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Interesting question, give me some minutes to look into it. Are you sure that's all there is to it ?
Ah, figured it.
It's quite simple really.
Let b1-5 be the 5 books in question and v1-10 be the values of each 10 weightings (I'm too lazy to write them down - v1 = 108, v2=112,v3=113 and so forth).
Let's write a system of equations now: b1+b2=v1 b1+b3=v2 b1+b4=v3 b1+b5=v4 b2+b3=v5 b2+b4=v6 b2+b5=v7 b3+b4=v8 b3+b5=v9 b4+b5=v10
Scratch that in the definition, b1-5 represents the weight of each of those 5 books. Regardless, now we have a system with 10 equations and 10 unknown variables which should be solved easily.
5C2=5!/(2! * (5-2)!)=5!/(2! *3!)=10 yes, just checking that there are ten possible combos:)
Easiest way to do it is to take 3 equations out of that bunch that use exactly 3 different variables. So let's say, these ones: b1+b2=v1 b1+b3=v2 b2+b3=v5 See ? 3 equations that only use 3 different variables. The system is solving itself now. Add the first two together: 2b1+b2+b3=v1+v2 And replace b2+b3 from the third equation 2b1+v5=v1+v2 And now you have b1=(v1+v2-v5)/2
Okay, didn't know that it was supposed to be a system of equations problem! I was almost prepared to do it manually) I think I kind of get the jist of it now. Thank you so much!
And from then onward, you can get b2 out of the first equation, and then b3 out of the second one and so forth.
Well, that's the beauty of mathematics! I didn't know either but once you start writing down the question in mathematical language things get so much clearer. The beauty of it is that there may very well be another way of solving it that we haven't figured yet. For the time being, this will do though.