Numbers 1-9 in the square so that each row, column and diagonal adds up to 15 (SEE NOTE)* Quiz by Ubbiebubbie

Can you name the numbers 1-9 in the square so that each row, column and diagonal adds up to 15 (SEE NOTE)*?

created by Ubbiebubbie

Enter a number in the box below
Correctly named numbers will show up below
Click any empty # or to answer for that location
ALERT!: Entering some wrong answers will end the quiz
Each row, column and diagonal will add up to 15. Only numbers 1-9 will be used, and there are no repeated numbers.
****IMPORTANT**** The first number has been provided. As there are two possible ways to solve I will eliminate one. The number in the bottom left corner is larger than the number in the top right corner.
Also try: Magic Square II
This quiz has not been verified by Sporcle

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There are 17 comments for this game.
(Warning: comments may contain spoilers)

Magic Square Quiz

by Ubbiebubbie

Created Dec 15, 2012 in Just For Fun
Game Plays 4,400

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	needapausebutton:	Dec 16th, 2012 at 02:31 GMT		7 points
	@Ubbiebubbie: There is another way to enter the numbers: 4 3 8 9 5 1 2 7 6

	Ubbiebubbie:	Dec 16th, 2012 at 02:38 GMT		2 points
	And now I have a sad face.

	Ubbiebubbie:	Dec 16th, 2012 at 02:40 GMT		3 points
	I think I fixed it.

	needapausebutton:	Dec 16th, 2012 at 02:57 GMT		4 points
	^ No, don't have a sad face, I wanted to alert you of that as soon as the quiz came out so you'd have a chance to adjust it. The more "prominent" you can make your Important notes, the better off you'll be (maybe add an asterisk at the end of the "Can you name..." instructions).

	Ubbiebubbie:	Dec 16th, 2012 at 03:05 GMT		3 points
	I am smiling now, and I have accentuated the note.

	needapausebutton:	Dec 16th, 2012 at 03:05 GMT		1 point
	I just sent you a related e-mail.

	hazelorb:	Dec 16th, 2012 at 17:45 GMT		10 points
	By telling us that a 3 does not go in the square right next to the 4, you basically told us to just use the transpose as the solution which I feel devalues the puzzle. Is there some way to let people enter the other number in each square? It doesn't detract from the minefield too much.

	charliep1973:	Dec 21st, 2012 at 08:06 GMT		-1 points
	What is wrong with my solution? Top 4 2 9 - Centre 8 6 1 - Bottom 3 7 5 ???

	charliep1973:	Dec 21st, 2012 at 08:08 GMT		4 points
	Oh, I just noticed that the diagonal without the 4 in it doesn't make 15. I thought it was too easy! :-)

	Ubbiebubbie:	Jan 21st, 2013 at 12:22 GMT		4 points
	Thanks for the CP, vikZ!

	vikZ:	Jan 21st, 2013 at 14:57 GMT		4 points
	No problemo!

	beforever:	Jan 21st, 2013 at 15:25 GMT		2 points
	Can any mathematicians or logicians run through how you would do in this strategically. I worked out that 1 and 7 can only go... you know, 'there' and 'there'. I feel like I lucked out, though, beyond that - and managed to just fluke it right on my first 'guess'. What's the 'real' way of working it out?

	Ubbiebubbie:	Jan 21st, 2013 at 16:12 GMT		1 point
	^ http://en.wikipedia.org/wiki/Magic_square There's a lot of history and math behind them. I won't bore you with the details.

	caramba:	Jan 21st, 2013 at 20:08 GMT		6 points
	@beforever: The game hint tipped me off as to what goes below 4, then I figured 5 had to be central, as the remaining numbers all pair up nicely into 10s that would be placed around the central 5. After that, it all kind of fell into place.

	Maritimer:	Jan 22nd, 2013 at 17:17 GMT		5 points
	I agree with @hazelorb. By telling us that the square next to the 4 is not a 3, it immediately implies the square below it is a 3 since the two solutions are transposes (mirror images along the diagonal that runs from top left to bottom right, for non-mathematicians). Might I suggest the following clue? "The number in the bottom left corner is bigger than the number in the top right corner."

	Ubbiebubbie:	Jan 22nd, 2013 at 20:07 GMT		2 points
	@Maritimer: I like it! Thanks!

	sub:	Feb 1st, 2013 at 19:04 GMT		1 point
	there's a fairly simple mathematical way to find out the middle element. Say you name the squares in order (from left to right, top to bottom) a,b,c...i, with a being the top left square and i the bottom right. Then you have the two diagonals: a+e+i=g+e+c=15, so a+i+g+c+2e=30. But a+g+d=c+f+i=15, so a+g=15-d, c+i=15-f, so 30=a+i+g+c+2e=15-d+15-f+2e, from where you get that d+f=2e. But d+e+f=15, so 3e=15 -> e=5.