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 2020-11-22 

Christmas (2020) is coming!

Showing all comments about the post Christmas (2020) is coming!. To return to the blog post, click here.

Comments

Comments in green were written by me. Comments in blue were not written by me.
@A: Aha, I should have thought of this! Puzzles completed :-)
(anonymous)
×1                 Reply
@Dean: you can also go through each answer one at a time and change a digit; if the number of wrong answers goes up then your answer for that question was correct
A
×1   ×1              Reply
@Dean: Yes, I'm planning to change how that bit works. Check back tomorrow or the next day for a more fun finish!
Matthew
   ×1              Reply
A bit harsh that we can’t tell exactly which answers are wrong! I won’t have time to revisit every puzzle - and its kind of less fun redoing something that is already correct... :-(
Dean
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@Marty: Yes, I'm in the middle of correcting the clues page to add these details back
Matthew
   ×1              Reply
Nice puzzle set as always. For those of you using the "plain/printable format", it does drop 1 key piece of info on clue 12 that can lead to an unclear clue: the digits are already in the right order. If you allow for rearrangements, there's 2 possible strings that satisfy all the clues I think.
Marty
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I couldn't get 19, so I brute forced Santa's location using what I already knew and then used that to brute force 19. Ha! (it turns out I was off on 19 by 34 - I still don't know where I went wrong)

This year's calendar was great. Thanks :)
Scott
×1   ×1              Reply
No good idea for the 21st, so I brute forced it, and chances are a messed it up are high!
Dave
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Loved the puzzle today. Can't tell if I made it more complicated than it needed to be, but either way it was fun!
Louis
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@alex: Well spotted. I'll attempt to fix it...
Matthew
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There seems to be an issue with the text formatting of 18, if you click hide this puzzle and re-open 18 the text loses it formatting and becomes:
The expansion of \((x+y+z)^3\) is
$$x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3xy^2 + 3y^2z + 3xz^2 + 3yz^2 + 6xyz.$$
Until you refresh the page
alex
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@Eric: Thanks, fixed it
Matthew
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there's a typo in the puzzle for the 18th: you wrote y^2 and z^2 terms instead of ^3

(also I agree that the puzzle for 16th was a banger)
Eric
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Nice one today (16 December) :)
Did he?... did he really?... starts to look like it... yes he did! :D
Gert-Jan
×3   ×6   ×3   ×1   ×2     Reply
@Moss: You were correct: as you were writing this, I was correcting the puzzle. It should now be correct
Matthew
         ×1        Reply
Hi Matthew

If I'm wrong, don't bother to answer, but I suspect there might be an error in the wording of Number 11.
I don't think it's possible to solve it right now.


Cheers
Moss
                 Reply
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