Browse Teasers
Search Teasers

## Diagonal Possibility

Math brain teasers require computations to solve.

 Puzzle ID: #22792 Fun: (2.76) Difficulty: (2.3) Category: Math Submitted By: chidam11 Corrected By: shenqiang

Fill the eight spaces below with numbers from 1 to 8. No number that comes before or after any other number should be placed in an adjacent space, either horizontally, vertically or diagonally.

(For Example 2 should not be next to the number 1 or the number 3)

The arrangement of the space is
* __ __ *
__ __ __ __
* __ __ *

## What Next?

See another brain teaser just like this one...

Or, just get a random brain teaser

If you become a registered user you can vote on this brain teaser, keep track of
which ones you have seen, and even make your own.

 umhlobo May 10, 2005 Unless I'm mistaken, you could have worded the question better. You seem to have left out the horizontal limitation, so one could have a number of other solutions e.g. 1,2 in top row 5,6,7,8 in middle row and 3,4 in bottom row ... and a bunch of combinations of these where there are no consecutive numbers vertically/diagonally next to each other. umhlobo May 10, 2005 Unless I'm mistaken, you could have worded the question better. You seem to have left out the horizontal limitation, so one could have a number of other solutions e.g. 1,2 in top row 5,6,7,8 in middle row and 3,4 in bottom row ... and a bunch of combinations of these where there are no consecutive numbers vertically/diagonally next to each other. markmonnin May 10, 2005 Well, he's right, but I assumed the horizontal limitation and go the correct answer. darthforman May 10, 2005 didida May 10, 2005 That was a great one!!! I actually got the answer!! hydrogen May 14, 2005 You seem to have left out the horizontal but a good one jarm May 14, 2005 i was lost from the start cmt1214 May 21, 2005 cool onlyeeyore Jun 06, 2005 The horizontal must have been changed in the edit so no worries there. I got it EASY until I found out that my setup was wrong due to the way it looks in the puzzle (hexagon with one additional space to the right). When I started to look at the answer I saw the setup was different, quickly hid it, and went to solve it again. This time it was fun figuring it out! Great teaser, though I would recommend some way to get the spaces lined up properly: ^^__ __ ^^ __ __ __ __ ^^__ __ ^^ Carats seem to work great! onlyeeyore Jun 06, 2005 I see the dilemma here... the carat idea worked great when I initially typed it! Oh, well! dimez_00 Jul 14, 2006 i found another solution *1 5* 3 8 2 7 *6 4* chidam11 Aug 05, 2006 No that is not possible as 1 is diognal to 2. stil Oct 27, 2006 Testing, testing- does this transfer from windowed input to fixed form with less distortion? (The O's would be replaced with digits; the X's would be considered nul or blanks. XOOX OOOO XOOX stil Oct 27, 2006 Another test - 0's not O's. X00X 0000 X00X (user deleted) Nov 11, 2006 Good one. This probably teaches good strategies of distribution. (user deleted) Jan 31, 2007 Here is another solution: 13 5796 24 chidam11 Feb 01, 2007 That's not it....You have placed 9 instead of 8. If 8 comes it would be beside 7... mbunap Jul 20, 2007 After seeing the difficulty score, I thought this one would take a while, but people must have been overthinking it. It took me three tries over the course of about 3 minutes to get the correct lineup. Fun. javaguru Dec 10, 2008 3 minutes? Try 15 seconds. It's immediately obvious that either 1 or 8 has to go in the middle of the hexagon, leaving only one place for the adjacent number. Repeat that for the other central location with the other digit with a single neighbor and the puzzle essentially fills itself out. How does this make the top three list of "fun" problems or get the difficulty rating it has? Paladin Mar 13, 2009 The top row must be either 35 or 53 or 46 or 64 The middle row must be 7182 or 2817 The bottom row must be either 35 or 53 or 46 or 64 (of course it cannot use the same 2 numbers as the top row and the order of the middle row sets the order of the top and bottom rows to keep the 3 from being diagonal to the 2 or the 6 from being diagonal to the 7) So there are 2 * 2 * 1 = 4 combinations: 53 & 2817 & 64 35 & 7182 & 46 46 & 7182 & 35 64 & 2817 & 53 These 4 combinations are the only possible solutions