2006 AMC 12B Problems/Problem 9
Problem
How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
Solution 1 (Alcumus)
Let the integer have digits , , and , read left to right. Because , none of the digits can be zero and cannot be 2. If , then and must each be chosen from the digits 1, 2, and 3. Therefore there are choices for and , and for each choice there is one acceptable order. Similarly, for and there are, respectively, and choices for and . Thus there are altogether such integers.
Solution 2
Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
and so on.
So, the answer is .
Solution 3
The last digit is 4, 6, or 8.
If the last digit is , the possibilities for the first two digits correspond to 2-element subsets of .
Thus the answer is .
Solution 4
The answer must be half of a triangular number (evens and decreasing/increasing) so or the letter B. -
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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