A certain number has four digits,...
Let A = 1st digit
Let B = 2nd digit
Let C = 3rd digit
Let D = 4th, last, digit
So the number = 1000A + 100B + 10C + D
the sum of which is 10.
A + B + C + D = 10
If you exchange the first and last digits, the new number...
The new number = 1000D + 100B + 10C + A
...will be 2997 larger.
So
1000D + 100B + 10C + A = 1000A + 100B + 10C + D + 2997
1000D + A = 1000A + D + 2997
-999A + 999D = 2997
Divide through by 999
-A + D = 3
If you exchange the middle two digits of the original
number, your new number will be 90 larger.
The new number = 1000A + 100C + 10B + D
...will be 90 larger.
So
1000A + 100C + 10B + D = 1000A + 100B + 10C + D + 90
100C + 10B = 100B + 10C + 90
-90B + 90C = 90
Divide through by 90
-B + C = 1
This enlarged number plus the original number equals 2558.
1000A + 100C + 10B + D + 1000A + 100B + 10C + D = 2558
2000A + 110B + 110C + 2D = 2558
Divide through by 2
1000A + 55B + 55C + D = 1279
So we have this system of 4 equations in 4 unknowns:
A + B + C + D = 10
-A + D = 3
-B + C = 1
1000A + 55B + 55C + D = 1279
Solve this system by either substitution, elimination,
or matrices or a combination of those, we get
A = 1, B = 2, C = 3, D = 4
So the desired number is 1234.
Edwin