Difference between revisions of "2007 AIME II Problems/Problem 6"
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Revision as of 23:32, 4 July 2013
Problem
An integer is called parity-monotonic if its decimal representation satisfies if is odd, and if is even. How many four-digit parity-monotonic integers are there?
Solution
Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of because of the given conditions. A clear pattern emerges.
For example, for in the second column, we note that is less than , but greater than , so there are four possible places to align as the second digit.
Digit | 1st | 2nd | 3rd | 4th |
0 | 0 | 0 | 0 | 64 |
1 | 1 | 4 | 16 | 64 |
2 | 1 | 4 | 16 | 64 |
3 | 1 | 4 | 16 | 64 |
4 | 1 | 4 | 16 | 64 |
5 | 1 | 4 | 16 | 64 |
6 | 1 | 4 | 16 | 64 |
7 | 1 | 4 | 16 | 64 |
8 | 1 | 4 | 16 | 64 |
9 | 0 | 0 | 0 | 64 |
For any number from 1-8, there are exactly 4 numbers from 1-8 that are odd and less than the number or that are even and greater than the number (the same will happen for 0 and 9 in the last column). Thus, the answer is .
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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