Difference between revisions of "1992 AIME Problems/Problem 2"

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== Solution ==
 
== Solution ==
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First, we line up the nine digits that may be used: <math>\displaystyle 1,2,3,4,5,6,7,8,9.</math> Note that each digit may be present or may not be present. Hence, there are <math>\displaystyle 2^9=512</math> working numbers.
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Subtracting off the one-digit numbers and the null set, our answer is <math>\displaystyle 512-10=502.</math>
  
 
== See also ==
 
== See also ==
 
* [[1992 AIME Problems]]
 
* [[1992 AIME Problems]]

Revision as of 19:12, 26 July 2006

Problem

A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?

Solution

First, we line up the nine digits that may be used: $\displaystyle 1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $\displaystyle 2^9=512$ working numbers.

Subtracting off the one-digit numbers and the null set, our answer is $\displaystyle 512-10=502.$

See also

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