Guest Posted February 13, 2011 Report Share Posted February 13, 2011 Can you place 10 coins in such a way that they lie in 5 straight lines and on each line there are 4 coins! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 It's Conan!!! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 no. consider the case where you have all 10 coins along the same line. you would have 6 lines of four. if you remove one coin from each end, you have 4 lines of four, and there is no way to make a 5th line with the two coins. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 (edited) .......1 ..2..3..4..5 .....6....7 ........8 ....9.....10 1. 2345 2. 1369 3. 147 10 4. 268 10 5. 5789 Edited February 13, 2011 by sayalzah Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 Okay. imagine a 3X3 table. leave the top left tab empty, and put one coin in all the rest. we have 8 coins. now, add a coin to the middle tab, on top of the other coin, and an other one on top of the coin in the right bottom tab. so we have : 1: middle vertical line- 1+2+1=4. 2: left vertical line: 1+1+2=4 3:second horizontal line: 1+2+1=4 4:3rd horizontal line: 1+1+1=4 5: diagonal line- from the top right to the bottom left- 1+2+1=4 is that alright? shall I add a diagram? (is it even possible...?) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 The problem I was having was I was creating the 5 lines of 4 but also a couple extra lines of 3 or 2 coins length. After a few tries I found a way to position the coins in such a pattern that the lines are all 4 coins long because the coins must be in the immediate spaces to the up/down/left/right/4 diagonals to be considered adjacent. My solution is as follows: 1 2xxx7 3(56)8 4xxx9 xxxx10 Note: Coins 5 and 6 are stacked on each other and there are no partial lines (i.e lengths<4 coins) due to the adjacent coin dilemma Therefore the lines are: 1,2,3,4 2,5,6,9 3,5,6,8 4,5,6,7 7,8,9,10 Quote Link to comment Share on other sites More sharing options...
0 Smith Posted February 13, 2011 Report Share Posted February 13, 2011 (edited) ..........1 .......2.....3 ......... 4 ....5 6.... 7 8 . ...9............10 * 1, 2, 5, 9 * 1, 3, 8, 10 * 3, 4, 6, 9 * 2, 4, 7, 10 * 5, 6, 7, 8 Edited February 13, 2011 by Smith Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 ..........1 .......2.....3 ......... 4 ....5 6.... 7 8 . ...9............10 * 1, 2, 5, 9 * 1, 3, 8, 10 * 3, 4, 6, 9 * 2, 4, 7, 10 * 5, 6, 7, 8 This solution requires curves to count as straight lines to be correct. The question specifically states straight lines. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 5 coins on the points of as star, join the points. The intersection of these lines is a pentagon, place a coin on each of these points. This gives 5 lines of 4 coins each Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 (edited) you can just have 1 2 3 4 5 ---------------------------6 7 8 9 10 then put it near a mirror! the reflection will have what you just asked for! Edited February 14, 2011 by roboberry Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 (edited) so we get something like this: 1 2 3 4 5 6 7 8 9 10 Edited February 14, 2011 by sana88 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 (edited) ___3 4927 _1__ 0_58 ___6 I was able to get maximum 4 lines 3786 4927 4156 3210 Edited February 14, 2011 by MayankGupta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 My solution would be to put each coin on point or "intersection" of a 5-point star. 5 points, intersections, 5 lines of 4 coins. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 ans is a star with five arms 1 2 3 4 5 6 7 8 9 10 five lines are -1 3 6 9,1 4 7 10,2 6 8 10,5 7 8 9,2 3 4 5 allignment in the diagram is not that right but u can draw them straight. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 Okay. imagine a 3X3 table. leave the top left tab empty, and put one coin in all the rest. we have 8 coins. now, add a coin to the middle tab, on top of the other coin, and an other one on top of the coin in the right bottom tab. so we have : 1: middle vertical line- 1+2+1=4. 2: left vertical line: 1+1+2=4 3:second horizontal line: 1+2+1=4 4:3rd horizontal line: 1+1+1=4 5: diagonal line- from the top right to the bottom left- 1+2+1=4 is that alright? shall I add a diagram? (is it even possible...?) Please add the diagram if you find no difficulty in that Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 YES star is the correct answer but try there is one more(CLOSE TO STAR)PLEASE TRY TO FIND IT Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted February 14, 2011 Report Share Posted February 14, 2011 Couldn't quite figure out PaperCat's instructions but they lead me to this arrangement. 1 1 112 211 Each number stands for the number of coins in that location. The five straight lines are the left and right verticals, the two lower horizontals and the diagonal from lower left to top right. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 14, 2011 Report Share Posted February 14, 2011 VERY NICE TRY CURR3nt Thanks for your effort Quote Link to comment Share on other sites More sharing options...
0 Smith Posted February 14, 2011 Report Share Posted February 14, 2011 (edited) In response to my first post... This solution requires curves to count as straight lines to be correct. The question specifically states straight lines. I disagree - perhaps my graphic is too rudimentary to correctly represent the straight lines, but I can easily draw the 5 lines implied by my solution without the use of curves. The only question is whether the additional line segments created by two points in various places are allowed within the wording of the puzzle. Edited February 14, 2011 by Smith Quote Link to comment Share on other sites More sharing options...
0 Smith Posted February 14, 2011 Report Share Posted February 14, 2011 In response to my first post... I disagree - perhaps my graphic is too rudimentary to correctly represent the straight lines, but I can easily draw the 5 lines implied by my solution without the use of curves. The only question is whether the additional line segments created by two points in various places are allowed within the wording of the puzzle. Brumster, perhaps you assume that my solution involves stacked coins as yours does. This was not the case. Mine is more akin to the 'star' solutions I see elsewhere. Quote Link to comment Share on other sites More sharing options...
0 curr3nt Posted February 14, 2011 Report Share Posted February 14, 2011 1 2 3 4 5 6 7 8 9 0 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 15, 2011 Report Share Posted February 15, 2011 1 2 3 4 5 6 7 8 9 0 YEAH!! thats the only other way i was talking about.smith got it. curr3nt again thanks for extra efforts Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 25, 2011 Report Share Posted March 25, 2011 Brumster, perhaps you assume that my solution involves stacked coins as yours does. This was not the case. Mine is more akin to the 'star' solutions I see elsewhere. Makes sense. It's hard to do diagrams on this site. It took me 5 minutes to get mine right. I like yours. It seems to be the best way without stacking. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted March 25, 2011 Report Share Posted March 25, 2011 see attached file..... Quote Link to comment Share on other sites More sharing options...
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Can you place 10 coins in such a way that they lie in 5 straight lines and on each line there are 4 coins!
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