. Let's call this total T. The weight of a single bale can be gotten by subtracting the weight of the other 4 from T, which can be done by subtracting the sum of the right two pairs. If we subtract a+b=110 and c+d=118 for instance, we are left with the weight of e. Now in that example, Q does not give the right 2 pairs in his example, if by 'right' he meant correct and not a direction. The hint from the teacher to the tutor to me to you on this 'correct' choosing of pairs is that you only know for sure that the lightest pair-weight is the lightest 2 bales, and the heaviest pair-weight is the heaviest 2 bales. Last is, all solutions are integers. Ready...set...solve!!!! A=54 B=56 C=58 D=59 E=62 Quote Link to comment Share on other sites More sharing options...

CraigD Posted December 16, 2007 Report Share Posted December 16, 2007 If you’re allowed to rearrange the constant side of the given equations, rather than change them, a+b=110, a+c=114, a+d=115, a+e=118, b+c=112, b+d=113, b+e=116, c+d=117, c+e=120, d+e=121 gives solution: a=56, b=54, c=58, d=59, e=62. I brute forced this, letting the computer try all possible rearrangements, so can’t report any great intuition in how I rearranged the constants. While being such a brute force junky may allow my intuition to atrophy, it’s reliable, and allows me to state with confidence that the only possible solutions for any rearrangement of the constants in the system of equations are 54, 56, 58, 59, 62. There are, of course, 5! = 120 possible rearrangements. Here’s the MUMPS program and data that committed this act of brute force:f N=2:1:(X," ") s (X," ",1,N)=(X," ",2,N)_" "_(X," "),(Y," ",N)=(Y," ",N)+1#N q:(Y," ",N) ;XNPERM: next permutation f K=1:1:(M(1)," ")-1 f A=1:1:(M(""),-1) i '(((M(A)," ",1,K-1)," 0")),(M(A)," ",K) f B=1:1:(M(""),-1) i A-B s S=-(M(:)," ",K)/(M(A)," ",K) i S f J=1:1:(M(1)," ") s (M(:D," ",J)=(M(A)," ",J)*S+(M(:)," ",J) ;XGJSOLVE: Gause-Jordan solve the matrix m M=M0 f A=1:1:(M(""),-1) s (M(A)," ",(M(A)," "))=(M0((X," ",A))," ",(M(A)," ")) ;XRESET: rearrange rightmost row of matrix s K=(M(1)," ") f A=1:1:(M(""),-1) i '(((M(A)," ",1,K-1)," 0")),(M(A)," ",K) q ;XISNOK: check for inconsistency k M f I=1:1 r R,! q:R="" s:R["""" R=(R,"""",2) s M(I)=R ;read matrix 1 1 0 0 0 110 1 0 1 0 0 112 1 0 0 1 0 113 1 0 0 0 1 114 0 1 1 0 0 115 0 1 0 1 0 116 0 1 0 0 1 117 0 0 1 1 0 118 0 0 1 0 1 120 0 0 0 1 1 121 m M0=M k XS s XS=0,X="1 2 3 4 5 6 7 8 9 10",Y=0 f x XRESET,XGJSOLVE,XISNOK s:' XS=XS+1,XS(X)="" w:' XS," ",X,! x XNPERM ;find all consistent arrangements Quote Link to comment Share on other sites More sharing options...

Turtle Posted December 16, 2007 Author Report Share Posted December 16, 2007 If you’re allowed to rearrange the constant side of the given equations, rather than change them, ... I still plan to find out more of the context of how the teacher presented the problem, but our solutions agree so I'll give my process after getting the new info. It is not so much rearranging to solve as it is solving to rearrange. First, I summed the list: 110+112+113+114+115+116+117+118+120+121=1156Because each bale is used 4 times, 4A+4B+4C+4D+4E=1156 (my tutor friend had this part early on, but I led us off on the system-of-equations route.:doh:) So, A+B+C+D+E=289 We know because A+B=110, and 110 is the smallest value of pair weights, that A & B are the lightest bales. (I think the students were supposed to figure this out on their own. :eek::)) Further we know that because D+E=121, the heaviest pair weight, that D & E are the heaviest bales. We now solve for C by summing A+B+D+E=110+121=231, and subtracting that from the sum of all, 289, we have 289-231=58. So C=58. Then, I reasoned since A & B are the smallest two and D & E the largest two, then C is the middle weight. So any pairing to give 121 has to have both terms greater than 58. There are only 2 possibilities: 60+61 & 59+62. But, if you let E=61, then E + C =119 and that weight's not on the list, so the answer is D=59 and E =62. Similarly I found A & B, then arranged the rest of the equations with the constant list. :cup: :thumbs_up Quote Link to comment Share on other sites More sharing options...

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