Question 1085289
A certain number has four digits,...
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Let A = 1st digit
Let B = 2nd digit
Let C = 3rd digit
Let D = 4th, last, digit
So the number = 1000A + 100B + 10C + D 
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the sum of which is 10. 
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A + B + C + D = 10
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If you exchange the first and last digits, the new number...
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The new number = 1000D + 100B + 10C + A
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...will be 2997 larger.
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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
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If you exchange the middle two digits of the original 
number, your new number will be 90 larger.  
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The new number = 1000A + 100C + 10B + D
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...will be 90 larger.
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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
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This enlarged number plus the original number equals 2558. 
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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</pre>