V x V magic square puzzle

[QuikSand]

I saw this published in a magazine... thought it might make a worthy puzzle for some of you here. Hope you enjoy...

The Simple Part...

Create a 5x5 Magic Square Puzzle, such that:
-the sum of each row and column is equal to the same value
-the sum of each major diagonal is the same value
-the sum of the four corners and the center square is the same value
-and for this particular puzzle, no number is used more than twice in any row or column

The Twist...

Use Roman numerals to fill in each box

And the grid should remain a magic square, with all the properties above, even if it is turned upside down.

[BrianD]

For those of us new to the magic square puzzles. Do we have to use all values from 1-25 for the squares?

[QuikSand]

Originally Posted by BrianD

For those of us new to the magic square puzzles. Do we have to use all values from 1-25 for the squares?

That is sometimes a rule for creating a magic square, but it is not a rule here, for fairly obvious reasons (e.g. XIV upside-down is not a legible numeral).

[Wolfpack]

Obviously, that restricts the universe to I, II, III, IX, X, XI, XIX, and XX for numbers under 25....

[Maple Leafs]

I'm guessing my solution of putting "I" in each square will not be accepted...

[Huckleberry]

That was both cocky and funny.

7.3

[QuikSand]

Originally Posted by Maple Leafs

I'm guessing my solution of putting "I" in each square will not be accepted...

Perhaps this is why this wasn't posed as a "puzzle" when I originally saw it, but rather as a "hey, look at this neat magic square."

Alas... filling the whole grid with Is, Xs, or some clever but simple combination of both isn't what we have in mind here. There is a solution that, as nearly as I can tell, never has any numeral appearing more than twice in any column or row.

I'll add that as a condition above, for clarity's sake.

[Maple Leafs]

Originally Posted by QuikSand

Alas... filling the whole grid with Is, Xs, or some clever but simple combination of both isn't what we have in mind here. There is a solution that, as nearly as I can tell, never has any numeral appearing more than twice in any column or row.

Killjoy.

[Pumpy Tudors]

Originally Posted by QuikSand

(1) the sum of each row and column is equal to the same value
(2) the sum of each major diagonal is the same value
(3) the sum of the four corners and the center square is the same value

(I numbered the three conditions for clarity)

Please pardon my ignorance about magic squares here, but are the values for each of these three conditions the same, or can the value from condition 1 be a different value from condition 2?

[Barkeep49]

Magic squares all have the same value so 1-3 should all produce the same sum. I am guessing, however, that this sum might be different when reversed.

[spleen1015]

Originally Posted by Wolfpack

Obviously, that restricts the universe to I, II, III, IX, X, XI, XIX, and XX for numbers under 25....

And if you can't use numbers more than twice, then this is impossible. 5x5=25, but with those as the only options, you only have 16 numbers to use.

We''re missing something.

[Pumpy Tudors]

Nobody ever said that we couldn't use a number above 25, did they?

[MIJB#19]

C turned upside down still makes a C.

[MIJB#19]

Something like:

X I I X C
I X C X I
X I I C X
C X X I I
I C X I X

Nah, that sounds too easy...

Edit: I know the diagonals are not correct yet, but this is ment as an example...

[MIJB#19]

I	I	X	X	C
I	X	C	I	X
C	I	X	I	X
X	X	I	C	I
X	C	I	X	I

I got to be missing something,,,

[QuikSand]

Originally Posted by MIJB#19

C turned upside down still makes a C.

Maybe in Dutch it does. Not here.

[QuikSand]

Originally Posted by Pumpy Tudors

Nobody ever said that we couldn't use a number above 25, did they?

No, I never said that... nor did I say there was a limit of using a specific number two times in the whole puzzle. Just in any given row or column.

Originally Posted by spleen1015

And if you can't use numbers more than twice, then this is impossible. 5x5=25, but with those as the only options, you only have 16 numbers to use.

We''re missing something.

Actualy, it's more like you're creating rules that aren't there. Actually, it's a lot like that.

[MIJB#19]

Originally Posted by QuikSand

Maybe in Dutch it does. Not here.

Crap, I was thinking of mirroring, instead of the 180 degree turn.

[twothree]

09 10 09 11 11
10 11 11 09 09
09 11 10 09 11
11 09 09 11 10
11 09 11 10 09

That would be one solution when represented in roman numerals. It looks like you need to use the numbers 9,10, and 11. "IX" "X" "XI" as those will "flip" easily and making adding up to 50 easy.

[QuikSand]

I'm starting to see why the original offering was just a "here's a cool magic square" rather than an actual puzzle.

As for twothree (that name, I am oddly drawn to it...) -- your puzzle certainly fills the rules that I set forth above, well done.

The original square that inspired this puzzle had a significantly wider range of numerals represented within... when I see it again tomorrow, I'll try to post another detail or two. But as I recall, it ised the numbers I, II, IX, X, XI, XIX and XX... but it, too, may have basically been an offsetting system of converse numbers, like the IX/XI system above.

[twothree]

09 19 11 10 01
11 01 19 09 10
01 09 10 11 19
10 11 01 19 09
19 10 09 01 11

Yeah, you could substitute a 19 and a 1 for one of the 9's and 11's in each row and column. To produce a magic square that works in roman numerals and does not contain any duplicates in either row, column, diagonal, or corner plus center. And it can be turned upside down with no problems.

[twothree]

LOL. I just noticed it also works turned upside down using the numbers the way I wrote them out. 01,09,10,11,19 becomes 10,60,01,11,61.

Though, 10,60,01,11,61 would not work as roman numerals, however.

[QuikSand]

The magic square that I originally saw published:

Uses each of the numerals I, II, III, IX, XI, XIX, and XX (but not X) at least once.
No numeral is used more than five times.
The sum of each row, column, and diagonal is 44.
...and the square is not built on the sort of "symmetry" principle that twothree is using above.

I suspect the value of this as a puzzle is probably nil at this point, but I'll wait to post the original magic square just in case anyone remains interested in this.

[twothree]

02 19 01 19 03
19 11 03 02 09
03 01 19 20 01
11 11 01 01 20
09 02 20 02 11

That would be one solution given the additional rules. Knowing that the sum is 44 helps in finding a solution. I just listed all possible sums of 44 following the rules as stated above:

20+19+3+1+1
20+19+2+2+1
20+11+11+1+1
20+11+9+3+1
20+11+9+2+2
20+9+9+3+3
19+19+3+2+1
19+11+11+2+1
19+11+9+3+2

Then just made up a square that would work. I tried putting 20 as the center square, but couldn't get it to work, so I had to settle for 19 before brute force presented a solution.

[QuikSand]

The original magic square is as follows:

II  IX  XI  II  XX
XI  II  XX  II  IX
XX  III IX  XI   I
IX  XI   I  XX  III
II  XIX III IX  XI

[twothree]

Originally Posted by QuikSand

The original magic square is as follows:

II  IX  XI  II  XX
XI  II  XX  II  IX
XX  III IX  XI   I
IX  XI   I  XX  III
II  XIX III IX  XI

II  IX  XI  II  XX
XI  II  XX  II  IX
XX  II  IX  XI  II
IX  XI  II  XX  II
II  XX  II  IX  XI

So, basically a square of symmetry using 20+11+9+2+2. And, then substitute (19+11+9+3+2) and 2 of (20+11+9+3+1) on a column and vertical row to get the original published square.