# Determine the number of 6 digit positive integers such that at least one pair of consecutive digits differs by an odd number.

I noticed that for an odd digit, then there are 4 choices for an even digit. For an even digit, there are 5 choices for an odd digit. There are 5 locations possible in the integer. What would be a good approach here?

Hint: look for examples of $$6$$-digit integers where consecutive digits never differ by an odd number.
• There are $10^6$ possibilities (10 digits including 0 to be placed in 6 positions) subtracted by $4^7$ for all even digits (including 0) subtracted by $4^7$ for all odd digits (including 0). Is that it? I am just learning the topic so thank you for the help. – dragonking Apr 2 at 1:12