Recreation Another game....

ok this game requires strategy. Look at the attachment, there are 8 boxes. What you should do is place numbers : 1 to 8 in each box. The rule is that each adjcent number should not be placed next to each other. That means 5 and 6 should not be aside, below or above ..(etc) each other...
So to make it easier for you:
numbers that shouldnt be next to each other are:
1&2
2&3
3&4
4&5
5&6
6&7
7&8

So while trying to figure out this problem, time yourself
I did it in 2 minutes ^.^;

GL

 

There is a way, believe me You just need to know which numbers you put in the middle.

 

Did you make this up tooma? It can be done. Why you ask simple logic.

I will explain all the box groups.

Number 1.
Where the three is at can be placed with any number that has two numbers by it (4,5,6, and 7). You will still get the same affect. One number will not touch (4 in this case) and the other will even at a diagonal, next to, and up or down, (5 in this case).

Number 2.
The best logic would be to place 1 and 8 then in the middle boxes. Because these are the only numbers that have one number touching them. In order to have 2 and 7 not touch they would have to be on one of the two free boxes. Either way you put it. In this case I did 8 on top and 1 on bottom. So in the free boxes it would have to be 2 on the first one. And 7 on the second one. So now we have 2,8,1, and 7 going in a line down the middle. Lets start with the number 3 (or 6) easiest to do. 3 (or 6) cannot touch diagonally. So in order for this to work then 3 will have to go on the two boxes left on the bottom (6 on either two on the top). If 3 and Six are next to each other then it cannot be done because 4 and 5 would touch.

Number 3.
This just goes and changes so 3 and 6 are on opposite sides. 5 cannot go below 6 no matter what. So the best way would be to put 5 by 3. Please see number 4.

Number 4.
Is one way to solve it if you follow all logic

Number 5.
Is the second way to solve it using 2,8,1, and 7 in a row.

There is only 2 more ways to solve this. Just switch it to 7,1,8, and 2. Apply the same logic as the others.

That wasn't to bad and kind of fun.

Please look at the attached file.

Here is another puzzle we should all be able to do.

NEXT PUZZLE:

Nine boxes. Going 3 X 3. Place 1 thru 9 in each box. But every row, column, and the 2 diagonals (x) has to equal 15.

Kind of like how you win in tic tac toe all those have to equal 15.

 

Basher, my calculus professor gave us his "special treat" to us
.
ok I finally got it. You confused me when you wrote 27 instead of 15. I was telling myself "I must be doing something wrong" :glazed:

So if anyone has a type of question that is similar to Basher and mine, please share

 

Basher had posted the answer to that first one. So its up to you if you want to do it and see how you do in each puzzle. Or we could turn this like the "Guess who's pic is this" thread.
We'll its up to you guys, anything goes....

 

Tooma is there something with that answer you posted? for some reason I only get half.

 

werid, I can see the whole attachment
ok here is the answer

[8][1][6]
[3][5][7]
[4][9][2]

[] = box =/

 

I may have missed something between the changes of the grid here but... isnt 3,4,5,6,7 all adjacent to each other in your answer there? I got one of those as a easter egg puzzle... how the hells a kid meant to do those?

 

Angel I was answering Basher's puzzle

 

Good work tooma.

Here is another Problem please look at the attachment.

There are nine circles total. Each circle goes from A to I. You must change all letters into numbers. The A circle is given to you. 4 equals A. Each time a circle over laps another it is the sum that belongs to those circles. Please note that some circles do lap three times.

To solve findout what letter does 1 to 9 go to and the rest of the numbers that are in the drawings.

Sorry for the bad drawing. My comp has only Art

Sorry had to edit it. I had everything right when I wrote but I messed up A. A equals 4. Now you should be able to solve it.

 

Basher, are you sure you havent missed any other rule, because if there isnt a limit to each circle. Then you could put any number and it will turn right.

If you edited the attachment, it did not have any effect. I could still see the old attachment.
:glazed: I guess you didnt...

 

Hey! I solved the first puzzle in about 10 seconds by just slapping numbers on, and thought it was so easy. But when Basher explained "the answer" where adjacent numbers couldn't be diagonal to each other either, I realized my answer was wrong cuz Toomaa never said anything about diagonals in the first place! Not fair!

 

I forgot to say diagonal, but i did mention it in the attachment ^__^;;; sorry for causing all this confusion :dizzy2:

 

As long as the first circle (A circle) has a 4 in it you should be ready to go.

Only rules are this.
9 circles labeled A, B,C, D to I. Each circle has to be replaced with a number from 1 to 9. You can only use a number once. So that the end result is circle A=4. circle b=?, circle C=?, etc. (The number for circle A has been given to you.)

The circles do over lap each other. The over lap part of the circle is what those two (or three circles) equal.

Exp.
Circle A + Circle C = Over lap of Circles A and C
Circle A = 4
and Over lap of Circles A and C= 6
then 4 + C = 6
Using this math you should be able to figure out what circle C number (equals). then try and figure out the rest.

It isn't to hard atleast I don't think so.

Just remember that sometimes the circles over lap 3 times.
Like Circle D + Circle E + Circle F = Over lap of Circles D, E, AND F.