1999 AIME Problems/Problem 5
From AoPSWiki
Problem
For any positive integer 
, let 
be the sum of the digits of , and let 
be 
For example, 
How many values of do not exceed 1999?
Solution
For most values of 
, will equal 
. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take 
as an example,
And in general, the values of will then be in the form of 
. From 
to 
, there are 
solutions; including and there are a total of 
solutions.
See also
| 1999 AIME (Problems • Resources) | ||
| Preceded by Problem 4 | Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
