Difference between revisions of "1992 AIME Problems/Problem 2"
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So, there are nine digits that may be used: <math>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>2^9=512</math> potential ascending numbers, one for each [[subset]] of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9\}</math>. | So, there are nine digits that may be used: <math>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>2^9=512</math> potential ascending numbers, one for each [[subset]] of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9\}</math>. | ||
| − | However, we've counted one-digit numbers and the [[empty set]], so we must subtract them off to get our answer, <math>512-10=502.</math> | + | However, we've counted one-digit numbers and the [[empty set]], so we must subtract them off to get our answer, <math>512-10=\boxed{502}.</math> |
{{AIME box|year=1992|num-b=1|num-a=3}} | {{AIME box|year=1992|num-b=1|num-a=3}} | ||
Revision as of 22:45, 4 September 2011
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
Note that an ascending number is exactly determined by its digits: for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.
So, there are nine digits that may be used: 
Note that each digit may be present or may not be present. Hence, there are 
potential ascending numbers, one for each subset of 
.
However, we've counted one-digit numbers and the empty set, so we must subtract them off to get our answer, 
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
