Question

An integer is called parity-monotonic if its decimal representation $a_1 a_2 a_3 \ldots a_k$ satisfies $a_i \lt a_{i+1}$ if $a_i$ is odd, and $a_i \gt a_{i+1}$ if $a_i$ is even. How many four-digit parity-monotonic integers are there?

Answer

Answer: 640

Concepts:digits parity multiplication principle

Difficulty rating: 2390

Small Hint:

Build the number from right to left, counting how many digits may immediately precede a given digit

Big Hint:

For every digit $d,$ exactly $4$ digits can precede it: the odd digits below $d$ together with the even digits above $d.$ The digit $0$ never qualifies.

Solution:

A digit $a_i$ may immediately precede $a_{i+1} = d$ exactly when $a_i$ is odd and less than $d,$ or even and greater than $d.$ Checking each $d$ from $0$ to $9,$ this always allows exactly $4$ digits: for example, $d = 0$ allows $2,$ $4,$ $6,$ $8;$ $d = 4$ allows $1,$ $3,$ $6,$ $8;$ $d = 9$ allows $1,$ $3,$ $5,$ $7.$ (Raising $d$ by $1$ trades odd choices for even ones, keeping the total at $4.$ ) Note that $0$ is never an allowed predecessor, since $0$ is even but exceeds no digit.

So choose the last digit $a_4$ in $10$ ways, then each of $a_3,$ $a_2,$ $a_1$ in $4$ ways; the leading digit is automatically nonzero. The count is $4^3 \cdot 10 = 640.$

2007 AIME II Problem 6

Answer: 640

Difficulty rating: 2390

Small Hint:

Big Hint:

Solution:

Problem 6 in Other Years