Question

Form a 3x3 magic square with constant 36
using consecutive numbers from 8 to 16​

Answer 1

Well, since the numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 the sum of all the 9 numbers is:

45, and, so, the sum in every column or row must be 15.

If you include the requirement for the diagonals then the number in the middle must be the 5…

Let’s use letters for the 9 cells:

a b c

d e f

g h i

The sums that include the cell in the middle are:

d+e+f = 15

b+e+h = 15

a+e+i = 15

c+e+g = 15

Notice all the 9 different letters are in those equations.

Let’s sum it all:

(a+b+c+d+e+f+g+h+i ) + 3e = 15*4 = 60

And the sum from a to i is 45.

So, 3e = 15 … so e = 5.

Now, just try.

What if a = 1? → i = 9

And what if we also force the b=2 ?

Then c = 15 - 2- 1 = 12 … Not possible.

Then b must be 5 or higher, but 5 is in the middle, so b must be 6 or 7 or 8.

Try 6:

1 6 8

* 5 *

* * 9

The last column exceeds 15…

Try b = 8

c = 6 and also the last column exceeds 15

So, no corner can be 1.

Then, the only possible place for the 1 is in b, or d or f or h.

If we try b=1 a = 6 → c = 8 and i = 4

6 1 8

* 5 *

* * 4

Then:

6 1 8

7 5 3

2 9 4

Answer 2

Answer:

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