Think of a unit of measurement.
Now rearrange its letters and read them out loud to form a sentence (an example would be u r a b, meaning "You are a bee"). Make sure the sentence compliments someone on their appearance.
Now, what is the word? What is the sentence?
Answer
It could also be a
quart
Because
You are a cutie! (u r a q t)
I found this question and became curious, can anyone tell me the answer and prove it, i know it seems fairly simple but just thought an explanation of this would make an interesting case.
Flip a fair coin repeatedly until you get two heads in a row (HH assuming H indicates head and T indicates Tails). On average, how many flips should this take? What if we flip until we get heads followed by tails (HT)? Are the answers the same?
Answer
Case 1: Waiting for HH
$A$ = average nb. of flips to get HH after T or nothing.
$B$ = average nb. of flips to get HH after H.
A is 1 flip, plus B if you get H and continue, or plus A if you get T and restart:
$A = 1 + {1\over2}B + {1\over2}A$
B is 1 flip, and you are done if you get H, or plus A if you get T and restart:
$B = 1 + {1\over2}0 + {1\over2}A$
This resolves to
$B = 1 + {1\over 2}A$
$A = 1 + {1\over 2} (1 + {1\over 2} A) + {1\over2} A$
$A = 1 + {1\over 2} + {1\over 4} A + {1\over 2} A$
$A = 6$
The expected number of flips to get HH is 6.
Case 2: Waiting for HT
$A$ = average nb. of flips to get HT after T or nothing.
$B$ = average nb. of flips to get HT after H.
A is 1 flip, plus B if you get H and continue, or plus A if you get T and restart:
$A = 1 + {1\over2}B + {1\over2}A$
B is 1 flip, and you are done if you get T, or plus B if you get H and wait for T again:
$B = 1 + {1\over 2}0 + {1\over 2}B$
This resolves to:
$B = 1 + {1\over 2}B$
$B = 2$
$A = 1 + {1\over 2}2 + {1\over 2}A = 2 + {1\over 2}A$
$A = 4$
The expected number of flips to get HT is 4.
Comparison
The difference comes from the fact that in the first case, if you get the wrong 2nd result, you start all over, while in the second case you continue with one good result.
This is my first puzzle, but strongly influenced in style by the type of puzzles you see monthly at Puzzled Pint. If you are fortunate enough to live in a city where they are hosted, I would really recommend going. Depending on how this one goes, it is planned to be the first of a series along the (deliberately clichéd) theme of a treasure hunt.
This story begins like any good treasure island story, with a cliché!
Your widowered father is aging, and the family have decided to move him to a retirement home so he can get the care he needs. He has been living by himself for the past decade and has accumulated a lot of various items - to be brutally honest, mostly junk. Being the most limber you volunteer to clean out the attic. While scrambling around - being carefull not to put your foot through the ceiling - you come across an old backpack which contains a collection of paraphernalia one might expect to see for an old-timey exploerer, such as a gas lamp, rope, hard hat, and a scrap of paper. The paper catches your eye. You pocket it and head down for an explanation.
"What's this?" you ask.
"That's an old treasure map." He says, somewhat matter-of-factly.
"Go on..."
"It is back from when I was in the navy. We found this map, found the island too. Scoured it as best we could, but only had a couple of weeks we could spare. It was abandoned, but we certainly weren't the first there. Rumour was that a treasure of some fantastic nature was hidden somewhere on it. I was putting together a few things to go back and search for it, but that is when your mother passed and my priorities shifted a bit. Then I guess I just lost the enthusiasm."
You examine the map closely. It contains a coded message which might give some clue as to location or nature of the treasure. Or it might be meaningless twaddle, but it is a good a place as any to start. 
Question: What is the message?
The handwriting on the map is not great, so a transcription of the coded phrase is given below if required.
B BBDAABBBCC CDDDAAA CCADDC
Answer
The message is:
X MARKS THE SPOT
The way to reveal it:
Each subset of letters leads to a location. A single A is A1, AA is A2, etc. Each respective square on the map refers to a letter: A1 is P(hilips), A2 is R (Arr!), A3 is E (Echo) and so on.
Previous puzzle of this kind: #1
A bird without wing, body, feet or tail,
An invention am I who can effortlessly sail.
I am unlike my lowly brothers who only deliver mail;
Surging over the winds I am majestic without fail!...Alas, that is a lie, I do have one shortcoming
In order to be useful to you I cannot be made thin.
Thanks a lot, you, you darned and cursed thing!
Now there are barriers I shan't break, oh, races I'll never win!You say that there's a Wikipedia article for anything?
Find mine, then, and declare yourself this puzzle's king!
Hint 1:
My stability was an issue that my creators did see
That is the reason why they taught me to sweep.
Hint 2:
UFOs have a generic shape;
I do too, though not from space
Answer
Are you?
https://en.wikipedia.org/wiki/Flying_wing
A bird without wing, body, feet or tail, An invention am I who can effortlessly sail.
Aircraft, plane or drone
I am unlike my lowly brothers who only deliver mail; Surging over the winds I am majestic without fail!
...Alas, that is a lie, I do have one shortcoming In order to be useful to you I cannot be made thin.
Needs to have room for engines, fuel, cabin inside the wing.
Thanks a lot, you, you darned and cursed thing! Now there are barriers I shan't break, oh, races I'll never win!
Will never break the sound barrier. Too thick to hold the equipment.
You say that there's a Wikipedia article for anything? Find mine, then, and declare yourself this puzzle's king!
Here is a series of movies that have something in common other than being thoroughly entertaining.
Find the next movie in this series.
Answer
Next in sequence is
"Ace of aces" or "Ace Ventura" (Ace)
The sequence can also be extended in the other direction:
Ten Things I Hate About You
Nine Months
The Hateful Eight
The Magnificent Seven
The Six Billion Dollar Man
I saw this puzzle many moons ago and so I assumed it would already be on here somewhere. I have looked and not found, though, and so I present to you a puzzle that baffled me when first I saw it.
What number replaces the question mark in the last example and why?
3531 → 0
3597 → 1
3891 → 3
9397 → 2
3869 → 4
2128 → 2
3786 → 3
3506 → 2
2700 → 2
7212 → 0
7205 → 1
2990 → 3
9503 → 2
1057 → 1
1160 → 2
1995 → ?
Answer
It is
2
You count the amount of
enclosed areas in the numbers
When Augustus de Morgan (a mathematician who was born and died in the 19th century) was asked about his age, he replied: “I was x years old in the year x².” What year was he born in?
Could such a strange lot have befallen someone who was born and died in the 20th century?
Answer
The only perfect square in the 19th century is:
1849
Which would make him
43 years old on that date, born in 1806
For the 20th century:
It is not possible.
The only perfect square is 1936.
However, that would make him 44 years old and born in 1892, which is in the 19th century.
My first is part of your digestive system, but you sometimes use only half of me.
My second is sometimes called what people sometimes call the shark hanging in the Smithsonian.
My third is my mom, once removed from Matilda and re-moved again.
My fourth is a greeting to my friend who is named after the end.
My fifth is a really fast run.
My sixth is in the number three.
My seventh can replace my first and my last.
My eighth is the last 2 of my dad's essay topics.
What is my all?
Note: I edited some of the clues for clarification.
Hint:
The answer is NOT punctuation marks. There is an actual obvious answer once you know the answers.
Hint 2:
In the hint for my third, the re-moved again part has significance.
Hint 3:
At least two of the clues are the same.
Hint 4:
My second is made of parts...
My third is re-moved again.
My sixth is not in #3
My seventh requires the others.
Answer
My first was answered by @A.D.
A colon
My second was answered by @GentlePurpleRain in the comments
A hyphen (a high fin)
My third was technically half-answered by @A.D
A tilde then re-moved (or shifted) is a "`"
My fourth was answered by @A.D.
A hyphen
My fifth was also answered by @A.D.
A dash
My sixth was answered by @Leslie P.
A hyphen
My seventh was not answered
It's a hyphen... it can be used like a pair of commas in a phrase within a sentence or in the same sense with parentheses
My eighth was answered by @frodoskywalker
End Parenthesis
My all is:
:-`----)
This is a problem I once found in a math textbook a tutor used with me.
A group of explorers is trying to get from their base camp to another camp which is eight days' walk through the desert.
However, each explorer can only carry five days' worth of supplies. How many explorers need to venture out to make sure that at least two of the explorers make it to the second camp and the rest make it safely back to base camp before being stranded without supplies?
This problem is somewhat similar to the camel transporting bananas puzzle, but this time there are multiple people who can exchange supplies between each other.
Answer
Six explorers, according to the following schedule:
Day 1: 6 explorers start with 28 rations and end with 22. Two return with 2 rations (1 per).
Day 2: 4 explorers start with 20 rations and end with 16. One returns with 2 rations.
Day 3: 3 explorers start with 14 rations and end with 11. One returns with 3 rations.
Day 4: 2 explorers start with 8 rations and end with 6.
Day 5: 2 explorers start with 6 rations and end with 4.
Day 6: 2 explorers start with 4 rations and end with 2.
Day 7: 2 explorers start with 2 rations and end with 0.
Could "almost" do this with 5 and just let them get a little hungry but the six is needed just for one meal. Note that there's no need to load everyone to max at the beginning since 4 explorers can only carry 20 meals.
UPDATE
As pointed out in comments, the solution above assumes that they eat at the end of the day (where day 8 should be "start with 0 and arrive hungry"). Below follows the rather similar solution for the more reasonable scenario of eat at the start or during the day.
Day 1: 8 explorers start with 40 rations and end with 32. Two return with 2 rations (1 per).
Day 2: 6 explorers start with 30 rations and end with 24. Two return with 4 rations (2 per).
Day 3: 4 explorers start with 20 rations and end with 16. Two return with 6 rations (3 per).
Day 4: 2 explorers start with 10 rations and end with 8.
Day 5: 2 explorers start with 8 rations and end with 6.
Day 6: 2 explorers start with 6 rations and end with 4.
Day 7: 2 explorers start with 4 rations and end with 2.
Day 8: 2 explorers start with 2 rations and end with 0.
This solution is more symmetrical, which feels better somehow.
Inspired by the famous puzzle (origin unknown to me)
Fill in the blanks in the following sentence using the same word OR its split versions (not anagrams).
A _____ surgeon, was _______ to operate on his patient because there was _____.
My version: Fill in the blanks in the following sentences using the same word or its split versions (no anagrams)
Detective Inspector Susan Collins thought ______, the butler, _______. It ______, the _______.
The word/s (and its split versions), in the above sentence could be dictionary words, acceptable slangs, names, proper nouns etc. No anagrams. There are no repeated words in my answer
The answer to the famous original puzzle
Notable, Not able, and No Table
Answer
Technically R Dye has got the answer. My answer (as Rubio pointed out) has 4 difffernt word versions
Detective Inspector Collins thought, W A Sherman, the butler, was her man. It was Herman, the washerman.
Of course R Dye deserves credit.
This puzzle is part of the "Piece de Resistance" series. Go back to Part 1 (Ace) for the story.
Ace Two Three Four Five Six ...
Aha! This one is of landscape orientation! Let's see...
Hint 1:
Title is hint.
Hint 2:
Threaten chickens to be devils
Answer
Suggested answers for all 16 cryptic clues follow below. Note that several of the clues employ 'printer's devilry' techniques, as suggested by '666', the mark of the beast from Revelation, appearing in the title.
Originally upset, Peter is awake (2)
UP (‘originally’ as in ‘first letters of’ Upset, Peter, meaning awake)
Ultimate veto in switch position (2)
ON (‘ultimately’ as in ‘last letters of’ vetO iN, meaning switch position)
They opposed stating threats (3)
EVA (printers devilry: They oppose dEVAstating threats) - solved by @JS1 in comments
Intend to wrap up alibi and lead mystery (3)
AIM (‘wrap up’ as in ‘ first and last letters of’ AlibI + ‘lead’ – first letter – of Mystery, meaning intend)
Rub, ail, sit! (3)
ION (printers devilry: RubIO Nails it!)
Stalker’s flag (4)
JACK (double meaning: Jack, as of ‘and the beanstalk’ and ‘union jack’)
Light unit and lead to domain (4)
LAND (‘Light unit’ + AND, meaning domain)
History within toothpaste (4)
PAST (‘within’ reveals ‘toothPASTe’, meaning history)
Exclamation reversed purpose of shelter (5)
HOUSE (‘OH’ reversed + USE (purpose), meaning shelter)
The fir among the trees (5)
Either STONE (printers devilry: The firST ONE among the trees) - suggested by @hdsdv and @JS1 in comments
Ring and veil uncovered reveals actress (5)
OLIVE ('ring' = O + an anagram of 'veil' clued by 'uncovered', i.e. LIVE, giving a female name which can be suggested by 'actress')
Mazed heart before an area (4)
ZONE (‘heart’ as in ‘central letter of’ ‘maZed’ + ONE (an, singular), meaning area)
Chicken! A searching over grass (5)
(best guess at present) PEACE (printer's devilry: Chicken PEA CEase arching over grass) - solved by @JS1 in comments
Settle public awareness (5)
LIGHT (double meaning, as in 'settle', 'land', 'dismount' and also in 'public awareness', 'public knowledge', 'bringing to light') - solved by @JS1 in comments
O pal! Dim tangled qualification (7)
DIPLOMA (anagram of ‘o pal dim’ clued by ‘tangled’, meaning qualification)
Deformed mint gel is changing form (7)
MELTING (anagram of ‘mint gel’ clued by ‘deformed’, meaning changing form)
Taken all together, the 16 solutions then form:
A connecting wall to resolve. The four answers are then connected as a group themselves, giving the final answer to the puzzle.
Resulting solution:
TIME - ZONE (time zone), ON (on time), PAST (past time), UP (time up)
HIGH - AIM (aim high), LAND (highland), DIPLOMA (high diploma), LIGHT (highlight)
POT - ION (potion), JACK (jackpot), MELTING (melting pot), STONE (potstone)
GREEN - OLIVE (olive green), PEACE (Greenpeace), HOUSE (greenhouse), EVA (Eva Green).The final connection between the four groups is:
TEA - TIME (teatime), HIGH (high tea), POT (teapot), GREEN (green tea).
If you're interested in starting the treasure hunt from the very beginning, check out the prologue!
Though your methods of solving the required riddle were dubious at best, you were still able to come up with the correct question to ask the concierge in plenty of time for your departure the next day. You were provided with a marked car (courtesy of Lamborghini, of course) and told your departure time would be 7:00 am. Restlessly, you made it through the night, finding yourself standing at the door of your car at 6:59 am. After a few seconds of thumb-twiddling, you hear a click, and the car door remotely unlocks. You swing it open and hop inside, finding a note on the dashboard.
Picking up the note, you read:
Dear [insert username here],
Congratulations on solving the last little brainteaser I left you! Unfortunately, I've erred, and the next clue isn't ready yet. Sorry about that. While you wait, please enjoy these neat factoids about Europe, since you'll be here for a little while yet!
Yours truly,
Bailey M
Puzzlemaster for the Treasure Hunt 'Round the World
You sigh. More waiting? But, for treasure, it will be worth it. Flipping over the letter, you read on the back:
20 Fun Europe Facts
- Europe is named after the Phoenician princess "Europa", from Greek mythology.
- Romania is the oldest country in all of Europe. It hasn't changed its name since 681 AD!
- The Netherlands has the world's densest rail system, with 113.8 kilometers of rail per 1,000 square kilometers.
- Over 50% of Greece's population lives in its capital, Athens.
- The most visited place in the entirety of Europe is, not suprisingly, Disneyland (located in Paris).
- Greenland, an island owned by Denmark, is the second-largest island in the world, surpassed only by Australia.
- Croissants, the famous French pastry, were actually invented in Austria.
- The national airport in Brussels sells more chocolate than any other location on Earth!
- Finland has the most McDonald's per capita in all of Europe.
- Hate mosquitos? Move to Iceland, where there are literally none.
- If you're getting bored of Europe, have no fear - Africa is only a 30-minute ferry ride from Spain.
- Europe is the second-smallest continent, but has the third-largest population, surpassed only by Asia and North America.
- The longest town name in Europe is Lanfairpwllgwyngyllgogerychwyrndrobwllllantysiliogogogoch. I'll wait for you to try to pronounce that.
- The most common name for a pet pig in France? Napoleon.
- It is forbidden to flush your toilet after 10 PM in Belgium, for some reason.
- The largest recorded colony of ants in the world is 6,000 kilometers long, and can be found in northern Italy. I promise I'm not sending you there.
- The Mediterranean Sea is the only sea in Europe that researchers believe has always been filled with water.
- The London Eye in (surprise) London is the world's tallest Ferris Wheel, standing at 135 meters tall.
- The entirety of Europe by land area is still smaller than Canada!
- An estimated 10% of Europeans were conceived in Ikea beds - that's about 73.3 million people, for those counting.
As you sit there, reading the provided facts and attempting to pronounce Lanfairpwllgwyngyllgogerychwyrndrobwllllantysiliogogogoch in your head, you can't help but think it's strange that the Puzzlemaster would slip up like this...after all, he's been preparing this Treasure Hunt for years, hasn't he?
Where is your next destination?
The story continues in the next part, Treasure hunt 'round the world! (clue 4)
Answer
This is tough, because as you all know Anything can signify anything.
For the facts:
- Europe is named after the Phoenician princess "Europa", from Greek mythology.
TRUE (1)
- Romania is the oldest country in all of Europe. It hasn't changed its name since > 681 AD!
FALSE (0)
- The Netherlands has the world's densest rail system, with 113.8 kilometers of rail per 1,000 square kilometers.
FALSE (0)
- Over 50% of Greece's population lives in its capital, Athens.
FALSE (0)
- The most visited place in the entirety of Europe is, not suprisingly, Disneyland (located in Paris).
TRUE (1)
- Greenland, an island owned by Denmark, is the second-largest island in the world, surpassed only by Australia.
FALSE (0), because considered as largest island according to Wiki.
- Croissants, the famous French pastry, were actually invented in Austria.
TRUE (1), especially according to English Wiki page.
- The national airport in Brussels sells more chocolate than any other location on Earth!
TRUE (1)
- Finland has the most McDonald's per capita in all of Europe.
FALSE (0)
- Hate mosquitos? Move to Iceland, where there are literally none.
TRUE (1)
- If you're getting bored of Europe, have no fear - Africa is only a 30-minute ferry ride from Spain.
TRUE (1)
- Europe is the second-smallest continent, but has the third-largest population, surpassed only by Asia and North America.
FALSE (0)
- The longest town name in Europe is Lanfairpwllgwyngyllgogerychwyrndrobwllllantysiliogogogoch. I'll wait for you to try to pronounce that.
TRUE (1), despite the fact that the initial L is missing.
- The most common name for a pet pig in France? Napoleon.
FALSE (0)
- It is forbidden to flush your toilet after 10 PM in Belgium, for some reason.
FALSE (0)
- The largest recorded colony of ants in the world is 6,000 kilometers long, and can be found in northern Italy. I promise I'm not sending you there.
TRUE (1) - the fact says "and can be found in northern Italy" which is definitely true (it doesn't state that's it can only or as a total be found in Italy)
- The Mediterranean Sea is the only sea in Europe that researchers believe has always been filled with water.
FALSE (0)
- The London Eye in (surprise) London is the world's tallest Ferris Wheel, standing at 135 meters tall.
FALSE (0)
- The entirety of Europe by land area is still smaller than Canada!
FALSE (0)
- An estimated 10% of Europeans were conceived in Ikea beds - that's about 73.3 million people, for those counting.
TRUE (1)
Given this fact sheet we get 10001011011010010001
Now we have a lot of options for such binary looking chunk of data:
As is 10001011011010010001
is binary representation for integer
571025which doesn't give me a hint to any place on earth for now. Factorizing doesn't reveal anything either.
6-bit
Base64 (with "0"-fill)
100010 110110 100100 010000
i 2 k QBase 32 => hdkh
5-tupels: 10001 01101 10100 10001 with the interesting fact, that 1 and 4 are equal.
Bacon with b=1
=> SOWS
Bacon with a=1
=> PTMP
Baudot with 1 repr. on-state
=> SC-S
Baudot with 1 repr. off-state (you never know)
=> HXFH
CCITT-2
=> ZFHZ could be an abbreviation for "Züricher Fachhochschule, Zürich"
MTK-2
=> ZPSZ
5-bit with A=1 encoding (= ASCII + 64)
10001 01101 10100 10001
17 13 22 17
Q M V Q=> nope
4-bit with A=1 encoding (= ASCII + 64, even if a bit strange since we only get to 15)
1000 1011 0110 1001 0001
8 11 6 9 1
H K F I A=> sounds like Sofia or Haifa but isn't identical.
It could mean HKF International Airport. Unfortunately there is no IATA code for HKF.
There would be HKD which would mean question 11 should be false, but it's definitively true.
Same holds for HKG (Hongkong) (question 12 should be true but is definitely false).
Interesting fact about Hongkong: it's also referred to as HKIA, but what are we gonna do with the F?
There's a seaport code with HKF but that doesn't fit to IA as "International Airport".
UN/LOCODE has HK for Hongkong, but there's no subcode for FIA.
ASCII alone doesn't really fit to 20 bit binary. BCD, Aiken et. al. don't work.
And so shall it be: looks like there are different types of Bacon code or one can simply add a Caesar shift by -1 to the results above and you get:
Bacon(2) with a=1 => OSLO
So I would head to to the capital of Norway and see what's coming next.
"Do sudoku answers always have a single minimal clue set?" No. Solved Sudoku puzzles may have more than one set of disjoint clue-sets. And those clue-sets can vary in difficulty, so one might be easy and another diabolical, even though their 81-number answers are identical.
Gareth McCaughan poses an interesting question:
I wonder what the maximum number of disjoint clue-sets that determine the same solution is. Obviously no bigger than
floor(81/17)=4, given that 17 is the minimum number of clues to make a unique solution; I bet 3 is easy but 4 might be difficult or impossible. (On the handwavy grounds that 81 is nearly5*17, my guess is that 4 is possible.)
So, is it 2, 3, or 4?
Answer for this puzzle is one word. Have fun!
Highlighted parts are where I realised I have clued it wrong... Sorry for all this inconvenience!
Hint:
usefulness level 1
The way of deciphering has been hinted in this post
usefulness level 2
athin should have used another colour instead of yellow. That would make things easier, hopefully...
This puzzle is a part of Ooohh, it all makes sense now! :D
Answer
After solving the nonogram (from @athin)
The title,
along with the binary found by solving the nonogram, clues Gray codes.
Breaking the code:
We can construct the Gray code by starting with [0, 1], and each layer we can flip the entire list, append 0 to the front of each element in the original list, append 1 to the front of each element in the new list, and then concatenate the lists. So [0, 1] becomes [00, 01, 11, 10] for instance. For 5 bits, we get the list
(Ok, maybe I interpreted the word "flip" too literally).
Anyway, I also wrote down the letters A-Z down the side starting 00000 (which makes sense given our code text contains some 00000's). Decoding the Gray code line by line gives the answer VANADIUM.
Label the vertices (or nodes) of this graph with positive integers so that any two nodes are joined by a edge (or line) if and only if the corresponding integers have a common divisor greater than 1 (i.e. they are not relatively prime). Do so in such a way that the total sum of the 13 numbers is minimum.
Answer
My best (manually)
With a sum of
79623
Update:
An exhaustive search of all 28.74 billion distinct arrangements with the lowest primes in the center has confirmed that this is the optimal solution. The C code for this search can be viewed here.
This is a continuation of the question A balance with three pans, detecting the lightest pan (find the one heavier ball). It was told to me by a friend, Markus Götz, who put it online here: Deviating Ball Puzzles (pdf).
The three-pan balance
Imagine a balance with not two, but three pans. Weightings using the balance follow these rules:
Let's call this the "lightest-pan-detection-rule" (LPDR).
The problem
You are given n balls, two of which are heavier (not necessarily of the same weight).
1) What is the largest n, so that the heavier balls can be identified with 2 weightings?
2) What is the largest n, so that the heavier balls can be identified with 3 weightings?
The normal (non-heavy) balls are all of the same weight. You are to identify the deviating balls by using the balance a maximum number of weighings stated in the puzzle, weighing only the given balls. You are also to present a method to identify the deviating balls, and explain why the value of n is the largest.
Answer
Updated complete answer
$n=3$ requires 1 weighing.
Put one ball on each pan. The one normal-weight ball will rise and the other two are the heavy ones. There are no other possibilities with $n=3$.
$n=4$ requires 2 weighings.
There are ${4\choose 2} = 6$ possibilities, and each weighing can only distinguish 4 states, so 1 weighing cannot be sufficient for $n=4$.
Put one ball on each pan and keep the 4th ball aside. If one pan rises, then the other two balls are the heavy ones and you are done. Otherwise, you must have weighed 2 light balls and 1 heavy ball, so the 4th ball is the heavy one. Swap it with one of the others. If you swapped it with a light ball, then the other light ball will now rise and you are done. If no ball rises, then you swapped the heavy ball with the other heavy ball. The two balls you swapped are the heavy ones.
$n=5$ cannot be done in two weighings. I could try to prove it, but doing so doesn't answer the question asked, so I won't.
Interestingly, $n=6$ can be done in two weighings. ${6\choose 2} = 15 < 16 = 4^2$, so in theory it could be possible. In practice, it is possible.
Put two balls on each pan, numbering and labeling such that pan A has (1,2), pan B has (3,4), and pan C has (5,6). After noting the result, move the odd balls to the next pan, so now A has (5,2), B has (1,4), and C has (3,6).
If no pan rose on the first weighing, then both heavy balls were on the same pan, making the other two pans of two light balls each weigh the same. On the second weighing, the balls are separated. So if the heavy ones are (1,2), then on the second weghing, pans A and B will each have a heavy and a light ball, and pan C will have two light balls, so pan C will rise. Likewise, (3,4): pan A would rise and (5,6): pan B would rise.
If one of the pans rose on the first weighing, then the two heavy balls were split. So, if pan A rose, then the possibilities are (3,5), (3,6), (4,5), and (4,6). On the second weighing, the results of those would be: (3,5): pan B rises, (3,6): no pan rises, (4,5): pan C rises, (4,6): pan A rises again.
Of the 16 possible pan rising combinations, the one one that cannot occur is no pan rising either time.
Since ${7\choose 2} = 21 > 16$, it clearly is not possible to differentiate $n=7$ in two weighings.
Update
For three weighings, I note that ${11\choose 2} = 55 < 64 = 4^3 < 66 = {12\choose 2}$, so the theoretical maximum for three weighings is 11 balls. This maximum cannot be achieved.
Proof that it is possible to find the heaviest two balls out of $n=9$ balls with three weighings:
First weigh (1,2,3) vs (4,5,6) vs (7,8,9).
Then weigh (1,8,6) vs (4,2,9) vs (7,5,3).
Then weigh (1,2,6) vs (4,5,9) vs (7,8,3).
Each of the ${9\choose 2} = 36$ possibilities gives a different set of answers. Here's a partial enumeration:
(1,2): NCN (1,3) NBB (2,3) NAB
(1,4) CCC (1,5) CBC (1,6) CNN (1,7) BBB (1,8) BNB (1,9) BCC
(2,4) CNC (2,5) CAC (2,6) CCN (2,7) BAB (2,8) BCB (2,9) BNC
(3,4) CAA (3,5) CNA (3,6) CBB (3,7) BNN (3,8) BBN (3,9) BAA
Any other combination is a rotation of one of the above and would also yield a unique pattern.
For $n=10$, if you put 3 balls on each pan, there are $18 > 16$ possibilities where no pan rises, 9 where both heavy balls are on one pan, and 9 where one heavy ball is left out and the other is unknown.
If you put 3 balls on pans A & B and 4 balls on pan C, then if pan A rises, you have 24 possibilities: 12 where one heavy is on B and one on C, 3 where both are on B, and 9 where one heavy is on A and one on B, but the one on B is heavier than the one on A (and the one on A weighs less than 2 normal ones). Note there are also 24 possibilities where pan B will rise, but 9 of those are overlap, but with the relative weights of the heavy balls switched. There are likewise 15 possibilities where no pan rises: 6 for both on pan C and 9 where one is on A and one is on B and the two heavy ones weigh the same. There are even 9 possibilities where pan C rises: one heavy ball on each of A & B, and both weigh more than two normal balls.
If you put 2 balls on each pan, there are 33 cases where no pan rises; 3 where both heavy balls are on the same pan, 6 where neither heavy ball is weighed at all, and 24 where one heavy ball is weighed and the other is not.
If you put 2, 2, and 3 balls, and pan A rises, then you have 17 possibilities: 1 where both heavy ones are on pan B, and 16 possibilities where the heaviest one is on pan B and the other heavy one is any of the other eight.
In summary, you cannot solve for $n=10$ with three weighings. Since the difficulties lie with not being able to distribute all the balls equally, adding an 11th ball will not make it easier. And, as stated above ${12 \choose 2} = 66 > 64 = 4^3$, so you cannot possibly weigh 12 or more in three weighings.
Thus $n=9$ must be the maximum for three weighings.
I think this is the most appropriate place to share this. I haven't figured out the answer yet, but I think I'm close. I couldn't find anything like this on the Internet, so I'm posting it here.
There are three travelers. They are inside a cave. Deep down in this cave, there are three doors, each door containing at least one diamond, and there are a total of 9 diamonds. Each traveler picks a door and gets the diamonds behind it.
The travelers loot their diamonds, and, before exiting the cave, all three must say simultaneously if they can deduce the number of diamonds that each of the other two have. They never lie, and all of them said they couldn't deduce it.
After these statements were said, one of the travelers realized that now he knows the answer.
So the question is:
How were the diamonds split between the three travelers? (the order doesn't matter)
I was going to write what I've managed to do so far, but I think it isn't a good idea to give away too much information even if its concealed by spoilers. This riddle was suggested by my discrete mathematics professor. It isn't homework and she didn't mention where it came from.
This puzzle is part 16 of Gladys' journey across the globe. Each part can be solved independently. Nevertheless, if you are new to the series, feel free to start at the beginning: Introducing Gladys.
Dear Puzzling,
Today I got a glimpse of what things may have looked like on Earth millions of years ago.
I'm debating whether I should give you a hint about today's puzzle or not. If it was anyone else I would try to point them into the right direction, but you, dear Puzzling, can probably work it out on your own. After all, I assume you have at least some interest in geography if you've followed my journey this far.
Wish you were here!
Love, Gladys.
1. Loop of rope / Shakespeare's pentameter
2. Technically just outside Las Vegas / Chameleon or gecko
3. Ground spice / E-book reader
4. String instrument / Time travel film
5. Inexpensive / Llama relative
6. Fossilized resin / Native American or South Asian
7. Brownish colour / Immigrants
8. Vagabond / Extreme stupidity
9. Riddermark / Very young child
10. Lanzhou's province / Worshipped at Delphi
11. Accepted in Greek and Irish shops / Light phenomenon
Gladys will return in "A burglar's sunglasses, a lady's odyssey".
Answer
Gladys is in
FOSSIL GROVE
The grid:
Loop of rope / Shakespeare's pentameter
LASSO / IAMBIC
Technically just outside Las Vegas / Chameleon or gecko
STRIP / LIZARD
Ground spice / E-book reader
CUMIN / KINDLE
String instrument / Time travel film
CELLO / LOOPER
Inexpensive / Llama relative
CHEAP / ALPACA
Fossilized resin / Native American or South Asian
AMBER / INDIAN
Brownish colour / Immigrants
KHAKI / ALIENS
Vagabond / Extreme stupidity
NOMAD / IDIOCY
Riddermark / Very young child
ROHAN / INFANT
Lanzhou's province / Worshipped at Delphi
GANSU / APOLLO
Accepted in Greek and Irish shops / Light phenomenon
EUROS / AURORA
The middles ...
are capital cities, if you take some letters from the left and right. So:
SOFIA, Bulgaria
TRIPOLI, Libya
MINSK, Belarus
OSLO, Norway
APIA, Samoa
BERLIN, Germany
KIGALI, Rwanda
MADRID, Spain
HANOI, Vietnam
SUVA, Fiji
ROSEAU, Dominica
Square Spin History: #1>#2
New Rules
This puzzle introduces two new square types:
Unmovable squares (Un)
Replaceable squares (Re)
Plus the concept of ambiguity!
Ambiguity of squares
Certain square types are defined as being ambiguous. When a square type is ambiguous, it can only be seen in the initial image of the problem! This does not mean that it only exists for the first move, ambiguous squares exist throughout the whole problem! Ambiguous squares will appear to be regular coloured squares in the desired image.
Unmovable squares
These squares contain a cross and they prohibit all moves that end up displacing it from its position.
Here is an example of an invalid move
However, this is a perfectly valid move
Replaceable squares
This type of square contains a circle and it has a fairly complicated condition attached to it. It prohibits all moves that don't replace it with the same colour of square when the move changes its position upon completion. Oh, and this square type is ambiguous!
Here is an example of a valid move
There are some pitfalls to be aware of! For example, this requires two moves
Recap of Previous Rules
- Basic rules from #1
Two mountain climbers scaled a tall and dangerous rocky mountain using teamwork and wits. They had set up a camera to take photos of the climb at regular intervals however, the camera corrupted all of the photos except for two. Luckily the two photos had caught the start and end of their journey. Can you figure out what happened along their journey? 
Get a 7x7 grid to work with here or alternatively get a 7x14 grid here
Answer
I'm at 11 now, by using the circle on the right you can make your life a little easier.
ORIGINAL POST: I'm at 12. The big 4x4 spin seems unavoidable, and everything else is in service to it. Feels like everything is as efficient as it can be, but I wouldn't be surprised if I missed something... nothing I did feels overly clever, and I'd expect something like that in a particularly crafted puzzle such as this. The gray circle on the right does nothing and that gives me pause.
This is the fourth in a series Lights in the Dungeon puzzles.
previous (3rd) <--|--> next (5th)
The five by five dungeon was only a little harder than the previous ones. You're feeling pretty good about it and you're even up for a little taunting. "Alright, Hooty, where the six by six?" you ask as you approach the cloaked figure waiting for you at the exit.
"Six by six?" asks the Lord, feigning ignorance. "I'm afraid I don't know what you talking about. You see, Fakky Jim and I have been working on your next challenge while you were busy rearranging tiles and walking through the dungeon, flipping lights like a good little monkey. Your next challenge is... a bit more grand. If you succeed, I promise to tell you the real reason I've brought you here."
Intrigued, you walk over to Fakky Jim and receive a scroll with the rules. You can tell they've been re-written from scratch as they were getting a bit illegible with all the strikethroughs and revisions from the previous puzzles.
- You will enter a dungeon in which you must visit every room and turn off the light in the middle, an easy operation to perform.
- As soon as you exit a room, every door to that room slams shut and locks. You can not re-enter that room through any means.
- The map below shows the current design of the dungeon with each room bordered by red and labeled with a alphabetic character.
- You may rearrange the rooms however you like before you enter but you cannot change the perimeter of the dungeon. In this case, that means you must maintain the exact stairstep shape that is eighteen rooms tall and somewhere between seven and eleven rooms wide.
- You cannot rotate or flip rooms. You may only translate them.
- The dungeon must have exactly one entrance and one exit.
- There must be exactly one possible path through the dungeon. Obviously, any path is reversible if you simply enter from the other end but that does not count as a separate possible path.
Surprised, you look up at Fakky Jim and ask, "What happened to the collapsing floors?" The Grand Vizier just stares at you. You're not sure what you expected from a cat, magic though it may be. You turn to Lord Hooty and repeat the question.
"Ah," he starts, a sorrowful look upon his face, "we changed that after losing the cleaning crew. It was a tragedy, really. They were only doing their job. At least they were tasty so that's a silver lining."
How can you rearrange the dungeon rooms so that you can win the game?
There are probably multiple solutions. The first valid answer will be accepted.
AAAAAAA
AABBBBB
CCCCCCC
DDDDDDEE
EFFFFFFF
FFFFFFFF
FFFFGGGGG
GHHHHMMMM
MMMMMMMMM
MMMMMMMMMM
MMMMMMMMMM
MMMMMMMMNN
NNNNNNNNNNN
NNNNNNNNNNN
NNNNNNNNNNN
OOOOOOOPPPP
PPPPPPPPPPP
PPPPPPPPPPP
For reference, here are the 16 possible rooms. (Note that this is all possible rooms. For this puzzle, you must use exactly the rooms shown in the map above.)
Answer
I started from the bottom-right corner and tried to get rid of all the Ms and Ns first, so I came up with:
Or as text:
ACPPPBCPAGPPMMPBHBOMMPAGHHMMPPPPPOMMPFGGEDMMPFGNOPMMFCPPPFPMMMMFHBEPMMMMFADFCMMMMPFBCFFMMMMPAEFDFMMMMPFNNNDMMMANCANNNCMMFNGMFNNNOMFPFOMANNNNOFNDPMFNNNNNNNNODANNNNNNNNNN
Inspired by this question: Which country is INDIA in?
Japan is in India.
New Caledonia is in Lima.
Greenland is in Quebec, but only during the winter.
Portugal is in November in the winter too.
Beijing is in the Hotel.
How is this possible?
Like in my second part, I'm searching for the name of an anime. Knowledge from the anime is required so maybe, if you havent seen it, you cant figure it out. I hope you have fun :)
So Mr. ... youre really that small? ok, it just distracted me. Lets begin with our job interview. You are 21 years old. What are your hobbys?
Well I do all kinds of stuff, during my time stayed abroad I came across thieves and world destructing plans. I eventually caught some of them myself, but I never charged money for it. In retrospect I should have, it could have been useful for a bike. Well thats what I do most of the time.
Ehh, ok. This is an employment offer for a professor. You won't be the next Indiana Jones here.
Bonus: Should the person take the offer?
Answer
Wild Guess Alert!
Pokemon (gotta catch 'em all!)
So Mr. ... you're really that small? ok, it just distracted me. Lets begin with our job interview. You are 21 years old. What are your hobbys?
Pokemon is 21 years old, but Ash Ketchum is always 10.
Well I do all kinds of stuff, during my time stayed abroad I came across thieves and world destructing plans. I eventually caught some of them myself, butI never charged money for it. In retrospect I should have, it could have been useful for a bike. Well thats what I do most of the time.
Ash stays in several different regions, where he encounters Team Rocket (thieves) and world destructing plans (he's come across several.) He never charges money, and the bike in the original Pokemon games is the most expensive item in that game.
Ehh, ok. This is an employment offer for a professor. You won't be the next Indiana Jones here.
Each Pokemon region has a professor
Should he take the offer?
yes because all of the Pokemon professors are named after trees, and ash is a tree!
My friends Peter and Sam are excellent mathematicians and always think strictly logical. Yesterday I told them: "I have secretly chosen two integers $x$ and $y$ with $1\le x\le y\le 9$. I have told their sum $s=x+y$ to Sam and their product $p=xy$ to Peter." Then the following conversation developed.
What are the values of $x$ and $y$?
Answer
To solve this I listed all possible solutions and then removed each entry that couldn't be it after the question. The thing is, a person knows what value x and y if there is only one entry in that list with that given s or p. Saying that you don't know the solution implies that I can remove all items from the list that only have a single entry with that s or p. Here is the list:
xy s p
11 2 1
12 3 2
13 4 3
14 5 4
15 6 5
16 7 6
17 8 7
18 9 8
19 10 9
22 4 4
23 5 6
24 6 8
25 7 10
26 8 12
27 9 14
28 10 16
29 11 18
33 6 9
34 7 12
35 8 15
36 9 18
37 10 21
38 11 24
39 12 27
44 8 16
45 9 20
46 10 24
47 11 28
48 12 32
49 13 36
55 10 25
56 11 30
57 12 35
58 13 40
59 14 45
66 12 36
67 13 42
68 14 48
69 15 54
77 14 49
78 15 56
79 16 63
88 16 64
89 17 72
99 18 81
After question 1 I can remove 11, 12, 13, 15, 17, 25, 27, 35, 37, 39, 45, 47, 48, 55, 56, 57, 58, 59, 67, 68, 69, 77, 78, 79, 88, 89 and 99.
That leaves us with:
xy s p
14 5 4
16 7 6
18 9 8
19 10 9
22 4 4
23 5 6
24 6 8
26 8 12
28 10 16
29 11 18
33 6 9
34 7 12
36 9 18
38 11 24
44 8 16
46 10 24
49 13 36
66 12 36
After question 2 we remove 22, 49, 66. we now have:
xy s p
14 5 4
16 7 6
18 9 8
19 10 9
23 5 6
24 6 8
26 8 12
28 10 16
29 11 18
33 6 9
34 7 12
36 9 18
38 11 24
44 8 16
46 10 24
After question 3 we eliminate 14, we now have:
xy s p
16 7 6
18 9 8
19 10 9
23 5 6
24 6 8
26 8 12
28 10 16
29 11 18
33 6 9
34 7 12
36 9 18
38 11 24
44 8 16
46 10 24
After question 4 we eliminate 23, we now have:
xy s p
16 7 6
18 9 8
19 10 9
24 6 8
26 8 12
28 10 16
29 11 18
33 6 9
34 7 12
36 9 18
38 11 24
44 8 16
46 10 24
After question 5 we eliminate 16, we now have:
xy s p
18 9 8
19 10 9
24 6 8
26 8 12
28 10 16
29 11 18
33 6 9
34 7 12
36 9 18
38 11 24
44 8 16
46 10 24
After question 6 we eliminate 34, we now have:
xy s p
18 9 8
19 10 9
24 6 8
26 8 12
28 10 16
29 11 18
33 6 9
36 9 18
38 11 24
44 8 16
46 10 24
After question 7 we eliminate 26, we now have:
xy s p
18 9 8
19 10 9
24 6 8
28 10 16
29 11 18
33 6 9
36 9 18
38 11 24
44 8 16
46 10 24
After question 8 we eliminate 44, we now have:
xy s p
18 9 8
19 10 9
24 6 8
28 10 16
29 11 18
33 6 9
36 9 18
38 11 24
46 10 24
Conclusion:
If Peter didn't know it now, he would eliminate 28 now which means that x = 2 and y = 8 is the answer. Sam knows this too because he thinks the same like Peter (and me).
We all know the Simpsons' little greyhound.
But what is Bart's pet saying in this picture?
Answer
It must be:
Eat My Shorts!
He's repeating anagrams of Bart's catchphrase. What a clever dog, that Santa's Little Helper!
For the record, Deusovi answered first (+1), verifying that they were all anagrams, but didn't come up with the one which most fits the puzzle.
He might also be barking "HANDWOVEN AT CAMO" ;)
The left and right numbers are linked. What should the last number on the right be, and why?
\begin{align} 135759&: 1 \\ 151364&: 4 \\ 255075&: 9 \\ 279422&: 36 \\ 292620&: 91 \\ 348777&: 135 \\ 398067&: 147 \\ 417894&: 265 \\ 459431&: 279 \\ 478926&: 307 \\ 609941&: 363 \\ 689245&: 435 \\ 814576&:\space? \\ \end{align}
Before anyone bothers, no right is not simply a polynomial of left (of course we could certainly fit one, but there is another way).
Hint
There is information somewhere in the community wiki post that may prove useful
Hint 2
The mapping from left to right is many to one and there are infinite possibilities for right.
Two more entries are
\begin{align}0&: 1\\1&: 0\\\end{align}
Hint 3
Nothing special about these, they are really just some more examples...
Since there were none, here are some with $8$s on the right: \begin{align}157388&: 8\\272813&: 18\\276276&:28\\291384&: 88\\\end{align}
To have a cube $\gt1$ on both sides, left must have at least $7$ digits, here are a few: \begin{align}1259712&: 27\\224755712&: 729\\559476224&: 1000\\1427628376&: 6859\\\end{align}
Answer
What's the $\bf ?$ in $~ \bf 814576 : \, ? ~$ ? $\require{begingroup} \begingroup \let \SSS \tiny \let \SS \scriptsize \let \S \small \def \T {\small\sf} \def \Q #1{{ \bf ? { \S\raise2mu ( } #1 { \S\raise2mu ) } }} \def \R #1{ \phantom{\Q0} \llap{ \bf #1 \kern 7mu } } \def \x #1{{\SSS\kern1mu\raise2mu \times \S\kern5mu 3 \kern1mu\SS\raise7mu #1}} \def \p { \kern8mu{ \S \raise1mu + }\kern8mu} \def \D { \kern6em } \def \E #1{ \D \llap{ #1 \kern4mu { \S\raise1mu = } \kern6mu } } $
$ \kern-12mu \E{ \bf ? } 514 $
Essential $\bf 1 \over 10\raise-4mu{\small\,3}$ of the story
$ \E{ \Q{\T digit} } $ how many enclosed regions, typically loops, in the drawing of the single $\T digit$
$ \E{ \Q{\T digits} } $ decimal value of the ternary number formed
$ \D \raise-6mu\strut $ when each $\T digit$ of $\T digits$ is replaced by $\Q{\T digit}$
$ \E{ \Q{814576} } \Q 8 \x 5 \p\Q 1 \x 4 \p\Q 4 \x 3 \p\Q 5 \x 2 \p\Q 7 \x 1 \p\Q 6 \x 0 $
$ \E{} \R2 \x5 \p\R0 \x4 \p\R1 \x3 \p\R0 \x2 \p\R0 \x1 \p\R1 \x0 $
$ \E{} 201001\raise-4mu{\small\,3} $
$ \E{} 514\raise-4mu{\scriptsize\,10} $ $\endgroup$
Middle $\bf 1 \over 10\raise-4mu{\small\,3}$ of the story
Notice the following retrieval from the community evidence locker. It originated from using various radices for the numbers being puzzled and displaying the only two radices with suggestive patterns.
left : right binary ternary
135759 : 1 1 1
151364 : 4 1.. 11
255075 : 9 1..1 1..
279422 : 36 1..1.. 11..
292620 : 91 1.11.11 1.1.1
348777 : 135 1....111 12...
398067 : 147 1..1..11 1211.
417894 : 265 1....1..1 1..211
459431 : 279 1...1.111 1.11..
478926 : 307 1..11..11 1.21.1
609941 : 363 1.11.1.11 11111.
689245 : 435 11.11..11 121.1.
Watch what happens, starting with the dots, when ...
... the middle two columns are ignored and left-column digits without enclosures are dotted out.
left right ternary
.....9 1
....64 11
...0.. 1..
..94.. 11..
.9.6.0 1.1.1
.48... 12...
.9806. 1211.
4..894 1..211
4.94.. 1.11..
4.89.6 1.21.1
60994. 11111.
689.4. 121.1.
No wonder the ternaries had so few $\small\tt 2$s, a peculiarity that had been noted early without direct result. Only the digit $8$ has two loops and thus transforms into $\small\tt 2$, while each of four digits$-0 \, 4 \, 6 \, 9-$has just one loop (or enclosed region) and transforms into $\small\tt 1$.And no wonder that a few ternary numbers were exactly as wide as the left-column numbers and that none were wider, which was not noted along the way.
Final $\bf 1 \over 10\raise-4mu{\small\,3}$ of the story
How would anyone notice the above relationship? By following up on clues, such as those just mentioned, and allowing for dumb luck coincidence.
The records room at Puzzling HQ has a dossier on The Case of the Really, really, really hard sequence, which was reported just 11 days after the present puzzle and hinges on an eerily comparable ternary modus operandi. Sure enough, that case was cracked by the poser of the present puzzle, in barely 86 minutes and 48 seconds. To see a new puzzle that uses a similar device must have been precious! And then to write a solution that flaunts the very key to their own unsolved puzzle.
The present puzzle's poser's pathological ternary obsession did not go unnoticed, nor uncontracted, by the detective assigned to this case. Sooner or later, well, much after the hint ...
... $0 \kern2mu {:} \kern1mu 1$ revealed that two wildly different numbers, $135759$ and $0$, can produce the same degenerate result, $1$, it was time to consider that individual digits of the puzzling numbers might be nullifying each other or even being ignored. The binary and ternary patterns above seemed like the simplest leads to follow when embarking on this. (Not) one of the (first) possibilities to touch on, if only to shake up thought patterns, was the graphical way of looking at digits in that other case.
And it happened to work. Tickle me pink.
Dear PSE users and moderators,
I’m new here in PSE, but I really need your help. There was this person who gave me a black envelope consisting 10+1 pages of puzzles, and also a scribble saying: “Find our favorites and you will be accepted to join our ‘pyramid cult’. Feel free to ask for help from your beloved friends on PSE. They will surely guide you into all the truth.” I’m also a newbie on grid puzzles, so, could you please give me any hint to solve these? It’s getting harder and harder later on..
- athin
Jump to the first page: #1 Numberlink | Previous page: #5 Slitherlink | Next page: #7 Fillomino
Rules:
- Draw a line to make a single loop.
- Lines pass through the centers of cells, moving in perpendicular direction with one of the cell sides, or turning. The loop never crosses itself, branches off, or goes through the same cell twice.
- The numbers show how many shaded cells there are in the direction of the arrow.
- The loop does not pass through the shaded cells or the cells with numbers, and shaded cells are not adjacent to their side.
Special thanks to chaotic_iak for testing this puzzle series!
Answer
First, some fairly trivial deductions:
Note that any "dead-end" triangle cannot be used, and so it must be shaded. This gives all of these:
And then if any of these cells were shaded, they would create a dead end cell next to them:
Then, a few less trivial deductions:
A single segment in the bottom left can be drawn. Then, consider the cells marked in red and purple:
Consider the [2↗] clue: one of the two remaining cells it points to is shaded. If it is the white cell, the lower red cell is shaded; if it is the red cell, the lower white cell is shaded. Either way, one of the two red cells is shaded, and so (because of the [3↖] clue in the lower right corner) the purple cells are all unshaded.
Finishing off the lower right corner:
With that knowledge, we can barely squeeze in the remaining required cells for the 4 clue:
And then we can keep making straightforward deductions from that to resolve the corner completely.
Using the last clue:
The only remaining clue is the 3↖ that's not in the corner. To allow enough space, one of the two shaded cells must be against the wall:
If either of the two endpoints in the center of the puzzle goes to the cell marked "c", it must then continue up and left to the cell marked "a" (or otherwise it would complete a loop too early). But one of those two cells must be shaded because of the 3 clue. So neither of those endpoints goes to the "c" cell, and therefore that cell is shaded.
And now the loop can be fully resolved:
Taking the letters on shaded cells, we see that the pyramid cult's favorite places are LOCHANS (a Scottish word for small lakes).
When you figure out that the guy speaking Gibberish meant
I am Hagey
you go back to the restaurant the following day, and hope for the best. When you walk in you notice the man who said the Gibberish in the first place but not the other man. After you order your food and start eating you hear the door bell chime and see the other man walk in and sit down. You hear him say something but it's not all that clear...
...something to do with, coffee? Anyways after he says that he starts spouting random numbers all over the place and then leaves. As soon as you hear him talking you got your notepad ready and after he finished you got this:
78 79 87 32 84 72 65 84 32 87 69 32 75 78 79 87 32 89 79 85 32 65 82 69 32 80 82 79 70 69 83 83 79 82 32 72 65 71 69 89 32 87 69 67 65 78 32 71 73 86 69 32 89 79 85 32 84 72 69 32 66 79 88 190
It's up to you to solve it. Can you do it?
Hint:
What the man said before he said the numbers was: "did you know you can program with coffee"?
Answer
Decrypting the cipher we get
using decimal to ASCII: "NOW THAT WE KNOW YOU ARE PROFESSOR HAGEY WE CAN GIVE YOU THE BOX."
There are a few mistakes in the encoding: there's a space missing between "WE" and "CAN", a space missing between the two "83"s that make up the Ss in PROFESSOR, and the last character (a period) should be encoded as 46. The rest matches up with the intended plaintext.
OP points out in comments that the intended encoding was actually
Javascript keycodes. Most of them match up with their ASCII counterparts, but the period is in fact encoded as 190 in this encoding.
One arm of a weighing scale is twice as long as the other.
Suppose you want to create a set of weights so that any object weight from 0.5 and its multiples up to 40 pounds can be exactly balanced there by placing a certain combination of these weights onto that scale.
What is the fewest number of weights you need, and what are their weights?
This question is different from the existing one. This one has a twist: the length of the arms and it must be from 0.5 and all its multiples.
Answer
If the object is in the pan with the long arm, then we have $$object+(some\ weights) = (some\ other\ weights)/2$$ If the object is in the other pan then we have the similar $$object+(some\ weights) = (some\ other\ weights)*2$$ The solution I have chosen only uses the first, i.e. the object is only ever on the long-armed side.
I use the seven weights $1$, $1$, $5$, $5$, $25$, $25$, $125$. The pattern obviously extends, being the powers of $5$, each weight occurring twice.
As proof that these weights work, here are all possibilities:
0.5 = 1 /2
1 = 1+1 /2
1.5 +1 = 5 /2
2 +1 = 5+1 /2
2.5 = 5 /2
3 = 5+1 /2
3.5 = 5+1+1 /2
4 +1 = 5+5 /2
4.5 +1 = 5+5+1 /2
5 = 5+5 /2
5.5 = 5+5+1 /2
6 = 5+5+1+1 /2
6.5 +5+1 = 25 /2
7 +5+1 = 25+1 /2
7.5 +5 = 25 /2
8 +5 = 25+1 /2
8.5 +5 = 25+1+1 /2
9 +5+1 = 25+5 /2
9.5 +5+1 = 25+5+1 /2
10 +5 = 25+5 /2
10.5 +5 = 25+5+1 /2
11 +5 = 25+5+1+1 /2
11.5 +1 = 25 /2
12 +1 = 25+1 /2
12.5 = 25 /2
13 = 25+1 /2
13.5 = 25+1+1 /2
14 +1 = 25+5 /2
14.5 +1 = 25+5+1 /2
15 = 25+5 /2
15.5 = 25+5+1 /2
16 = 25+5+1+1 /2
16.5 +1 = 25+5+5 /2
17 +1 = 25+5+5+1 /2
17.5 = 25+5+5 /2
18 = 25+5+5+1 /2
18.5 = 25+5+5+1+1 /2
19 +5+1 = 25+25 /2
19.5 +5+1 = 25+25+1 /2
20 +5 = 25+25 /2
20.5 +5 = 25+25+1 /2
21 +5 = 25+25+1+1 /2
21.5 +5+1 = 25+25+5 /2
22 +5+1 = 25+25+5+1 /2
22.5 +5 = 25+25+5 /2
23 +5 = 25+25+5+1 /2
23.5 +5 = 25+25+5+1+1 /2
24 +1 = 25+25 /2
24.5 +1 = 25+25+1 /2
25 = 25+25 /2
25.5 = 25+25+1 /2
26 = 25+25+1+1 /2
26.5 +1 = 25+25+5 /2
27 +1 = 25+25+5+1 /2
27.5 = 25+25+5 /2
28 = 25+25+5+1 /2
28.5 = 25+25+5+1+1 /2
29 +1 = 25+25+5+5 /2
29.5 +1 = 25+25+5+5+1 /2
30 = 25+25+5+5 /2
30.5 = 25+25+5+5+1 /2
31 = 25+25+5+5+1+1 /2
31.5 +25+5+1 = 125 /2
32 +25+5+1 = 125+1 /2
32.5 +25+5 = 125 /2
33 +25+5 = 125+1 /2
33.5 +25+5 = 125+1+1 /2
34 +25+5+1 = 125+5 /2
34.5 +25+5+1 = 125+5+1 /2
35 +25+5 = 125+5 /2
35.5 +25+5 = 125+5+1 /2
36 +25+5 = 125+5+1+1 /2
36.5 +25 +1 = 125 /2
37 +25 +1 = 125+1 /2
37.5 +25 = 125 /2
38 +25 = 125+1 /2
38.5 +25 = 125+1+1 /2
39 +25 +1 = 125+5 /2
39.5 +25 +1 = 125+5+1 /2
40 +25 = 125+5 /2
40.5 +25 = 125+5+1 /2
41 +25 = 125+5+1+1 /2
...
93.5 = 125+25+25+5+5+1+1 /2
I'm not sure this is optimal, but it is at least a generic solution that can be extended to any weight range.
EDIT:
Electric_monk's answer shows that you can use the same number of weights but with a lower maximum weight for the range given in the problem.
Keeping all the weights on the short arm, opposite the object, you have the traditional solution with weights $1$, $2$, $4$, $8$, $16$, $32$, $64$. This allows one to weigh any object up to $127/2=63.5$ in weight with seven weights. So for the range given in the problem, there is no need to use a subtractive system of weights yet (unless there is some more optimal answer with 6 weights only). It seems therefore that it would have been better for a larger range to have been used, e.g. $0.5$ to $80$.
EDIT2:
Oray was the first to announce that
6 weights is enough for the given range of object weights.
But I found something similar.
The weights $1$, $1$, $5$, $5$, $25$, $50$ suffice. Almost the same configurations can be used as in my 7 weight solution, except that when 25+25 was used before, you now use the 50, and instead of -25 + 125/2 we use (25+50)/2. The heaviest object that this set can handle is 43.5.
You can also use $1$, $1$, $5$, $6$, $28$, $56$ to get objects up to 48.5 pounds, but the various arrangements are rather more haphazard.
“The Two Sheriffs” puzzle was already discussed here Two Sheriffs and Eavesdroppers. This puzzle has only one difference: we have seven suspects instead of eight.
Two sheriffs in neighboring towns are on the track of a killer, in a case involving seven suspects. By virtue of independent, reliable detective work, each has narrowed his list to only two. Now they are engaged in a telephone call; their object is to compare information, and if their pairs overlap in just one suspect, to identify the killer.
The difficulty is that their telephone line has been tapped by the local lynch mob, who know the original list of suspects but not which pairs the sheriffs have arrived at. If they are able to identify the killer with certainty as a result of the phone call, he will be lynched before he can be arrested.
Can the sheriffs, who have never met, conduct their conversation in such a way that they both end up knowing who the killer is (when possible), yet the lynch mob is still left in the dark?
Original problem is takken from: Mathematical Puzzles, a Connoisseur's Collection, Peter Winkler. The solution for seven suspects belongs to Yoav Kallus.
Answer
Assign each suspect a number $1-7$. Define the xor of two numbers between $1$ and $7$ as follows: write them in binary, then xor them together bitwise. For example, $3 \oplus 6=011\oplus 110=101=5$.
Some observations:
$a\oplus b$ is different from both $a$ and $b$.
If $a\oplus b=c$ and $d\oplus e=f$, then the sets $\{a,b,c\}$ and $\{d,e,f\}$ have an element in common. This can be verified by checking the only sets of three numbers where two of them xor together to make the other are {1,2,3},{1,4,5},{1,6,7},{2,4,6},{3,4,7},{3,5,6},{2,5,7}, and that any two of these intersect.
The converstion: Lets say Alice has {a,b} and Bob has {a,c}. Alice announces $a\oplus b$ and Bob announces $a\oplus c$. If Alice and Bob say the same thing, then they know they have the same pair, and hang up. If not, there are two possibilities:
Why the sheriffs know the killer: It's easy to see why they do in case 1, since they each heard the other's pair. In case, 2, when Alice hears Bob say $b$, she knows $b$ is not the killer, so it is $a$. Same thing happens for Bob.
Why the lynch mob doesn't: Suppose that they hear the numbers $a_1$ and $b_1$, followed by the pairs ${a_2,a_3}$ and ${b_2,b_3}$. It could be the case that Alice had the pair ${a_2,a_3}$ and Bob has the pair ${b_2,b_3}$, in which case the killer would be the intersection of these sets (this would be case 1). However, it could also be the case that Alice had $\{a_1\oplus b_1,b_1\}$ and Bob had $\{a_1\oplus b_1,a_1\}$. Here we are in case 2, where Alice and Bob just randomly chose to say $\{a_2,a_3\}$ and $\{b_2,b_3\}$. No matter what sets they randomly announced, by bullet point 2, they will always intersect, so this is really indistinguishable from case 1.
For example, here are two situations where the Sheriffs have the same conversation, but the killer is a different person.
Alice has {1,2}, while Bob has {1,5}. Alice says 3, Bob says 4. Since 4 is not in {1,2} and 3 is not in {1,5}, we are in the first bullet point. They then each announce their lists, {1,2} and {1,5}. They now both know that the killer is 1.
Alice has {4,7}, Sheriff 2 has {3,7}. Alice says 3, Bob says 4. Each number is in the other's list, we are in the second bullet point. They know now the killer is 7. Alice randomly announces one of the pairs {1,2} or {5,6}, since these are the pairs which xor together to make 3 (besides {4,7}). Let's say she chooses {1,2}. Similarly, Bob randomly says one of {1,5}, or {2,6}, and he randomly chooses {1,5}.
