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Question 641671: the sum of the digits of a three-place number is 16. if the digits are reversed and the resulting number is added to the original number the sum is 1049. if the resulting number is subtracted from the original number the difference is 297. what are the numbers?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! the sum of the digits of a three-place number is 16.
if the digits are reversed and the resulting number is added to the original,
number the sum is 1049.
if the resulting number is subtracted from the original number the difference
is 297.
what are the numbers?
:
Let x = the 100 digit
Let y = the 10's
Let z = the units
then
100x+10y+z = the original number
:
Write an equation for each statement, simplify as much as possible:
:
"the sum of the digits of a three-place number is 16."
x + y + z = 16
:
"if the digits are reversed and the resulting number is added to the original
number the sum is 1049."
(100x+10y+z)+(100z+10y+x) = 1049
combine like terms
100x + x + 10y + 10y + z + 100z = 1049
101x + 20y + 101z = 1049
:
"if the resulting number is subtracted from the original number the
is 297."
(100x+10y+z)-(100z+10y+x) = 297
remove bracket
100x + 10y + z - 100z - 10y - x = 297
combine like terms
100x - x + 10y - 10y + z - 100z = 297
99x - 99z = 297
Simplify, divide by 99
x - z = 3
x = (z+3)
:
Multiply the 1st equation by 101, subtract the 2nd equation
101x + 101y + 101z = 1616
101x + 20y + 101z = 1049
---------------------------subtracting eliminates x and z, find y
81y = 567
y = 567/81
y = 7 is the 2nd digit of the numbers
:
Back to the 1st equation
Replace x with (z+3) and y with 7
(z+3) + 7 + z = 16
2z + 10 = 16
2z = 16 - 10
2z = 6
z = 3
Find x
x = z + 3
x = 3 + 3
x = 6
then
673 is our original number
:
:
See if that checks out in the statement:
"if the resulting number is subtracted from the original number the
is 297."
673 - 376 = 297, confirm our solution of 673
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