Make them sum to 10,000
Haskell, 99 97 82 81 bytes

-16 bytes thanks to Delfad0r (taking inputs as a list, using abusing that we don't need to deduplicate -> n can always be in [0,4] &amp; using a clever combination of input-format and ap)!



filter((==1e4).sum).mapM([read.(take n&lt;&gt;(c:).drop n)|c&lt;-['0'..'9'],n&lt;-[0..4]]&lt;*&gt;)


Try it online!