Maria's Puzzle Book [Break]

[VyseGolbez]

Now for the first batch of puzzles:

Puzzle 01

Difficulty: 2

The Misarranged Pearl-Bracelet

Simon is a man of beauty. He often crafts several accessories himself according to his sense of beauty. That sense however is only appreciated by few, as it is the beauty of numbers and logical systems that fascinates Simon. Just a few days ago, one of his aquaintances criticized a design for a bracelet Simon made. This aquaintance suggested to swap two neighbouring pearls in that design, resulting in the one pictured above. Knowing Simon’s sense of beauty and that only two neighboured pearls were swapped, how did the original design look like?

Puzzle 02

Difficulty: 3

The Lottery

A company celebrated their 10th anniversary. Part of this celebration was a lottery where every employee was allowed to draw one lot. There were only winning lots and the prizes were one projector, wine and several towels. Mrs. Ainsworth got lucky and won the projector, Mr. Bailey won one of the towels. The probability of this occuring is exactly 10%. So how many bottles of wine were offered as prize?

Puzzle 03

Difficulty: 1

The Rainbow Triangle

The above triangle is made up of several small triangles. Some of them were already colored with four colors. Color the remaining triangles according to the following rules:

In every bigger triangle made up of four small triangles no color may show up more than once.
Only the four colors that you see above may be used.

For this puzzle a picture of your solution is sufficient, you don’t need to write a step-by-step solution.

Puzzle 04

Difficulty: 3

Anniversary in Ladder Town

Ladder Town is called that way because it looks like a ladder from the sky. Ladder Town is currently celebrating its 500th anniversary. In this town there are two main streets that are parallel to each other. These main streets are connected by a total of 13 side streets. All these streets can be traversed in both directions. The department store of Ladder Town is near the town entrance, in the first side street. The festival grounds for the anniversary are on the city plaza which is located on one of the main streets between the fifth and sixth side street. The home of Mr. Smith is located at the end of the town, in the 13th side street. Now Mr. Smith realized that he still had to buy some things so he hopped into his car and went straight to the department store. On the way back he has some more time though, so while he traverses in the direction of his home on the main streets, he may decide to drive into one of the side streets since the town is so nicely decorated because of its anniversary. He also wants to pass by the festival plaza and then drive into the next side street because there’s a friseur and he needs a new hair cut. He doesn’t change direction on this side street. After that he continues his travels like before, but he takes at least one of the following four side streets. Eventually he reaches his home.
So, how many different paths could Mr. Smith have taken on his way back with the information given above?

Puzzle 05

Difficulty: 1

Multiplying Trees

Pictured above is a nice pine tree made up of matches. It is possible to make two trees out of this one tree by moving only one match. Can you do it?

For this one, a picture of your solution is sufficient as well.

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As you can see, these first five are more on the easy side, as the highest difficulty is 3. You have time to send your PM until next Saturday (that’s the 18th February 2017) 8 pm UTC. You don’t have to send all five solutions in one PM, so you can send multiple PMs, like say you solve Puzzle 02 on monday and send a PM and then solve Puzzle 05 on wednesday and send a second PM. You are also free to edit your solution until the deadline hits, so while that is technically only one try, you can correct yourself if you happen to notice at a later time that you made an error.

[VyseGolbez]

Regarding Puzzle 04:

he may decide to drive into one of the side streets

Just to clarify: This means that he might choose at each junction to take the side street. This was a bit of a poor wording on my part, thanks @Karifean for the question that pointed this out. Thankfully, most people haven’t tried to solve this one yet.

[VyseGolbez]

So time’s up, let’s see how everyone has done through this first set. For the solutions, if you see the keyword “Addendum”, anything after that was written by me.

Puzzle 01 Solution (provided by Blackrune)

Now at first I thought this was some complicated sequence of 1s and 0s or 1s, 2s, and 3s. I also tried making it symmetrical or respecting the fact that in a bracelet the beginning and end are technically connected too. Then I got inspired. Turns out the solution was much simpler once you see it.
Numbers. Beauty.

Prime numbers.
If you switch the seventh and eight pearl, all the dark blue pearls mark the prime numbers in the sequence.
2, 3, 5, 7, 11, 13, 17, 19

This puzzle was solved by @Blackrune and @CounterShadowform. @Karifean, you didn’t even try it; you didn’t already give up, did youuuuu?

Puzzle 02 Solution (provided by CounterShadowform)

The formula for the probability of the intersection of Mrs. Ainsworth winning the projector and Mr. Bailey winning one of the towels, regardless of both order drawn and who else got what prize, is
https://d2qdztz5tk5zxi.cloudfront.net/original/2X/0/03aaa1cf319fd33174c708e6d3b49b8ba8f989ea.png
in which n is the total number of prizes, n_projectors is the number of projectors offered (which we know to be equal to 1), and n_towels is the number of towels offered. The number of wine bottles, n_wine, must be equal to n − n_projectors − n_towels.

Since there are a natural number of total prize items and 10 has only two prime factors, one of the above fractions contributing to P must equal 1/2 and the other 1/5. We know that, since n_projectors = 1, one of the two numerators must equal 1. The other, since it must account for only towels in the presence of one or more wine bottles, cannot exceed n − 1 in order to allow the wine room to exist among the non-projector prizes. In addition, the two denominators must be two consecutive positive integers, being equal to n and n − 1, and the larger of them cannot be any less than 4, since there are one projector, at least two towels, and at least one wine bottle; the larger also cannot exceed 10 or else the product would necessarily be less than 1/10. The only working solution is n = 5 and n_towels = 2, for which 1/5 * 2/4 = 2/5 * 1/4 = 1/10 = 10%. Plugging these into the formula for n_wine, n_wine = n − n_projectors − n_towels, we find n_wine to be 5 − 1 − 2 = 2; therefore, two bottles of wine were offered.

Addendum: Interestingly enough, the above formula also works if you take n = 6 and n_towels = 3, as 1/6 * 3/5 = 1/10 as well. In that case, n_wine would still be n_wine = 6 − 1 − 3 = 2 though, so while the number of employees and towels is unknown, the number of wine bottles is 2 in either case. @Blackrune is the only one who saw that second case in his solution.

This puzzle was solved by @Blackrune, @CounterShadowform, and @Karifean.

Puzzle 03 Solution (provided by Karifean)

https://d2qdztz5tk5zxi.cloudfront.net/original/2X/d/d84d60e89494be8db5f3c0abcc37e68fa86bafa2.png

You said we didn’t need to document our M.O., but I’ll do it anyways. You can solve it by following two rules: 1. if there are 3 colors already represented in a bigger (4 part) triangle, the last part is the unrepresented color. 2. if there are 2 colors already represented in a bigger (4 part) triangle, check to see if either of the two remaining colors would create a rule violation in the adjacent bigger triangle; if yes, place both colors so that no violation occurs. Do this iteratively until everything is colorized.

This puzzle was solved by @Blackrune, @CounterShadowform, @Karifean, @Restkastel, and @Exalt.

Puzzle 04 Solution (provided by CounterShadowform)

If Mr. Smith, starting from the department store, initially heads to the main road not next to the festival grounds, he must take an odd number of side streets to reach that side. If he takes the road from which the grounds are visible, on the other hand, he must take an even number of side streets to stay on that side. There are four side streets before the plaza, and the numbers of combinations can simply be added together for the two possible starting positions of this leg. Starting away from the plaza, he may take either one or three side streets out of the four, so assuming he goes that way he has (∑(n=1, 2, 4 nCr (2n-1)) possible ways to get to the plaza. Starting the other way, he may take zero, two, or all four, for a total of ∑(n=0, 2, 4 nCr 2n)) possible paths. These two values can then be added together to produce a final value for the number of paths for this leg.

Once Mr. Smith reaches the festival grounds, there is only one given path from there to the end of the sixth side street. Since he takes at least one of the following four side streets, all of the 2⁴ paths through that area are valid except the one defined by taking none of the side streets; therefore, that leg has 2⁴ – 1 possible paths. For the final two side streets before turning into the thirteenth, any combination of turns and non-turns is valid, so those two have 2² possible paths through and/or past them.

Therefore, the total number of possible ways home is

which evaluates to 960.

Addendum: The first part of this formula (the part with the sum) can also be expressed as 2⁴, as he has the choice between two options for the segments between 1st and 2nd, 2nd and 3rd, 3rd and 4th, and 4th and 5th sidestreet, but is always on the same side for the segment between 5th and 6th sidestreet. Both @Karifean and @Blackrune saw that in their solutions, however they made errors along the way, which is why this solution is featured.

This puzzle was solved by @CounterShadowform.

Puzzle 05 Solution (provided by Karifean)

Obviously since there are only four matches the only way there could be two pine trees would be to have them be made up of two matches each. Overlap would’ve been the other possibility, but this picture was described to have “one pine tree” despite the top part of the pine tree technically forming a second pine tree, so using overlap is obviously an invalid answer. We already have one complete two-match pine tree at the top so all that’s left to do is move one of the bottom matches towards the other to complete the second.

This Puzzle was solved by @Blackrune, @CounterShadowform, @Karifean, and @Restkastel.

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Everyone’s score for both this round and overall (as this was the first round):

1. @CounterShadowform with 10 medals
2. @Blackrune with 7 medals
3. @Karifean with 5 medals
4. @Restkastel with 2 medals
5. @Exalt with 1 medal

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Now for the next batch of puzzles:

Puzzle 06

Difficulty: 4

Professor Johnson’s evil game

Professor Johnson is bankrupt, and that two weeks before his son’s birthday. He hasn’t bought a present yet, so how to get money fast? Then he remembers that he could use his superior intelligence to exploit a lot of money from the unsuspecting Professor Davis, one of his early colleagues and someone who always was a bit gullible. So the next day Professor Johnson meets Professor Davis at a cafe the latter frequents and starts a conversation with him: “Oh hello, old colleague, what a coincidence meeting you here! Might you be interested in a little gamble, now that we’re growing older and older? It’s one where you can win a lot of money with a high probability.” Professor Davis replies: “Sorry, but I’m always sceptical if someone promises me money or a tonic against hairloss.”
“Oh come on, let me at least explain the rules. For old time’s sake.”
"Okay, fine."
That’s what Professor Johnson was waiting for. So he started his explanations: “You know, under 99 different numbers, there is usually one that is divisible by 99, isn’t there?”
“If those numbers are picked at random, I would agree.”
“Goooood, very good. And you would also agree that it doesn’t matter how many digits these numbers have, wouldn’t you?”
“Again, if those numbers are big enough and truly random, then yes, I’d agree.”
“Perfect, then let us take random numbers with 19 digits, which start with a 1 and are otherwise only made up of 3s and 4s.”
“And where do you plan to get such random numbers?”
“Oh we just use a die. We start our number with a one and then roll a die 18 times, adding a 3 if the die shows a number less than four, and adding a 4 if the die shows a number bigger than three. Then, if the resulting number is divisible by 99, you win, ,and if it’s not, I win.”
“But that would be largely unfair, as you’d win 98 times before I win once on average.”
“Right, right, but it wouldn’t be as unfair if I’d wager 9900€ against 100€ from you, wouldn’t you agree?”
“Maybe, but I still doubt I could win a big amount of money with a high probability.”
"Fine, I didn’t want to bet a measly 9900€ but a whopping 2500000€ against your 100€, and that each game and you can decide when we stop. Besides, I don’t want this to be a fair game, I want you to win with a high probability!"
Eventually Professor Davis agrees to the terms. While he knows that Professor Johnson is a sly man, he doesn’t remember ever being lied to by him, so if Professor Johnson states that he can win with a high probability, then that means that he can win with a probability larger than 50%, of that Professor Davis is certain.
So the two Professors play 100 times, and 100 times Professor Johnson wins, who has therefore won 10000€ of which he can buy a present for his son. The game obviously wasn’t fair, but who really had the better odds, and why?

Puzzle 07

Difficulty: 2

Too many sales in New Lemonsdale

Like many other small towns, New Lemonsdale has some weird festivals. One festival this town celebrates focusses on selling lemonade. But this year there are way more people selling lemonade than the preceding ones. And since only those that sold lemonade in the last year are supposed to sell this year as well, there must be several people selling these refreshments illegally. So the police called all salespersons to the market square of the city. While they can’t check firsthand if someone is selling legally or not, as most legal sellers either don’t have their papers with them or don’t possess such papers in the first place, they had the idea to ask the sellers about the legality of their colleagues, as the sellers do know whom of them is selling legally and who isn’t. So the salespersons, 84 in number, were tasked to form a giant circle and state wether their right neighbor is selling legally or not. Begrudgingly, while the legal sellers always answered truthfully, the illegal ones assured their illegal colleagues their legality and denied the legality of any legal sellers. Now, when everyone in the circle gave their judgement, all people that stated their neighbor to be illegal were tasked to leave the circle. The rest should then close the circle and repeat this procedure. In the end, 12 people remained in the circle. Now it slowly dawned to the police that every legal seller always said thee truth while every illegal seller always lied. So they were wondering what they could conclude from that. Then, detective Airhead mentioned that while he can’t say who is illegal and who isn’t, he at least can tell how many illegal sellers there are at minimum. Can you tell that too?

Puzzle 08

Difficulty: 2

Multiplying trees XXL

Another match task. But this time there is a lot more asked of you. Out of the pine tree pictured above make eleven. You may only move the matches, not break them in any way.

Puzzle 09

Difficulty: 5

First think, then gift

Mr. Gift is a very busy man and thus has sadly no time to care for his daughter Anna. He wants to excuse that by overwhelming her with presents on her birthday every year. This year he has prepared 8 presents, one larger than the other. Those presents are all of differing sizes. The boxes Mr. Gift prepared to send the presents are of such dimensions that not only the present itself fits inside, but also all smaller boxes together. Because of that, Mr. Gift contemplates if he sends the 8 boxes individually or just the biggest box with the other 7 boxes inside it. Then he realizes that there are several more possibilities of putting boxes in other boxes. After all, he can put boxes inside boxes which themselves are in another box and so on and so forth. How many different possibilities does Mr. Gift have of sending those gifts to his daughter?

Puzzle 10

Difficulty: 3

Finding the balance

The mobile pictured above is in perfect balance. The horizontal sticks all have the same weight. All the strings weigh nothing. For a single stick to be in balance, the weight to the left and right must be equal. The exact position of the weights can be ignored, so it doesn’t matter how near or far an object is to the central string from which a stick is hanging. The same forms weigh the same. At one position you see a questionmark. Which form must be there for this mobile to be in balance?

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So, happy puzzle-solving! You have time to send your PM(s) until next Saturday (25th February) 8 pm UTC. And if you’re someone who hasn’t sent in solutions for last week, you can still sent solutions for this week, or any following weeks for that matter. While it’s going to be harder to reach the overall first place the longer this competition is running, you could still be featured as a solution or even reach the first place of a round.

[Karifean]

“Didn’t even try it”? I spent way too long trying, I just didn’t submit a solution because I never came up with a satisfactory one. shrug

[Blackrune]

Oh damn I was sooo close in 4. Should have triple-checked.
Well, new round new luck.

[VyseGolbez]

I spent way too long trying

Okay, I admit, imagining you going “It’s useless, it’s all useless!” is way more entertaining

[pictoshark]

Solved Puzzle 10. Probably won’t do the others.

It’s not that bad.

[VyseGolbez]

And time’s up! So solutions for Puzzles 06 to 10:

Puzzle 06 Solution (provided by Karifean)

A number is divisible by 99 if and only if it is divisible by 9 and 11.

A number is divisible by 9 if the sum of its digits is divisible by 9. There are very few numbers divisible by 9 in the described format. For it to happen, the number must either have one 3 in it, or 10 3s in it, with the number of 4s being either 8 or 17.

However, the number must also be divisible by 11. A number is divisible by 11 if the alternating sum of digits is divisible by 11. This means in our case that the odd and even numbered digits must add up to the same number. There is no way to make the difference between the even and odd digit sums higher than 10 after all.

In other words, the number must consist of one 3 and 17 4s, and the 3 must be situated on an odd-numbered digit. Leaving us with only 9 possible numbers to go with:

1434444444444444444
1444344444444444444
1444443444444444444
1444444434444444444
1444444444344444444
1444444444443444444
1444444444444434444
1444444444444444344
1444444444444444443

The number of candidates is 2^18, or 262144, and since the chance distribution is uniform, that means we have a 9 / 262144 chance, or about 1 in 29127 chance of getting one of these numbers.

This means Professor Johnson has the better odds in the game. He would’ve had to bet about €2912711 against Davis’ €100 for the game to be fair.

Addendum: Now the interesting thing is that Professor Johnson did not lie. The game wasn’t fair, and Professor Davis had a chance greater than 50% to win a large amount of money. If his goal would be to win 200000€, his chances aren’t too bad. We assume that Professor Davis doesn’t play endlessly until he wins 200000€, but stops if he either won a game or lost 2300000€. But how big is the probability for Professor Davis losing 2300000€? Since he loses 100€ every time he loses, he would have to lose 23000 games without winning one so that he doesn’t go home with a net total of 200000€ or more. The probability of him winning an individual game is 9 / 262144, as Karifean pointed out. The probability of him losing is 1 - (9 / 262144), which is ca. 0.999966, or 99,9966%. The probability of him losing 23000 consecutive games is (1 - (9 / 262144))^23000, which is ca. 0,454, or 45,6%. This means that his chance of winning one of those games, and therefore winning at least 200000€, is 54,6%.

This Puzzle was solved by @Karifean and @CounterShadowform.

Puzzle 07 Solution (provided by Karifean)

Obviously, a simple minimum number of illegal sellers is (number of sellers this year (84) - number of sellers last year (unknown number)).

Besides that, perhaps it would be an idea to illuminate exactly how the ring of salespeople changes with every iteration; a salesperson remains in the ring if and only if their neighbor is part of the same group as them. As such, it may make sense to visualize the salespeople in the ring not as individuals, but as groups. To illuminate what I mean, let me make an example.

Say we have a group of 12 salespeople, distributed as first 3 legal, then 2 illegal, then 4 legal, then 3 illegal salespeople. Naturally, the last person in each of these groups will claim their neighbor to be illegal, while everyone else will claim their neighbor to be legal. So after the first iteration where everyone who claimed their neighbor to be illegal steps back, the group turns into first 2 legal, then 1 illegal, then 3 legal, then 2 illegal. Next iteration, the single illegal is eliminated, merging the two legal groups, so the groups are then 1 legal, 0 illegal, 2 legal, 1 illegal, or in other words, 3 legal, 1 illegal. Finally, in the last iteration, we end up with 2 legal salespeople.

Could we have been able to tell we’d end up with 2 legals from the start? Actually yes! If you may notice, in every iteration, the same number of legal and illegal salespeople is eliminated. As should be apparent from the grouping model, it is impossible for this to not be the case, because 1 person from every group of neighboring legals/illegals is eliminated every time.

So the number of eliminated salespeople must be half legal, half illegal. We know that 84 salespeople were reduced to 12, so 72 salespeople were eliminated. 36 of those are legal, the other 36 are illegal. The 12 remaining salespeople could be either legal or illegal, but we can say one thing for sure: At minimum, 36 of the salespeople are illegal sellers.

This Puzzle was solved by @Karifean and @Blackrune.

Puzzle 08 Solution (provided by Karifean, Blackrune, and me)

You’re probably wondering why several people are credited for this one. That’s because there are multiple different solutions and one of them wasn’t found by a participant (which coincidentally was the one I expected to see). All solutions have in common that the number 11 is formed, and not 11 trees.

https://d2qdztz5tk5zxi.cloudfront.net/original/2X/2/21e0f0d91cf83d8d5b72f3c510b3d910591cb505.png

This Puzzle was solved by @Blackrune, @Karifean, and @King_Titanite_XV.

Puzzle 09 Solution (provided by Karifean)

Every box can be either separate, or in any of the bigger boxes. For example, Box 1 (the smallest) can be by itself, in Box 2, in Box 3, in Box 4, Box 5, Box 6, Box 7 or Box 8 - eight possibilities. Conversely, Box 8 only has one possibility, it has to be by itself. The number of possibilities for the boxes inbetween are 7 to 2 respectively. Any of these possibilities can coexist with any other, and it is exhaustive - there is no possible way for the boxes to be arranged that is not covered by this.

So the number of possibilities is 8765432*1, or 8!, which makes 40320 possibilities.

This Puzzle was solved by @Karifean.

Puzzle 10 Solution (provided by pictoshark)

Allow the weight of a stick to be v
Allow the weight of a square to be x
Allow the weight of a circle to be y
Allow the weight of a triangle to be z
Allow the weight of a mystery object to be a

From the bottom left going from left to right and then up, each Mobile can be written as the following equations.

[1] x+y=z

[2] v+x+y+z=2y+z

[3] 4x+y=3y+a

[4] 2v+x+3y+2z=v+4x+4y+a

Equation [2] can be written as:

v+x+y+z=y+y+z

By taking away y+z from both sides of the equation you wned up with

v+x=y

Which can be rearranged to get:

v=y-x

By subbing this into Equation [4] we get:

2y-2x+x+3y+2z=y-x+4x+4y+a

Which simplifies down to:

-x+5y+2z=3x+5y+a

Which can be rearranged to get:

2z-4x=a

Equation [3] now!

4x+y=3y+a

Our known value of a in terms of x and z can be subbed in here to get:

4x+y=3y+2z-4x

Which can be rearranged to get:

8x-2y=2z

Halfing it gives you:

4x-y=z

AND NOW

LET’S LOOK AT EQUATION [1]!

x+y=z

x+y must equal 4x-y as both of them equal z

x+y=4x-y

Which can be rearranged to get:

2y=3x

Halfing it gives you:

y=1.5x

And noooooow back to this line

4x-y=z

Our known value of y in terms of x can be subbed in here to get:

4x-1.5x=z

Which simplifies down to:

2.5x=z

Remember this?

v=y-x

Our known value of y in terms of x can be subbed in here to get:

v=1.5x-x

Which simplifies down to:

v=0.5x

To be honest, there was no need for me to work out v just now. BECAUSE THIS PUZZLE SOLVED ITSELF LONG AGO

We got a little something a while back when we fiddled with Equation [4].

A known value for a in terms of x and z.

2z-4x=a

Our known value of z in terms of x can be subbed in here to get:

5x-4x=a

x=a

Which means our mystery object has the same weight as the square.

And the triangle and the circle have drasically different weights to the square.

Therefore, the mystery object is a square, as it was either a square, a triangle or a circle, and it had the same wieght as the square but different weights to all the other possible options.

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I can check my work using Equation 4, ihi.

2v+x+3y+2z=v+4x+4y+a

v=0.5x
y=1.5x
z=2.5x
a=x

2(0.5x)+x+3(1.5x)+2(2.5x)=0.5x+4x+4(1.5x)+x

x+x+4.5x+5x=0.5x+4x+6x+x

7x+4.5x=11x+0.5x

11.5x=11.5x

Yaaaaaaaaaaaaaaaay

This Puzzle was solved by @Karifean, @Restkastel, and @pictoshark.

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Now for score:

Current round:

1. @Karifean with 16 medals
2. @Blackrune with 4 medals
3. @CounterShadowform with 4 medals
4. @Restkastel with 3 medals
5. @pictoshark with 3 medals
6. @King_Titanite_XV with 2 medals

Overall:

1. @Karifean with 21 medals
2. @CounterShadowform with 14 medals
3. @Blackrune with 11 medals
4. @Restkastel with 5 medals
5. @pictoshark with 3 medals
6. @King_Titanite_XV with 2 medals
7. @Exalt with 1 medal

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And of course the 3rd batch of puzzles:

Puzzle 11

Difficulty: 3

The completely impossible triangle

Above you can see a triangle made up of several small points. Someone posed the challenge to connect those points using the following rules:
The points should be connected in one go.
The lines connecting two points always have to have the minimum possible length.
If a point was already visited, it can’t be visited again.
At each point the direction has to change.
Above you see a small line that follows these rules. However, doing this for the entire triangle is impossible. Why?

Puzzle 12

Difficulty: 4

Lighter buns

The bakery Sweet Stuff is famous for its buns, which always weigh exactly 100 gramm. That is thanks to the head baker. However, he also has an apprentice that doesn’t always manage to bake buns weighing exactly 100 gramm. Today, they have 10 baskets each containing somewhere between 100 and 200 buns. 9 of these baskets are filled with buns baked by the head baker, therefore every bun in those baskets weigh exactly 100 gramm. The tenth basket is filled with buns baked by the apprentice, who only managed to get the weight correct for five buns, the others all weigh only 99 gramm. Of course, those can’t be sold, but both don’t know which basket is the one with the apprentice’s buns. So they have to weigh the buns somehow. We assume that the head baker knows that his apprentice baked five buns weighing 100 gramm and baked all others weighing 99 gramm. We also assume that he possesses a scale that can weigh even up to 100 kg precisely, although this scale has a little error always showing at least 1 and at most 5 gramm less than what it should show. How can he identify the basket with the apprentice’s buns in only one weighing process?

Puzzle 13

Difficulty: 3

Needle Blessing

Mrs. Pinegreen still remembers the last christmas. Now this isn’t because of the meal or the presents or because of her family. No, it’s thanks to the christmas tree she had that year. This tree had this habit of losing its needles. On the first day it wasn’t so bad, as it only lost one needle, but on every following day the tree lost twice as much needles (sometimes also one more or one less) than on all preceding days together, until it lost all 100000 needles. Mrs. Pinegreen always cleaned up the needles on every day. So how many needles did she have to clean up on the 8th day?

Puzzle 14

Difficulty: 4

Back and Forth and Round and Round

Now for a staple for math puzzles: Below you see a row of numbers where you get to the next by using some sort of principle. Your job is to determine which number comes next.

1
2
4
8
16
122
433
677
1543
8585
11716
78332

Puzzle 15

Difficulty: 5

Professor Johnson is lost at sea

Professor Johnson went on vacation to the island Tangentia. The coastline of this island is straight for miles on end. He also employed an assistant for his yacht, with which he departed from this coast. He explicitly asked of that assistant to be a seasoned man of the sea who is dependable. As only one person applied for the job, sailor Yarn, Professor Johnson had no choice but to employ him. Now, when they departed, Professor Johnson decided to take a nap, however, sailor Yarn, who should take the wheel, fell asleep as well. Now the two were lucky to not just sink by ramming into a random coast. When Professor Johnson woke up and saw his sailor sleeping, he woke him asking: "Can you tell me for what I even pay you?"
Sailor Yarn replied: “What? Huh? What’s the matter?” - “The last two and a half hours you had the wheel, so can you tell me where exactly we are?” - “One moment, boss, let me check.” With those words he checked the digital fuel tank gauge. “Okay, looking at the amount of fuel left, we traveled pretty much exactly 100 miles. That means we are at most 100 miles away from the coast. I don’t know in which direction the coast is, though.” - “And you think I do? Think of something, damn it!” - “I might have an idea how we can get to the coast without using a compass, because looking at the thing, it seems to be broken as well.”, grinned sailor Yarn. “We only need to traverse 730 miles to get to the coast without fail. I simply need to activate my navigational system which I have in my pockets. While it can’t tell us where we are or where which cardinal direction is, for that I would have to have activated it at the beginning of our journey, it still allows us to navigate in relation to our current position. This, a sheet of paper, a ruler, a pair of compasses and a pencil would be enough to realize my plan. With that we would be able to traverse straight lines and even circle arcs.” - “I have these things in my cabin, but we only have enough fuel to travel another 660 miles. So think of a better plan!” - “I’m very sorry, boss, but I’m not able to think of a plan only traveling 660 miles.” - “By all good things!”, cursed Professor Johnson, “Then I’ll have to think of something.” Of course the genius Professor Johnson was able to think up a plan to savely reach the coast with the amount of fuel they had left and without knowing where north was. How did he navigate to accomplish this feat?

---

And again, happy puzzle-solving. You have time for a week again, so send your PM(s) until next Saturday, the 4th March, 8 pm UTC.

[pictoshark]

To be honest, there was no need for me to work out v just now. BECAUSE THIS PUZZLE SOLVED ITSELF LONG AGO

I hope I never change.

[VyseGolbez]

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:

1. @Blackrune with 14 medals
2. @Karifean with 3 medals
3. @pictoshark with 3 medals
4. @Wonderlander with 3 medals

Overall:

1. @Blackrune with 25 medals
2. @Karifean with 24 medals
3. @CounterShadowform with 14 medals
4. @pictoshark with 6 medals
5. @Restkastel with 5 medals
6. @Wonderlander with 3 medals
7. @King_Titanite_XV with 2 medals
8. @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.

---

Bonus

https://www.youtube.com/watch?v=1vF9Jg0XKOs

[Blackrune]

Of course the fun stopped when simple geometry got involved.

[pictoshark]

Difficulty: 6

Wow. Neat.

[VyseGolbez]

Do you not like Geometry?

But look, not all of them are about that.

Questions that came up:

Puzzle 17: 3 larger weights? I assume this means heavier weights?

Yes, with larger weights I meant heavier weights.

Puzzle 18: Are the doors soundproofed?

We’ll consider the doors and the building soundproofed, so you have to ask a question to find out which door to take.

Puzzle 20: I need to know how old you can be before you count as in school or kintergarten for this. Also can we assume no-one got held back? I know people that got held back six years in school and I’m in the UK which makes that veeeeeeery rare.

Ah, I feared this might be very different depending on your country. In this puzzle, “being in school” is equivalent to being 6 years or older and “being in kindergarten” is equivalent to being 3 years or older.

[VyseGolbez]

Since I haven’t gotten any solutions yet and you said that you had a busy week when I asked on discord, I’m going to give all of you another week to solve the current batch of puzzles. But I ask you to please use the additional time I’m giving you. This means the new deadline is Saturday 18th March 8 pm UTC.

[VyseGolbez]

You see, the ring around the chalice and the chalice itself don’t have the same volume.

In this statement, does “the ring around the chalice” refer to the whole interior of the lower segment, or just the area not also inside the upper segment? (Or possibly something else entirely?)

Ah, so the wording was too unclear. Let me make it clear with a profile of the described object:

A and B are the two volumes this puzzle refers to.

---

EDIT: So while I got a few answers this week, the amount is still a bit too low, as I have only gotten one answer to Puzzle 17, one answer to Puzzle 18, and one answer to Puzzle 20. From what I’ve heard, this was again a time problem, not an interest problem, so I am willing to give another week. So that gives you time until Saturday 25th March 8PM UTC. Remember that with the answers I’ve gotten until now it is technically possible to ask for a hint for Puzzle 19 right now. Hope this week isn’t as busy as the last two for you.

[VyseGolbez]

Ugh, still haven’t gotten enough solutions…

This time, I’m going to ask you to fill a poll regarding each puzzle, just so I know where everyone stands. But first, I need to write an addendum and an apology regarding Puzzle 17:

As it turns out, the sample solution I had for that one is wrong. Shocking, I know. It would be correct if the number 854 itself would be impossible to reach, but the puzzle text I gave you clearly states the opposite. A solution still exists, however, if you see the “numbers from 1 to 854 can be reached” as a vacuous truth, with the possibility of 855 being reachable as well still intact. For that one, I however have no idea how to reach said solution without using a brute force approach. Therefore, I am going to allow two different solutions: You can either try to find a solution assuming that the witch can only weigh everything from 1 to 853, or you can try to find a solution assuming that the witch can weigh everything from 1 to 854, possibly including 855. Bonus points if you do the latter and can present me a method that doesn’t rely on brute force.

Now for the polls I talked about. They are basically the same for each puzzle, but I’d still ask all of you to fill each of them, and I mean all of you, so I’ll tag everyone who has collected medals: @pictoshark @Blackrune @Karifean @Restkastel @CounterShadowform @Exalt @King_Titanite_XV @Wonderlander

Puzzle 16

• I haven’t looked at the puzzle, and don’t plan to do so
• I haven’t looked at the puzzle, but plan to do so
• I have looked at this puzzle, but don’t have a solution
• I have already sent a solution to this puzzle

Show results

Puzzle 17

• I haven’t looked at the puzzle, and don’t plan to do so
• I haven’t looked at the puzzle, but plan to do so
• I have looked at the puzzle, but don’t have a solution
• I have already sent a solution to this puzzle

Show results

Puzzle 18

• I haven’t looked at the puzzle, and don’t plan to do so
• I haven’t looked at the puzzle, but plan to do so
• I have looked at the puzzle, but don’t have a solution
• I have already sent a solution to this puzzle

Show results

Puzzle 19

• I haven’t looked at the puzzle, and don’t plan to do so
• I haven’t looked at the puzzle, but plan to do so
• I have looked at the puzzle, but don’t have a solution
• I have already sent a solution to this puzzle

Show results

Puzzle 20

• I haven’t looked at the puzzle, and don’t plan to do so
• I haven’t looked at the puzzle, but plan to do so
• I have looked at the puzzle, but don’t have a solution
• I have already sent a solution to this puzzle

Show results

So I’ll give another week again (Saturday, 1st April, 8PM UTC), and depending on how the polls go, I’ll do further stuff during the week.

[Blackrune]

I’m just stumped on the others at the moment, though I guess there’s always a chance inspiration will suddenly strike…
Maybe it’s time to give out some small hints so they don’t go to waste~?

[King_Titanite_XV]

I took looked at to mean “tried to solve” here. Sorry for not reading properly

[VyseGolbez]

So okay, seems to look like hints are needed. I’ll put them in boxes as well, just in case someone among you doesn’t want hints (wouldn’t want to ruin it):

Puzzle 16

All volumes that are needed here can be expressed by combining cones of various sizes. As the volume of a body is proportional to the third power of a length on that body if similar bodies are viewed, you don’t even need the formula for the volume of a cone.

Puzzle 18

I said in the puzzle itself what questions aren’t allowed. As hint I’m going to say what specifically is allowed: You can use compound subjects (i.e. combining statements with “and” and “or” and the like), so “Is it correct that the sky is blue and the grass is green?” would be a valid question.

Puzzle 19

All is not as it seems. For this one, I’ll say look for homonyms of the word “Square”

Puzzle 20

Try to think of ways of how to break down a big number into several factors. You no doubt did that in school at some point.

[VyseGolbez]

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.

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Score

Current round:

1. @Karifean with 11 medals
2. @Blackrune with 3 medals
3. @CounterShadowform with 3 medals

Overall:

1. @Karifean with 35 medals
2. @Blackrune with 27 medals
3. @CounterShadowform with 17 medals
4. @pictoshark with 6 medals
5. @Restkastel with 5 medals
6. @Wonderlander with 3 medals
7. @King_Titanite_XV with 2 medals
8. @Exalt with 1 medal

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For now this little competition will go on a break. I urge you all to use the time to read Higurashi.