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Difference between revisions of "2007 AIME II Problems/Problem 6"

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==Solution 2==
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==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(<math>0</math> can't be at the front and do digit is less than <math>9</math>). There are <math>4</math> options to add no matter what(try some examples if you want so the recursion is <math>S_n=4S_{n-1}</math> where <math>S_n</math> stands for the number of such numbers with <math>n</math> digits. Since <math>S_1=10</math> the answer is <math>\boxed{640}</math>.  
 
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(<math>0</math> can't be at the front and do digit is less than <math>9</math>). There are <math>4</math> options to add no matter what(try some examples if you want so the recursion is <math>S_n=4S_{n-1}</math> where <math>S_n</math> stands for the number of such numbers with <math>n</math> digits. Since <math>S_1=10</math> the answer is <math>\boxed{640}</math>.  
  

Revision as of 12:40, 4 June 2017

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

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. 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

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 do digit is less 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 $\boxed{640}$.

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 $4^{k-1} \cdot 10 = 4^3\cdot10 = 640$.

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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