Question

The middle digit of a three digit number is half the sum of the other two digit the number is 7 more than 35 times the sum of its extreme digits the new number obtained by reversing the digits is 396 more than the original number find the original number

Answer 1

this is your answer


let a = 100's digit
let b = the 10's
let c = the units
then
100a+10b+c = the original number
:
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
:
The number is 20.5 times the sum of its digits.
100a+10b+c = 20.5(a+b+c)
:
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
:
in the first equation, b = .5a + .5c, replace a with (c+6)
b = .5(c-6) + .5c
b = .5c - 3 + .5c
b = c - 3
:
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