If I didn't worry about the sums being distinct I only had to worry about the last digit. This made a p&p straightforward for me. There are 3*4*3*7*9*5=11340 ways of choosing a number from each set. I essentially made a tree-diagram with overlapping branches. And went in order of sets 5,4,6,2,1,3 to get the larger sets out of the way first. Results: 1131 end in 0 1137 end in 1 1136 end in 2 1132 end in 3 1132 end in 4 1135 end in 5 1135 end in 6 1133 end in 7 1136 end in 8 1133 end in 9

As for Ady's comment. I don't see much in S5 except I JUST NOTICED I MADE A BIG MISTAKE. I MISSED THE 19. EVERY DIGIT IS IN THAT SET AND I COULD HAVE COMPLETELY IGNORED IT. To do it my way anyway.

There's no obvious reason to ignore S5 if you're looking for 'distinct' sums. Although 1260 is the product of the other set sizes.