I think the answer is 4011 for reasons that follow.
First we construct a 4x4 table:
|
1 |
2 |
3 |
4 |
A=5 |
|
|
|
500 |
B=6 |
|
|
60 |
|
C=1 |
|
|
1 |
|
D=3 |
|
|
|
3000 |
We use the given references (A4=500 to place the number 500 in row A column 4, etc.).
Note that A4+B3+D4+C3=500+60+3000+1=3561, the given number.
The rules are that in any column, row or diagonal we cannot have more than one of the same order of ten. So, for example, in column 4 we've entered 500 and 3000, that's hundreds and tens, so the other two cells must contain ones or tens.
Let's start filling in the missing cells. Row A must contain 5000, 500, 50 and 5. Column 3 must contain all the orders of 10. So A3=5000 and D3=300. Also B4=6 and C3=10. We end up with:
|
1 |
2 |
3 |
4 |
A=5 |
50 |
5 |
5000 |
500 |
B=6 |
6000 |
600 |
60 |
6 |
C=1 |
100 |
1000 |
1 |
10 |
D=3 |
3 |
30 |
300 |
3000 |
If we add the numbers in each row we get 5555, 6666, 1111 and 3333.
If we add the numbers in each column we get 6153, 1635, 5361 and 3516.
Finally we use the coordinates to add up: A2+B4+C2+D4=5+6+1000+3000=4011.