Hide answer & extension
There is 1 number whose digits are all non-zero and whose digits add up to 1 (1).
There are 2 numbers whose digits are all non-zero and whose digits add up to 2 (2, and 11).
There are 4 numbers whose digits are all non-zero and whose digits add up to 3 (3, 12, 21, and 111).
There are 8 numbers whose digits are all non-zero and whose digits add up to 4 (4, 13, 31, 112, 121, 211, and 1111).
The amount of numbers is doubling each time. You can justify this by noticing that every number whose digits are all non-zero and whose digits add to \(n+1\)
can be made from a number adding to \(n\) by either adding 1 to the final digit or appending a 1 onto the end of the number.
Therefore, there are 128 numbers whose digits are all non-zero and whose digits add up to 8.
Extension
There is 1 number whose digits are all non-zero and whose digits add up to 1.
There are 2 numbers whose digits are all non-zero and whose digits add up to 2.
There are 4 numbers whose digits are all non-zero and whose digits add up to 3.
There are 8 numbers whose digits are all non-zero and whose digits add up to 4.
There are 16 numbers whose digits are all non-zero and whose digits add up to 5.
There are 32 numbers whose digits are all non-zero and whose digits add up to 6.
There are 64 numbers whose digits are all non-zero and whose digits add up to 7.
There are 128 numbers whose digits are all non-zero and whose digits add up to 8.
There are 256 numbers whose digits are all non-zero and whose digits add up to 9.
Are there 512 numbers whose digits are all non-zero and whose digits add up to 10?