SOLUTION: how to solve the sum of the digits of A 3-digit number is 16 if the digits are reversed and the resulting number added to the original number the sum is 1049 if the resulting numbe
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Question 732844: how to solve the sum of the digits of A 3-digit number is 16 if the digits are reversed and the resulting number added to the original number the sum is 1049 if the resulting number is subtracted from the original number the difference is 297 what is the number? Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! First condition
x+y+z=16................................(1)
Second condition
100x+10y+z+100z+10y+x=1049
101x+20y+101z=1049............................(2)
third condition
100x+10y+z-(100z+10y+x)=297
99x-99z=297
/99
x-z=3
x=z+3.......................................(3)
substitute x in (1) & (2)
z+3+y+z=16
y+2z=13...................(4)
101(z+3)+20y+101z=1049
101z+303+20y+101z=1049
202z+20y=1049-303
202z+20y=746..................(5)
1.00 y + 2.00 z = 13.00
Total value
20.00 y + 202.00 z = 746.00
Eliminate y
multiply (1)by 202.00
Multiply (2) by -2.00
202.00 y 404.00 z = 2626.00
-40.00 y -404.00 z = -1492.00
Add the two equations
162.00 y = 1134.00
/ 162.00
y = 7
plug value of y in (1)
1.00 y + 2.00 z = 13.00
7.00 + 2.00 z = 13.00
2.00 z = 13.00 -7.00
2.00 z = 6.00
z = 3
y=7,z=3,therefore x= 6
100*6+70+3
673 is the number
