Question 464206
16 is not a "digit" in base 10; i.e. all of your addends must be single-digit.



There are several ways to approach this problem. First we can divide all the numbers by 2; so instead of using even digits 2,4,6,8, we can use the digits 1,2,3,4 and try to get 13.


After this it is a bit tricky. We could visualize it as a permutation problem where we have a bunch of balls and "dividers" to split up 13 into six numbers. However it gets difficult when we have to subtract all the cases where there is an addend of 10 or more. Here I will just list them out, by order of largest digit:


111244
112234
122224
112333
122233
222223


There are no sets of six with largest digit 2, as the sum would be at most 2*6 = 12. So there are six ways (assuming order of digits is irrelevant).