This round was not very productive. In fact, this time one puzzle wasn’t solved by you guys. While it was tried by one person, they didn’t get the correct solution.
Puzzle 11 Solution (provided by Karifean)
Let us examine the very top of the triangle and attempt to find a way to connect all dots in this corner in a line that A. conforms to all rules described in the problem and B. does not begin or end in this cutout.
As should be plainly obvious, the green lines are mandatory. There is no way around them if we want to include the point in the far corner and still want both ends of the line make it out of this corner. As a direct consequence, the red lines cannot be in the path, as they violate the rule about the line changing directions at every point. As a result, both ends of the line are forced to take the blue lines to continue their path. But therein lies the problem - both blue lines end up at the same dot, ending the line completely.
From this we can derive that it’s impossible to create such a line in a corner without having its beginning or end there. Connecting all dots in a corner is relatively simple if we do have a beginning or end there, as for example one of the blue lines can simply be left out. However, unfortunately, we have three such corners in our triangle, and we only have 1 beginning and 1 end. Meaning that in order to create a line that connects all dots, we absolutely need to fill out a corner without a beginning or end. But, as we’ve just established, this is impossible without closing off the path immediately. Thus, there simply cannot be a line that connects all dots without violating any rules.
This Puzzle was solved by @Karifean, @Blackrune, @pictoshark, and @Wonderlander.
Puzzle 12 Solution (provided by Blackrune)
If all buns on the scale have proper weight, the expected offset is between 1 and 5 grams.
Once there’s one wrong bun, the offset would be between 2 and 6. Since that overlaps the expected margin of error for perfect buns, we need to make sure it’s at the very least shifted enough to not overlap.
We need 5 wrong buns for the scale to show between 6-10grams less than expected.
However, there are also five perfect buns in the apprentice’s basket, so in order to guarantee there are at least five wrong buns, we need to take at least 10 buns from each basket. With that, the offset from the wrong basket should show between 6 and 15 grams less.
(Why 15? Because there is the possibility that we pick more than 5 wrong buns from the apprentice’s basket, which means we can miss up to 10 grams from that.)
But of course just taking 10 from each and putting it all together wouldn’t really help much.
We’d just be 6-15 grams short with no information gained.
We need to take different amount of buns from each basket to make any sort of progress here.
Since we have to again make sure there ar eno overlaps, we need to shift the range by 10 each time.
So we take
10 from the first basket (if wrong buns are present in this one, the scale will show between 6-15 grams less.)
20 from the second basket (if wrong buns are present, scale will show between 16-25 grams less.)
30 from the third (scale will show between 26-35 grams less)
40 from the fourth (36-45 grams less)
etc.
We just put them all together, and can then decisively tell from the amount of missing grams which basket contained the wrong buns, all in one go.
Addendum: You could even take no buns from the tenth basket, in that case if the wrong buns are present there, the scale would show 1-5 grams less.
This Puzzle was solved by @Blackrune.
Puzzle 13 Solution (provided by Blackrune)
At first glance it might looke like it’s hard to predict how many needles the tree would lose, seeing how it can randomly lose one more or one less.
Of course nothing is random here, there is only one sequence of losing needles that enables the tree to lose an exact total of 100000 needles.
I’ve determined this through a tedious process of trial&error. Each day except the first one has 3 ‘settings’, +1 needle, -1 needle, or neutral.
It’s also obvious rather quickly that the whole thing takes around 12 days before you’re either past 100000 needles or would inevitably pass it on the next day.
So for each day, I first tried -1 needle, then went through the whole process always adding +1 needle on subsequent days until the 12th. If I didn’t manage to pass 100000 needles, that day can not be a day with the -1 needle setting. If I do, it could be.
Vice-versa for +1 needle, and when in doubt, I checked neutral with the appropiate ‘speeding’ or ‘braking’ depending on what I’d already established.
Anyway, long story short, the correct sequence is this:
1st: 1
2nd: (1*2) -1 = 1
3rd: ((1+1)*2) -1 = 3
4th: ((1+1+3)*2) = 10
5th: ((1+1+3+10)*2) +1 = 31
6th: ((1+1+3+10+31)*2) -1 = 91
7th: ((1+1+3+10+31+91)*2) +1 = 275
8th: ((1+1+3+10+31+91+275)*2) -1 = 823
9th: ((1+1+3+10+31+91+275+823)*2) -1 = 2469
10th: ((1+1+3+10+31+91+275+823+2469)*2) -1 = 7407
11th: ((1+1+3+10+31+91+275+823+2469+7407)*2) = 22222
12th: ((1+1+3+10+31+91+275+823+2469+7407+22222)*2) +1 = 66667
1+1+3+10+31+91+275+823+2469+7407+22222+66667 = 100000 <- checks out
Therefore the tree lost 823 needles on the eight day!
Addendum: One could save the trouble of the trial and error process by starting at the end and dividing that number by 3, the rest of that division telling if one more or one less needle has fallen. If the rest is one, then one more needle has fallen, if the rest is two, then one less needle has fallen. For example, at the end of the 12th day, 100000 needles have fallen overall. That divided by 3 is 33333 rest 1, therefore 333332 + 1 = 66667 needles have fallen on that day alone. On the 11th day, 33333 have fallen overall, divided by 3 is 11111, therefore on the 11th day, 111112 = 22222 needles have fallen. On the 10th day, 11111 needles have fallen overall, divided by 3 is 3703 rest 2, however since we can’t add 2, we have to instead say that on that day 3704*2 - 1 = 7407 needles have fallen. Etc. etc.
This Puzzle was solved by @Blackrune.
Puzzle 14 Solution (provided by Blackrune)
Curse you for making me go through so many possible approaches to this.
Anyway, some things were obvious. For example, the number always doubled as long as they were still in the single digits.
This suggested that whatever method is used is somehow linked to the number of digits, because the jump from 16 to 122 was suddenly much more massive.
Eventually I noticed that flipping the digits and multiplying by two gave me 122.
Anyway, that still wasn’t quite what was happening here.
Switching the digits of the subsequent numbers around eventually gave me the right idea.
The title of the puzzle was also a clue.
Basically, you need to add two numbers together.
You get one of these by going one one step backwards and the other one by going one step forward from the first digit and then read the number (almost) normally from there. Here’s the twist: You kind of have to see the digits as being in a circle, so going ‘back’ from the first digit means you start at the final digit and then go forward from there. (meaning the original first digit is the second digit there)
For the other number, you just read from the second digit onwards and then add the original first digit at the end.
For single digits any step just makes you end up at the same digit again, giving you the original number both times. Which is why it always just doubled.
1
2
4
8
16
In the case of 16, both forwards and backwards essentially amounts to starting at 6, making both numbers 61.
61+61 = 122
Then we get:
221+212=433
334+343=677
776+767=1543
5431+3154=8585
5858+5858=11716
17161+61171=78332
83327+27833=111160
I rest my case.
This Puzzle was solved by @Blackrune.
Puzzle 15 Solution (provided by me)
The coast is at most 100 miles away from the yacht. The coast is straight over hundreds of miles, so it can be thought of as a tangent to a circle around the yacht with a radius of 100 miles. We don’t know which of the infinite tangents of this circle is the coast. So we have to traverse 100 miles in any direction from the get go before we can make any observations if we have already reached the coast. If we would then traverse the circle, we would have to travel a distance of 100 miles * (1 + 2π). This would be appr. 728 miles, which is more than we have fuel for. While the chance is high to reach the coast before using up all of the fuel, it’s not 100%, and this is a life or death situation, so a better solution would be in Professor Johnson’s best interest. And we can indeed cut a few corners, as we don’t necessarily need to visit every point of the circle, we only need to cross every tangent of the circle. If at the beginning we don’t go only 100 miles in one direction, but a bit further, then we already cross a whole bundle of tangents coming from the left and right of the circle.
If we travel 100 miles to point A and then continue to point B and then continue on the outmost tangent to point C, then while we traveled a good bit longer than if we would have if we had gone to C from A on the circle, we also don’t need to travel the part of the circle between C’ and A, as we already visited all tangents between those points. Furthermore, once we reach point D, we can save even more distance by going straight to the tangent in point C’ from D, as that way we cut all tangents between D and C’.
But is this strategy short enough to not exceed 660 miles? Let us imagine the circle encased in a square. Then this square has the length 2 if we see 100 miles, the radius of the circle, as one length unit. If we choose B so that B is a corner point of this square, then the distance between the starting position and B is √2 and the distance between B and C is 1. If we continue from C on the circle to D, which is half of the circle’s circumference, then this distance is π and the distance between D and the tangent in C’ is 1 again. The lengths of these parts added up and multiplied by 100 equals ca. 655,58 miles.
This Puzzle was solved by no one.
Score
Current round:
- @Blackrune with 14 medals
- @Karifean with 3 medals
- @pictoshark with 3 medals
- @Wonderlander with 3 medals
Overall:
- @Blackrune with 25 medals
- @Karifean with 24 medals
- @CounterShadowform with 14 medals
- @pictoshark with 6 medals
- @Restkastel with 5 medals
- @Wonderlander with 3 medals
- @King_Titanite_XV with 2 medals
- @Exalt with 1 medal
Here’s the next batch of puzzles:
Puzzle 16
Difficulty: 3
Creative Flowerpot
An artist designed a new artsy flowerpot for a flowershop. He started with a 1 meter high, hollow cone made out of thin metal. Then he put it on a table and sawed the lower third of this cone off. He took this lower third and put it up-side-down, so that the wider opening was pointing upwards, on top of the rest of the cone. At the bottom end, where both parts now had the same diameter, he soldered the two parts together. After that, he soldered a metal stick on the tip of the cone, and on top of that he put a metal disk in the form of a circle, so that disk and stick were perpendicular to each other. Now, one had at least two possibilities to use this object. One could put it on the metal disk and use it as a vase, in this form it kind of looks like a chalice, or one could put it the other way around and cultivate various herbs. As this object was laying around in the shop’s display window, Simon, who you might remember is a math fan, saw it and was a bit annoyed: You see, the ring around the chalice and the chalice itself don’t have the same volume. If this artist had separated the cone at a different height, these two parts would have had the same volume. At which height should he have separated the cone?
Puzzle 17
Difficulty: 5
The weighing arts of the berry witch
On the marketplace of a small town there is a witch that sells various kinds of berries. For weighing these, she uses different kinds of weights, 8 total, and a beam balance. When Mrs. Smith was buying berries there she saw that all weights are square numbers, namely 1, 4, 9, 25, 36, and 3 larger ones. Now she was wondering: “But how can you weigh with these weird weights? What if, say, I would order 773 gram blueberries?” - “Kihihihi, no problem, I can weigh every weight between 1 and 854 gram with only one weighing process!” Mrs. Smith can’t really believe that and thus proclaimed: “Okay, then I would like to have 773 gram blueberries!” The berry witch indeed weighed 773 gram with only one weighing process. Mrs. Smith thought that this is true number magic and thus accepted the witch. However, this witch’s magic is only a small cry compared to the magical feat that you can accomplish: With the given information you can say how many of her weights the witch used. So how many did she use?
Puzzle 18
Difficulty: 6
Truth or Lie
The townhall of Lyingsfort has two entrances. Behind one of them is a dragon. This dragon is called Olga and is the quick to anger secretary of the mayor. The other door is the entrance to the city’s choir. In front of those two entrances is an inscrutable person. He is the only person that knows behind which door the beast lingers and behind which door the musical event takes place. But like every inhabitant of Lyingsfort, he either always lies or always speaks the truth. Of course it is unknown if he is a notorious lier or not. And yet it is sufficient to ask him only one question that can be answered with Yes or No to know for sure behind which door the choir meets. Which question would you ask this person? This question may not contain people or things that actually don’t exist, and conditional questions are forbidden as well. So questions like “Suppose you had a collegue that always says the opposite to you, …” are not allowed.
Puzzle 19
Difficulty: 4
Squaring the tree
Out of the tree pictured above you have to form at least 20 squares that all have to have the same size. This is no error, you are reading this correctly! You may move all matches, but you are not allowed to break any of them.
Puzzle 20
Difficulty: 3
My goodness, did they grow up
Two housewifes were having a little conversation about this and that. Here is a little excerpt:
“You know, 52 years ago life was a lot harder…”
“But I wasn’t even alive 52 years ago!”
“Oh right, you’re quite a bit younger than me. By the way, how are your children?”
“Imagine this: If I multiply their ages, I get the number 1260, and all of them are already in school!”
“But didn’t you tell me earlier that the product of their ages is 715 and that all of them are in school or in kindergarten?”
“Oh yeah, but this was some time back, they obviously got older now. I also have to leave, you see, Monika’s birthday is tomorrow, and I still need to buy a present.”
“Oh really? How old does she get?”
“Come on, you should be able to reach the answer to that question yourself.”
How old does Monika get?
I hope you are still enjoying yourselves. The time limit stays the same, one week. Solutions have to be sent in by the next Saturday, 11th March, 8pm UTC.