Commander Posted August 25, 2015 Share Posted August 25, 2015 Link to comment Share on other sites More sharing options...
Acme Posted August 25, 2015 Share Posted August 25, 2015 The four solutions in spoiler. The red line in the top left solution joins the digits sequentially and in occult practice is known as a 'magic path'. In use, the path alone would be inscribed on an amulet or such-a-matter to achieve whatever the intended magic effect is. Each of the four solutions has a unique magic path, though I left out 3 of them as they did not satisfy my magic effect intentions. 2 Link to comment Share on other sites More sharing options...
Commander Posted August 26, 2015 Author Share Posted August 26, 2015 (edited) The four solutions in spoiler. The red line in the top left solution joins the digits sequentially and in occult practice is known as a 'magic path'. In use, the path alone would be inscribed on an amulet or such-a-matter to achieve whatever the intended magic effect is. Each of the four solutions has a unique magic path, though I left out 3 of them as they did not satisfy my magic effect intentions. Well done Acme ! You have done it quite quickly ! +1 from my side Are there any more solutions ? We observe that no other number than 1 or 8 can occupy the central spots. Therefore there won't be any other solutions except variations with 1 & 8 at these spots. Edited August 26, 2015 by Commander Link to comment Share on other sites More sharing options...
Roamer Posted August 26, 2015 Share Posted August 26, 2015 1 & 8 determine where 2 & 7 come both 3 & 5 and 4 & 6 go together on one side, where 6 and 3 lay away from the 2 or 7 so just those 4 solutions. 2 Link to comment Share on other sites More sharing options...
Commander Posted August 27, 2015 Author Share Posted August 27, 2015 (edited) 1 & 8 determine where 2 & 7 come both 3 & 5 and 4 & 6 go together on one side, where 6 and 3 lay away from the 2 or 7 so just those 4 solutions. Perhaps these are the 16 Unique Patterns possible : Edited August 27, 2015 by Commander Link to comment Share on other sites More sharing options...
Acme Posted August 27, 2015 Share Posted August 27, 2015 Perhaps these are the 16 Unique Patterns possible : patterns3.jpg No Commander; just the four I gave and that Roamer explained. The four correct that I gave and that you repeated I have outlined in Blue and I have added red lines to show your errors. 1 Link to comment Share on other sites More sharing options...
Commander Posted August 27, 2015 Author Share Posted August 27, 2015 No Commander; just the four I gave and that Roamer explained. The four correct that I gave and that you repeated I have outlined in Blue and I have added red lines to show your errors. Hi Acme, Yes, I see the Errors in my Patterns and I am surprised how I missed these obvious violations. Sorry, I did not verify thinking that there won't be a clash by switching interchangeably 3 & 5 and 4 & 6 and missed the effect of 2 with 3 and 6 with 7 which led to my Errors. Link to comment Share on other sites More sharing options...
Acme Posted August 27, 2015 Share Posted August 27, 2015 Hi Acme, Yes, I see the Errors in my Patterns and I am surprised how I missed these obvious violations. Sorry, I did not verify thinking that there won't be a clash by switching interchangeably 3 & 5 and 4 & 6 and missed the effect of 2 with 3 and 6 with 7 which led to my Errors. No worries Commander. I made a similar error last week with the light bulb puzzle that TimeSpaceLightForce posted. I have also looked at your recently posted knight's tour puzzle thread and the one on [presumably] division, but I haven't responded because I don't understand the division problem and the knight's tour is more-or-less a programming problem these days. Thanks for some fun though. 1 Link to comment Share on other sites More sharing options...
Commander Posted August 27, 2015 Author Share Posted August 27, 2015 No worries Commander. I made a similar error last week with the light bulb puzzle that TimeSpaceLightForce posted. I have also looked at your recently posted knight's tour puzzle thread and the one on [presumably] division, but I haven't responded because I don't understand the division problem and the knight's tour is more-or-less a programming problem these days. Thanks for some fun though. Acme : Thanks and it’s nice to share interesting Puzzles ! The Knight’s 16 Moves and 17 Moves on the Six-Square Chess Board is not that tough once you are given that there is a solution. We do division calculations as explained below. As an example 123 x 456 = 56088 and a whole division of 56088 by 123 is carried out in the following steps. 123 ) 56088 ( 456 492 ---- 688 615 ---- 738 738 Take enough digits starting Left of 56088 greater than 123 and therefore we take 560 and see that 560 / 123 is 4 x123 = 492 plus reminder 68 So we write 4 as the first result digit 492 below 560 and 68 below as shown We bring down 8 which is the next digit of 56088 to make 688 Then similarly 688 / 123 = 5 x 123 = 615 plus remainder 73 is shown next bringing the last digit 8 alongside 73 to make 738 and the next step of 6 x 123 = 738 ends the division without a remainder. A Puzzle can be made out of this we can make the following pattern calling it ‘Three 3s’ letting only the 3s to be visible and # stands for any digit except 0 and $ stands for any digit including 0. #$3 ) #$$$$$ ( #$$ #$$ ---- #$$ #$$ ---- #3$ #3$ The solver has to find the possibility of replacing #s and $s suitably to recreate all digits of the ORIGINAL DIVISION SUM above. The exposed three 3s may or may not give enough logical clues for this and also there may be more than one solutions. The Seven 7s Problem is similar to this but a larger perfect division. There could be more than seven 7s in the Solution but these 7s exposed are sufficient to logically work out the other digits and the Solution is Unique ! Link to comment Share on other sites More sharing options...
Commander Posted August 28, 2015 Author Share Posted August 28, 2015 1 & 8 determine where 2 & 7 come both 3 & 5 and 4 & 6 go together on one side, where 6 and 3 lay away from the 2 or 7 so just those 4 solutions. Yes, you are right ! +1 for you 1 Link to comment Share on other sites More sharing options...
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