Solutions - 2020

These are the solutions to the Christmas Lab Calendar 2020.
The solutions are actually based on the solutions I received from participants. Thank you so much, you made my December great with your enthusiasm and creativity.
If you are looking for the calendar itself, find it here:

1st of December:

Thanks a lot to everyone who submitted an answer! 😍

A lot of you got it right, and some of you came up with extremely creative ‘alternative’ answers!

The PhD student will definitely consider all the suggestions to how she can get rid of all those wires with wifi and other wireless connections! She may also talk to the technical service about putting the aircon connection OUTSIDE of the lab. She will also consider how a ‘crossing’ is defined in a three-dimensional lab!

However, the PhD student was thinking of a layout like the this (or the mirror-version) when she made the claim.

She is also happy if you suggest to pull the cable under the optical tables, and if you suggest higher order of spirals, like this one (Thanks Anne ?)

2nd of December:

Thank you for all the answers!

A few of you wanted the postdoc to empty all the boxes for screws (or apples and bananas).

However, the correct answer is that the postdoc, who knows that the boxes are wrongly labeled, only needs to look at a single screw from the box labeled as mixed.


Sebastian W. summarizes what the Postdoc needs to do:


Draw from the box with label „mixed screws“ – you get an imperial screw

You know „mixed screws“ can only contain imperial screws now (because the current label is wrongly)
You therefore know „metric screws“ cannot contain imperial screws but must be the mixed one
You then know that „imperial screws“ must contain the metric screws


Draw from the box with label „mixed screws“ – you get an metric screw

You know „mixed screws“ can only contain metric screws now
You therefore know „imperial screws“ cannot contain metric screws but must be the mixed one
You then know that „metric screws“ must contain the imperial screws


I hope the postdoc relabels the boxes right away, otherwise the next person is a bit screwed too! (thanks to everyone who has made this joke 😉 )

3rd of December:

I’m so impressed – and maybe a bit worried – by how much effort you put into answering the riddles! I learn a lot from your answers. I’m sorry if this is stealing away your time from important matters.

The puzzle from yesterday can be solved in many ways, and I have gotten the most excellent answers to how many slides the master student had when he discussed his slides yesterday. The answers ranges from two numbers yelled through the lab over snips of code to a full solution.

Firstly, the master student has 4 slides, of which 3 contains results.

As Jens pointed out, it is actually a Diophantine equation, and the sequence of solutions can be looked up in The On-Line Encyclopedia of Integer Sequences:
For the full solution, the secretive Ann O’Nymous shared this:

Let the stack have N = n + m slides, where n is the number with results and m the number without results.
Then the probability of picking 2 distinct slides with results is n(n-1)/[N(N-1)], which should be 1/2.
That means, we must find pairs (n,m) of integers that solve the equation 2n^2 – 2n = (n+m)^2 – (n+m). That is n = m + 1/2 + 1/2 sqrt(8m^2 + 1).

If we know a solution m_i, then the next one m_{i+1} follows the rule:

m_{i+1} = 3 m_i + sqrt(8 m_i^2 + 1).

This can be proved:

I had the three numbers N = n + m with n = m + (1 + sqrt(8m^2+1))/2.
Obviously, we have a solution whenever sqrt(8m^2+1) is an odd number.
This means, I can introduce a constant d with 2m^2 = d(d-1). [1]
This allows me to write n = m + d — we have a solution whenever there is a pair of integers m, d that satisfy 2m^2=d(d-1).
Finally, the recursion relationship m –> 3m + sqrt(8m^2+1) can be recast as m –> 3m + 2d – 1 = M.

We now do a proof by induction.

Induction start:
The pair of integers m=1, d=2 obviously satisfies 2m^2=d(d-1).

Induction step:
Let the pair m,d be integers that satisfy 2m^2=d(d-1).
Now, we find 2M^2 = 2(3m+2d-1)^2 = (4m+3d-1) (4m+3d-2).
Finding this decomposition is a bit fiddly and requires (naturally) the induction assumption 2m^2=d(d-1).
We thus have found that the new M satisfies the identity 2M^2 = D (D-1), with D = 4m + 3d – 1. M and D are integers, because they are sums of integer multiples of the integers m and d.

End of proof.


This formula will give all solutions.

1) We notice that the solutions come as families: Once you know one solution, you can generate infinitely many larger ones. Furthermore, we can estimate what the distance between solutions within a family is:

If m is a solution, then M = 3m + sqrt(8m^2 + 1) is the next one, so the ratio between subsequent elements within the same family is in the range 3+sqrt(8) <= M/m <= 6.

2) Earlier, we found that the recurrence within a family can be formulated as:
M = 3m + 2d – 1, D = 4m + 3d – 1.
As it turns out, we can simply invert this matrix equation to find:
d = 3D – 4M – 1, m = 3M – 2D + 1 = 3M – sqrt(8M^2 + 1).
Note that every solution M,D implies a _SMALLER_ solution m,d, which again are guaranteed to be integers by construction. More precisely, we can show that for M>0 the smaller solution is in the range M/6 <= m <= M/5, so a positive solution can never be “back-traced” to a negative one.

3) We can now prove that there is only one family of solutions:
We know that there is a family with solutions m0=0, m1=1, m2=6.
Assume that there is a second family with a solution M, then because of 2) we could generate smaller solutions from it. Because each such step would reduce the magnitude of the solution, but would not lead to negative numbers and we would at some point end up with a solution between 0 or 6. However, we can easily see by brute force that all solutions in this range belong to our already known family.

Therefore, there is no second family.

[1] Introduction if d:
8m^2+1=c^2 ==> 8m^2 = c^2-1 = (c+1)(c-1), where c must obviously be odd,
because c^2 is the even number 8m^2 plus one. So it can be represented as
c = 2d-1, therefore 8m^2 = 4d(d-1) or 2m^2=d(d-1).

Thank you, Ann! ?

4th of December:

Thank you for coming up with so nice answers to this maybe slightly silly question!

As Jens points out, we are dealing with a nontransitive set of dice.

Both Chris and Vladimir make an amazing analogy that I had not come up with myself: It is like rock, paper, scissors. And it is not nice to play rock, paper, scissors against someone who knows what you will choose! In this case, however, there is also an element of probability, and the better dice of a given pair wins 5 out of 9 games in that pair.

Some participants went further to consider a battle between all three cubes. Dietmar even made this very nice table


When looking at all possible combinations of three dices thrown it turns out that dice #3 rolls the lowest number 11/27 while dice #1 and #2 loose 8 times each.

Assuming the goal is to get the highest number then it turns out that dice #3 is the best choice (11/27 versus 8/27 for the others).

So depending on what is required to be the winner, dice #3 is the worst or best if all dice are in play!

I also greatly enjoyed the hints on how to sort the screws. Sadly the current corona restrictions do not allow anyone to hire bachelor students at the moment, so the sneaky PhD will probably just sort them while she waits for some code or some of her experiments to finish running.

5th of December

 I’m sure Julia is super happy that you helped her figure out how many fractions her not so talented student can solve using the unconventional method!

A lot of you shared a snippet of code and the answers (this is also how I solved this problem)

Ann O’Nymous summarizes that the 176 solutions for a, b, c, d being integers with 0 < a,b,c,d < 10, falls into 3 classes:
Type 1 trivial: a = c = 0, c and d arbitrary (81 cases),
Type 2 trivial: a = b and c = d (81 cases),
14 nontrivial solutions:
(1 / 2) * (5 / 4) = (15) / (24)
(1 / 4) * (8 / 5) = (18) / (45)
(1 / 6) * (4 / 3) = (14) / (63)
(1 / 6) * (6 / 4) = (16) / (64)
(1 / 9) * (9 / 5) = (19) / (95)
(2 / 1) * (4 / 5) = (24) / (15)
(2 / 6) * (6 / 5) = (26) / (65)
(4 / 1) * (5 / 8) = (45) / (18)
(4 / 9) * (9 / 8) = (49) / (98)
(6 / 1) * (3 / 4) = (63) / (14)
(6 / 1) * (4 / 6) = (64) / (16)
(6 / 2) * (5 / 6) = (65) / (26)
(9 / 1) * (5 / 9) = (95) / (19)
(9 / 4) * (8 / 9) = (98) / (49)

Chris further points out that if Julia’s student has a 1.4% chance to be right for 0 < a,b,c,d < 10 and a 0.013% chance for 0 < a,b,c,d < 100 – probably not enough for the student to pass the final exam, but Julia will teach him or her how to multiply fractions correctly ?

6th of December:

Oh man, I was in the best mood on Sunday thanks to your INCREDIBLE answers to this bit of trickery! I am so sorry that some of you apparently spent so much time on this question! ⌛

As many of you pointed out, I actually asked you to solve the well known Three Utilities Problem. If you have never heard of this, check the above Wikipedia page out. You should also take a look at this page: (thank you Chris!) where it is neatly proven that…

The Three Utilities problem has no solution in 2D.

 Buuuuut! Many of you points out that the PCB has two sides. Someone also suggested a multi-layer PCB.

This is a small gallery including some of the amazing solutions I received:

Mogens and Sebastian W put the last cable under the bread board.


Jens, Peter, and Ann puts the cable through one of the connectors

The ‘just make a wormhole’-solution is one of my personal favorites (thanks Jens!), but I will also highlight the answer from Mikkel The Masterstudent, who points out that the PhD should just use a breadboard like this ‘cute half size breadboard’ (check the link, that is actually the description!)

7th of December

This question was maybe a bit lame compared to the previous questions, but the number of replies to this one was surprisingly high compared to the Sunday question. If you find my questions too hard, let me know!

Vladimir shares this solution:

Since ‘The postdoc sits next to the ytterbium PhD’ and ‘The rubidium PhD sits next to the ytterbium PhD’, the first three in a row are the postdoc (P), the ytterbium PhD (Y), and the rubidium PhD (R). Left are the master student (M) and the supervisor (S). Since ‘The master student sits … not next to the rubidium PhD’ and ‘The rubidium PhD sits… not next to the supervisor’, there is no one else after the rubidium PhD, and both M and S should sit on the P-side of the P-Y-R-row. Since ‘The postdoc sits… not next to the master student.’, it must be supervisor sitting next to the postdoc. The final answer is M-S-P-Y-R.
This (and the mirror-version with Rubidium PhD – Ytterbium PhD – Postdoc – Supervisor – Master) is the only solution.

8th of December

Thank you for all the funny answers! 🤩

I’m sorry that some of you had to give up on this one, especially because the solution is very obvious once you have seen it!

Peter summarizes it like this:

 First a solution, which I guess is the standard one:
Die 1: 0-1-2-3-4-5
Die 2: 0-1-2-6-7-8
The two dice give you 12 different faces. The simplest analysis would say you need 14 digits: 10 (0-9) for the single digit and 4 (0-3) for the 10-digit.
This means you need to make/show two of these digits to be superfluous.
the first one is simply the 3 on the 10-digit group. As only 30 and 31 are days, we can use the same dice as for 03 or 13.
The second superfluous digit is the interesting one, where the paths diverge:
Solution 1: as each die can be turned around in any way 6 and 9 are the same number just upside down of each other. Hence, we can leave out the 9, and only need to have 0-2 on each die, filling up 3-8 on the remaining sides. This solution is given above.
Solution 2: it is not specified that we always need to use both dice to represent the day. In that case, 1st-9th of a month only require one die, while the number 0 is only used in 10th, 20th, and 30th. Thus, we can remove the superfluous 0 from Die 1 in the above solution, and instead replace it with the missing 9.

9th of December

I personally enjoyed solving this puzzle when I first encountered it, but I was honestly a bit worried that it would be too easy! Yet, you came up with great answers!

Dietmar solved the puzzle by considering what the weight of the lightest of the heavy boxes and the heaviest of the lightest boxes:

The heavy ones on average carry 56/700 of the total weight, the light ones on average 51/700 of total weight. This means that the heaviest light box must carry <56/700 (otherwise it would belong to the heavy ones) and the lightest heavy box carries >51/700. The heavy and the light boxes together carry 428/700 of the total weight, so 272/700 must be packed in other boxes.

The weight of those boxes must be between 52/700 and 55/700. 272/55=4.9…, so exactly 5 boxes do the job. Using more would result in having at least one of them fulfilling the “light box” criterion.


It can be written even more compactly, and the ever-fabulous Ann O’Nymous simply writes

The mass of each box is in the range
1/4 * 32/100 >= m >= 1/4 * 68/100 * 3/7
12.5 <= 1/m <= 13.725
Assuming that the supervisor cannot use their Jedi powers to create fractional boxes, the number must be 13.
To summarize: The supervisor has packed 13 boxes!
As some of you point out he has indeed distributed his books and papers very nicely into the boxes such that they are almost identical in weight. Equally heavy, that is!


 10th of December

Wow, thank you for all the amazing Eulerian paths! Did you know that I asked you to draw an Eulerian path ( I definitely didn’t, but Jens kindly pointed this out! I learn a lot more than expected from doing this calendar.

I will just show a few of the solutions that I’ve received!

This one from Hannes is very accurate, even takes the beam dump into account! But I’m a bit disappointed that you didn’t find the time to actually build it, Hannes 😉

This one from Fermi2 was recreated on a key lab-supply!
The bonus question of course also got a lot of attention (I eveb got a video from the Mother Of Master-Student!), but only one sharp eyed participant saw the bonus-bonus question!
HOWEVER! I keep returning to one answer. Dear all, lean back and enjoy this amazing, aesthetically pleasing and somehow oddly hypnotic solution from Vladimir! Imagine the soft music in the background for these!

11th of December:

For this question I have no cool fun facts, sorry about that! Luckily the question from the 12th contains a lot of cool stuff, so give this a quick glance and keep scrolling 🤶

This question also did not seem to cause you trouble! It is however a bit ambiguous since it has two solutions.

Peter makes this summary of this exercise:
Let us notate the numbers of parts as n1 (amplifiers), n2 (adaptor), and n3 (jumper). Correspondingly, the prices will be called p1, p2, and p3.
Then, the solutions are:
1) n1=5, n2=1, n3=94;
2) n1=0,n2=20, n3=80;
The two equations we get from the first statement are
p1=10, p2+2p3=4, p2=2p3+2.
The last two equations for the prices can be easily combined to obtain
p2=3, p3=1/2.
As p3 is the only single-item price below the average price of 1 Pound per item, it must be greater than 0 (in case we need to divide at some point).
Substituting n3=100-(n1+n2) into the price-weighted equation, and reordering, we get
(p1-p3)*n1+(p2-p3)*n2 = 9.5*n1+2.5*n2 = 100*(1-p3)=50
n1 = (50-2.5*n2)/9.5 = (100-5*n2)/19.
As both n1 and n2 are positive integers, the numerator must be 19*5*N (N integer greater or equal to 0, factor of 5 as it will be divisibile by 5 and 19 is not).
The only two cases of N are 1 and 0, which yields
n2=1 -> n1=5 -> n3=94
n2=20 -> n1=0 -> n3=80.
Vladimir comes to the same result, writingThere are two solutions: 1) N_amp = 0; N_ad = 20, N_j = 80; 2) N_amp = 5; N_ad = 1; N_j = 94. Both satisfy the conditions of the problem, but the last one seems to be the one the author of the question was looking for 😉

Vladimir, you see right through me!
So maybe we will not have to follow the advice of Fermi 2 who suggested that we can live without the long-leadtime amplifiers by licking the MCP cable to detect our ions 😜

12th of December

Wow, again I’m amazed by your brilliance!

I admit that the question was a bit tricky, since I did not specify which cards the PhD wants to use for the card-square. As we will see, there are multiple solutions.

Thus, as Dietmar writes, the PhD has 11 cards with a sum of 77, he needs 9 with a sum of 63. Obviously, he has to remove 2 cards with a sum of 14. So the 7 must stay and plays a “central” role in the following: The card in the middle contributes 4 times to the desired sum of 21, so it must be the 7. The lowest number then, which is either 2 or 3, can not sit in one of the 4 corners since only two combinations of remaining cards sum up to 21.

Before we see the multiple solutions, I’d like to mention, that this is not just a normal question, but a magic question! Jens points out that this question is about magic squares (, and writes:
The question is equivalent to asking what the magic square of order 3 is, because magic squares stay magic under addition. The magic constant for order 3 is 15, so we just need to add 6/3 = 2 to each entry of the magic square of order 3:

[2,7,6]] + 2 on all entries = the above magic square, still staying magic 😀
Jens also points out, that using this selection of cards, the above solution is unique up to a rotations and reflections, and that in contrast to magic squares, magic triangles do not have unique (up to rotation and reflection) solutions, but multiple ones.
So! Let’s see a small selection of solutions!
First, Dietmars multiple solutions:

Thomas considers what cards satisfy the condition of summing to 21 most often and then kick out some of the other cards.
Well, uniqueness has already been mentioned by Jens! But thanks, Thomas for bringing up the exercise for the reader, I can very much relate 😉
Sebastian W makes this most wonderful drawing!
The triangle is also magic! ( Thanks Jens 😊
For the triangle, Sebastian W has been busy photoshopping!
If you want more magic, Jens recommends this page:
Now I know what I will be reading once I’ve prepared the question for the 16th!

13th of December

Apologies for the delay of this solution! We had a small cake-related incident of ‘Oh shit, I did not start the recording 🤦‍♀️– Is the icing all broken?’.

And the icing was all broken.

If you scroll to the bottom you will see the solution with the tiny left-over pieces. Maybe I will redo the video, but the first retake – from today (Thursday) with a donut – was not a success.


For this problem, I got a lot of interesting solutions!

The first I would like to mention is from someone who taught me math in my first year at university, who answered:

The lowest number of cuts is 3.

This is not the correct answer, sorry!


The cake can be cut into six pieces with only two cuts!

Mikkel and Anne solves the question in a similar way, and Anne has also labeled the pieces already in preparation for Christmas vacation:

Christoph and Raphaël, and Vladimir have a different way of solving the question! They use the three-dimensionality of the cake!

Dietmar came to the same conclusion and even explained how to do the cutting: With a cake-cutting string!

I found the above very amusing, but WHAT ABOUT THE ICING?

When asked about whether every piece could have icing, Dietmar answered:

Cut away the legs, rearrange the three pieces and cut all of them in 2 pieces with a single cut, the probability of getting icing is doubled 🙂

And this is the perfect answer in my opinion! Check it out (this is done with left-over cake, sorry that it is not a perfect horse-shoe!)


Anne, I guess your brothers will be more happy if you cut it like this! 🤶

14th of December

You did not seem to find this riddle so challenging, and neither did the postdoc in the drawing. Vladimir sums up:

from the first two student guesses and that each one has one correct guess it means tow possibilities: a) 1 = Heisenberg and 3 = Planck, or b) 5 = Heisenberg and 2 = Planck. If we consider option a), then 1 = Dirac is incorrect in statement 4, therefore the other one should be correct: Pauli = 4. This is also present in statement 1, therefore the other guess should be incorrect (that picture 2 showed Bohr). The same incorrect guess is present in statement 3, therefore the other one should be correct: Dirac = 3. But this spot is already taken by Planck, therefore option a) is invalid.

Let’s check option b). Spot 2 is occupied, therefore Bohr cannot be there. Thus, the remaining options in statements 3 and 5 should be true: Dirac = 3 and Pauli = 4. The remaining spot 1 is left for Bohr.

Answer: 1 = Bohr, 2 = Planck, 3 = Dirac, 4 = Pauli, 5 = Heisenberg.

Chris and Sebastian have solved the question with truth tables:


Dietmar asks an excellent question: Why are the pictures placed like this?

I will be inspired by Thomas and ask this as a question for the reader! 😉

15th of December

It makes me a tiny bit sad that none of the chemists (or those beyond chemistry!) answered this question. It is otherwise a quite nice question based on true events.

Peter writes that he almost forgot to answer because this question was so easy!


But here the easy solution comes. Thomas writes (translated from Danish)


The combined flow rates are
1/3+1/6+1/10+1/15 beakers/min = 2/3 beakers/min.

Takes 90 seconds to empty the beaker assuming that the liquid is leaving only through the four holes.


Maybe a tiny bit of liquid was spilled over the top in the middle of the panic, but the correct answer is indeed 90 seconds!


Also, the nice chemists were actually only laughing at the postdoc and the PhD and promised that the fume hood is also available in the future 😍 – Probably because the PhD and postdoc left a nice note and a drawing of a little cat.


16th of December

The coding master student certainly had a lot of dictionaries!

Peter writes: Don’t be mad at me, but I think, there are again two solutions for the riddle. 😇
It depends on how you interpret “After the superfluous dictionaries have been removed, the remaining dictionaries have 3 elements more than before the dictionaries were deleted.”.
If this is meant as the new number of entries compared before the whole procedure, or only the last step, where the master student removed 15 dictionaries. Ironically, both variations yield a valid solution of integer numbers of dictionaries and entries per dictionary.

Don’t worry, Peter, I will definitely not be mad!

Let’s see the two possible solutions.
Sebastian W writes:
Let’s call the number of dictionaries in the initial state d and the number of elements per dictionary in the initial state n. We know that:

10*n / (d-10) = 1

from the first operation of deleting 10 dictionaries as well as

15*(n+1) / (d-25) = 3

From the second operation, which we can regroup to:

I: 10*n = d – 10
II: 5*n = d – 30

Plugging one in the other, we get that


And therefore know that after deleting a total of 25 dictionaries we have in the final state:

25 dictionaries with
7 elements each

The other solution is summed up by Jens:
Initially he has 100 dictionaries with 9 elements each and 900 elements in total.
He removes 10, so he has 90 dictionaries left and needs to redistribute 10*9 = 90 elements into the remaining 90 dictionaries. Each dictionary gets 90/90=1 more element and has now 10.
He removes 15, so he has 75 dictionaries left and needs to redistribute 15*10 = 150 elements into the remaining 75 dictionaries. Each dictionary gets 150/75=2 more elements and has now 12. [makes sense: now each remaining dictionary has overall 3 more elements, and 75*12 = 900.]

How lovely to have TWO nice solutions 😀

17th of December

Fun little challenge! Optimal connections are very important.

I received these two tries from one participant but I’m sad to say that this is in fact worse than the version, the PhD is trying on the screen (to which the participant agreed. Points for trying, though! 😉)

The correct answer looks like this:

This pretty solution was drawn by one of my favorite university-dropouts but did not come with any explanation.

It is however a neat little optimization problem, and Vladimir shared his thoughts (here accompanied by the neat drawing done by Chris 😉 On this drawing, b = x.)


The idea is optimize for x in the layout above. X = 0 gives the path shown on the screen, with a length of L = 2*sqrt(2)*a ~ 2.83a. But for any arbitrary x it is equal to:

L = x + 4*sqrt[(a/2)^2 + ((a-x)/2)^2] = x + 2*sqrt[a^2 + (a-x)^2]

In order to find extremum, one can take a derivative and equal it to zero. Thus, optimum x is:

1 + 2*(1/2)*(-2*(a-x))/sqrt[a^2 + (a-x)^2] = 0;

4*(a-x)^2 = a^2 + (a-x)^2;

X = a – a/sqrt(3);

The total path length for this x is then

L = a*(1-1/sqrt(3)) + 2*sqrt(a^2 + a^2/3) = a*(1-1/sqrt(3) + 4/sqrt(3)) = a*(1 + sqrt(3)) ~ 2.73a

And this is of course correct! This result can be generalized to rectangular layouts, and is some sort of two-dimensional minimal surface problem. I found this quite cool:


Raphaël and Christoph pointed out that this actually resembles a honey-comb lattice (which I would not have come up with myself, so thank you guys!) – honey-comb lattices are your favorite lattices, right?

Again, minimal surfaces 😀


18th of December

Thanks to everyone ready to help the supervisor organize his calendar.

Sebastian W writes:

Here’s the updated schedule, I hope all the people involved in it are okay with changing their plans because of the supervisors pattern-liking behavior:


Administration – Talk – Teaching – Administration – Supervision – Talk – Supervision – Teaching


and Chris answers that:

A possible pattern is N-S-T – S-A-N-T-A. As I used N for the new category because T was already taken I even created the word SANTA (which was totally on purpose xD)

The PhD student is connecting equipment in the lab.

  • The aircon (1) must be connected to a controller (A).
  • Some instrument on table (2) must be connected to the computer (B )
  • The lasers on table (3) must be connected to the UPS (C)

Is the PhD student right when she claims that she can make the three connections without crossing the cables?

How can she do this?



Sebastian and Chris are of course completely right! I wonder what NST is short for 🤔

The solutions they have found are mirror version of each other, and actually this is the only way to reorder entries from four categories (assuming that the ignoring the mirror version) – given, of course, that you do not consider how the categories are distributed. I, for one would not put teaching as the last entry in the calendar!


This question is by the way actually a version of Langford’s cubes (colored cubes, same idea), and for 4 different categories (or colors) there is only one unique solution (and the flipped version).


Jens nailed the bonus question, writing:

Bonus answer: There is no solution for n=5 and n=6. One can see this since at these lengths, the max length of the block outside is 3 or 4, and one is forced to fit too many numbers inside and outside (after looking for a while 🙂 ).


For n=7, there are the 26 solutions

“71316435724625”, “71416354732652”, “73161345726425”,
“74151643752362”, “72632453764151”, “72462354736151”,
“72452634753161”, “73625324765141”, “17125623475364”,
“17126425374635”, “57141653472362”, “57416154372632”,
“27423564371516”, “57236253471614”, “57263254376141”,
“37463254276151”, “51716254237643”, “41716425327635”,
“36713145627425”, “26721514637543”, “23726351417654”,
“24723645317165”, “52732653417164”, “35723625417164”,
“62742356437151”, “35743625427161”

(and another 26 flipped).


Jens also found the solutions for n=8, and shared them (Thank you, Jens! I do actually really enjoy all these numbers in my mailbox 😊).

You can also get the answers from OEIS:


As a bonus info, the number of solutions for 24 categories (or pairs of colored cubes) was found in 2005 after 3 months of computation time by C. Jaillet, M. Krajecki, and A. Bui. There are 46,845,158,056,515,936 solutions (and another 46,845,158,056,515,936 flipped 😉 ) (see C. Jaillet, M. Krajecki, and A. Bui, Parallel tree search for combinatorial problems: a comparative study between openmp and mpi, Studia Informatica Universalis, vol. 4, no. 2, pp. 151–190, December 2005)


If you want to know more, check out the surprisingly short wiki-page!

19th of December:

 What area has been marked off on the optical table?

This exercise can be solved the pedestrian way, by cutting and counting. This is shown here by Chris:


The area is 30,5 laser-table units


Ann has also found the correct area, and points out a very intriguing general solution:

The general way is easy: just cut the part of the table out with a saw and weigh it 😝

(That works actually for any shape and is surprisingly accurate. Actually into the 1970s the standard method for measuring e.g. a hysteresis curve on an oscilloscope was to place a sheet of paper on the scope, trace the curve, cut it out as well as a square of known size from the same piece of paper and weigh both. I had to do that in one of my undergrad lab courses and it blew my mind.)


The cut-and-measure is indeed mind-blowing, but one could also, as Ann pointed out in a subsequent email:

1) Count all squares that are completely inside the contour.

2) For each partially contained square add 1/2 per line segment that passes through it.


Which is exactly the general way of calculating areas on a lattice grit, and it can be rewritten to a neat little formula known as Pick’s theorem ! Jens goes a bit further than Pick’s theorem in his answer:

In general, the shoelace formula ( is useful for Polygons, but when the vertices lie on lattice points we can use Pick’s theorem that states that the Area is A = B/2 + i – 1, where B is the number of vertices on the perimeter and i is the number of vertices inside the polygon. I count B = 29 and i = 17.


With B = 29 and i = 17, Pick’s formula yields an area of 30.5.

20th of December

This three-in-one calendar entry is one of my personal favorites!


Will Christoph be able to find exactly – no more, no less than – 6 out of 7 cards being in the correct envelope? Some of you said yes, but Ann cracked it, writing:

If six cards match their envelopes then the last one also has to match. So the probability of exactly 6 matches is zero.

Sorry for those of you who bothered with combinatorics 😅



The moon! 🌝 The fact that I got more than one email saying ‘I had not really thought about the moon before’ made me so happy! 🌒


Chris made this neat illustration showing the problem and pointed out that we need to do a projection of the half circle unto a plane.


It turns out, that the shape of the crescent moon is a circle with an ellipse missing. Curious until you’ve thought about it 😀

Maybe I will update this answer, so consider stopping by at some other time!



Jens wrote an answer which made me snort with laughter (rather embarrassing, actually).

These curves, which have a constant width, are called *drum roll* curves of constant width ( ). A simple one is the Reuleaux triangle, but one can get crazy with Reuleaux polygons shapes constructed out of these.

I enjoyed this for example:

You need to watch the video Jens linked. Right away! Maybe twice, even! 😉

21st of December:

Biking in the rain and wind? A well-known (but maybe not so mathematically interesting) problem.


Sebastian W writes

Going to the university, she needs 6/12 = 1/2h = 30 min.

Going back, she needs 6/18 h + red light stops = 20min. + 7*55s = 26min 25s

So she’s still faster on her way back.

Young Peter shares this neat (sorry, Danish) solution:

The bonus question was very nicely answered by Vladimir:

Regarding bonus question, the speed of the bottom point of the wheel is always zero (unless PhD student practices drifting 😉), and the top point is double the speed of the bike, that is, 36 km/h. And all the above is true for the Earth frame of reference 😊

Though I need to add that drifting is not unthinkable in Danish conditions.

22nd of December

I have been completely confused about the days lately 🤷‍♀️, so I could very well relate to this question!


Jens gave full explanation:

I’m going to say today is Friday, the older daughter was right when she said today is Friday but wrong when she said Tuesday and Wednesday; and the younger daughter was right when she said yesterday was Thursday, but wrong when she said Saturday and monday.


The day after tomorrow should then be Sunday, which is when the movie should play.


For the more important question, namely WHAT IS THE MOVIE, I’ve gotten those answers:

  • The movie is (of course) “Der Wilderer vom Silberwald” (Germany, 1957) starring Rudolf Lenz.
    he ever-green Home Alone (followed up by Home Alone 2, and then this little youtube gem )
  • My mama guessed Love Actually 😍 and then she and my sister sat down and watched that.
  • Jens said: “Julefrokosten” is my answer. This is the most Danish looking thing I could find around a reasonable time on Sunday 27th. This can’t be wrong – I feel as certain as having no idea at all! I mean – I’m certain. But not really.
    The last was accompanied by this image of what was actually shown Sunday the 27th:

However, Den Eneste Ene is a much better movie!

One of the subjects of the riddle apparently sighed: another Cinderella story! 😒🙄

I’m still waiting for the truth, though!

23rd of December

Cookies! Awesome! Too bad they are all gone.

The mother of the master student was incredibly efficient, and baked…. Nico enlightens us:

x is the total number of cookies baked by the very nice mum of the Master student. So after Monday the amount of cookies y left is

x – 6/7 x -1/7 = y   (1)

So on Tuesday the master student and the sister we have the same thing happening and z cookies are left:

y – 6/7 y – 1/7 = z    (2)

So now the dad enters the stage and as he seems to know the math very well eats the right amount of cookies such that only one cookie is left for the master student:

z – 6/7 z – 1/7 = 1  (3)


Solving this is left now as an exercise for the reader. Just kidding. Solving equation 3:

z = 8

Take that solution and solve equation 2 leads to 57 cookies where left on Monday evening. So at Monday morning we had 400 cookies according to equation (1).



Peter points out that this iteration is simply a geometric series:

C_(n-1) = a*C_n +b -> C_0 = a^n C_n+b* (a^n – 1)/(a – 1). As our C_n=1=b here, it is a perfect geometric series of 400 = 7^3 + 7^2 + 7^1 + 7^0.



This has also been noted by Jens:

This can be stated as a recurrence relation

7a(n)-a(n-1)==-1, where a(d) = 0.

when it takes d days (or iterations) to finish all the cookies.

The general solution is

a(n) = 1/6 (7^(d – n) – 1)

Surviving 4 iterations gives 400, 57, 8, 1, 0. What if we want the cookies to remain for 10 days?

As I mentioned this is a rapid cookie decay, so mom better baked a lot. We need 47079208 cookies to keep them for 10 days!

47079208, 6725601, 960800, 137257, 19608, 2801, 400, 57, 8, 1, 0


This unfathomable number of cookies is quite… unfathomable!


24th of December:

This was the final entry of the calendar (at least for 2020), and I must admit that I felt rather satisfied with myself when I pressed the send-button.

I personally found this last exercise tricky, yet very entertaining. If you have not yet given it a go, I really recommend it 😀

Not many people answered this one (no surprise 😉 ) but I guess it was a question of timing.


Here I share the answer from Sebastian W (this answer made me so happy btw! This riddle in particular is rather well suited for grandparents!):

Now for the last solution. After combined work of the whole family around the Christmas table, we have come up with the following:


The math student joins at 14:32 from the office, eats cookies, wears a black shirt and his gift is a sweater.


The biology student joins at 14:31 from the living room, eats an orange, wears a green shirt and his gift is the vaccine.


The chemist joins at 14:28 from the bedroom, eats Studentenfutter, wears a blue shirt and his gift is kitchen equipment.


The PhD joins at 14:29 from the uni, eats rye bread, wears a grey shirt and his gift are books.


The engineer joins at 14:30 from the kitchen, eats marzipan, wears a red shirt and is happy without any gifts.


Sebastian, I totally agree!


Also, our master student is really great, isn’t he? He solves every problem we throw at him!

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