# Create a 3x3 table with a specific rule

Take 9 distinct numbers from [0 to 9], then put the numbers to a 3x3 table, so :

Each cell = Last digit of (sum of 2 numbers in the same row of the cell + sum of 2 numbers in the same column of the cell).

If the numbers are a,b,c,d,e,f,g,h, and i, so :

[a,b,c]
[d,e,f]
[g,h,i]

a = last digit of sum (b,c,d,g)
b = last digit of sum (a,c,e,h)
c = last digit of sum (a,b,f,i)
d = last digit of sum (a,g,e,f)
e = last digit of sum (b,h,d,f)
f = last digit of sum (c,i,d,e)
g = last digit of sum (a,d,h,i)
h = last digit of sum (b,e,g,i)
i = last digit of sum (c,f,g,h)


Note : There is only 1 answer if we ignore answers from permutation of columns, permutation of rows, and reflections of diagonals. from total 72 solutions (6x6x2).

• Any motivation behind asking this question? – Matsmath Oct 14 '16 at 11:36
• @Matsmath : Motivations? For reputations and fun. Why do you asking this? – Jamal Senjaya Oct 14 '16 at 15:15

The numbers could be

a=0, b=1, c=9, d=3, e=4, f=2, g=7, h=8 and i=6

So the matrix looks like

[0,1,9]
[3,4,2]
[7,8,6]

Reasoning

I started with the "0" as it is the lowest number - the sum of b,c,d,g would have to be either 10 or 20 - 10=1+2+3+4 (0 taken already), but it was impossible to complete the table, so there's the 20. Then I took the lowest and highest to complete the row. Tried the same with the column, but 2 and 8 didn't work, so next try was 3 and 7. The rest was just inserting values that didn't break the first steps. :)