Two Three-Digit Numbers Added to Make Four-Digit Sum. No Number Can Repeat Itself.

Question

A friend posed this question to me a few days ago, and I just haven't been able to get it off of my mind.

"There are ten numbers from 0 to 9. You can use any number you like in the box, but the number cannot repeat itself."

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+[][][]

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[][][][]

If you can provide the way in which arrived to the answer, that would be great. I would really like to learn.

Answer 1

1)    246 + 789 = 1035   
2)    249 + 786 = 1035   
3)    264 + 789 = 1053   
4)    269 + 784 = 1053   
5)    284 + 769 = 1053   
6)    286 + 749 = 1035   
7)    289 + 746 = 1035   
8)    289 + 764 = 1053   
9)    324 + 765 = 1089   
10)   325 + 764 = 1089    
11)   342 + 756 = 1098    
12)   346 + 752 = 1098    
13)   347 + 859 = 1206    
14)   349 + 857 = 1206    
15)   352 + 746 = 1098    
16)   356 + 742 = 1098    
17)   357 + 849 = 1206    
18)   359 + 847 = 1206    
19)   364 + 725 = 1089    
20)   365 + 724 = 1089    
21)   423 + 675 = 1098    
22)   425 + 673 = 1098    
23)   426 + 879 = 1305    
24)   429 + 876 = 1305    
25)   432 + 657 = 1089    
26)   437 + 589 = 1026    
27)   437 + 652 = 1089    
28)   439 + 587 = 1026    
29)   452 + 637 = 1089    
30)   457 + 632 = 1089    
31)   473 + 589 = 1062    
32)   473 + 625 = 1098    
33)   475 + 623 = 1098    
34)   476 + 829 = 1305    
35)   479 + 583 = 1062    
36)   479 + 826 = 1305    
37)   483 + 579 = 1062    
38)   487 + 539 = 1026    
39)   489 + 537 = 1026    
40)   489 + 573 = 1062    
41)   624 + 879 = 1503    
42)   629 + 874 = 1503    
43)   674 + 829 = 1503    
44)   679 + 824 = 1503    
45)   743 + 859 = 1602    
46)   749 + 853 = 1602    
47)   753 + 849 = 1602    
48)   759 + 843 = 1602    

Here's the brute-force Ruby program I used to enumerate these:

(2..10)
  .to_a
  .permutation
  .map { |a, b, c, d, e, f, g, h, i| ["#{a}#{b}#{c}".to_i, "#{d}#{e}#{f}".to_i, "1#{g}#{h}#{i}".to_i] }
  .select { |a, b, c| a < b && a + b == c }
  .map { |a, b, c| "#{a} + #{b} = #{c}" }