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avatar+198 

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?

 Oct 20, 2019
 #1
avatar+2390 
+2

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

 Oct 20, 2019
 #2
avatar+198 
+2

Your better at modular arithemtic than you think

 Oct 20, 2019
 #3
avatar+2390 
+2

I actually used guess and check in this problem XD XD, I only understood the format meaning of modulus I actually barely even know how to apply it.

CalculatorUser  Oct 20, 2019
 #4
avatar+23324 
+3

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.

 

laugh

 Oct 21, 2019
 #5
avatar+2390 
+1

Thank you for the modulus lesson!

CalculatorUser  Oct 24, 2019

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