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Thank you Matthew. 23rd was my favourite puzzle as the cuisenaire rods helped me and I worked with my son to get a final answer. Happy New Year.

loving the new and harder types of puzzles this year :)

@Seth Cohen: Even with those hints I just can't seem to get this one!

I really like 22, and will be using it with my top set Year 10s when I do circle theorems next term :)

I love doing your puzzles, your advent ones as well as the Chalkdust Crossnumbers - thank you!

I solved the puzzles but I think I had to ignore Clue 17 for it: wondering why Clue 17 is false, as it doesn't match the criterion given by Clue 9?

Thanks for another super fun holiday math puzzle! I look forward to this every year!

@Ryan: Got it! I like your method -- just keep eliminating square numbers until you're left with what you need.

I still wanted to figure out why my original method was wrong. And it finally dawned on me:
My mistake was not realizing that my answer of 241 was just a lower bound. The value of n needed to be AT LEAST 241, because my analysis said that 241 needed to be divided out. But any number >241 would also do the job of dividing out 241. So I needed to think about higher numbers too.

Another great puzzle, Matthew, thank you!

@(anonymous): Hi Seth, sorry, I forgot to put my name on my post. I hope it was useful!

@Seth Cohen: Hi Seth,

Your analysis about the multiplicity on primes under 250 is key.

One other thing that helped me is I wrote out '500! x 499! x 498! x 497! x ... x 2! x 1!', stared at it, played with different ideas, and eventually saw that I could rewrite it by grouping together pairs of factorials, which I'll detail in the next paragraph.

I was thinking about how to group that expression into squares, and I eventually lucked out and saw I could do this rewrite: 500! x 499! x 498! x 497! x ... x 2! x 1! = 500 x (499!)^2 x 498 x (497!)^2 x ... x 2 x (1!)^2. This opened up the floodgates for me. I was able to find *an* answer for n. I then used the same analysis you proposed and proved it was the *smallest* answer for n. I hope this helps!

(Oops, forgot to put my name in the comment about puzzle 12.) My thoughts for puzzle 12 are below, covered up:

I thought about prime factors. For the full expression to be a square, all prime factors need to appear an even number of times. So to find n, I can see what prime factors in the numerator appear an odd number of times, and divide them out.

Looking at the prime numbers less than 500, any that are >=250 will appear an even number of times in the numerator. For example, 251 will appear in the 251! term, the 252! term, all the way up to the 500! term. That's 250 appearances, which is an even number.

But what about less than 250? Let's look at 241. That will appear in 241! up to 500! (260 times), but it will also appear in 482! up to 500! (19 times), because 482=241*2. So 241 appears 260+19=279 times, which is an odd number. So 241 needs to be divided out. Likewise with numbers less than 241, like 239. I didn't count the number of appearances of all numbers below 241, but I figured that if n=241, the denominator being 241! will divide out all the numbers that need to be divided out. But that didn't work.

So what am I missing? Any hint would be appreciated! Thanks!

The only puzzle I haven't been able to solve is 12, though I was able to get the final solution without that one because I was close enough that it was just a guess between two options. But I want to solve 12! I thought I had the critical idea, but my answer was wrong. Maybe I could get a hint? I'd love to do a spoiler tag here to explain my original thinking, but I'm not sure how to do that...maybe I'll try a couple things below.
test

(anonymous)

                 

I think this is at least my 9th year solving the math(s) advent calendar, and as always I enjoyed it a lot! Thanks so much, Matthew!

Really enjoyed this years puzzles, Matt! 12 and 14 were favorites as well as the logic "meta-puzzle".

Thanks for another great year Matt! I agree about the step up in difficulty. I found the final logic puzzle, in particular, very difficult, and definitely got frustrated at times! I even tried to resort to a brute force solution in Python since there are only 10,000 possible solutions, but there's a trap baked in related to the clues referencing themselves that made this not work out for me! In the end, I ended up solving it using good old manual reasoning with a bit of trial and error :-)

I agree, this year was a step up in difficulty, which I quite enjoyed! My favorites this year: 2, 12, 16, 21, 22, 23.

Thanks for another great advent calendar, Matthew!

This year felt much more difficult than previous years. Definitely not complaining, especially with how day 23 felt completely impossible until it wasn't -- that feeling is why I love these sorts of puzzles. Cheers!

@Aaron: Wow thank you Aaron, that was just the idea I needed. Very fun.

@Reza:

I didn't use a recurrence. When dividing into three areas, one of the the lines must span the entire grid creating two areas. Then one more line is used to split one of the just created areas into three areas. I just figured out how many ways there are to place the first line, then how many ways there are to place the second. There are two main scenarios you need to deal with:
1) Two horizontal or vertical lines
2) One horizontal and one vertical line

Hope that helps

I would appreciate a hint for puzzle 23. I am struggling to derive a recurrence relation or formula. How to find the pattern for number of ways?

Thanks again Matthew. It was great fun.

My favourite so far of all your advents, a good mix of the trivial and the tricky - day 12 took me far too long until the answer came to me!

Thanks Scroggs!

Wonderful fun as always. Some fiendish ones this year!

I've just completed all the puzzles in the mscroggs.co.uk Advent calendar! My favourite puzzle was the function inequalities.

This year's puzzle was a roller coaster unlike last year's. Do not assume anything is true except what can be deduced. Can't wait to see what tomorrow's clue is.

Whoaaaa just saw the clue for the 24th. This is gonna be fun to unpack! Definitely a step up from previous years!

Thanks @Matthew: and @Allan Taylor: for the tips!

@Seth Cohen: Consider a truth table covering the four T/F combinations. Which one is consistent?

This clue is a really clever anchor for establishing some truths (I mean, could you trust a clue that says "x clue is true"?). The single consistent solution is really clever. Thanks Matthew, this was a fun challenge.

Hey Matthew. I have a request. I know that, after we enter 24 answers, it says below the grid if they're all correct. I'm wondering if might be possible for that message to say WHICH answers are wrong, if any. If we have any wrong, it's not hard for us to suss out which ones are wrong: just go to a day, change the answer for that day, and see if the count of incorrect answers changes. That's not hard to do, but it's tedious. So I figured I'd check if the message could do that work for us. No problem if not, but figured I'd ask!

PS. Trying to wrap my head around the clue on the 23rd. Its logic is messing with my brain.

9, 16 and 17 seem to give contradictory clues, perhaps 17's clue should have 'or' instead of 'and' ?

The clue for today's puzzle just made everything make way more sense.

Looking forward to it, as always!