rookie1ja Posted March 31, 2007 Report Share Posted March 31, 2007 Write Numbers - Back to the Geometry Puzzles Write the numbers from 1 to 8 into the squares, so that the squares with consecutive numbers do not touch (neither edges nor corners). This old topic is locked since it was answered many times. You can check solution in the Spoiler below. Pls visit New Puzzles section to see always fresh brain teasers. Write Numbers - solution Link to comment Share on other sites More sharing options...
Guest Posted May 23, 2007 Report Share Posted May 23, 2007 Here's another solution. 1 Link to comment Share on other sites More sharing options...
Guest Posted May 23, 2007 Report Share Posted May 23, 2007 strikefiend, the corners of the 4 and 5 are touching. Link to comment Share on other sites More sharing options...
kingofpain Posted May 25, 2007 Report Share Posted May 25, 2007 I think the trick is that the middle 2 squares touching a lot of squares have to be 1 and 8 since they have only 1 adjacent number which allows you to put more numbers around them... once you start there, the puzzle virtually solves itself...am I making sense? Link to comment Share on other sites More sharing options...
Guest Posted May 30, 2007 Report Share Posted May 30, 2007 Here's another solution. Link to comment Share on other sites More sharing options...
Guest Posted May 30, 2007 Report Share Posted May 30, 2007 How about this 6 4 2 8 1 7 5 3 Link to comment Share on other sites More sharing options...
Guest Posted June 5, 2007 Report Share Posted June 5, 2007 I think the trick is that the middle 2 squares touching a lot of squares have to be 1 and 8 since they have only 1 adjacent number which allows you to put more numbers around them... once you start there, the puzzle virtually solves itself...am I making sense? absolutely correct. since the 2 middle squares touch almost all the other squares, except one each, they need to be 1 and 8. then that means the ones on the left and right need to be 2 and 7 and the ones on the top and bottom are the remaining four numbers. any combination that follows this configuration and the edges/corners rule is a possible solution. Link to comment Share on other sites More sharing options...
Guest Posted June 22, 2007 Report Share Posted June 22, 2007 I thought I had a different solution but upon further analysis it didn't work... False alarm Link to comment Share on other sites More sharing options...
Guest Posted June 22, 2007 Report Share Posted June 22, 2007 I think the trick is that the middle 2 squares touching a lot of squares have to be 1 and 8 since they have only 1 adjacent number which allows you to put more numbers around them... once you start there, the puzzle virtually solves itself...am I making sense? i think u r making sense... but thats me. but i also think what u mean is that the two middle squares are touching all but one other square. and as u said 1 and 8 are the only ones that have 1 other adjacent # . so they therefore are the only #s that can fit into the middle spots. then of course u put the # that is adjacent to those #s in the only spot where each of the two middle spots can't touch Link to comment Share on other sites More sharing options...
Guest Posted June 22, 2007 Report Share Posted June 22, 2007 How about this 6 4 2 8 1 7 5 3 thats the same puzzle upside down Link to comment Share on other sites More sharing options...
Guest Posted June 22, 2007 Report Share Posted June 22, 2007 Here's another solution. thats the same puzzle flipped Link to comment Share on other sites More sharing options...
Guest Posted June 25, 2007 Report Share Posted June 25, 2007 oops Link to comment Share on other sites More sharing options...
Guest Posted June 25, 2007 Report Share Posted June 25, 2007 6 4 1 7 2 9 5 3 Link to comment Share on other sites More sharing options...
Guest Posted July 2, 2007 Report Share Posted July 2, 2007 4 6 7 1 8 2 3 5 Nobody else has posted this one so I thought I would. Link to comment Share on other sites More sharing options...
Guest Posted August 5, 2007 Report Share Posted August 5, 2007 I am new at this. Could this be another solution? 1 3 5 7 6 8 2 4 Link to comment Share on other sites More sharing options...
Guest Posted August 7, 2007 Report Share Posted August 7, 2007 5 3 2 8 1 7 6 4 I guess this is the soultion Link to comment Share on other sites More sharing options...
Guest Posted August 26, 2007 Report Share Posted August 26, 2007 There is actually only one answer to this problem, but there appears to be four. Because the design is symmetrical across 2 axis, the answer can be mirrored to create what appears to be an additional 3 answers. Link to comment Share on other sites More sharing options...
Guest Posted September 22, 2007 Report Share Posted September 22, 2007 I am new at this. Could this be another solution? 1 3 5 7 6 8 2 4 No - the 7 and 6 are adjacent. No consecutive numbers can touch. The only solution *must* have 1 and 8 in the center. Link to comment Share on other sites More sharing options...
Guest Posted September 23, 2007 Report Share Posted September 23, 2007 Here's another solution. -- 4 7 -- 6 1 5 2 -- 3 8 -- strikefiend, That is not correct .... 4 and 5 shared a point (meet through the diagonal). Link to comment Share on other sites More sharing options...
Guest Posted November 19, 2007 Report Share Posted November 19, 2007 heres another way 3 6 7 1 8 2 5 4 Link to comment Share on other sites More sharing options...
Guest Posted November 21, 2007 Report Share Posted November 21, 2007 1 6 7 3 8 4 5 2 And this one? it looks to mee that there are multiple solutions... greta Link to comment Share on other sites More sharing options...
Guest Posted December 18, 2007 Report Share Posted December 18, 2007 In Littlemisshotpink's solutiom 5 and 4 would be adjacent! Link to comment Share on other sites More sharing options...
Guest Posted December 21, 2007 Report Share Posted December 21, 2007 Here is the only one solution mirrored on 2 axis. Link to comment Share on other sites More sharing options...
Guest Posted January 10, 2008 Report Share Posted January 10, 2008 Here is the only one solution mirrored on 2 axis. That is correct, 8 and 1 should be in the center since they're the only two numbers with one adjacent number (1->2 and 8->7). This makes that between 1 and 2 the 8 is, and between 7 and 8 the 1 is. 4 numbers left, those are the easy pick. Solved this in less then a minute. Link to comment Share on other sites More sharing options...
Guest Posted March 10, 2008 Report Share Posted March 10, 2008 (edited) What about 13 7586 42 Edited March 10, 2008 by BlueDevil408 Link to comment Share on other sites More sharing options...
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