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

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<noinclude>== Problem ==
 
<noinclude>== Problem ==
</noinclude>An integer is called ''parity-monotonic'' if its decimal representation <math>a_{1}a_{2}a_{3}\cdots a_{k}</math> satisfies <math>a_{i}<a_{i+1}</math> if <math>a_{i}</math> is [[odd]], and <math>a_{i}>a_{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are there?<noinclude>
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</noinclude>An [[integer]] is called ''parity-monotonic'' if its [[decimal representation]] <math>a_{1}a_{2}a_{3}\cdots a_{k}</math> satisfies <math>a_{i}<a_{i+1}</math> if <math>a_{i}</math> is [[odd]], and <math>a_{i}>a_{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are there?<noinclude>
  
 
== Solution ==
 
== Solution ==
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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 <math>\displaystyle 4^{k-1} * 10 = 4^310 = 640</math>.
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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 <math>\displaystyle 4^{k-1} \cdot 10 = 4^3\cdot10 = 640</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:08, 16 July 2007

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.

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

See also

2007 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions
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