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determine all triplets
inWhere
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hi again,
This looks like another where trial is going to get you the solutions. Once a, b, and c pass a certain value, abc is just too large for the LHS to ever reach it.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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There's a handful I found mathematically (i.e. not trial-and-error). WLOG choose one variable to fix and solve for the other two, we'll go with
. To keep from tripping over ourselves on notation let's use x and y for b and c, as they will be the variables we're solving for. So we haveWe want to find a y such that x is an integer. The easiest way is to set the denominator to 1, so let's do that.
As long as y is an integer x will be too, so you can grab some triplets this way. That doesn't rule out other solutions, I'll let you know if I find any more.
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Worked on this some more and I think we can narrow it down to very few possibilities. Unnecessarily convoluted math is spoilered.
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Since the terms are sort of symmetrical if you are doing trials you can leave a few out. For example if you try (2,2,3) then you don't have to try (2,3,2) or (3,2,2), since every combination of addition and multiplication is used and in addition and multipliction order does not matter (they are commutative).
I guess I am saying that you only need to try different combinations, not permutations. Order does not matter.
At least that's how I see it.
If like me you sometimes have trouble remembering which is which between combinations and permutations, just remember this four word sentence:
A combination lock isn't.
I know, you have to let it sink in for a minute...
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Thinking about it some more, we can actually narrow our search even further. Remembering that we asserted a <= b <= c, this means that
must be true. We showed before that the RHS of this inequality is smallest when b is smallest, which in our case would be b = a. And the minimum value allowed for c is c = a, so if the inequality
is not true then there is no solution for that value of 'a'. Solving the inequality we get
This inequality does not hold for a >= 4, so we only need to search on a = 2 and a = 3 (note that this inequality does not hold for a = 2, but because of the quirk we noted earlier the RHS of the original inequality actually increases for a short while as b increases so it does allow for possible answers). This leaves us with something like 10 possible triplets to check, easily done with trial and error.
Last edited by TheDude (2011-10-20 15:12:19)
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