# Sum in a sudoku with sum total of 15 and 45

I have created a sudoku puzzle with the following restrictions:

• Each row and column sum to $$45$$.
• Each row and column in the nine $$3$$ by $$3$$ sub-grids sum to $$15$$.

Is such a sudoku unique?

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• "...where each row and column adds to a total of 45..." is redundant, since it is in the nature of a sudoku puzzle that its rows and columns sum to $45$. This is because each row contains the numbers $1$ through $9$, and $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. – Hugh Feb 11 at 5:04
• "where each (3*3) 9 individual squares of rows and columns total a sum of 15" >>> do you mean 45? – Kryesec Feb 11 at 5:33
• @Kryesec I've assumed that that means that each of the 3 by 3 sub-grids is a magic square with "magic constant" 15. In other words, each row/column in each 3 by 3 sub-grid sums to 15. – Hugh Feb 11 at 5:47

No.

Of course, there's rotation and reflection, but even beyond that you can swap any two rows within its group of three rows (eg swap row 1 and 2 or swap 4 and 6), and you can swap a group of three with one of the other groups of three (eg swap 1,2,3 with 4,5,6), and the same applies to the columns.

• I wonder if there's any kinda of transformation that would work here that doesn't work on sudoku in general. – Rawling Feb 11 at 12:00

I was beaten to a very similar answer, but you can create such a square based on the magic square below:

abc
def
ghi

Turn it into this:

abcdefghi
defghiabc
ghiabcdef
bcaefdhig
efdhigbca
higbcaefd
cabfdeigh
fdeighcab
ighcabfde

Since there are multiple possible magic squares, it's not unique. When you change the number at the very center (i), which you can, you get a different sudoku (not even a rotation or reflection).

One such arrangement of the 3x3 grid could be:

1 5 9
8 3 4
6 7 2

This would then be repeated along to the right, but rotating the order of the rows to create:

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

The same could be done to repeat downwards but rotating the columns this time:

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

However:

This creates one such possibility of your given solution, as each of the 3x3 sections could be placed in the top left and the rotations occur from there.