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You usually post a list of successful entries - will you be doing so this year (or can I just not find it). I have gotten a number of my students into doing this each year and I am keen to see if any are still doing it.

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I couldn't figure out a rule for 21/14 so I decided to count it, realized that I had only gotten 1/8 of the way through with the counting in 5 days, so I took the result at the time, multiplied it by 4, then brute forced it down. I figured one out after though. Good puzzles \|

Great year-end fun as usual! Will there be some kind of annotated solutions afterwards? Also hoping to use this space to discuss the solutions after the competition is closed. Especially interested in the neat solutions for 14 and 21 that some commenters alluded to. If allowed, would also like to share some insights for 9 which many seemed to struggle with (brute force not needed!) after the competition closes.

Another great collection of puzzles, thank you! 21 was, indeed, a challenge. Perms and combs got there quite quickly after a long time of not getting there :) Happy holidays everyone!

@Katharine: I only had one combo that worked. Once you wrap your head around the 19th and 20th, things should sort themselves out.

For 19th and 20th, yeah they warped my brain for a while too. The way I thought about it was this: if the 19th is true, can the 20th be true or false or is it a broken? (Remembering how "and" statements work with true/false: both have to be true for the full statement to be true.) And if the 19th is false, can the 20th be true or false or is it broken?

I've correctly solved each puzzle and have a combination for the pieces that satisfies every rule, but it didn't work. In past years' calendars, there seemed to be only one unique final solution, but this year there seem to be multiple options. Are we allowed to order more than one part from the same company?


Also, I think the clues on the 19th and 20th are breaking my brain. How do I know that I can trust anything that is said on December 20th?

@Alireza: For 9 there is definitely a neat formula. I would recommend either investigating this with grids of different dimensions and/or thinking about the ways the line passes from one square to another. 21 was not easy - there’s been some discussion already in these comments (including from me)

@Aisha: Thanks! I did think of that originally but even going back now I can’t seem to find a neat way to apply that approach (and even the OEIS doesn’t seem to give one for the sequence I get solving this for smaller cases). Would be interested to know how you went about this! As it happens, with a bit more thought I did end up finding a single (double-summation) formula that could be used to calculate this

@Alireza: 9: I didn't try to find a formula, but I made the grid in a spreadsheet, and for each cell, I calculated if the line would pass through that cell.

21: I thought about each group of even A's as a single unit, and thought about how many ways that group of A's could be slotted among the B's. The simplest version of this is the following. You have 2 A's and 9 B's. Treat the 2 A's as a single unit that can't be divided. How many ways can you slot that unit among the B's? Now do that for every possible configuration of even groups of A's. It seems like it'd be a ton of calculation, but it ends up being "only" like 30 cases or something.

Oh my goodness, this was so fun! Thanks for putting this together, I've never done it before but I think I'm gonna have to start doing it every year. 21 was a beast (my favourite though) and 19 was incredibly sneaky, but forshadowed excellently.

Thanks for putting this together. I just happened to see someone post about the challenge on Mastodon early in December and then quickly got hooked. I've really enjoyed all the puzzles.

I would appreciate any small hint on how to proceed.

9: Is there a neat formula for how many squares the line passes through in terms of the dimensions?

21: Should we find a relation between the number of lists of n and the number for n-1 or some other way?

@Aisha: Hm, I used exactly the same method for both of these days. I'm not sure if it lends itself to a nice clean formula, but it might! I'll share my approach after the contest is over!

@Blake: there is an elegant solution but it is very different from day 14. If a small hint is allowed:

recurrence relations!

@(anonymous): It’s certainly do-able, but whereas I was able to find a single simple summation formula for the day that asked for no odd consecutive amounts, this one has eluded me so just wondering if an elegant approach does exist

@Chad: What was your wrong answer for #15? I had that one wrong too at first, and it took a kind of sneaky "Oh of course!" insight for me to figure it out.

Thanks for putting this together--this brought me a lot of joy this year! I only got Day 15 wrong and I really don't understand why, but was still able to solve it all, which was very satisfying.

PS, I am sure there is an elegant way to do #21, but I also just did it brutishly in Excel.

Thanks for another great year of puzzles! I especially liked Dec 9. I didn't figure out an elegant way to calculate the answer all at once (is there a way like this?), but I did it in a spreadsheet, where I figured out how to have each cell calculate whether it'd be on the line. It was satisfying to see the grid filled with 0's and 1's, with the 1's carving a nice path from corner to corner.

@Blake: It's actually not that many different parts. I did it in a spreadsheet, and there were 30 things to calculate. But once I figured out how they should be organized and calculated, each one was very quick.

Another great puzzle calendar, thanks!
I eventually coded Day 21 and while I can certainly see how it could be broken down into a very large number of cases for the purposes of calculation, I am wondering if there is an elegant approach to this one?

The clue on the 20th is breaking my brain.

Question: For the Dec 15 puzzle, does a trivial solution in which the only odd factor is 1 count? Nothing in the text prohibits this currently, but want to make sure this was not just overlooked.

Many thanks for your puzzle again this year - I suspect I have run into a similar contradiction to Bob. It may be an interpretation issue. When you say that the answer is not a factor, do you mean any factor or one particular factor?

@Bob: I have noticed this too. I will try not to worry until the final day!

Nice puzzle today, I am confident of my answer because I have checked it with brute force. I am also confident that my answer to day 13 is a correct answer, but may be not the only correct answer, I do get a direct contradiction in the clue though. Runners should be bought from a shop that was excluded for runners in day 13.

In todays example should both digits in the row of 2 be zeros? If they should be ones then why?

@Matthew: Don't worry, we'll still do them all! Thanks!

@Dhruv: Though actually, now that I think about it, the largest possible value of m doesn't really make sense either. It doesn't matter what m is; the only thing that affects the final number is n. I'm going to wait for clarification from Matthew before solving this one.

@Seth Cohen: Yes I have found only value of n that works too. Plus, with this value of n, I seem to be getting a clash of sorts with another door.

For the 11th, do you mean the largest possible value of m? Unless I'm thinking about this wrong, there's only one value of n that works.

Thank You for the notification of this year's bounty from yourself.
Wish you a Merry Xmas and a Wonderful New Year !!!