Magic Squares

In a magic square of order 4, we are filling in numbers from 1 to 16.

So, sum of all the numbers=1+2+3....16=136

which gives us, sum of numbers in a row, column or diagonal= 136/4=34

$\begin{matrix} A & E \\ G & H \end{matrix}$ $\begin{matrix} F & B \\ I & J \end{matrix}$

$\begin{matrix} K & L \\ C & O \end{matrix}$ $\begin{matrix} M & N \\ P & D \end{matrix}$

let this be the magic square of order 4

then, A+H+M+D=34

C+L+I+B=34

A+E+F+B=34

C+O+P+D=34

adding all the above equations, we get

2A+2B+2C+2D+H+M+L+I+E+F+O+P=34+34+34+34

2(A+B+C+D) + (H+L+E+O)+(F+I+M+P)=4X34

2(A+B+C+D) + (34)+(34)=4X34

2(A+B+C+D)=2(34) WHICH GIVES US A+B+C+D=34.

there are 16 numbers from 1 to 16, it can be easily inferred that sum of all numbers (1+2+3+.......16)=136........since there are four rows so sum of each row or column is 136/4=34. Now accordingly we can arrange the numbers in the squares.....

7+10+11+6=34

wheres the fun in algebra? just sit there for an hour and make it up because its harder to do like that.

Label all the numbers in the magic square be $x_1$ , $x_2$ ,...., $x_{16}$ . The magic constant is $\frac{1}{2} n(n^2 -1)$ = 34. Then we have $x_1+x_2+x_3+x_4+x_{13}+x_{14}+x_{15}+x_{16}=68$ $(x_1+x_4+x_{13}+x_{16})+(68 - (x_6+x_7+x_{10}+x_{13})=68$ $(x_1+x_4+x_{13}+x_{16})+[68-\{68-(x_1+x_4+x_{13}+x_{16})\}]=68$ $2(x_1+x_4+x_{13}+x_{16})=68$ $x_1+x_4+x_{13}+x_{16}=34$ And from question, $\{x_1,x_4,x_{13},x_{16}\}=\{A,B,C,D\}$ . Therefore, $A+B+C+D=34$

I took the first four intergers, 1, 2, 3, and 4, added them together to make 10, took the last four intergers, 16, 15, 14 and 13, added them together to make 58. Found the mid point between 10 and 58 which is 34. :)

Lowest even numbers 2+4=6 highest odd numbers 13+15=28 Added together 6+28=34

add 16+15+14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 136 this will gives u the total sum of the solution. then this number will divided by 4 we will get 34.

I have a solution. We know each row adds up to the same number. So, the number should be 1+2+3+4+5+......+16=136. This number is divided by 4 which gives us 136/4=34(actually, I just sat and tried to fill all the grids but failed everytime but the sum of the numbers always added up to 34, my dad explained me the logic behind it) Now, we know that each row, column and diagonal adds up to 34. So, here, any two numbers in, let's say, a row adds up to 1/2 of 34 which is 17. I considered the two numbers to be A and B(first row) and C and D(last row). So A+B=17 and C+D=17. Adding, we get A+B+C+D=34.

To know more on magic squares see https://youtu.be/-czqJpGGQck. Jump to 14:50 for 4X4 Square.

We can divide up the square like this:

Let $s = A+B+C+D$ denote our desired sum of corners, and $r$ , $g$ and $b$ denote the sums of the red, green and blue shaded regions, respectively. From the problem description, we can directly infer that the total sum of the square $s + r + g + b = 1 + 2 + 3 + \cdots + 16 = 136$ . This also means that each line sums up to $\frac{136}{4} = 34$ . By adding the sums along all of the colored lines, we arrive at the following equation:

$(s + r) + (s + g) + (s + b) = 6 \cdot 34 = 204$

We can solve this equation by substituting the total sum: $(s + r) + (s + g) + (s + b) = 204 \\ 2 s + (s + r + g + b) = 204 \\ 2 s + 136 = 204 \\ 2 s = 68 \\ s = 34$

I thought that it might be only a verbal trick, so there are 4 numbers in each row, column or diagonal (4 squares) so i just arranged the numbers in order so the corner numbers will be (1+4+13+16=34) lol maybe i got lucky or smart i don't know please tell me if I'm wrong :D

Using Gauss' addition technique, the sum of 1 through 16 becomes (1 + 16) + (2 +15) + (3 + 14) + ... which translates to 17 x 8 = 136 Dividing 136 by 16 yields an average value per square of 8.5 therefore any four symmetrically placed squares will have a sum of 8.5 x 4 = 34

1 15 14 4
12 6 7 9
8 10 11 5
13 3 2 16 is a solution,so 1+4+16+13=34

Line one 04 14 15 01 Line two 09 07 06 12 Line three 05 11 10 08 Line four 16 02 03 13 This is magic square use 1 to 16 numbers

A magic square of order 4 is a matrix:

[a(i,j)], i, j = 1,2,3,4; with:

a(i,j) ∈ {1,2,3,...,4^2}; with the properties:

Σ{i = 1; 4 (a(i,j))} = M(4); j = 1,2,3,4;

Σ{j = 1; 4 (a(i,j))} = M(4); i = 1,2,3,4;

Σ{k = 1; 4 (a(k,k))} = M(4);

Σ{k = 1; 4 (a(k,(4 - k + 1)))} = M(4);

ΣΣ{(i = m; m + 1)(j = p; p + 1)(a(i,j))} = M(4); m = 1,2,3; p = 1,2,3;

Sum(n) = ΣΣ{(i = 1; 4)(j = 1; 4)(a(i,j))} =

Sum(n) = Σ{(k = 1; n^2)(k)} = n^2(n^2 + 1)/2; for n = 4:

Sum(4) = 4^2(4^2 + 1)/2 = 136; it can be show that:

M(n) = Sum(n)/n = n(n^2 + 1)/2; (MAGIC NUMBER), for n = 4:

M(4) = 4(4^2 + 1)/2 = 34;

if we denote:

A = a(1,1); B = a(1,4); C = a(4,1); D = a(4,4);

E = a(2,2); F = a(2,3); G = a(3,2); H = a(3,3);

we have:

ΣΣ{(i = 2; 3)(j = 2; 3)(a(i,j))} = M(4) = E + F + G + H; (1)

Σ{k = 1; 4 (a(k,k))} + Σ{k = 1; 4 (a(k,(4 - k + 1)))} =

A + B + C + D + E + F + D + H = M(4) + (M(4) = 2M(4); (2)

substituting (1) in (2), we have:

A + B + C + D + M(4) = 2M(4);

A + B + C + D = M(4) = 34. (ANSWER)

Solution 1: For a 4x4 magic square the sum of rows/columns/diagonals is [1+2+...16]/4 i.e. 34. This is going to be the sum of the numbers in any proper closed shape.

Solution 2: Now for a 4x4 magic square the jump number is 4. If we start with 1 then jump 4 numbers and land on 6. Then once again jump 4 numbers & land on 11. Then once again jump 4 to get 16. Now add those numbers i.e. 1 + 6 + 11 + 16 == 34. This has worked for me for a very long time!

Well the #ANSWER is 34 but for how to create a magic square of order 4 .

You can go to Fourth-order Magic Squares

Note that there are many types of magic squares having integers $1$ to $16$ that can be constructed. But the sum of the vertex numbers is always $\boxed{34}$ .