The worlds first riddle!

Colored balls: 2
First weighing: r1,g1 vs. r2,y1
Second weighing:
If first weighing balanced: r1 vs. r2
Otherwise: g1 vs. y2
The possible results of each weighing are L (left side heavier), R (right side heavier), E (equal).
The heavy balls that result from the possible combined outcomes are:
E,L - r1,g2,y1
E,R - r2,g1,y2
L,L - r1,g1,y1
L,E - r1,g1,y2
L,R - r1,g2,y2
R,L - r2,g1,y1
R,E - r2,g2,y1
R,R - r2,g2,y2

No one understands





Goodness knows whether or not anyone will solve Trya's question about how many words can be made from the letters SAMPLE.

sample:
6+6*5+6*5*4+6*5*4*3+6*5*4*3*2+6*5*4*3*2*1=1956

As you may know, Kuphus polythamia, is the longest living bivalve known to man… it is not know if wimmins know of a longer shipworm.


Well shiver me timbers; Aaagh Mark returns for talk like an archetypal pirate day.

Good to see ya matey. I think you know Ekename – a comely wench believed to have been involved in the great A2K Whipped Cream Scandal and became very slippery as a result… or is he the 300 pound steel worker from Milwaukee with seis bambinos living in a double-wide. Yarrr, I can only dream of such luxury.


Goodness knows and so does Mark the answer to the ‘Sample’ question, and to be honest, although I don’t want to blow my own trumpet (I ain’t double jointed you know) so do eye.

N = p(6.1) + p(6.2) + (6.3) +(6.4) + (6.5) + (6.6) = 6+30 + 120 +360 +720 + 720 = 1956

Lang may yer lum reek young Mark… well compared to me and other Dinosauria from taxonomic, morphological and ecological standpoints.


Colored balls:

I cannot fault yore logic and I have:

Two weighings suffice.

Weigh one red and one green ball against one yellow and the other green ball.
If the weights are equal, then the red and the yellow differ in weight.

A weighing between these two balls allows us to deduce the weights of all other balls.

If the red-green combination was heavier than the yellow-green combination in the first weighing, then the green ball in the red-green combination is certainly heavy and the other green is light.

Now take the red from the red-green combination and the yellow from the yellow-green combination. Weigh these together against the remaining red and the remaining yellow.

The only interesting case is when this weighing is "equal". Then, the red from the red-green combination must be heavy and the yellow from the yellow-green combination must be light.




Buenos dias, como estas Eke, U wanna shake ya booty: https://www.youtube.com/watch?v=MXVx6yJQbn8


We just have time, but how do we measure forty-five minutes when you only have tres (3) identical fuses, each of which takes an hour to burn?

We have matchsticks with us, flintlocks, cutlasses and a barrel of rum.

NOTE: The fuses burn non-uniformly.

So, for example, the two halves of a fuse might burn in 20 minute and 40 minutes respectively (or any other time).



So how do pirates know if ye be a pirate or knot?






A: Ye think, therefore ye ARRRR!!!!!

fuses:
time 0m: Light both ends of fuse A and one end of fuse B.
time 30m: Fuse A has burned out and fuse B is half burned. Light the other end of fuse B.
time 45m: Fuse B has burned out, and you still have fuse C to ignite your explosive!

This is timely. I recently extended the sequence of the number of achievable times with N fuses: https://oeis.org/A283075 (scroll down to the bottom). a(13) took 12.5 hours to compute and used 249GB on a machine with 256GB.

This reminds me of the time I was at tending a bar in Wisconsin and three bi-valves mussel up and come in to suck it and see.

They become unhinged at the very thought of the three likely looking clams pole dancing and already half full to the gills.

The lads proposed a transylvanian lottery where the girls get shells equal to the sum of the lowest of two numbers chosen randomly from 1 to 14 and the boys get shells equal to one number, the highest number randomly chosen,

Each team bets one shell per game , rough or smooth 50/50, until the other team had lost all its shells to determine who pays for the mineral water and what with much roister, boister and oyster they left but I still wonder who paid in all probability?.

I am so sorry, there appears to have been an alphabetical malfunction, in that I did not say you were a hippopotamus. I said, the hippocampus is the place for short-term memories while the cortex is home to long-term memories and the results, published in the journal Science, showed that memories were formed simultaneously in the hippocampus and the cortex.


I hope that explains why I support Scotland becoming Canada’s 11th province.
Come on, I’m serious, after all Scotland is closer to Newfoundland than Hawaii is to California… and don’t forget, more than half of Canada's prime ministers have claimed Scottish heritage.


“This is timely. I recently extended the sequence of the number of achievable times with N fuses: https://oeis.org/A283075 (scroll down to the bottom). a(13) took 12.5 hours to compute and used 249GB on a machine with 256GB.”


Holey moley! That was timely indeed and as for-

“a(13) took 12.5 hours to compute and used 249GB on a machine with 256GB.”

That is absolutely amazing, I am flabbergasted by your perspicacity, perspicuity and decipherability. I welcome the time when you will be inducted into the A2K Hall of Fame and presented with the Golden Keys to the executive bathroom.

I would however wish to point out that from the example shown in your link:

“…when the first rope's flames have reached the middle…”

But as each rope burns at a nonconstant rate, the ‘middle’ is no indicator of time. Or am I missing something? Just sayin’.


Eke beautifully wrote, “This reminds me of the time I was at tending a bar in Wisconsin and three bi-valves mussel up and come in to suck it and see...”

Man, I was just across the street in the Harley-Davidson museum – dagnabbit if that don’t beat all and the answer is plain to sea: https://www.poetryfoundation.org/poems-and-poets/poems/detail/43914





The greatest common divisor (gcd) of two or more integers is the greatest integer that evenly divides those integers. For example, the gcd of 8 and 12 is 4 (usually written as gcd(8,12)=4). Two integers are called coprime (or “relatively prime”) if their gcd is equal to 1.

So a reasonable question to ask is…
Given two randomly chosen integers a and b, what is the probability that gcd(a,b)=1?

If no prime divides n, then n=1, by the fundamental theorem of arithmetic. So we want to find the probability that no prime divides gcd(a,b). Let a and b be two randomly chosen integers and p a fixed prime. If p divides both a and b then gcd(a,b)≥p.

We are interested in the probability that this does not occur for any p.
Now I need to say what I mean by “randomly chosen”. It’s very handwavey, but we will calculate the probability for an integer chosen uniformly at random between 1 and N, and then let N tend to infinity. We won’t actually do that, but that’s what we’re thinking of.

Every pth integer is divisible by p, so the probability that p divides a and b is 1p2, and the probability that it divides at most one of a or b is 1−1p2. That’s a good start, but what we want is this probability over all primes. The probabilities for each prime are independent of each other, so you can just multiply them all together. Therefore the probability that gcd(a,b)=1 is Pr(gcd(a,b)=1)=∏p prime(1−1p2).


Make mine a Harvey Wallbanger ;0)





In this thread, there are 100 members. Each of whom is either crazy or sane.

However, if you choose any two members at random, at least one is crazy.

How many sane members are there?

"I would however wish to point out that from the example shown in your link:

“…when the first rope's flames have reached the middle…”

But as each rope burns at a nonconstant rate, the ‘middle’ is no indicator of time. Or am I missing something? Just sayin’."

You're absolutely right. That should be stated differently. That wasn't my doing.

sane members: at most 1

Assuming three random numbers are chosen (ladies get sum of lowest two, guys get highest). On average, the women's shells will outnumber the men's as often as the men's outnumber the women's.

probability that gcd(a,b)=1

The number you seek is 6/pi^2.

Congratulations Mark you found the sane one – but who is it!!!


“The number you seek is 6/pi^2.”

Touché mon ami.

I shall be seeking an immediate refund from the Universidad San Pablo de las Salinas from whom I purchased my Masters in applied mathematics for $19.99.

Clearly this diploma is not working and I will now have to rely on an even cheaper version from Harvard covering: Differential equations; approximation theory (broadly construed, to include representations, asymptotic methods, variational methods, and numerical analysis); and applied probability.

At least I can say I did have breakfast in Annenberg Hall (formerly Alumni Hall) back in the day… <sigh>


I apologize for the absence of Eke due to a sore head - which is due entirely to Harvey Wallbanger - so I have suggested she move the bed further from the wall!!!


I wonder iffin’ y’all could help me identify Harvey…

All I know is he is one of a set of triplets, always tells lies and needless to say they all look alike.

The other siblings are Harvard who is also an inveterate liar and Howard.

Their Mom told me that they went out wearing red, white and blue bowling shirts, but didn’t know who was wearing what color.

However the good news is that one of them is attending a senior management meeting in the A2K kindergarten today and you can ask ONE question.

Due to time restraints this question has been limited to no more than Trois (3) words.

BTW: You have no idea who will turn up, but I hope it is Howard as he is the only one who always tells the truth.

So what question are you gonna ask?

In the meantime I am going to have a siesta as I am experiencing various unpleasant physiological and psychological effects following the excessive consumption of ethanol.

Call me an old hillbilly and you can STFU…

However, I have now been informed that STFU isn’t necessarily the first acronym that comes to peoples mind when I mean: Stand Tall For (the) Union; but I find something profoundly poignant in this week’s news from the U.K. that the dear old newspaper The Guardian is actively considering moving its headquarters from north London back to its birthplace in Manchester.

If the paper does indeed return home, the locals will find it completely unrecognisable as the great, crusading liberal organ that started life in their magnificent city back in 1821, before it set off in 1964 to lose its fortune in London.

The modern equivalent of the prodigal’s father was C. P. Scott, who edited the Manchester Guardian (as it was originally called) for 57 years from 1872, becoming its owner in 1907 when he bought it from the estate of the son of its founder, a cotton merchant.

In the 19th century, the Manchester Guardian was the voice of manufacturing and free trade, a stalwart champion of small government. With its eye constantly on promoting economic growth, the paper took a particularly dim view of workers’ strikes, arguing that the raison d’etre of trade union leaders was to avoid settling disputes, since ‘they live on strife’.

In the words of its working-class competitor, the Manchester and Salford Advertiser, the Guardian was ‘the foul prostitute and dirty parasite of the worst portion of the mill-owners’.

According to David Kynaston’s masterly history, Austerity Britain 1945-1951, the paper even opposed the creation of the NHS. This was on the morally repugnant grounds that free healthcare for all would ‘eliminate selective elimination’, leading to an increase in the numbers of congenitally deformed and feckless people. In other words, the Guardian wanted the weak to be left to die.

To which I say; HDF – Halt Die Fresse.


Now I know y’all want to know where Mark and Eke have gone and although I am sworn to secrecy, I had my fingers crossed behind my back, so it don’t count - right!

Well, when Mark found out Eke liked golf… he asked if she would like to play a round!!!

Mark has a good approach, so if he was to score a hole in one… I’ma sure we will be able to solve this puzzle:


One of the best kept A2K secrets is its private golf course – I know, I was shocked too – anyhoo, they have this odd system to book your tee-off times, whereby each minute members would come and put in one numbered golf ball apiece into a massive empty chute.

The first minute starting at 12noon (Mountain Standard Time (MST)), the first member came and put in a ball (number 1).

The second minute, 2 members put in one ball apiece (2 and 3). A minute later, 4 members put in 1 ball apiece…

This pattern continued until the chute was full at exactly six post meridiem.

At what time was the chute half empty?

tee times:
1 minute before 6:00 p.m. However, that was one big chute. More balls were placed into it than there are atoms in the universe.

Oh please, when I say, Have a ‘Gute fährt’, it has no scatological reference to anything other than ‘have a good journey’ in German.

Mine Gott what were you thinking!


G’day Mark, I hope you and yours enjoyed the weekend and what were you thinking?

“1 minute before 6:00 p.m. However, that was one big chute. More balls were placed into it than there are atoms in the universe.”


Man I hear that, A2K membership is expanding exponentially due to the fact that you can join… but you can never leave. (In my defense I did say it was ‘massive’)

It is all part of the circle of life and can be witnessed at the A2K selection process for Moderators when they find a Trump supporter…

There are n A2K members in a circle, numbered 1 thru n. Going around the circle, every second person is removed from the circle, starting with person number 2, 4, and so on.

Can you prove the last person standing is not a Republican, or at least show that the number of the last person remaining in the circle can be obtained by writing n in binary, then moving the leftmost 1 to the right?

So for example, with n = 13 persons (1101 in binary), the last person is number 11 (1011 in binary).



Note: This conundrum has been sponsored by the leftwing Bernie Sanders woz robbed circle of friends.

Should you wish to sponsor a puzzle, please write your own epitaph on the back of a ‘Benjamin’ and leave in reception – Thank You and y’all have a nice day.

Seriously, do you know your Homo’s from your Hetero’s?


Laughing So Hard I Dropped My Taco and My Sombrero Fell Off!



Try to remember the simple way:

Homonyms: spelled the same – sound the same (can – can)

Homophones: same sound – different spelling (week – weak)

Homographs: spelled the same – different sound (wind – wind)

Heteronyms are a type of homograph. Heteronyms are words that are spelled the same, but have different pronunciation.

Still confused?


For example: Oh tell v hotel, and some of you may know the A2K Hotel outside Springfield, it has 1000 rooms. Initially, all rooms are marked with - signs.

From days 1 thru 1000, the chief moderator toggles marks on some of the rooms: from + to - and from - to +.

On the i-th day, the signs on rooms that are multiples of i get toggled.
On the 1001-th day, all rooms marked with + signs are opened.

Which rooms are these?

I am incandescent with sanctimonious ignominy over the fact that the Brits have purloined our very own indigenous peoples word ‘mugwump’!!!

Forshame on you Limey’s as the term originates from the Native American Massachusett language, spoken in part of New England in pre-colonial times. Mugumquomp meant a war leader, from which mugquomp – meaning an important person or kingpin – derived.

American settlers appropriated the word mugwump for the American language, with its most notable use being in the 1884 US presidential election, but it was used prior to the vote.

This meaning led to 'mugwump' being the label given to a group of Republican political activists who rejected party allegiance because of corruption, to support the Democratic candidate Grover Cleveland, who was seen as a reformer.

The word has also been used in literature. Roald Dahl used it to describe a group of caged monkeys in his children’s book The Twits. The mugwumps are forced by the Twits to do tricks, at the threat of a beating. The mugwumps later outwit the Twits and escape.

And this is how they did it…

You have 50 mugwumps, 50 Twits and 2 rooms. You can distribute the mugwumps and Twits in any way you wish between the two rooms, but all must be used.

Then one of the rooms is chosen at random and one person will be chosen from that room at random.

How would you maximize the chance its a mugwump?
What is the probability of doing so?


Who won the 1884 election?



BTW a word of caution to all Democrats thinking of escaping the Trump machine – A movie ticket in Toronto will set you back $18. And for a $8.99 bottle of California wine expect to be up-charged to around $20. There is no three-buck chuck in Canada!

rooms: squares because they have an odd number of divisors
mugwumps: a single mugwump in room 1; 49 mugwumps and 50 twits in room 2; probability of mugwump is 74/99

Holey guacamole, Mexico supplies 82% of avocado shipments into the US and I know Mark has a massive fan club and following, which is why I am as concerned as anyone over the disappearance of whatshername, but I think this quote from Jespah the A2K Chief of Staff may be responsible -

“We know you want to make A2K a better place. But eek, leave the malformed topics alone.”

So there you have it –

Now what does Mark think…

“mugwumps: a single mugwump in room 1; 49 mugwumps and 50 twits in room 2; probability of mugwump is 74/99

Lets do the math and see iffin he is right:

Let’s calculate the total probability.

P( mugwump ) = P( Room 1 ) * P( mugwump in room1 ) + P( Room 2 ) * P( mugwump in Room2 )

P( mugwump ) = 0.5 * 1 + 0.5 * 49/99

P( mugwump ) = 0.7474

Thus, we end up with ~75% chance of picking a cute mugwump and can confirm Mark’s findings.


Now what about the A2K hotel:

Mark says, “squares because they have an odd number of divisors”


Can you believe this guy!

A room is toggled as many times as the number of divisors it has.

For example, room number 24 is toggled on days 1, 2, 3, 4, 6, 8, 12 and 24.

Now, divisors come in pairs like 24 = 1*24 = 2*12 = 3*8 = 4*6. So the total number of divisors is even, except when the number is a perfect square, in which case the total number of divisors is odd.

Since all rooms are initially marked with - signs, only those rooms that have an odd number of divisors have + signs eventually. These are room numbers 1, 4, 9, 16, 25, 36, and so on.

I can confirm you can.



Jes and Eek are playing a game. They are teammates, so they will win or lose together. Before the game starts, they can talk to each other and agree on a strategy.

When the game starts, they go into separate soundproof rooms — they cannot communicate with each other in any way. They each flip a coin and note whether it came up Heads or Tails. (No funny business allowed — it has to be an honest coin flip and they have to tell the truth later about how it came out.)

Now Jes writes down a guess as to the result of Eek’s coin flip; and Eek likewise writes down a guess as to Jes’s flip.

If either or both of the written-down guesses turn out to be correct, then both win as a team. But if both written-down guesses are wrong, then they both lose.

Can you think of a strategy they can use that is guaranteed to win every time and does not involve ‘flipping me the bird’?





For those who are fans of Baseball we can see what the Boston Beaneaters* Charles Radbourn thinks –

https://en.wikipedia.org/wiki/File:Old_Hoss_Radbourn_finger.jpg

Seriously, the Boston Braves used to be called the Beaneaters!!!


No seriously, how much wood could a woodchuck chuck
If a woodchuck could chuck wood?

coin flipping: One of them will guess what she flipped. The other will guess the opposite of what she flipped. Exactly one of them will be correct.