2007 AIME II Problems/Problem 6

Problem

An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd, and $a_{i}>a_{i+1}$ if $a_{i}$ is even. How many four-digit parity-monotonic integers are there?

Solution 1

Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of $a_1,\ a_2,\ a_3$ because of the given conditions. (Also note that 0 cannot appear as 0 cannot be the first digit of an integer.) A clear pattern emerges.

For example, for $3$ in the second column, we note that $3$ is less than $4,6,8$ , but greater than $1$ , so there are four possible places to align $3$ 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 $4^{k-1} \cdot 10 = 4^3\cdot10 = 640$ .

Solution 2: Recursion

This problem can be solved via recursion since we are "building a string" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications( $0$ can't be at the front and no digit is greater than $9$ ). There are $4$ options to add no matter what(try some examples if you want) so the recursion is $S_n=4S_{n-1}$ where $S_n$ stands for the number of such numbers with $n$ digits. Since $S_1=10$ the answer is $\boxed{640}$ .

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2007 AIME II Problems/Problem 6

Contents

Problem

Solution 1

Solution 2: Recursion

See also