MODULAR ARITHMETIC II

Don't care about the white squares, just care about the grey squares.

We can fill the horizontal grey row with 1 2 3 4 5

The remaining four grey squares will contain 4 numbers from the list 1 2 3 4 5

Since is 2 (mod 6)

we just guess and check

1  + 2  + 3 + 4 + 5 = 15

That is the sum of the center horizontal row.

We then can guess and check the sum of the vertical row plus 15 to see if it fits 2 mod ( 6)

Lets say the center square is 1?

Then 29 / 6 remainder is 5

center square is 2?

Then 28 / 6 remainder is 4

Center square is 3?

then 27/6 remainder is 3

Center square is 4? 

then 26/6 remainder is 2

Center squar eis 5?

25/6 remainder is 1

We see by guess and checking that the answer is 4, four should go into the center square

Your better at modular arithemtic than you think

Suppose you want to fill the 5x5 square below with the following restrictions:
(1) The numbers 1-5 must appear in every row and column;
(2) The sum of the numbers in gray boxes is 2 (mod 6).

What number must go in the center square?

Let the number go in the center square \(= \mathbf{x}\)

Let the sum of the numbers \(1+2+3+4+5 = 15\)

Let the sum of the numbers in gray boxes \(= 15 + (15-x)\)

Then:

\(\begin{array}{|rcll|} \hline 15 + (15-x) & \equiv & 2 \pmod 6 \\ 30-x & \equiv & 2 \pmod 6 \quad | \quad 30 \pmod 6 = 0 \\ 0-x & \equiv & 2 \pmod 6 \\ -x & \equiv & 2 \pmod 6 \quad | \quad \cdot (-1) \\ x & \equiv & -2 \pmod 6 \\ x & \equiv & -2+6 \pmod 6 \\ x & \equiv & 6-2 \pmod 6 \\ \mathbf{x} & \equiv & \mathbf{4 \pmod 6} \\ \hline \end{array}\)

The number 4 must go in the center square.