Question 986202
Spent a lot of time on this problem, unable to get integer solution when the remainder is 8, changing the remainder to 12 will solve it

let a = the 100's digit
let b = the 10's digit
let c = the units
:
Write an equation for each statement
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twice the middle digit of a three-digit number is the sum of the other two.
2b = a + c
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 if the no. is divided by the sum of its digit, the answer is 56 and the remainder is 12.
{{{((100a+10b+c-12))/((a+b+c))}}} = 56
here we can replace a+c with 2b from the 1st statement, so we have
{{{((100a+10b+c-12))/(3b)}}} = 56
100a + 10b + c - 12 = 56(3b)
100a + c = 168b - 10b + 12
100a + c = 158b + 12
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 If the digits are reversed, the number becomes smaller by 594.
100c + 10b + a = 100a + 10b + c - 594
Combine like terms
100c - c + 10b - 10b = 100a - a - 594
99c = 99a - 594
simplify divide by 99
c = a - 6
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Back to the first equation 2b = a + c, replace c with a-6
2b = a + a - 6
2b = 2a - 6
Simplify, divide by 2
b = a - 3
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Back to the 2nd eq: 100a + c = 158b + 12, replace b with (a-3), c with (a-6)
100a + (a-6) = 158(a-3) + 12
101a = 158a - 474 + 12 + 6
101a - 158a = -474 + 18
57a = 456
a = 456/57
a = 8 is the first digit
then
b = 8 - 3
b = 5
and
c = 8 - 6
c = 2
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Find the no. 852
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You can check this for yourself in each statement.