Quick sum extension – answers
December 9, 2007 am31 3:52 am
Using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly once each, the sum of any list of 1- and 2- digit numbers we create will be a multiple of 9. Several commenters demonstrated this convincingly here.
“Pseudonym” (probably not his real name), suggested several extensions, and I added a few of my own:
- What are the maximum and minimum sums, using only 1- and 2- digit numbers?
- Can every multiple of 9 between those 2 numbers be created?
- If we waive the limit on the number of digits in each addend, verify that all possible sums are still multiples of 9.
- Now, having waived the limits on the number of digits, what are the maximum and minimum sums?
- How many multiples of 9 between that max and that min can be created?
- Can we play the same set of games successfully in base n, using the digits 0, 1, 2, … , n-1 ? Eg, in base 6, using the digits 0 – 5, can we make a list of one and two digit numbers that add to 100 (base 6) ?
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JD2718 
The minimums will always be easy: promoting any digit to a tens place (or to any higher place, in any base) will increase the value of the sum, so the minimum in base
is always 
, which in our case is 45.
Question 5 seems like the hardest — I haven’t investigated much, but I’m curious where, how and why the ability to increase by 9 every time breaks down.
It might be interesting to look at #5 from the other side: what if we had only 10 digit numbers, we would go from 1234567890 to 9876543210. Almost a billion multiples of 9, but we only catch 9! of them….
I am going to look at max(only n digit) vs min(n+1 digits). I think the first skip might occur there.