Question 146473
You're looking for a number ABC.
You know that A can vary from (0,9).
You know that B can vary from (0,9).
You know that B can vary from (0,9).
You know that 
1.{{{A-B=B-C}}}
because the difference between the hundreds digit and its tens digit is equal to the difference between its tens digit and its units digit.
You also know that 
2.{{{A+B+C=9}}}
because the sum of the digits is 9.
If we re-arrange eq. 1, you get,
1.{{{A-B=B-C}}}
1.{{{A+C=2B}}}
If we re-arrange eq. 2, you get,
2.{{{A+B+C=9}}}
{{{A+C=9-B}}}
Since the two right hand sides are equal they are equal to each other.
{{{2B=9-B}}}
{{{3B=9}}}
{{{B=3}}}
Now you have the middle number. 
Using eq. 1, you get 
A is either greater than 3 or less than 3 from B. 
So it can either be 6 or 0.
C is also either greater than 3 or less than 3 from B.
So it can either be 0 or 6. 
ABC is then either 630 or 036. 
Usually 036 does not count as a 3 digit number.
ABC=630.

I'm not sure what the 1,2,3,4 choice are. Is that the first number.