Difference between revisions of "1999 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| − | For any positive integer <math>x_{}</math>, let <math>S(x)</math> be the sum of the digits of <math>x_{}</math>, and let <math>T(x)</math> be <math>|S(x+2)-S(x)|.</math> For example, <math>T(199)=|S(201)-S(199)|=|3-19|=16.</math> How many values <math>T(x)</math> do not exceed 1999? | + | For any positive integer <math>x_{}</math>, let <math>S(x)</math> be the sum of the digits of <math>x_{}</math>, and let <math>T(x)</math> be <math>|S(x+2)-S(x)|.</math> For example, <math>T(199)=|S(201)-S(199)|=|3-19|=16.</math> How many values of <math>T(x)</math> do not exceed 1999? |
== Solution == | == Solution == | ||
Revision as of 17:51, 16 March 2008
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 7 to 1999, there are 
solutions; including and there are a total of 
solutions.
See also
| 1999 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
