Well, I think it’s about time to close this round and give you the solutions.
Puzzle 16 Solution (provided by me)
The volume of a body is always proportional to the third power of a length on this body, if similar bodies are viewed. Our body is a cone and we use its height as the length on the body. If we use a bad approach, we will however get to an equation of third grade. While these equations are theoretically solvable, the formulas needed are not very well known. Following is a solution that only uses simple school math. The complete cone has the height n. We separate this cone just like the artist, so the the top part has the remaining height m. The volume of this top part therefore has the volume m^3. The bottom part has the volume n^3-m^3. If we put the smaller cone inside the ring, we have to further subtract the volume of the part of the smaller cone that is inside the lower third of the big cone. This volume is m^3-(2m-n)^3. Therefore the volume of the ring is n^3-2m^3+(2m-n)^3. You probably noticed that we don’t work with the actual volumes of the bodies themselves. For those you usually also need a factor which is different depending on what type of body you calculate the volume of. In the case of a sphere, this factor would be 4/3 π . Since we are however going to compare the volumes of similar bodies to each other, or rather, equate those, these factors are irrelevant and would just be subtracted on both sides of the equation. Equating those two volumes gives us m^3= n^3-2m^3+(2m-n)^3. This can be transformed to 5m^3-12m^2n+6mn^2=0.
Nothing speaks against putting n to 1 and getting a value for m from there. After all, it is said that the original cone has the height of 1 meter. Then we get 5m^3-12m^2+6m=0. Dividing this by m gets us 5m^2-12m+6=0. Then dividing it by 5 gives us m^2-(12/5)m+6/5=0
As one of those is greater than 1 and is therefore geometrically impossible, 
is the only possible solution, and the cone must be separated at a height of 
which is roughly 29 cm.
This Puzzle was solved by no one.
Puzzle 17 Solution (854 can't be reached provided by me, 854 can be reached provided by Karifean)
854 can’t be reached:
A first consideration: Of course the berry witch can put weights in both sides of the balance, which means that she can not only weigh sums of the weights, but also differences. Would she for example want to weigh 20 gram, she would put the 25 gram weight in one side, and the 1 and 4 gram weights in the other, and fill that side with berries until the balance is in balance. If we were to calculate all possible sums and differences of the first five weights, we would see that some weights can’t be measured yet:
18, 42, 43, 54, 55, 59, 63, 67, 68, 72.
All other weights between 1 and 74 can be measured. Our next weight should now have the maximum weight of 2*74, since we get to 74 with the first five weights and would reach the next number by subtracting these weights from the new weight. With this view, the most fitting square number would be 144. However, 18 would still be impossible to depict with that, and we want a maximal row of depictable numbers. Taking that into consideration, 64 would be the next weight. This one does not only allow us to weigh 18 gram, but also all other numbers mentioned above. However, now we get a new chain of impossible to depict numbers, namely the above numbers raised by 64, or 82…
This is that way because we still can’t form a matching summand to add to 64 with the other weights. If we continue in this way, the other two weights are 196 and 576. To get to 773, we definitely need the two heaviest weights, which brings us to 772, missing 1 gram, so that weight must also be used on the same side, bringing us to the solution that 3 weights were used. Now, this is not a mathematical proof, only a possible thought process of how to get to the solution.
854 can be reached:
I brute-forced this one. Wrote a program that creates all possible triples for the remaining weights, checked if they could reach all grams from 1 through 854, and then, once it found one, checked how 773 can be weighted with these weights.
The weights I found for the ‘heavier’ weights are 144, 196 and 625. They can weigh 773 grams by putting 1 and 4 on the same side as the thing you’re weighing and 625, 144 and 9 on the opposite side. So in conclusion she used 5 of her weights.
This Puzzle was solved by @Karifean.
Puzzle 18 Solution (provided by me)
You can ask the following question: "Is it correct that you are either a liar of that the choir meets behind the right door?"
Let’s look at a table that tells us what he says in which case:
We see, that, independent of him being a liar or not, he answers Yes if the choir meets behind the right door, and says No if they meet behind the left door.
This Puzzle was solved by @Karifean.
Puzzle 19 Solution (provided by me)
A square number can also be called a square, and here you see the square of the size 1 depicted 6 times as a roman numeral. Furthermore, 1^1 is also depicted 5 times, which equals 1 and is therefore a square of the same size, which already brings us to 11 squares. Then there’s 1^1^1 4 times, and so on and so forth, bringing us to a total of 21 squares of the same size.
This Puzzle was solved by no one.
Puzzle 20 Solution (provided by CounterShadowform)
715 has only three prime factors—5, 11, and 13—showing that the housewife has exactly three children and that their ages as reported earlier must have been 13, 11, and 5. Since birthdays can occur throughout the year and it is unknown when this conversation takes place, the children’s ages might not have increased by the same amount, but the maximum difference in increases (for integer ages) can’t be more than one year. We already know that the children’s current ages multiply to 1260, so the only possibility for the set of their current ages is 15, 12, and 7. If Monika’s birthday is tomorrow (and assuming that time of birth is not being factored in—otherwise someone could turn a new age later on the day of the conversation), then her age will be the next to update. Since the difference between two children in number of birthdays since the earlier conversation can’t be more than one, and the middle child has had one fewer birthday since that conversation took place, the middle child must be Monika. As the middle child is currently 12 years old, this means that Monika is turning 13.
This Puzzle was solved by @Blackrune and @CounterShadowform.
Score
Current round:
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@Karifean with 11 medals
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@Blackrune with 3 medals
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@CounterShadowform with 3 medals
Overall:
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@Karifean with 35 medals
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@Blackrune with 27 medals
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@CounterShadowform with 17 medals
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@pictoshark with 6 medals
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@Restkastel with 5 medals
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@Wonderlander with 3 medals
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@King_Titanite_XV with 2 medals
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@Exalt with 1 medal
For now this little competition will go on a break. I urge you all to use the time to read Higurashi.
