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6 NUMBERS must be identified from the following combinations, which are suggestion of the 6 unidentified numbers,with a clue given is the total number of good numbers in the combination (tgn) and total good numbers in good position(tgp). Find the the secret 6 numbers using the clues below.
285963 tgn3 tgp2
695395 tgn2 tgp0
105280 tgn3 tgp0
394514 tgn3 tgp0
725983 tgn3 tgp1
183945 tgn3 tgp1
894166 tgn2 tgp0
635412 tgn3 tgp1
355462 tgn2 tgp1
461794 tgn1 tgp0
909079 tgn1 tgp0
Last edited by triple (2015-08-20 08:24:15)
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hi triple,
Welcome to the forum.
This sounds a bit like the peg game (two players) of Mastermind. One person sets up four pegs of various colours and hidden from view. The other tries combinations to guess the four and gets told how many colours are right and how many in the correct place. You score this by counting how many tries it took and then swapping roles.
Solving yours may take a while.
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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Hi triple;
Do you know if this problem has a solution? I can't find one.
I agree with Bob's suggestion that this appears to be a Mastermind-like puzzle and have gone about trying to solve it accordingly. All possible (I think) options fairly quickly run into a contradiction.
Last edited by phrontister (2015-08-23 17:51:05)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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I can't find one either. I'll check my working ... if no errors then I can prove this.
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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Hi Bob;
Here's a revised version of the proof I wrote in post #5 (now deleted), which contained an error in logic. Also now included are images of my workings in Excel.
I've numbered the combinations from #1 to #11 to help with my notation.
Yellow cells in the 'Working grid' denote test numbers. Orange cells and 'x'es in the proof denote eliminated numbers. Red cells denote contradiction locations (only one, even if more than one).
Below, I try to show that the numbers 9, 7 & 0 in combination #1 are invalid, by which it follows that the puzzle has no solution.
9 : If #1 = 9xxxxx, xx9xxx or xxxxx9, then #2 = xxxx9x. 0,1,4,6,7 are eliminated (from tgn=1), so #4 = x35xx2. #3 also has all of #4's 3,5,2, contradicting its tgn2.
7 : If #1 = xxxx7x, then #2 = xxx7xx. 0,1,4,6,9 are eliminated (from tgn=1), so #5 = 8xxxxx...with less than its tgn2.
To eliminate 0 I used the following proofs (a), (b) & (c) for when #1 = x0xxxx or xxx0xx to eliminate 4, 6 & 1 (9 & 7 already being eliminated):
(a) If #2 = 4xxxxx or xxxxx4, then 1,6,7,9 are eliminated (from tgn=1) and #8 = 3x45xx or 3xx5x4. #6 then = xx3x45, eliminating 8. But #5 needs 8 for its tgn2.
(b) If #2 = x6xxxx, then 1,4,7,9 are eliminated (from tgn=1) and #8 = 3xx5xx, contradicting its tgn3.
(c) If #2 = xx1xxx, then 4,6,7,9 are eliminated (from tgn=1) and #8 = 3xx51x. #6 then = 1x3xx5, eliminating 8. But #5 needs 8 for its tgn2.
Last edited by phrontister (2017-02-27 00:28:16)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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