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Ionacarr, odd digits are 1,3,5,7,9 (as normal) and B,D,F, which represent 11,13 and 15. So the only choices for each of the thee cells in 5dn are B, D or F, with two of those repeated.
Similarly, if any of the odd numbers (the last property) end in a letter they must end in B, D or F.

You cannot assume that such numbers contain only decimal odd numbers before conversion. For instance FFD contains all odd hex digits, but FFD is decimal 4093, which has only two odd digits.

ionacarr, it's worth persevering and by taking Dryden's tips and looking at earlier posts you should start to make progress.
I found the best site for hex conversions was "List of binary, octal, hexadecimal and decimal numbers by Bob Rutherford" Remember too that each of the numbers 2 to 9 will appear once only. It's then just a case of using logic and trawling through the properties systematically again and again.

I'll have a go, then, but having done a few numericals before, I know that 'again and again' usually means 'hours and hours'.

How is it possible at the outset to say that 2 to 9 appear once only? I can see that instructions b and c lead in that direction but surely instruction a allows for the first peg removed to be one of them as well?

I answered my own question about half way through the solve. After four hours I now have a grid that's complete apart from two cells, but I know what will (eventually) be going into them.

I shall persist in the hope of proper PDMs and proper satisfaction, of the kind that for some of us does depend on good old words.

Struggling with this as every time I get to fill in 10dn I can never find any satisfactory solution for it. Would anyone be kind enough to put me out of my misery and just tell me what it is?

Hi aquestion, are you remembering that when it is divisible by 7, 10 and 64 these numbers are hexadecimal so in decimal 7, 16 and 100?

Ah, that explains it! Thanks very much!