Solutions 2021
1st of December:
Cable-tieme indeed!
The PhD student wanted help to make a closed loop of reopenable cable ties - some of you did not believe this, but it is a thing, and it is super geil!
Also, the amazing Anne informed me that 'Bioinformaticians like to call small sequences k-mers. So we have cable-mers in this case.'
Let's see how we can connect these cable-mers and solve this challenge!
If the answer is too long, just scroll and see the figures 😉 But The Calendar Team loves these answers!
A lot of you pointed out that (here with the words of Thomas G:)
'You need n-1 cable ties to tie together n chains, and another one to make it closed (assuming that "closed" means annulus-shaped). Since there are 9 chains, one could be inefficient and pick 9 ties at random and then stick everything together with these.
However...', Thomas continues, and then he gives the right answer.
The right answer is 7. Jens describes it with words:
8, 7, 6, 6, 5, 5, 5, 4, 3; step 0
16, 6, 6, 5, 5, 5, 4, 2; step 1
23, 6, 5, 5, 5, 4, 1; step 2
30, 5, 5, 5, 4; step 3
36, 5, 5, 3; step 4
42, 5, 2; step 5
48, 1; step 6
49; step 7
At least one step less than 8, which one would get by just cutting open a curled up chain of length 49 and cutting them into the appropriate pieces (cut open + 7 cuts -> 8 steps). The key is that in the last step, the chain whose ends to be joined to form the ring are joined by a chain which vanishes. This can be reached in many ways (the solution is not unique). E.g. she could've used the chain of length 7 and join all others pairwise or she could've taken one each out of four of the chains that don't have length 3 and use the one with length 3 to do the last 3 moves (with the chain of length 3 vanishing in the last move).
Fermi2 sent this little mathsterpiece to us
7 cables ties should be opened, and no less.
The explanation comes in two parts.
* Part one: if you have n chains and p already-opened cables ties, on which condition can you make a single loop without opening any new cable tie?
If p>=n then it is fine: you have enough ties to link the n chains together; and you have some extra opened ties that you can add freely.
Now we can prove that it is not possible if p<n, and you can do that recursively on n:
The sentence to prove is: "With n chains and p opened ties with p<n, you can't form a single loop".
** If n=1 and p=0, you have no opened cable tie and you are not allowed to open any, so you are screwed.
** Let's assume that the sentence above is true for a given n, and consider a situation where you have n+1 chains and p opened ties. If p=0 it is again game over, so we can assume that p>=1. You can then take one opened cable tie to link two of your chains. You now have n chains and p-1 opened ties. The sentence is true for n chains, so if p-1<n you can't make a single loop. Therefore at the beginning, if you have p<n+1 it is also impossible, so the sentence is true for n+1 chains as well.
** By recursion the sentence is true for any integer n.
* Part two: With part one, we can rephrase the problem: What is the minimal number p of links to open so that we end up with n<=p chains?
We don't have a winning strategy for the general case. In our particular case, there is no way to open 6 ties or less and end up with the same number of chains left (a single closed tie is also a chain!): if you want to keep 6 chains or less, you have to open at least 12 ties (the ones of the three smallest chains). And here there is a way of opening seven ties and ending up with seven chains: you have to open every tie of the two smallest chains. With these seven ties, you can link the seven remaining chains and create a single loop :)
Disclaimer: No cable tie were harmed, wasted, or left lying in the lab while solving this riddle.
The Head of The Calendar Team had to snort laughing at the disclaimer.
Finally, the graphical explanations are maybe even more straightforward.
Sebastian W simply explores all options:
And Ed shows how to connect to make the 💗.
2nd of December
Mogens solves this problem by going backwards:
"Then six where lext" = 6
"and for the second board I took one-third of the ones in that box you have there" = 6*1,5
"You gave me three for my first board" = 9+3
"the PhD remembers, 'and then one-third of the remaining ones went into the box that I had to build'." = 12*1,5
"True, and I gave two to the guys next-door the other lab, they also needed some" = 18+2
"Then I gave one-third of the remaining ones to the student helper" = 20*1,5
"and the Calendar Team took nine" = 30+9
"First, I took one to replace that broken one" = 39+1 = 40
Thus, there must have been 40 pieces in the box from the beginning.
True! The right answer is 40.
Though the chronology of the question was admittedly not entirely clear.
Jens adds: The way the question sounded, I was at first a bit confused about the time at which the Bachelor student removes a third. Assuming She is last to modify the stock from the text seemed reasonable a posteriori (the only allowed integers up to 100 satisfying the second last statement are {13, 40, 67, 94})
The Calendar Team also really likes the curiousity about the 9 RF components, but it seems like Fermi2 figured it out:
The Calendar team took 9: 30 are left. We guess that it was to build a large antenna to broadcast worldwide the daily riddles of the Calendar.
What else could we need those components for??
3rd of December:
Cedric tips the calendar team (thanks a lot 💖)
Julia finds a nice solution to the riddle of the professor. This one was actually not found by so many of you!
Dietmar did this - WITH PENS 🥰
Dietmar borrowed the pens from the secretary - who even joined 💖
Sebastian W got really creative with question C.
Michael even sent a new puzzle!!!! How do we turn the horse by moving only one stick?
4th of December:
the "obvious" solution is
192
384
576
Now, for other possible combinations: Since I'm not a theoretical (they will for sure find an impressive but complicated solution) but rather an experimental physicist I tried the following "Ansatz":
Let's have a look at possible combinations of last digits, here they are:
..1 ..2 ..3 ..4 ..6 ..7 ..8 ..9
..2 ..4 ..6 ..8 ..2 ..4 ..6 ..8
..3 ..6 ..9 ..2 ..8 ..1 ..4 ..7
The first digit combinations in principle can only look like these:
1.. 1.. 1.. 2.. 2.. 2.. 2.. 3..
2.. 2.. 3.. 4.. 4.. 5.. 5.. 6..
3.. 4.. 5.. 6.. 7.. 7.. 8.. 9..
1
2
3
is not possible, since the lowest 100 number would be 145, times 3 it's 435.
Now, combining first digit combinations with possible last digit combinations not so much combinations are left.
I came up with only two more solutions, here they are:
273
546
819
and
327
654
981
Easy!
So the first thing to note about this puzzle is, that only the first number needs to be considered to traverse all possibilities. Second notice is that only all numbers (including) between 100 and 333 need to be checked for the lowest value as lower values would not yield enough digits and higher values would yield to many.
The obvious idea is then to permute the digits. Since the carried over values need to be correct, there is only one way to rearrange "219" which is "192". This leads to the triple:
192 -> 384 -> 576
Cedric has solved this with a single(!) line of code. Much appreciated 🤩
And Jens, the problem-crusher adds a bit of bonus-bonus for us:
There is one option that breaks the rules! You want nines in here? Nine says nein and disguises herself as a zero:
267, 534, 801
Thank you all 💖
5th of December
Assuming that the inner circle has area of 1 (arbitrary area units), the rings surrounding it should have areas 3, 5, 7, 9.
The only way to realize the scenario is 1+3+5 = 9
Bonus: Once we have one more ring added (area 11), then there is another possibility
1+3+7 = 11.
BBonus: Adding two more rings (area 13,15), we can have two solutions with only consecutive rings being hit.
1+3+5 = 9
3+5+7 = 15
Unit area, I like that! I also like the approach of The Fermis:
First let's play the favourite trick of the theoreticians and set pi = 1 so that the area of the rings are integers. We can also set hbar = c = 1 but that's not mandatory.
The problem-chewing Jens points out that this can be solved by considering Pythagoran triplets:
... one just looks for Pythagorean triples at first.
We want \pi r1² + \pi r2² = \pi r3²
and divide by \pi to see that we are looking for Pythagorean triples.
Answer:
(3,4,5) is the lowest Pythagorean tripe and matches the available circles. So he hit 3 and 4 (one or more times) while she hit 5 (one or more times).
Max considered what 'accurate' means.
Bonus-Bonus Question
====================
We now only think about sets a_i and b_i of consecutive numbers. We also note that we know, that 6 rings can't be a solution, since we have already found all possibilities.
We briefly note that, if the condition that the colleague needs to hit her rings "accurately" must be interpreted as "she only hits one ring anyway", the solution is simple. We just add next smallest ring to the original solution A(1)+A(2)+A(3)+A(4)=17=A(8). This would be fulfilled with the 8th ring.
There is however another solution that only requires seven rings, building on top of the previous logic. Without further giving any reason for completeness we can now just give an example:
A(2) + A(3) + A(4) + A(5) = 3 + 5 + 7 + 9 = 24
A(6) + A(7) = 13 + 11 = 24
Since we already now the original solution, which hits only consecutive rings, this adds another one: The PhD students hits rings 2,3,4 and 5 and his colleague hits rings 6 and 7.
This does not prove that there are no other solutions with 7 rings, but this is unnecessary for the question.
Here follows the solutions from Ed, Paul and the PhD student who is pretty good at magnet dart!
6th of December:
Here, honestly, was a mishap.
I'm sorry on behalf of The Calendar Team! This is what happens when a new riddle is found for an already drawn question!
Thanks for all the solutions to a meaningless question.
Ed fixed my problem by allowing the student to run the other way around. I really liked this!
Thanks, Ed!
However, correct answer to everyone who came up with a number, and sorry again for wasting your time on calculating unphysical quantities!
7th of December:
Let's see some of the funkiest solutions! Here comes solutions from Jens, JPJ, Dietmar, and Ed.
Now, let's dive a bit deeper!
Paul writes:
a) There is no other possibility to fill out the corner at the top right
and not making the remaining length of the diagonal line to short for
anything else than to place the first shape and as soon as it is done by
the same argumentation we also have to place the second one below. At
this point for the third and fourth it becomes interesting, as they can
only be placed if the original shape has exactly the geometry which is
shown by the unit-shape S.
b) There are many possibilities, so here just one of them.
When I was finished I started to think if there is a general pattern,
and there is! Exactly quadratic numbers are possible, I'm happy to share
my thoughts with you :-) :
Prove by induction: n^2->(n+1)^2
If we always re-scale everything, such that the shapes we want to cut
out have the size of the unit-shape S, then we can simply add another
column on the left and row on top (shown in green). These can be
constructed by using a line of length 3n+2 as shown on the right and
choosing the appropriate of the three possibilities to crinkle it into
an L-shape, which nicely fits on the shape on the left.
It is not possible to get any other number:
The length of the left edge x has to be the sum of attached smaller
shapes, so it can be written as x=p+q*sqrt(2) for some integral numbers
p and q. As the total number of small shapes x^2=p^2+2*q^2+2*p*q*sqrt(2)
has to be integral, we can deduce that either p or q has to be zero. If
p is zero we have to place q small shapes with their diagonal at the
left edge, which will either fail at the upper or lower corner. Thus q
is zero and we can cut out p^2 shapes.
The Fermis are also on this track. Here is the proof from Paul and the drawing from the Fermis. Beautifully fractally.
If we do not demand pieces to be identical in size, one can cut the trapezoid into three parts of the same shape (and different sizes), as shown in panel c.
Using this approach we can divide the initial piece into any number N>2 of the same shape parts. Indeed, upon cutting a single trapezoid into 3, we increase the number of parts by 2. Starting from three parts and cutting some parts into three, we can get any odd number eventually. Panel d shows how we get 9 parts. Starting from the 4-part division in panel a and cutting some parts into 3, we can obtain any even number of parts.
8th of December
since I have no posts at home (really, no joke!) I had to use egg and coffee cups :-)
(sadly no picture attached)
The PhD student tell the Postdoc with a grin in his face:
'Are you kidding me? What kind of example are we giving to the bachlor student about serious work here? Of course it's impossible!'
Then he does not even bother to try with the 11 posts since he quickly understands that the Postdoc is tricking him again. However with the 12 posts and 5 flips per move, he comes up with this answer in 4 moves (u: pointing up, d: pointing down, underlined the ones that were flipped):
0) uuuuuuuuuuuu
1) ddddduuuuuuu
2) ddddudddduuu
3) dddduduuuddu
4) dddddddddddd
Max and co. went all in, look-up table and all! Did you ever hear about Dijkstra's theorem? I didn't!
Flipping the posts in the first picture is possible in move by simply flipping the upside down posts once (also in three if one does two dummy moves)
Flipping the posts in the second picture is not possible
Flipping 11 posts is also not possible
Bonus: This requires a minimum of four moves
Working:
First, we completely ignore the position of the posts.
When two posts are flipped at the same time, and we look at the total number of posts in the correct orientation (A) we find that there are only two cases (call the upside-down orientation B):
Flipping over two posts in the same orientation changes this overall number by +/-2 (AA->BB)
Flipping two posts of opposite orientation changes this overall number by 0 (AB->BA)
It might not be possible to do either case depending on the actual situation (i.e. one can't flip 2 over if there is only one). However we see, that the total number of posts only changes by an even number. Therefore it will never be possible to flip all posts to the correct orientation if the initial number of posts in correct orientation has a different parity than the total number of posts (i.e. odd number + even number always stays odd).
Therefore it is clear that the first case can work (1 is initially correct, 3 total) and we find the obvious solution to simply flip the two outer posts in the first step. [If we really want to do it in three steps for whatever reason, we just flip the first two posts twice, and then outer two posts once]
The situation in the second case can't work, since 2 are initially correct, while there are still3 in total.
The case for eleven posts (all correctly oriented A, need to be inverted B) and flipping four at once is extremely similar. In each step we can:
Flip over four or zero posts of the correct orientation --> changes number in correct orientation by +/-4 (AAAA->BBBB)
Flip over three or one posts of the correct orientation --> changes number in correct orientation by +/-2 (AAAB->BBBA)
Flip over two posts of the correct orientation --> changes number in correct orientation by 0 (AABB->BBAA)
Again the total number only changes by an even number. Since we start with an even number of upside down posts (0) and want to have all posts (11) upside down, this is not possible.
Bonus:
For this question we investigate the directed graph, which shows from which position (number correct = A, number upside down=B) we can end up in which other positions.
In the table we can look up any state (i.e. A=8,B=4) and then go down to find the colored squares which map to each of the possible moves (color coded in the smaller table on the right).
I.e. we find that by using any move(!) of the form "ABBBB"->"BAAAA" we end up in the A=11, B=1 situation.
We now apply a variant of Dijkstra's algorithm to find the shortest path from the initial situation in step 0 to any other situation by simply looking at all next possible situations and writing the next highest integer there, if no "path" has already been found. I.e. we find that we can go from A=12,B=0 only to A=7,B=5. As there has been no shorter path to this situation we label it with the total number of steps so far (1) and repeat from there.
In the end the shortest path form the start can be traced backwards. It requires four moves in total. A possible shortest way would be:
Flip 5 A (ends in situation A=7, B=5)
Flip 3 A and 2 B (ends in situation A=6, B=6)
Flip 3 A and 2 B (ends in situation A=5, B=7)
Flip 5 A (ends in situation A=0, B=12)
9th of December:
One solution we found is to cut the big cake in two equal pieces (for example by cutting along the diameter of the disk, which you can construch with a ruler and a compass, the only tools of the geometrician), and cut the middle cake in 32 pieces of area 1 (you can do that by cutting the cake in 2 then each piece in 2, etc. until you have 2^5 = 32 pieces. You can also do that with only a ruler and a compass).
The two first persons can share the big cake, the third can take the entirety of the small cake and 7 of the 32 pieces of the second cake (area 18 + 7 = 25), and the last person can take the last 25 pieces of the second cake (area 25). By the way you don't need to cut all the individual 32 pieces and make a mess out of it, but just 8 of the good size so that you have 7 to give to the third person :)
JPJ and R solves the problem like this, and after this comes the solution from Alex:
Paul solves the problem, and attach this amazing picture! Thanks, Paul!
The solution from Marian is extremely nice. He writes:
Nice question. Sadly due to corona our Master students didn't bring any cake but 6 cm diameter is a really small cake !!
But the solution if one assumes that the icing was done during the cake was cooling in the baking form is straight forward.
6cm stays as a whole 9cm^2
the 8 cm cake has to be cut into two pieces on with 12.5 cm^2 icing and one with 3.5 cm^2 icing.
the 10 cm cake has to be cut in two pieces with 12.5cm ^2 and 12.5cm^2 icing.
The best cuts are made in the 10 cm in the middle, and the 8 cm cake 0.9758 cm from the outside.
If one is only is only allowed to do 2 cutes but can cut all three cakes because they are sitting very close to each other we have an alternative solution.
The Fermis also have a nice solution for this one, including a MUST WATCH VIDEO! We had it already last year, but watch it again!
Now you could argue and say it's not very fair, because the four different persons don't get the same amount of pieces. And this is absolutely true, as the following video proves:
https://youtu.be/gnArvcWaH6I?t=200
10th of December:
so for this solution is will call the number of coffees that each of the students had before the bespoke coffe-break x.
Then after the coffee break the non-decaf-drinking student will have enjoyed another number of coffees I'll call Δ.
From the information we deduce that the decaf drinking student does not drink any additional normal coffee (so his count stays at x). The decaf-drinking student on the other hand will end up at a certain number y=x+Δ, such that "x will be half as much plus one". Or in equations:
x=1/2y+1=1/2(x+Δ)+1
We also need to make sure that y>x (otherwise we would not have any new coffee), this means Δ>0.
We can now find the number of coffees that were drunk during the coffee break:
Δ=x-2
and the total number of normal coffees drunk by both students in sum during the day Σ:
Σ=x+x+Δ=3x-2
So the answer is:
The students drank 3 times the number of coffees they had before the break minus 2 in total.
Some realistic values would i.e. be drinking 3 coffees each before the break and 7 in total after the break (1 new during the break seems plausible, and the wording "the decaf" implies that they only drink a single cup during the break each).
Jens writes
before the coffee break, assuming during the coffee break only one coffee is consumed, the two students had already 3 normal coffees. After the break, the decaf drinking student will have still 3 coffees in his system since decafs don't count, and the non-decaf drinking studen will have 4. 4/2+1=3 =-O.
Bonus:
For n consumed coffees during the coffee break (some people don't stop after n=1!), one has
y=x-n; (y has no offset since the decaf-drinking student says that she'll 'have decaf now', implying that there will be none non-decafs consumed)
y=x/2+1;
for x the number of coffees of the non-decaf drinking student and y the number of coffees of the-decaf drinking student. So, x=2n+2 and y=n+2.
Values n>3 are not recommended.
Thanks, Jens, for taking care of our health!
The coffee-responsible PhD student answers with coffee-ticks!!! Thanks, Ed 💖
You may note that the Head of The Calendar Team has adapted a new ticking-strategy!
11th of December:
Pizza? 5 cuts? 16 Pieces! Why? Because Pentagram Pizza ]:->
JPJ is of course correct!
(Here accompanied by the pentagram by Paul, because sorry JPJ, your handdrawn pentagram is not this shiny and devilish!)
Ed shares this nicely cut pizza
Team Marian finds this relation:
for 1 cut we have 2 slices for 2 cuts we have 4 etc. ... we can make a rule out of this.
cuts :0,1,2,3,4,5
slices:1,2,4,7,11,16
This means #slices =n^2/2+n/2+1
Jens, who last year blew the minds of the solution-readers by knowing the OEIS sequences very well, writes:
The problem is known as 'circle division by lines' or circle cutting problem. There is a nice page by Wolfram math:
https://mathworld.wolfram.com/CircleDivisionbyLines.html
on which the corresponding OEIS sequence is linked as well :)
And Paul advises us to be careful cutting pizza in public (in particular if you use the pentagram), since "this way, you might summon an angry Italien (Even worse if you also have pineapple on it)."
Alexabder and the Fermis are thinking outside of the (pizza) box!
Alex suggests that
Given that only straight cuts are allowed, one can achieve a number of 32 pieces of pizzas with 5 cuts, if you stack them on top of each other: The first cut makes 2pieces, those are placed on top of each other and are cut, that makes them 4 parts....
If you are even allowed to fold the pieces and then cut them, you could achieve in combination with stacking infinite pieces. Allowing one fold per piece would for example lead to 3 pieces after one cut, so five cuts could produce 3^5=243 pieces.
Sounds a bit greasy!
The Fermis have a different way of making infinity-pizza. They are not assuming straight cuts:
We have two solutions to cut the pizza. The first one is by using non-straight cuts, as illustrated in the figure attached (top left drawing), which leads to an unbounded number of pieces.
The more serious solution is illustrated below to the right and follows this rule: "each time you draw a line, it should intersect every already-existing line". A line can only intersect another line once, so this should be the maximal number of pieces we can get. For each line you draw, you add one piece plus as many pieces as you cross other lines.
Staring with one line that cuts the pizza in two, you then add a second line and two pieces, then a third line and three pieces, etc. So the maximal number of pieces you can have with n cuts is (n^2 + n + 2)/2, which for n=5 gives 16.
Regarding the bonus question! So many good suggestions!
Jens says: "the upper right corner looks like an abstract drawing of a "balance for charges" or a device for "measuring charges" to me. An electrometer measures charges. Best guess on what should fit most closely to these observations: a 'Peltier-electrometer'. It might schematically even look a bit like that drawing. Let this be my final answer. Anyways a next best guess might be the even more abstract representation of a potentiometer: maybe the plus and minus represent a voltage drop and the small bump in the middle is the wiper"
Max suggests that it "could be Positronium? Not sure about the weighing scale though... (First I thought that would be a Coulomb balance, but that didn't start with P)"
Team Marian thinks: "it looks like a swing with two things sitting on it in an equilibrium. Therefore we had potential difference, potential equalization or proton weight."
Julia is certain: "There is a proton on a scale which is why I guess "proton mass" is what is drawn here."
Dietmar is onto this too, " It's a balance for protons which we usually call palance :-)"
The artist, when asked, was very pleased with these many good suggestions, and maybe he would even like to pick a different meaning for his lil drawing. Buuuut, he was a bit offended by Team Marian, who also wrote "We also thought maybe related to the work in your lab it could be a really bad drawn transistor sign and therewith a "photon gate"? But the + made no sense." He intended to draw a positron.
The sheet btw looks like this as of the 9th of Jan.
12th of December:
I really enjoyed this puzzle, because at first glance it is utterly puzzling!
Ivo writes: I figured the 12th of December answer would be 720 degrees, as the coin has to in total travel 2x the circumference. It’s of course important here, as mentioned, that the coins are the same type (and thus size!)
This - two turns - is the general vibe in the answers, though this is one of the few questions with WRONG answers!?!?! Dear Calendees, you seem to have been too busy during December!
Team Marian writes (I've edited group member status, we will not share who came up with the wrong answer)
Spontaneously our beloved group member said 360 degree, but of course he was wrong. So if one think like this that the center of the second coin has to travel 2r instead of 1r it absolutely makes sense that the second coin has to be turned twice = 720° to end up at the same spot position.
The Fermis write:
The coin does two entire turns, so 4pi in radiants. You can see it in the following way: If you have your two coins touching and you fix their centers, you can rotate them like gears. If you rotate until you end up with the same initial situation, each coin will have done a full turn (clockwise for the first, counterclockwise for the second). Now if you do the very same movement, but consider it in the framework of one of the coins, you get exactly the situation described in the riddle. Therefore the moving coin does 1+1 = 2 turns.
This you can also apply to a situation where the moving coin has a radius twice as small as the fixed coin: in the gear point of view, the small coin turns twice, the large coin turns once in the other direction, therefore in the riddle point of view, the small coin turns three times.
This is also why you have a difference between the solar day and the sidereal day :)
Dietmar provides this illustration:
Ed simply tested it out
Max (Team Max??) writes:
The bottom right circle (center point at A) depicts the coin which is stationary and the bottom left circle (center point at B) depicts the original position of the "rolling" coin. To get insight into the full rotation angle, we will track the location of point B, while the coin is rolling around the stationary one. A third circle around a center point C is showing the coin after it has been rolling about a certain arc-length a along the edge of the stationary coin and they now touch at a point E. Note that all circles have the same radius r.
We are interested to find the angle α (blue) which denotes the total rotation the rolling coin has to perform to get there. This angle can be found by tracking the new location of the original touching point D, which we call F. This is measured in relation to an absolute coordinate reference given by a line through C parallel to the line AB. Since the coins can't "slip", the arc-length traveled along the stationary coin is also the arc-length traveled along the rolling coin, so b=a.
Since the radii of the coins are also identical, we find that a=rβ=b=rγ and therefore β=γ. (Note that if the two coins have different radii, this would be the step were that matters). The interpretation of the formula is, that the rotation due to the "rolling" in its own frame of reference is exactly identical to the rotation angle about the stationary coin.
But there is a superimposed motion due the changing location of the coin. We therefore have to find the angle δ between the line parallel to AB and the connection between A and C. As these are intersections of parallel lines, we directly find δ=β, since they are so-called alternate angles.
In total we find α=γ+δ=2β.
So the overall solution is, that the rolling coin turns twice as far as its rotation angle compared to stationary coin.
Or for full rotation this means, that the coin has to turn 720°.
and illustrates with this illustration:
13th of December
Uh, probability!
The consensus regarding this question was rather clear, but I DID have wrong answers!
Julia got it right, doing the combinatorics:
There are six options how to select two out of four people (1,2; 1,3; 1,4; 2,4; 2,4; 3,4 with 1,2=Dr 3,4=Postdoc ) and four out of those are a mixture of Dr. and Postdoc (1,3; 2,3; 1,4; 2,4). This leads to a probability of 4/6 = 2/3 that 1 Dr and 1 Postdoc are selected.
Team Marian asks a question which I will (very late!) forward to you all! Marian, maybe by next year they will have answered this ;)
Since we have Secret Santa coming up, we were wondering if it makes sense to include more people to reduce the likelihood of someone drawing themselves?! Should we include all of the people in the institute or should we do an exclusive ColdMolecule Secret Santa if we want the probability that someone drawing themselves stays small?
Max has done a counting-tree
One of you points out that it is clear why the professor was too slow in bringing up this point! He is working on the calendar puzzles during his meetings!
14th of December:
What a touching story, no? You thought so too, right? Because we got A LOT of answers for this one.
Most of them with a lot of 💖💖💖 such as this one from Ed
| Anna | Tobi | The Missed Friend <3 | Timm | Sarah |
Andrey nicely explains his line of thoughts:
The office we look for (hereafter office X) is in the middle of the raw of offices, which goes as Anna (A), Tobi (To), X, Timm (Ti), Sarah (S).
Here is the proof.
From the first and third statements, we find that Timm's office is in between X and Sarah's, which gives us the part X-Ti-S of the office arrangement.
From the second and third statements, we infer that neither A nor To go after S, so there is nothing there.
Now, from the first statement, it follows that To is next to X, giving To-X-Ti-S.
Finally, the second statement demands that we put A next to To, so A-To-X-Ti-S is the answer.
Paul draws his solution
The PhD student was really happy for all the help 💖
He actually managed to solve the problem himself and sent me a picture of having found the right office and a good friend!
15th of December:
Sneaky one! During the correction phase there was some internal discussion in The Calendar Team about what counts as a correct answer, because obviously, there will be times where the bachelor student and the PhD student has their left foot on the pavement at the same time. However, do they put this foot down at the same time?
Alexander is pretty clear:
They will never put their left foot down at the same point of time! After 8 steps of the PhD, the situation is exactly the same as in the beginning and in between they were not synchronised with their left feet.
But it is december, and The Calendar Team is generous! Correct solutions for everyone with a good answer 🤶💖 And a good answer is the observation that the cycle is repetitive, and they are back at the same point again where they started.
The fermis answer with a nice drawing
We think that the two students will never set their left foot at the same time. We made an experiment to see this by letting them walk on the beach and taking a photo of their steps from the top. The result is the illustration attached. It looks like the long-legged student puts the left foot on the groung exactly between two steps of the other student.
Max solves answers with a table:
They never put their left foot down at the same time. Here is a graphical depiction of the situation. One can see, that the cycle repeats after the PhD student made 4 and the Bachelor-Student made 6 steps. In the meantime there was one situation, when both put their foot down at the same time, but it's still the right foot for the PhD.
And the Fermis say
We think that the two students will never set their left foot at the same time. We made an experiment to see this by letting them walk on the beach and taking a photo of their steps from the top. The result is the illustration attached. It looks like the long-legged student puts the left foot on the groung exactly between two steps of the other student.
16th of December:
"To my opinion a torus does not help. No new routes will open up that could not be reached by fiddling a wire at the border of a regular board. The same holds true for Möbius. At least that's what my Möbius board tells me :-)"
Alexander solves the torus but concludes that the band is not possible... - read further to see if we agree!
Max takes us through his thought process
Even if I can connect the left and the right side, I can't do it. I get stuck here, where the left and the green wires would need to cross...
But note, that this is not a torus, but a cylinder. It changes if I really have a torus and the top and the bottom are connected. Then we can do this magic:
On a Möbius strip the solution is also a bit ambigous I fear. If I have a Möbius-Strip, the green and red wires cross on the backside of the PCB, but I wouldn't count this. As soon as I get the full way around the Möbius Strip the red and green wires are in the same orientation, so it is not possible to do it. If it is allowed to contact the elements from the back (which would make sense, since we lay wires there anyway, front and back loose their meaning on a Möbius Strip), then we can do this:
Team Marian solves the question like this:
I was super impressed by the clarity of Julias illustrations!
Jens finally revealed the story about the black hole from last year!
The black hole has dreamed of being 'more normal' all along... The green dashed line goes through the hole of the torus and 'around the back'.
As mentioned last year, the problem is the three-utilities problem ( https://en.wikipedia.org/wiki/Three_utilities_problem ). This year, we have a torus with Euler characteristic x = 0 and the bonus question asks about the Möbius strip, which fortunately also has x = 0. That may give us some hope that the additional freedom gained with these shapes also allows a solution for the Möbius strip: indeed there is. One can avoid a crossing by letting one pair of cables run along the twisted side of the Möbius strip such that the two cables 'circle' nicely around each other. A good drawing can be found on the page of the three utilities problem (above)
The Fermis write:
The riddle of yesterday was very nice! In fact, last year one of us (R.) got very frustrated not to find any solution to this problem, and even more to learn that there were none. He got on the Internet to find an explanation, and got entangled in the world of planar graphs. There was a nice explanation given by Euler's formula, but then of course crazy people mentioned the possiblility of drawing things on different surfaces.
And that's where the Euler characteristics enters.
It's a number that describes a surface
Or a solid or a manifold
In any high dimension
It's a topological invariant
You can count holes with it
That's the Betti number aspect of it <3 <3
Algebraic properties are plenty
The product one is the best
It makes booooom in your mind
With lots of sparkles
You can do homotopy, it's there
Homology, it's there
Topology, it's there
Cohomology still there
You can't escape it
Euler characteristics da best
Hem. Back to the riddle. The Euler characteristics of a plane is 2 and this makes the riddle of last year impossible. Now if you lower this to, say 0, then it is not impossible anymore (now we need to see if it becomes possible). And the characteristics of a torus and of a Möbius strip are exactly 0 :)
For the torus it's easy, see the drawing attached (the three points on the bottom are each linked to the three points at the top)
For the Möbius strip things got complicated. Drawing it was hard and didn't seem to work. So we started building some, and it didn't work either (see photo 'Fail'). And then we realized that a correct Möbius strip is not the one you build with normal paper, but the one you build with transparent material. And there it works! See the 2 photos 'Success'.
So we learned that a Möbius strip you build with regular paper is equivalent to a normal ribbon (and has Euler characteristics 2), and the true mathematical one has to be transparent (and Euler characteristics 0), so ALWAYS build Möbius strips with transparent material, like this:
https://www.youtube.com/watch?v=4mdEsouIXGM
here is someone who understood Möbius strips (watch all her videos it's great!)
10 minutes later a follow-up email arrived with this:
Oh and I just saw this one:
https://www.youtube.com/watch?v=CruQylWSfoU
She also masters tori, and brings the riddle to the next level: you can add a fourth network to the problem! [Add brain explosion sounds]
The Fermis also attach how to not do the band, and then how to do it right:
Finally, Paul writes a nice, correct answer, and then he adds:
Again I couldn't resist to add some animations (see attachment). I'm sure, the workshop would be happy to construct one of these boards, after just finishing splitting all these shapes ;)
Thank you, Paul!!
And now, the updated solution!
Paul provides even more insights!
When reading through the solutions there was another figure which suddenly came into my mind. It is basically the möbius-strip in its final form ... (drum noise) ... the klein bottle pcb-board.
Only disadvantage: The workshop has literally to rip reality apart to construct such a thing.
Don't worry, Paul, we don't ask the workshop to make these, if they are really important we outsource them to PCBPool or something, and they can handle this kind of jobs easily!
17th of December:
Dietmar simply writes:
Your working with the Rubidium D_2-line, and you enjoy it :-)
Oh Dietmar, you know me! And the theoreticians finally made the fits to my line works, thanks to the clever postdoc!
Max shows us how to calculate this value
From text we find that "their" result y is different from the PostDocs Result x so that
y-1000=1000-x
already including the information that y>1000 and x<1000.
Then the Postdoc tells us how to arrive at y starting from x:
y=5/3*x-80
Putting that into one equation:
5/3*x-80-1000=1000-x
===> 8/3*x=2080
===> x=780
So the correct result is probably 780.
This number is obviously very special, since it is not only a triangular number but also often quoted as the rough wavelength of the Rubidium D2 ground state transition ;)
Who doesn't love Rubidium? My most beloved Postdoc once quoted Bill Phillips: "Rubidium is God's gift to Bose–Einstein condensates". We may not have a condensate, though, but we like it anyway.
The ever-amazing Ed wrote:
The PhD students were distracted playing with the bouncy balls, so they did not notice that Rubo (probably magnetically levitating) was working hard on the board to solve the Riddle! He was very excited to find that the answer was one of his lucky numbers!! And he is deeply happy that the calendar team understands its hidden beauty :)
18th of December:
Ever spilled a lot of isoprop everywhere? I want to say no, buuuuut I revealed that this is exactly what happened last year around the 15th.
Now, lets check for half-full-ness without spilling!
Ivo says: draw a line on the side of the box to where the IPA level is. Then turn the box upside down - if the IPA was more than halffull, the level should now cover the line that you marked on the side!
Dietmar follows the same line of thoughts
lay the container on one side. Since it is semi-see-through one can mark the level of the liquid on the bottom of the container. Turn it by 180 degrees.
New level higher than the mark: more than half full
New level lower than the mark: less than half full
For a cylindrical container you do the same, but roll it by 180 degrees
Neat! Max draws it like this:
But what if we don't want to lay the bottle all the way down? Maybe we do not trust the lid totally. Sure, it should be tight, especially since it is a volatile and highly flamable liquid, but still.
Lukas suggests to
1) Choose a selection of your favorite 1/2" post and stack them up until it equals the height of the container. Now make sure that you can split the stack into pieces with equal length. This is your ruler to check if it is fuller than half.
He also suggest something something that actually leads to isopropanol everywhere...
2) Take some straight object and make a cross along the diagonals of one of the sides. At the crossing point, punch a hole and see if you get some isoprop out. But please but a beaker below to not waste any isoprop. ;)
Maybe no hole-punching, but we're onto something.
Julia nicely explains it:
I would try to tilt the container until I am either able to see the liquid at the bottom move to the side (then it is less then half full) or some liquid spills out of it before seeing the bottom (then it is more than half full).
Pretty brilliant from Julia!
The Fermis point out that "This method works for both rectangular and cylindrical container, as illustrated on the attached drawing. For the rectangular section you can even choose if the two corners you check are on the same rectangular face (the level of the water should follow the diagonal of this face for a half-full container) or if these two corners are along a big diagonal of the parallelepiped (the level of the water should then follow this big diagonal when it is half-filled)."
Sebastian W illustrates it nicely, and so does the Fermis (First Sebastian, then the Fermis)
19th of December
Julia makes it completely clear what is going on:
The ball will travell in cm:
80 (down) + 40 (up) + 40 (down) + 20 (up) + 20 (down) + 10 (up) + 10 (down) + ...
When doing this an infinity times this leads to 240 cm.
The ball flies
(80+40) + (40+20) + (20+10)+...=
120+60+30+...=
120*(1+1/2+1/4..)
This is the well known geometric series which converges towards 1/(1-1/2)=2
So, the ball flies 120*2=240 cm.
If the ball would be a tortoise, Achilles would never catch up!!!
Wait, what? Jens points out that this question is related to Zeno's paradox, or Achilles and the tortoise, and recommends this video: https://www.youtube.com/watch?v=Jwtn5_d2YCs
20th of December:
The art of permuting!
Let's see some solutions! Alexander submits quite a lot of solutions!
Sebastian shares a single solution.
And this is absolutely not cheating! This is efficiency!
But actually, there are 160 combinations satisfying this criteria as many of you point out!
Paul writes: There are 160 in total, but for each one 2^4=16 are basically the same, as they can be created by the following operations - Just listing these equivalent solutions will leave only ten (listed as shown in the figure on the right)
Permutation Sum
[ 1 3 9 7 2 4 6 8 5 10] 19
[ 1 4 6 10 2 5 3 9 8 7] 20
[ 1 4 7 9 3 6 2 8 10 5] 20
[ 1 4 9 5 3 8 2 10 6 7] 18
[ 1 6 7 9 2 3 8 4 10 5] 22
[ 1 9 7 6 3 5 8 4 10 2] 22
[ 2 3 7 9 4 5 1 10 8 6] 19
[ 2 3 9 7 5 6 1 8 10 4] 19
[ 2 4 9 6 5 7 1 8 10 3] 19
[ 3 2 10 6 4 7 1 9 8 5] 18
The ever-complete Jens sent me all the answers, and wonders: Not sure how long this keeps the professor busy. Doesn't he have to write a proposal or something?
Max also sent all the solutions (in a .txt file though) produced with a little snip of code. He writes: For the Bonus question my script found 1908 solutions (again including all rotations etc), so I will just print one here ;)
I will not show the solution from Max, instead the one from the Fermis!
1) because it is pretty, and 2) because it comes with some quite interesting text! Look after the text
For the bonus question we find many solutions, one of which is illustrated on the right. If you want to construct any solution, you just need to color the intersections as on the illustration and use a dice:
- For color red roll the dice and fill one of the two red intersections with the number you rolled. On the other red intersection write the number which is on the lower side of the dice (both should have a sum of 7!)
- For color blue roll the dice again, until you find a number you haven't written yet. Fill one of the blue intersections and fill the other one with the number on the lower side of the dice.
- For color green fill the two last intersections with the two remaining numbers (you can roll the dice a last time to decide which goes where).
That's our deterministic way of randomly solving this problem!
By the way, we found out that this drawing is very close to a cube: it is exactly the stereographic projection of an octahedron, which is the dual of the dice (see Bonus illustration, taken from https://demonstrations.wolfram.com/StereographicProjectionOfPlatonicSolids/). So actually you can construct any solution by throwing your dice only once: it has to land on one of its corner but once you manage to do that (we're sure wou can find a way), you can just project stereographically the centers of the faces of your dice with their number and link these numbers when the corresponding faces have an edge in common and you get a solution to the riddle :D
By the way, if you like stereographic projections (and crazy maths animations), we can recommend this movie:
https://www.youtube.com/playlist?list=PLHdeFhyhsVd_CJDxD9Y_Tj1bwo76MWbXI
It's a bit slow, but it's nice. And we think it will get much more viewers in Germany after legalization of some substances.
Yes, the bonus question had to be corrected, and with this came a little additional question. Is 6 the maximal number of intersections one can make between three circles, I wondered.
Luckily, Jens was ready to answer this question too:
Yes, one can avoid triple-intersections and yield for n circles 2*(n,2) intersections since each circle can intersect every other one at most 2 times [by (n,2) I mean the binomial coefficient]. For n=2: there are max. 6 intersections.
21st of December
I personally spent quite some time on this puzzle. It seems like you were all much faster! A lot of code-snips were emailed to The Calendar Email, but here I'll just share the ones with explanations
Dietmar first takes a look at the highest number possible and writes:
9876543210
According to Ed's coffee ticks rule we have
9+7+5+3+1=25 (the positives)
8+6+4+2+0=20 (the negatives)
However we change the digits, the positives can not be equal to the negatives, since 45 is odd. So, let's apply a different strategy, make the difference 11. We thus have to increase the positives by 3 (decrease the negatives by 3)
This can be done by arranging the digits like this:
9876524130
(9+7+5+4+3=28)
(8+6+2+1+0=17)
I have no proof that this is really the largest number one can make, but I'm sure it's the "Wanted" one :-)
To solve the riddle it was helpful for me two note the following points:
-The number of odd digits from 0 to 9 is odd, thus also the result of the alternating sum is always odd and cannot be zero.
-We can build the highest number from left to right, greedy each time taking the largest possible digit such that it is still possible to arrange the remaining ones in a way that the complete number is dividable by eleven.
From this point I just started doing trial and error:
987654...->This was to greedy! However we distribute the last numbers the highest sum we can attain is to small: 9+7+5+3+2-(8+6+4+1+0)=7. So we have to go one step bag.
98765... -> This works just perfectly: 9+7+5+4+3-(8+6+2+1+0)=11, so taking the largest numbers alternating both "bins" will give us the highest possible number:
9876524130 Hooray!
As a follow up a short explanation why the trick with the alternating sum works:
Let a(x) be the alternating sum as explained in the riddle, then we can prove the slightly stronger claim, a(x)=x mod 11, by induction, starting from a(0)=0 mod 11:
a(x)=x mod 11 -> a(x+1)=x+1 mod 11
Let now k be the number of carries (or number of consecutive nines from the right).
-If k is even, ie. we have to replace an even number of nines by zeros. These replacement exactly cancel out and and we only increase one digit at an even decimal place by one, thus a(x+1) = a(x)+1.
-If k is odd the most left one doesn't cancel out and so we need to subtract ten for replacing a nine by zero at an even decimal place and additionally subtract one for adding a one to a decimal place at a odd position. We get a(x+1)=a(x)-9-1=a(x)-10=a(x)+1 mod 11 which happens to be the correct congruence-class.
And Jens is on point:
the largest number is 9876524130; it can be divided by 11: 9876524130/11=897865830.
A brute force search finds this number. Seemed quite sad to do it that way given the nice fact you gave us on the divisibility. One can come up with a perhaps slightly better method using your tip. One looks at the partitions of 11 of length 5, since there are 5 pairs (x0-x1) + (x2-x3) +... and one wants to minimize the difference in the most significant digits (in the order x0, x1, ...). The partitions are
{1, 1, 1, 1, 7}, {1, 1, 1, 2, 6}, {1, 1, 1, 3, 5}, {1, 1, 2, 2, 5},
{1, 1, 1, 4, 4}, {1, 1, 2, 3, 4}, {1, 2, 2, 2, 4}, {1, 1, 3, 3, 3},
{1, 2, 2, 3, 3}, {2, 2, 2, 2, 3}
Now one can play around with the combinations of these partitions giving a pattern <big,big,big..., .... small+1,X,small> where we start with the smallest number and try to increase every second digit by the smallest amount while trying to have the largest digits in the beginning. I.e. we want something starting like 9876... (how far can we go down like that?) and something ending like ...a2b1c0 where we look for ...,a,b,c. The partition {1,1,3,3,3} matches this pattern best and gives the number 9876524130.
22nd of December:
Here we have entered the critical Christmas days and vacation time kicks in for the Calendar! Not so many answers, but high quality answers!
Julia and Sebastian W draws it:
Alexander formulates it shortly (but takes it to n boxes!)
2 Boxes: ¾
3 Boxes: 11/12
4 Boxes: 25/24
n Boxes: (1 + ½ + 1/3 + … + 1/n )/2
Alex doesn't go to n, but thinks outside (or on top of) the box
for 2 boxes, the outermost can have an overhang of 3/4.
for 3 boxes, the outermost can have an overhang of 11/12, if you are going for the stair configuration.
for 4 boxes, the outermost can have an overhang of 25/24, if you are going for the stair configuration.
But there is a way to arrange them in a manner to get more overhang: If you place two boxes beside of each other on one box, you can push this to the edge until the one box on top is totally over the edge, so for 3 boxes we get an overhang of 1.
If you then add one box exactly on top of e.g. the upper left box, the right box can achieve an overhang von even 5/4 over the edge.
Paul is also onto this point:
If we build a tower with four boxes we would end up with 25/24. The following arrangement beats this, but I have no idea if this is optimal.
We move the two intermediate boxes symmetrically out as far as possible that the top one is still able to hold the weight without being lifted. If we are exactly at this point then we a weight of 1/2 from above on the inner edge and 1 in the center.
We can apply this argument symmetrically to the most left box and deduce that this is the last point where it can pass the weight to the bottom box without falling of.
Summing the distances of the bottom and side box leads to 1/2+2/3=7/6=28/24 which is slightly more than with the tower.
Finally, depending how similar the ground shape is to a square we can gain an additional factor of up to sqrt(2) for both arrangements by rotating everything by 45°. Here a demonstration with some CDs from my parents :-)
Jens provides the correct answers, but he also provides some reading!
this problem is known as the book stacking problem. As mentioned in my answer for the 19th of December (remember, the bouncy ball and the infinite sum), this problem has the interesting solution that infinitely many books can be infinitely stacked outwards away from the table edge.
Anyways, I recommend the following resources:
https://mathworld.wolfram.com/BookStackingProblem.html
https://www.youtube.com/watch?v=Jwtn5_d2YCs (we already had this on the 19th, but we post it again)
https://www.maplesoft.com/support/help/maple/view.aspx?path=MathApps%2FTheBookStackingProblem
and particularly the last resource is nice to see how the general problem for n books arises :). The distances between the books in book length units are the summands of the harmonic series for the maximum book overhang.
The nice answer is:
23rd of December:
Bouncy balls! There has been a lot of bouncing since the 6th of December, and it doesn't seem to end anytime soon.
We are also deep in the Christmas hibernation, and this question received only five answers. In general we had two camps for this question.
Camp 'Let's go directly to 3D' and camp '2D rocks!'
Jens belongs in camp 3D and writes:
This question reminded me a bit of the four color theorem
https://www.youtube.com/watch?v=NgbK43jB4rQ
https://en.wikipedia.org/wiki/Four_color_theorem
The rules on the patterns allowed for the arrangement of the balls in this task was not completely clear to me in this task. I was wondering if some degree of tight-packing is required or not (like in the picture shown below), if there is a given number of balls or whether the pattern has to be periodic but was thinking these limitations on the types of answers are not wanted. So consider this pattern a guess on whether or not it will fall within the rules: A 3D tightest-packing configuration (at least the balls can fulfill the requirement of 'touching' in this configuration).
Team 2D doesn't care about tightest bouncy-ball packing. Paul (who of course also included a solution in 3D) found this solution:
24th of December:
I'm super amazed that I got five (5!) answers to this question! In the Christmas vacation. You rock!
It is maybe a bit much to show the table multiple times, so let's simply see the correct solution as shown by Alexander
And the table as shown by Max.
The Calendar team had a great Christmas break, btw, and found it difficult to select the solutions to share. Now, however, this selection has finally been uploaded, and The Calendar Team can proceed to an even more important task:
Selection of the winner! This will be posted as a news item, and on the calendar-page. Stay tuned 💖