Introduction:            

Hiii! Today I am writing a note again about the magic squares, but this time with a slightly more difficult type. Read it so you can see.

The task and solution:

Complete the magic square using each number from 11 to 19 exactly once. The sum of the numbers in every row, column, and diagonal must be 45. Find the unknowns A, B, C, Y, X and Z.

The numbers you must use are:
11, 12, 13, 14, 15, 16, 17, 18, 19

This is the task for now. You see a magic square, but with unknowns. Let's start solving it.

First, we need to find the sum of every diagonal, line, and column, which you know is always the same in magic squares.

We see that the only line, column, or diagonal where we can find the sum is one of the diagonals, where all the numbers are filled

18 + 15 + 12 = 45  
So the sum is 45.

Then we can write all the sums of the magic square:

18 + X + Y = 45

• Z + 15 + A = 45
• B + C + 12 = 45
• 18 + Z + B = 45
• X + 15 + C = 45
• Y + A + 12 = 45
• 18 + 15 + 12 = 45
• Y + 15 + B = 45 

After that, we can remove the used numbers from the numbers we need to use. The numbers that are left are:
11, 13, 14, 16, 17, 19

Now let's simplify the equations.

1.    18 + X + Y = 45
      X + Y = 45 - 18
      X + Y = 27

2.     Z + 15 + A = 45
       Z + A = 45 - 15
       Z + A = 30

3.     B + C + 12 = 45
       B + C = 45 - 12
       B + C = 33

4.     18 + Z + B = 45
       Z + B = 45 - 18
       Z + B = 27

5.     X + 15 + C = 45
       X + C = 45 - 15
       X + C = 30

6.     Y + A + 12 = 45
       Y + A = 45 - 12
       Y + A = 33

7.     Y + 15 + B = 45
       Y + B = 45 - 15
       Y + B = 30
     

The equations that I found are these:

X + Y = 27

Z + A = 30

B + C = 33

Z + B = 27

X + C = 30

Y + A = 33

Y + B = 30

And the only sums I see are 27, 30, and 33. I also have 11, 13, 14, 16, 17, and 19 to use. Now each of these three numbers is the sum of two of the other numbers.

Now let's find this sum. But wait... There are several possibilities. Let's write them all down: 

27
11 + 16 = 27 
13 + 14 = 27

30
11 + 19 = 30
13 + 17 = 30
14 + 16 = 30

33
14 + 19 = 33
16 + 17 = 33

And the way we can find them is to see if they fit in our equations. Let's see.

Before start:

Left numbers:
11, 13, 14, 16, 17, 19

First, I will find Y.

X + Y = 27

Y + B = 30

Y + A = 33

To find Y, we need to consider all the possibilities. Let's start.

Case 1

Let's say Y is 11.

First, I need 11 + X to cost 27. For that, X needs to be 16. I have 16 from the left number, so this one is possible.

Then I need 11 + B to cost 30. For that, B needs to be 19. I also have 19 from the left number, so this one is possible, too.

Finally, I need 11 + А to cost 33. For that, A needs to be 22, but the problem is that I don't have 22 in the left numbers. So Y can't be 11.

Case 2

Let's say Y is 13.

First, I need 13 + X to cost 27. For that, X needs to be 14. I have 14 from the left number, so this one is possible.

Then I need 13 + B to cost 30. For that, B needs to be 17. I also have 17 from the left number, so this one is possible, too.

Finally, I need 13 + А to cost 33. For that, A needs to be 20, but the problem is that I don't have 20 in the left numbers. So Y also can't be 13.

Case 3

Let's say Y is 14.

First, I need 14 + X to cost 27. For that, X needs to be 13. I have 16 from the left number, so this one is possible.

Then I need 14 + B to cost 30. For that, B needs to be 16. I also have 19 from the left number, so this one is possible, too.

Finally, I need 14 + А to cost 33. For that, A needs to be 19, and we have 19 in the left numbers, so Y can be 14!

Wait. I know we "found" Y, but we need to check the other cases to see whether Y could be something else.

Case 4

Let's say Y is 16.

First, I need 16 + X to cost 27. For that, X needs to be 11. I have 11 from the left number, so this one is possible.

Then I need 16 + B to cost 30. For that, B needs to be 14. I also have 14 from the left number, so this one is possible, too.

Finally, I need 16 + А to cost 33. For that, A needs to be 17, and I also have 17 in the left numbers. So Y can be 16, too!

Case 5

Let's say Y is 17.

First, I need 17 + X to cost 27. For that, X needs to be 10, but the problem is that I don't have 10 in the left numbers. I don't think I need to check whether the other equations fit, because this case is already finished. So Y can't be 17.

Case 6

Let's say Y is 19.

First, I need 19 + X to cost 27. For that, X needs to be 8, but I don't have 8 in the left numbers, so Y also can't be 13.

Now, from these cases, I understood that Y can be 14 or 16. Let's see the two possibilities.

CASE 1

Y is 14. When Y is 14, we already know some of the other unknowns:

Y = 14 

X = 13 

B = 16

A = 19

Now let's find the other unknowns using the equation. You can try to find it by yourself. I will only write the answer. 

Z = 11
C = 17

And the magic square becomes like this:

You know, there's another case to consider: if Y is 16, but I'll say that would also fit because the square is mirrored. If you want, you can write down all the cases, but I will write here only the other magic square:
 

End:
Finally! I think this was the longest note I've ever written.  I hope you learned something too, because I actually understood some new things while writing. Also, don't forget that if you read this note, you should read the other note for magic squares, which is Magic Squares - Fill The Square With Told Numbers and Told Sum. See you in my next note. Bye!