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Question 970206: the middle digit of a three digit number is half the sum of the other two digits. The number is 20.5 times the sum of its digits. The new number obtained by interchanging the digits in the unit's and hundred's places is more than the original number by 594. Find the original number
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! let a = 100's digit
let b = the 10's
let c = the units
then
100a+10b+c = the original number
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Write an equation for each statement, simplify as much as possible
:
the middle digit of a three digit number is half the sum of the other two digits.
b = .5(a+c)
b = .5a + .5c
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The number is 20.5 times the sum of its digits.
100a+10b+c = 20.5(a+b+c)
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The new number obtained by interchanging the digits in the unit's and hundred's places is more than the original number by 594.
100a + 10b + c + 594 = 100c + 10b + a
100a - a + 10b - 10b = 100c - c - 594
99a = 99c - 594
simplify, divide by 99
a = c - 6
We know that c has to be 7, 8, or 9
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in the first equation, b = .5a + .5c, replace a with (c+6)
b = .5(c-6) + .5c
b = .5c - 3 + .5c
b = c - 3
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If c = 9, then the number is 369
if c = 8, then the number is 258
if c = 7, then the number is 147
All of these numbers, when reversed, have a difference is 594?
Which of these satisfy the 2nd statement
369/18 = 20.5
258/15 = 17.2
147/12 = 12.25
Find the original number 369
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