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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?

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Hint: look for examples of $6$-digit integers where consecutive digits never differ by an odd number.

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  • $\begingroup$ Each digit is odd or each digit is even. Where should I go from here? $\endgroup$ – dragonking Apr 2 at 0:21
  • $\begingroup$ Count all possibilities for these. You should find that's pretty easy. $\endgroup$ – David Apr 2 at 0:48
  • $\begingroup$ 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. $\endgroup$ – dragonking Apr 2 at 1:12

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