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yeah I'm here \o/
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damn. That took me way to long. xD
In case you wondered: all the possible combinations of triplets:
http://rextester.com/PFIU21502
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Nice :) Save your code it may be useful soon.
But I don't think it works right... There shouldn't be 0-s in the last few solutions. And there are some combinations missing (i.e. 13+70+98=181).
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yeah noticed those flaws too, but was too lazy to iron those out since i found some valid solutions xD
but (13+70+98=181) is included, since 13+98+70=181 (2nd row) uses this combination already. (https://en.wikipedia.org/wiki/Commutative_property)
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made your code better, so that the solution is there immediately (btw first time using pascal)
Code
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nice one! looks way better than my clumsy try :) (not that good at programming^^)
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coding is the way :)
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(13, 146, 224), (98, 101, 184), (70, 90, 223) 383 each is another solution.
Mean value theorem saves more than half time.
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But the sum is having two of the same digit (3).
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I thought what you mean the same digit is
For i =/= j, a_i =/= a_j s.t. a_1+a_2+a_3 = a_4+a_5+a_6 = a_7+a_8+a_9
Nevertheless, I solved it too. :)
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Thank you for the gift. Looks like more math puzzles await me.
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