The thousands digit is 3, and the hundredths digit is 6.
Let the tens digit be a, and ones digit be b
3+6+a=b or 3+6+b=a or 3+a+b=6 or 6+a+b=3
That is,
9+a=b, 9+b=a, a+b=3, or a+b=-3
The last one is out, so we have
9+a=b, 9+b=a, a+b=3
The only solution for 9+a=b is a=0, b=9, which gives 3609
The only solution for 9+b=a is b=0, a=9, which gives 3690
The solution for a+b=3 are
a=0, b=3, which gives 3603
a=1, b=2, which gives 3612
a=2, b=1, which gives 3621
a=3, b=0, which gives 3630
So all 6 solutions are:
1. 3603
2. 3609
3. 3612
4. 3621
5. 3630
6. 3690
Edwin
