Quote: Originally Posted by Barman58
For the original question
Count down the digits on left hand 10-9-8-7-6 then add the right hand
6 + 5 = 11
Ya can't do that - you're starting with an arbitrary number, 10 in this case.
You could start with any arbitrary number to count down, say 25. Then the countdown would be:
25, 24, 23, 22, 21.
Of course we can't add 5 more fingers of the other hand to the 21, to yield 26...
The solution to get the correct amount of digits would be:
(n2 - n1) + 1 = amount of digits.
(25 - 21) + 1 = 5, just as the other (10) countdown:
(10 - 6) + 1 = 5...
Then and only then, we can add the 5 more to get the ? how-ever many...
I've seen snags like this with 'buffer-sizes', where the +1 wasn't done:
(Last_index - First_index) is always short by one,
even when using the 'c' base index of :
(4 - 0) = 4, but to get all
the items, we need the + 1...