Difference between revisions of "1999 AIME Problems/Problem 5"
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For most values of <math>x</math>, <math>T(x)</math> will equal <math>2</math>. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take <math>T(a999)</math> as an example, | For most values of <math>x</math>, <math>T(x)</math> will equal <math>2</math>. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take <math>T(a999)</math> as an example, | ||
<cmath>|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|</cmath> | <cmath>|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|</cmath> | ||
| − | And in general, the values of <math>T(x)</math> will then be in the form of <math>|2 - 9n| = 9n - 2</math>. From <math>7</math> to <math>1999</math>, there are <math>\left\lceil \frac{1999 - 7}{9}\right\rceil = 222</math> solutions; including <math>2</math> and there are a total of <math>223</math> solutions. | + | And in general, the values of <math>T(x)</math> will then be in the form of <math>|2 - 9n| = 9n - 2</math>. From <math>7</math> to <math>1999</math>, there are <math>\left\lceil \frac{1999 - 7}{9}\right\rceil = 222</math> solutions; including <math>2</math> and there are a total of <math>\boxed{223}</math> solutions. |
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:40, 4 July 2013
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,
![\[|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|\]](http://latex.artofproblemsolving.com/a/e/b/aeb23361465dcde0d545b66e70279e2ff9536f54.png)
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 • 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
