SMS

If all rows and columns add to the same sum, then this equation must be true:

$a+b=a+c=b+d=c+d$ .

If $a+b=a+c$ , then $b=c$ .

If $a+d=c+d$ , then $a=d$ .

We only need to know about two of the numbers, because we'll know immediately what the other two are!

Let's pick $a$ and $b$ . There are $10$ possibilities for $a$ , because it's an integer from 1 to 10. There are also $10$ possibilities for $b$ . This gives us $10 \times 10 = 100$ possibilities for the semimagic square.

Note that a+b=a+c, a+b=b+d. From this we know that b=c and a=d, so all we have to do is pick a and b, and we will automatically get a magic square. There are 10 ways to pick each number, giving us 10*10=100 possibilities.

When a+b=a+c, then b=c and a, b, c, d cannot be all distinct integers, which means an integer must be used at least twice. Consider two cases:

1. Case 1: When a=b=c=d, there are 10 ways to fill the magic square.
2. Case 2: When a=c, b=d, there are $10 \times 9$ = 90 ways to fill the magic square.

Since if a=b, then c=d, there are no other possible cases. Hence, the total ways to fill the magic square is 10+90=100 ways.

Look at row a + b and column a + c , if a + b = x and a + c = x , then b = c . The same relationship can be said about a and d .

This means that whatever integer we choose for a and b , we will choose for d and c respectively.

So, the problem reduces to finding the number of ways we can add 2 number, lets say a + b . We we have 10 choices for each number, and repetitions are allowed, therefore:

$10 \cdot 10 = \fbox{100}$

For the conditions to be satisfied $a=d$ and $b=c$ must be true.

$a$ and $b$ can take any value from 1 to 10.So total ways = $10 \times 10=\boxed{100}$

10+18+16+14+12+10+8+6+4+2=100

10 for {(1,1,1,1),(2,2,2,2)......(10,10,10,10)} 18 for {(1,2,2,1),(1,3,3,1)......(1,10,10,1),(2,1,1,2),(3,1,1,3).....(10,1,1,10)} 16 for {(2,3,3,2),(2,4,4,2),(2,5,5,2)....(2,10,10,2),(3,2,2,3),(4,2,2,4)....(10,2,2,10)} ................until 2 for {(9,10,10,9) and (10,9,9,10)}

Then sum up all possible ways.

clearly we must have c=b and d=a so the answer is 10^2=100.

Without loss of generality, let us assume that the grid is filled in with a, b, c, and d, as pictured in the problem. The sum of the top row is $a+b$ , and the sum of the left column is $a+c$ . In order for the square to be semi-magic, $a+c=a+b$ . Subtracting $a$ from both sides gives us that $b=c$ . The same process can be repeated to show that $a=d$ . Thus, there are 10 choices for the value of $a$ (also the value of $d$ ), and 10 choices for the value of $b$ (also the value of $c$ ). $10\times10=100$

To satisfy the above mentioned circumstances the diagonals must be same.

First, let one of the diagonals is consists of 1's.By this we get 10 distinct patterns.

Again, we consider the diagonal lines consist of 2's,3's,4's,.................10's.

In each case we get 10 distinct patterns.

so in total we have * 100 distinct patterns. *

see we have ten ways to fill a and 10 to fill b,now c and d have to take a value satisfying both conditions ie,sum of rows ans sum of column is same,so we have in all 10*10 such possibilities=100,hence the ans is 100

Segundo o enunciado, $a + d = b + c$ , com isso:

1 + 1 = 1 + 1

1 + 2 = 2 + 1

...

1 + 10 = 10 + 1

Ocorre o mesmo com 2, 3, ..., 9 e 10. Daí,

$10 \cdot 10 = \\ \boxed{100}$

There are 10 ways to fill same numbers in the squares.

For each row and column to have same sum, we can only have at most 2 different numbers in one arrangement. We must choose two numbers out of 10 numbers and the arrangement is important. Thus we have 10 x 9=90 more ways.

Total ways is 10+90=100

since a+b=a+c, b=c. Similarly, a=d. Two cases: a=b or a != b. If a=b just pick a number => 10 ways. If a =! b 10 choose two numbers, position matters. =10*9=90 more ways. So there are 100 ways total.

The requirements can be expressed as $b+a=a+c=c+d=d+b$ , which represents the sum of each column and row. From the first two expressions, $b=c$ . Using the second and third expressions, $a=d$ . This means there are two separate values in the grid that can be any qualified integer ( $1$ to $10$ ) independent of the other value. The number of subsets with length two for 2 sets of 10 is just $10\times10$ , or $100$ .

We know that: $S = a + b \wedge S = c + d \wedge S = a + c \wedge S = b + d$
Thus: $a + b = a + c \implies b = c$
and, similarly: $c + d = b + d \implies b = c$
Now we need to calculate the amount of ordered integers (x, y) where both can obtain 10 different values. Note: x represents both a and c, which is justified by the fact that a and c have to be equal. Similarly, y represents both b and d. Say the amount of options is N, then we get: $N = 10 \times 10 = \fbox{100}$