A good way to go about figuring this problem is denote Adam's house number by abc, where each of a, b, and c are numbers between 0 and 9. So, if Adam's house is abc, Ben's house is bc, and Paul's house is c. Then, we know that their sum is 912. So let's set this up:
OK. So, if we think about this problem like regular old addition, we have c+c+c would leave a remainder of 2 when divided by 10. In other words, c+c+c=3c would have to have a 2 in the ones digit. Numbers like this are 2, 12, 22, 32, 42, 52, 62, 72, 82, 92, etc. But since c is an integer (a whole number), 3c would also have to be a whole number. So in our list, we need to look at numbers divisible by 3. This leaves 12, 42, and 72. If 3c=12, then c=4. If 3c=42, then c=14, but c cannot be 14 because c should be between 0 and 9. Hence, c=4. Our problem now becomes:
OK, we know 4+4+4 is 12, so adding in the usual way makes us leave the 2 below the equal sign, and carry the one above the b, like this:
Alright. Now, we need to have b+b+1 have a one when divided by 10, so b+b+1=2b+1 could equal 11, 21,31,41,51,61,71,81,91,etc... But if 2b+1=21, then b would have to be 10. But b must be between 0 and 9, so we choose 2b+1=11, and so b=5. Almost done. Now our problem is
OK, but when we add 4+4+4, we get 12, so we carry the one to the fives. But 5+5+1=11, so we again need to carry the 1 over to a. Then, a+1=9, and so a=8.
Hence, our number for Adam's house is 854, Ben's is 54, and Paul's is 4.
Hope this helps.