Question 1103484
let a = the 100's digit
let b = the tens
let c = the units
:
Write an equation for each statement, simplify as much as you can
:
A three digit number is equal to 17 times sum of its digit.
100a + 10b + c = 17(a+b+c)
100a + 10b + c = 17a + 17b + 17c
100a - 17a + 10b - 17b + c - 17c = 0
83a - 7b - 16c = 0
:
 If 198 is added to the number the extreme digit get interchanged.
100a + 10b + c + 198 = 100c + 10b + a
100a - a + 10b - 10b +c - 100c = -198
99a - 99c = -198
simply divide by 99
a - c = -2
Rearrange
c = a + 2
:
 The sum of the first and third digit is 1 less than the middle term. 
a + c = b - 1
a - b + c = -1
Using these two equations
a - b + c = -1
a + 0 - c = -2
----------------Adding eliminates c
2a - b + 0 = -3
2a - b = -3
Rearrange
b = 2a + 3
:
Using the 1st simplified equation: 83a - 7b - 16c = 0
replace a with (2a+3); replace c with (a+2)
83a - 7(2a+3) - 16(a+2) = 0
83a - 14a - 21 - 16c - 32 = 0
83a - 30a - 53 = 0
53a = 53
a = 1
then
b = 2(1) + 3
b = 5
and
c = 1 + 2
c = 3
Find the number. 153 is the number
:
:
Check this in the first statement
"A three digit number is equal to 17 times sum of its digit.
153 = 17(1+5+3)
153 = 17(9)