Algebra (Linear Algebra)

Question

The sum of the digits of a three-place number is 14. If the digits are reversed and the resulting number is added to the original, the sum is 1171. If the resulting number is subtracted from the original number, the difference is 693. What is the original number?

Answer 1

Let x = 100's, y = 10's and z = 1's

x+y+z = 14 EQN1

100x + 10y +z + x+ 10y + 100z = 1171 = 101x+20y+101z EQN2

100x + 10y +z -x -10y -100z = 693 = 99x-99z EQN3

x = 693/99 +z = 7+z

Substitute x into EQN1

7+z+y+z = 14

y = 7-2z

substitute x and y into EQN2

101(7+z) +20(7-2z) + 101z = 707+101z + 140-40z + 101z = 1171

162z = 324

z = 2

y = 7-4 = 3

x = 7+2 = 9

The original number is 932

932+239 = 1171

932-239= 693