To demonstrate that it is not possible to construct a combination of 16 numbers that add up to exactly 256 in each row, column, and diagonal, we can use the parity method.
If we add up the numbers in a magic square of odd size (e.g. 3x3, 5x5, 7x7), the resulting sum is always an odd number. This is because each number on the main diagonal is counted twice, while each number on a secondary diagonal is counted only once. As there is an odd number of diagonals, the total sum will be odd.
On the other hand, if we add up the numbers in a magic square of even size (e.g. 4x4, 6x6, 8x8), the resulting sum is always an even number. This is because each number on a diagonal is counted only once, and there is an even number of diagonals. Therefore, the total sum will be even.
In the case of a 4x4 magic square, the sum of the numbers must be equal to 256. Since 256 is an even number, each row, column, and diagonal must add up to an even number. Therefore, if we add up the numbers on a main diagonal (which contains 4 numbers), we will get an even sum, as each number is counted twice. Similarly, if we add up the numbers on a secondary diagonal (which contains 4 numbers), we will get an even sum, as each number is counted only once. Therefore, the sum of the two diagonals will be even.
However, since the diagonals already add up to an even number, it is not possible for the rows and columns to also add up to an even number, unless some numbers are counted more than once. But since each number appears only once in a magic square, it is not possible to have an even sum in all the rows, columns, and diagonals. Therefore, it is not possible to construct a 4x4 magic square with a sum of 256 in each row, column, and diagonal.
To demonstrate this, we can use the following notation:
A1, A2, A3, A4 are the numbers in the first row;
B1, B2, B3, B4 are the numbers in the second row;
C1, C2, C3, C4 are the numbers in the third row;
D1, D2, D3, D4 are the numbers in the fourth row.
The sum of the numbers in each row, column, and diagonal can be represented as follows:
S1 = A1 + A2 + A3 + A4
S2 = B1 + B2 + B3 + B4
S3 = C1 + C2 + C3 + C4
S4 = D1 + D2 + D3 + D4
S5 = A1 + B1 + C1 + D1
S6 = A2 + B2 + C2 + D2
S7 = A3 + B3 + C3 + D3
S8 = A4 + B4 + C4 + D4
S9 = A1 + B2 + C3 + D4
S10 = A4 + B3 + C2 + D1
In order for it to be possible to construct a combination of 16 numbers that add up to exactly 256 in each row, column, and diagonal, all values of S1 to S10 must be equal to 256.
We can use the following equation to calculate the value of each sum:
Si = ai1 + ai2 + ai3 + ai4, where i is the number of the sum and ai1, ai2, ai3, ai4 are the four numbers that make up the sum.
We can rearrange the equation as follows:
ai1 = Si - ai2 - ai3 - ai4
We can use this equation to calculate the value of each number based on the values of the other three. For example, we can calculate the value of A1 as follows:
A1 = S1 - A2 - A3 - A4
We can use this same equation to calculate the value of all the other numbers.
However, if we try to apply this equation to all the sums, we will see that there is a conflict. For example, if we try to calculate the value of A1 using the sums S1, S5, and S9, we will get three different equations:
A1 = S1 - A2 - A3 - A4
A1 = S5 - B1 - C1 - D1
A1 = S9 - B2 - C3 - D4
These equations cannot all be true, because each number must appear only once in the combination of 16 numbers. Therefore, it is not possible to construct a combination of 16 numbers that add up to exactly 256 in each row, column, and diagonal.
To demonstrate that it is not possible to construct a combination of 16 numbers that add up to exactly 256 in each row, column, and diagonal, we can use the parity method.
If we add up the numbers in a magic square of odd size (e.g. 3x3, 5x5, 7x7), the resulting sum is always an odd number. This is because each number on the main diagonal is counted twice, while each number on a secondary diagonal is counted only once. As there is an odd number of diagonals, the total sum will be odd.
On the other hand, if we add up the numbers in a magic square of even size (e.g. 4x4, 6x6, 8x8), the resulting sum is always an even number. This is because each number on a diagonal is counted only once, and there is an even number of diagonals. Therefore, the total sum will be even.
In the case of a 4x4 magic square, the sum of the numbers must be equal to 256. Since 256 is an even number, each row, column, and diagonal must add up to an even number. Therefore, if we add up the numbers on a main diagonal (which contains 4 numbers), we will get an even sum, as each number is counted twice. Similarly, if we add up the numbers on a secondary diagonal (which contains 4 numbers), we will get an even sum, as each number is counted only once. Therefore, the sum of the two diagonals will be even.
However, since the diagonals already add up to an even number, it is not possible for the rows and columns to also add up to an even number, unless some numbers are counted more than once. But since each number appears only once in a magic square, it is not possible to have an even sum in all the rows, columns, and diagonals. Therefore, it is not possible to construct a 4x4 magic square with a sum of 256 in each row, column, and diagonal.
To demonstrate this, we can use the following notation:
A1, A2, A3, A4 are the numbers in the first row;
B1, B2, B3, B4 are the numbers in the second row;
C1, C2, C3, C4 are the numbers in the third row;
D1, D2, D3, D4 are the numbers in the fourth row.
The sum of the numbers in each row, column, and diagonal can be represented as follows:
S1 = A1 + A2 + A3 + A4
S2 = B1 + B2 + B3 + B4
S3 = C1 + C2 + C3 + C4
S4 = D1 + D2 + D3 + D4
S5 = A1 + B1 + C1 + D1
S6 = A2 + B2 + C2 + D2
S7 = A3 + B3 + C3 + D3
S8 = A4 + B4 + C4 + D4
S9 = A1 + B2 + C3 + D4
S10 = A4 + B3 + C2 + D1
In order for it to be possible to construct a combination of 16 numbers that add up to exactly 256 in each row, column, and diagonal, all values of S1 to S10 must be equal to 256.
We can use the following equation to calculate the value of each sum:
Si = ai1 + ai2 + ai3 + ai4, where i is the number of the sum and ai1, ai2, ai3, ai4 are the four numbers that make up the sum.
We can rearrange the equation as follows:
ai1 = Si - ai2 - ai3 - ai4
We can use this equation to calculate the value of each number based on the values of the other three. For example, we can calculate the value of A1 as follows:
A1 = S1 - A2 - A3 - A4
We can use this same equation to calculate the value of all the other numbers.
However, if we try to apply this equation to all the sums, we will see that there is a conflict. For example, if we try to calculate the value of A1 using the sums S1, S5, and S9, we will get three different equations:
A1 = S1 - A2 - A3 - A4
A1 = S5 - B1 - C1 - D1
A1 = S9 - B2 - C3 - D4
These equations cannot all be true, because each number must appear only once in the combination of 16 numbers. Therefore, it is not possible to construct a combination of 16 numbers that add up to exactly 256 in each row, column, and diagonal.