i made it look like a “I”, with two lines down the center. That way, there would be a line up top, and a line on the bottom, two vertical lines connecting the top and bottom, and the diagonals would be 5 and 6.
If you layout two sets of four coins horizontaly, with one set above the other. Then place a coin above the two end coins on the top row and a coin bellow the end two coins on the bottom row.
This gives you Two Horizontals, Two Verticals and Two diagonals each with four coins.
cool, there are more than two ways of doing it, but always using each coin in two rows with possible combination of (diagonal,diagonal), (diagonal, vertical),(diagonal,horizontal),(horizontal,vertical).
THE “I” AND THE “H” HAVE THE SAME DISPOSITION OF COINS, BUT THE WHOLE FORMATION ITS ROTATED, SO I COUNT THOSE TWO AS ONE, THE 5 TO 5 ITS TWO, AND THE FIRST I MADE AND HIS ROTATION MAKES THREE
The 2 answers I had in mind were the ‘I’ configuration, as several have pointed out above, and the ‘star’ arrangement, as Rosey and Suineg describe & parallax links to.
2 stacks of six coins each… I think the point of this puzzle was to think outside 2 dimentions. When you stack coins vertically you can count to four – three times in each stack.
OOOOOO
OOOOOO
There are many great answers to this! Well done everyone!
I agree with aaron hergert if you stack them there are many ways this can be achieved. Including making the six four coin rows with only eight coins.
Just form a box such as this however each coin has one coin on top of it while excluding the 4 unnecessarry ones:
i made it look like a “I”, with two lines down the center. That way, there would be a line up top, and a line on the bottom, two vertical lines connecting the top and bottom, and the diagonals would be 5 and 6.
Not sure, but how about these:
http://i225.photobucket.com/al...../coins.jpg
it is in the form of a H with a big middle part
0 0
0 0 0 0
0 0 0 0
0 0
2 diagonal
2 horrisontal
2 vertical
didn’t come out right!!!!!!!!!!!!!!!
just move the second ones on the top and bottom to the edge so it makes an H shape
If you count diagonals and columns as “rows,” then you can make an “H” with 6 rows of 4 coins each:
X X
X X X X
X X X X
X X
Oops, this system doesn’t count spaces.
X….X
XXXX
XXXX
X….X
one of the way
http://img222.imageshack.us/im.....wereb0.jpg
One is a six-pointed star. I’m working on another.
I think:
O O
OOOO
OOOO
O O
Oops, It diddnt come out as intended:
o–o
oooo
oooo
o–o
Try again : If this does not work. I am trying to make an ‘H’ with 2 horizontal components.
O O
O O O O
O O O O
O O
can’t figure this out, the closest I can come is to make a circle haha
Mine looks like an “H” with a double line going across. Same as the “I” but sideways.
If you layout two sets of four coins horizontaly, with one set above the other. Then place a coin above the two end coins on the top row and a coin bellow the end two coins on the bottom row.
This gives you Two Horizontals, Two Verticals and Two diagonals each with four coins.
http://i3.photobucket.com/albu.....uzzle1.jpg
http://i3.photobucket.com/albu.....uzzle2.jpg
I think this is an easy one.
the keyword being I
As in, I.
oooo
oo
oo
oooo
lol
oooo
..oo
..oo
oooo
ok better
the problem is solve in any configuration that you can used any coin in 2 of the lines: i think this is one solution too:
O O
O O O
O O
O O O
O O
i does not save the space wait a minute..
O…….O…..O
……………….
O……O……O
….O……O
O……O……O
………………..
O…….O……O
i hope this work
cool, there are more than two ways of doing it, but always using each coin in two rows with possible combination of (diagonal,diagonal), (diagonal, vertical),(diagonal,horizontal),(horizontal,vertical).
i made seven ups
but with 14 coins, i made up the problems again
take the two middle coins of the upper row and the lower row and you get 6 rows of 4 coins with 12 coins
ok, it’s easier when you use a rather liberal definition of the word row
its like this:
O………………O
O……..O…….O
…..O……..O
O……..O…….O
O………………O
ANOTHER SOLUTION I CALL 5 “TO” 5 IS THIS:
………O
O…O….O…O
…O………O
O…O….O…O
………O
THE “I” AND THE “H” HAVE THE SAME DISPOSITION OF COINS, BUT THE WHOLE FORMATION ITS ROTATED, SO I COUNT THOSE TWO AS ONE, THE 5 TO 5 ITS TWO, AND THE FIRST I MADE AND HIS ROTATION MAKES THREE
Joe- your 1st configuration was interesting; I think I follow, but just to be sure, can you explain what you did?
The 2 answers I had in mind were the ‘I’ configuration, as several have pointed out above, and the ‘star’ arrangement, as Rosey and Suineg describe & parallax links to.
Just stack the coins in the 12, 3, 6, and 9 o’clock positions.
http://i3.photobucket.com/albu.....zle1-1.jpg
Joe- that works too, nice.
i found another way:
………………….OO
………………….OO
………………..OOOO
………………O..OO..O
I THINK THIS WORK AND IS A DIFFERENT WAY.
0 0
0000
0000
0 0
Left and Right rows (vertical), two middle rows (horizontal), top left to bottom right row (diagonal), bottom left to top right (diagonal).
2 stacks of six coins each… I think the point of this puzzle was to think outside 2 dimentions. When you stack coins vertically you can count to four – three times in each stack.
OOOOOO
OOOOOO
There are many great answers to this! Well done everyone!
I agree with aaron hergert if you stack them there are many ways this can be achieved. Including making the six four coin rows with only eight coins.
Just form a box such as this however each coin has one coin on top of it while excluding the 4 unnecessarry ones:
0 0
0 0
I hav figured out the solution it tuk me quite a while but…i finally got da ansa here it is:
0 0
0000
0000
0 0