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INTPf/INTPx Trivia Contest 2019

Utisz

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Questions remaining: 0/50

Link to INTPx thread.

Hope you enjoy. Good luck!

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Quiz Complete! Scores here

Winner: lbloom (245 points)
Runner-up: Serac (240 points)
Third place: Penguinhunter (185 points)

Congratulations!


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Identify the misheard lyric:


10 points lbloom (INTPx): Have you ever seen the rain - Creedence Clearwater Revival

20 points C.J.Woolf (INTPx): I can clearly now Lorraine is gone


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If Chad is five
Then Eritrea is six
And if Eritrea is six
Then Kazakhstan is seven
Then Kazakhstan is seven
Then Kazakhstan is seven

One doubts they're going to heaven

But which countries are one, two, three and four?

25 points Penguinhunter (INTPx): (1) Equatorial Guinea, (2) Cameroon, (3) Uganda, (4) Sudan

Countries ordered by longest serving presidents:


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Name the songs mixed together here:


Okay, admittedly this one was undercooked. There are three songs.

The first and by far the most obscure one is William Basinski - d|p 1.2.

The musician who wrote the second song is in his seventies.

The third song is something you could make a nice pie with.

10 points scarydoor (INTPx): Third song is Aphex Twin - Rhubarb


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How many convex regular polygons (i.e., equilateral triangles, squares, pentagons, and so on) can be drawn from the dots in the following animation?


The dots should form the vertices of the polygon and the polygon should be preserved in each frame of the animation. There is no restriction on how many polygons one dot can be used in.

20 points Hephaestus (INTPx): four triangles, three squares



The star is regular, but not convex (plus the points are not its vertices but rather travel along it).

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The colour ▀▀▀▀▀ replaced the colour ▀▀▀▀▀ a year after it was first published. But what's so special about ▀▀▀▀▀ anyways? (Or more precisely: what's so mundane about it?)

20 points MarkovChain (INTPx): Cosmic Latte

▀▀▀▀▀ is the average colour of the universe, according to a team of astronomers from Johns Hopkins University (originally they had worked out that the average colour was ▀▀▀▀▀, but corrected this soon after).

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You've entered a triathlon that involves swimming, running and maths. The starting point (x) is on land while the finishing point (+) is in the water.


You can swim a constant 2km/h in water while you can run a constant 10km/h on land. The straight-line distance (the dashed line) from start to end is 10km, of which 5km is land and 5km is water. However, you don't need to follow a straight line. The shortest path from the starting point to water is 3km (the dotted line). How far would you travel if you were to take the fastest route from the start to the end?

20 points ibloom (INTPx): 11.069 km

Actually I had computed the wrong answer before posting the quiz, but anyways, there are a couple of possible methods:

1) Serac (awarding 10 points for the method): letting y be the distance along the shore where you enter the water, the distance on land is sqrt(9 + y^2) km, the time on land is sqrt(9 + y^2)/10 h, the distance in water is sqrt(9 + (8-y)^2) km, the time in water is sqrt(9 + (8-y)^2)/2 h. Hence the total time is t = sqrt(9 + y^2)/10 + sqrt(9 + (8-y)^2)/2 hours. Where the slope of that function is 0, you've got the minimum, so differentiate for dt/dy and solve for the value where it's 0.

2) You can also use Snell's law, which tells us that in the fastest route, the ratio of the sines of θa and θb (see below) is equal to the ratio of the velocities in both media (10/2 = 5).



Working it out is rough but gives the same result.

3) Code it out I guess and brute-force the minimum time :)

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This short novel is about a man who lives on an egg and pursues a relative of the man who lives alongside him. The book received little attention in the lifespan of the author.

20 points Penguinhunter (INTPx): The Great Gatsby

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Why is Xzibit so happy in this photo?


Rocco Castoro would surely get this one.

10 points Serac (INTPf): There's metadata in the photo suggesting he's on Victoria Island.


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From which musical work is the following taken?


25 points Serac (INTPf): Liszt: Liebestraum No. 3


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Which philosopher (edit: whose doctoral supervisor shared his first name*) can be credited with the following paraphrased quote?

I am quite sure of the fact that more people die as a result of virtuous foolishness than outright evil.

* Adding edit 22:23 UTC because question might be underspecified.

10 points lbloom (INTPx): Anatole France
10 points Penguinhunter (INTPx): Dietrich Bonhoeffer

This question was definitely underspecified (sorry). I edited to hopefully make the answer more unique.

Mostly known for contributions to the philosophy of science, he was also credited with coining a paradox that speaks to current tensions over political correctness and inclusiveness.

20 points lbloom (INTPx): Karl Popper

It seems to me certain that more people are killed out of righteous stupidity than out of wickedness.
Doctoral supervisor was Karl Bühler.

He coined the paradox of tolerance.

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Which national flag is missing from the sequence below?


Odds are we'll be adding another Mexican flag on the end of that.

35 points lbloom (INTPx): France (Michel Hazanavicius)

The flags indicate the nationalities (more specifically countries of birth) of previous winners of the Academy Award for Best Director:

  • Tom Hooper (England): The King's Speech
  • Michel Hazanavicius (France): The Artist
  • Ang Lee (Taiwan): Life of Pi
  • Alfonso Cuarón (Mexico): Gravity
  • Alejandro G. Iñárritu (Mexico): Birdman
  • Alejandro G. Iñárritu (Mexico): The Revenant
  • Damien Chazelle (United States): La La Land
  • Guillermo del Toro (Mexico): The Shape of Water
Also Alfonso Cuarón is a strong favourite for the 2018 award on Sunday.

Dey coming o'er here, takin' our Academy Awards for Best Director

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The alt-right say it likes to fill safe spaces. What is this mathematical structure called?


40 points Rolling Cattle (INTPf): Flowsnake (aka. Gosper Curve)

It's a space-filling fractal/curve.


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Name the three people who have contributed to this "face":


35 points Limes (INTPx):
  1. Bernie Sanders (first proposed by lbloom 5 points)
  2. Helena Bonham Carter (first proposed by Blorg 5 points)
  3. Mike Tyson (first proposed by Madrigal 5 points)


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Billy is a 2d being with perfect 1d eyesight who lives on an infinite 2d plane (in the positive x and y directions). Currently Billy is at the origin (0,0) of a plane with a point at (a,b) for all a and b whole positive numbers. The points on the plane are mathematical points, meaning that they have no width or height, but if Billy looks right at them, he can see blue. On the other hand, if he is not looking at a point (in the arbitrary distance), he sees white. (If he looks along the x or y axis or away from the positive quadrant he sees nothing.)

Here we see Billy looking at the point (1,1), seeing blue.


Can Billy ever see white from his current position?

20 points Rolling Cattle (INTPf): He is almost surely going to see white wherever he looks.

Points can be described by the gradient he's looking out at, which is the fraction y/x (for example, looking at gradient 2, he looks at the point (1,2); looking at gradient 1/2, he looks at the point (2,1)). When y and x are integers (whole positive numbers) he will see blue; in other words, when y/x is rational, he will see blue. On the other hand, if he looks out at a gradient of π (pi) for example, he will see white since π is irrational (he'll look just past, e.g., (7,22) since 22/7 is a very good approximation of π, but an approximation is not good enough).

Fun fact 1: Since the rational numbers are countably infinite and the irrational numbers are uncountably infinite, essentially between any two blue (rational) gradients, there are infinite white (irrational) gradients. Hence he will almost certainly see white looking out at arbitrary gradients.

Fun fact 2: The gradient which, roughly speaking, stays furthest away from any blue point is the golden ratio φ (aka. "the most irrational number": the hardest to approximate with a rational number)

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It's Billy time again!

If Billy sees a point (1,1), he cannot see the points directly in line behind it like (2,2), (3,3). We will call points like (1,1) visible and points like (2,2) hidden. Assume that the plane is actually finite extending to the point (n,n) (again for n a positive whole number). As n approaches infinity, the fraction of visible points converges to a particular value. What is that value?

20 points Rolling Cattle (INTPf): 6/π²

So Billy can see (1,2) but he cannot see (2,4), (3,6), etc., because (1,2) blocks him. We can generalise this by saying that a point (a,b) is visible if and only if a and b have no common factor other than 1: if and only if a and b are coprime. Now given two random positive integers a < n and b < n, the probability that they are coprime as n approaches infinity will be 6/π².

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In 1988 he played a character named after a star. In 1989 he played the same character as he would again in a later 1992 movie. In 2014 he played a character that played a character with some passing resemblance to this 1989/1992 character. In 2019 he will play the same character as he did in a 2010 movie. Who is he?

20 points Hephaestus (INTPx): Michael Keaton

Beetlejuice/Betelgeuse 1988, Batman 1989, Batman Returns 1992, Birdman 2014, Toy Story 3 2010, Toy Story 4 2019.

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Complete the meme:


20 points Hephaestus (INTPx): B

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What song was banned by many U.S. radio stations, perhaps partly due to what the songwriter allegedly did with a pencil (though certainly not related to its lyrics)?

20 points lbloom (INTPx): Rumble - Link Ray & His Ray Men:


The song has no lyrics but was still banned by many U.S. stations. He allegedly used a pencil to poke holes in the speakers of his amplifier.

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What is the smallest number (if any) of the form 1[0]ᵐ1[0]ⁿ1 that is prime, where [0]ᵐ denotes m 0's and [0]ⁿ denotes n 0's (m ≥ 0, n ≥ 0)?

For example, 111, 1011, 10101, 10001001, 10000000011, 10100000001, and so on, are all numbers of this form.

20 points Serac (INTPf): All are multiples of 3; specifically 3 × [3]ᵐ⁺¹[6]ⁿ7 (e.g., 111 = 3 × 67; 1011 = 3 × 367; 1000101 = 3 × 333367). Furthermore, 1001001 = 1000000 + 1000 + 1. Dividing each term by 3 leaves 1/3 as remainder for each, which sums up to 1.

As another explanation: any number (base 10) is divisible by 3 if and only if its cross-sum (the sum of all its individual digits) is divisible by 3. For example, the cross-sum of 249 is 2+4+9=15, which is divisible by 3, and hence 249 is divisible by 3. On the other hand 251 is 2+5+1=8, which is not divisible by 3, and thus nor is 251. The cross-sum of any number 111, 1011, 10001001, etc., is always 3, hence all such numbers are divisible by 3, and hence none of them can be prime.

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There is a letter in the English alphabet that if you write it down, then flip it 180° and write it down, then flip it 180° again and write it down, you arrive at a three letter word for something that roughly half of us could never officially be while staying in one piece. What's the word?

20 points Serac (INTPf): nun

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You and your identical septuplet daughters wake up on a desert island with no food, no water, and no idea how you got there. Things look grim, but then you come across the following circular raft with twelve symmetrically positioned seats:


There's only room for one daughter on each seat and no room for you unfortunately. Even worse, the raft is not very buoyant and needs to be perfectly balanced (with the centre of gravity on the midpoint of the raft) to not tip over. Luckily your daughters are identical septuplets of identical weight and if you put six of them on the raft, one in every second seat, it floats just fine. But if you try to put a seventh in one of the remaining seats, the raft becomes unbalanced and sinks on that side.

So you cast off six of offspring in the raft in the hope that they will float into a shipping lane and be rescued. Some time later you and your least favourite daughter die of exposure on the island. But actually, you became quite fond of that daughter, and began to wonder: could you have given her a better chance? Was there a way you could have put all seven daughters on the raft and kept it balanced?

20 points Serac (INTPf):


The blue seats are unoccupied; the rest are occupied. The simplest explanation is that the pink seats balance (2), the orange seats balance (2) and the green seats balance (3). So altogether they must all balance too.

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In which movie does the following staircase feature?


20 points lbloom (INTPx): Gattaca


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The following chess position is not possible to reach for a number of reasons:


Move the fewest possible pieces around to make the chess position one that is possible to reach. The moves do not have to be legal chess moves; you can pick up any piece and move it elsewhere to a vacant spot.

30 points Penguinhunter (INTPx): move Re1 to g8, move Kc4 to e8, move a3 pawn to b3

Unfortunately I made some mistakes designing the board, so while I thought it was possible in two moves, three moves are required. Furthermore, the answer is not unique. However, Penguinhunter offers the only valid solution thus far.

The reasons for the illegal position are:

  1. Both kings in check at the same time
  2. Double check on white king not possible (no discovered check possible)
  3. Double check on black king not possible (no discovered check possible)
  4. White has two pairs of doubled pawns indicating two captures but black is only missing one piece
  5. The black rook starting on h8 could not have gotten out of h7/h8 so long as the e7, g7 and h7 pawns have not moved (unless it were captured by a white knight and promoted back onto the board, but with only one black pawn missing, only either the second dark-squared bishop or the rook could have been promoted)
The solution proposed by Penguinhunter addresses these problems as follows:

Re1 to g8: Solves (2) and (5)

Kc4 to e8: Solves (1) and (3)

a3 pawn to b3: Solves (4)

The solution I originally had and was suggested by Penguinhunter, who corrected himself: move Re1 to g8, move d3 pawn to c3

This does not work, as Blorg points out 10 points because the doubled pawn that is left is on a3, a dark square, but the black piece left unaccounted for is the light-squared bishop, so the pawn could not have doubled on a3 taking that bishop. This solution does not work, nor does any solution with only two moves.

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From which musical piece is the following taken:


25 points jawdropper (INTPf): Ojalá - Silvio Rodriguez


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Assume that your favourite gas/petrol has 600g/L of carbon and that a litre of it weighs around 0.77 kg. How many kilograms of gas/petrol would you have to burn to create a kilogram of CO² (in an efficient engine; give or take 1%)?

20 points lbloom (INTPx): ~0.35 kilograms

About 0.35 kilograms of petrol (about 0.454 litres) would be needed to produce 1 kilogram of CO²: Though surprising, in CO², the atomic weight of C is ~12 and O is ~16, so the weight of CO² is ~34, 12/44 = ~27.3% carbon, and 32/44 = ~72.7% oxygen. When petrol is burnt in an efficient engine, about 99% of the carbon will mix with oxygen of the air producing CO². Hence most of the CO² weight actually comes from the oxygen in the air: petrol can produce almost three times its own weight in CO² when burnt.

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He was born in the 50's, but most of us now know him by another name. Probably the best-known movie he acted in was in 1982, but that's not what he's most famous for. He's a big fella, but not as big as some of the bigger fellas he had supposed run-ins with. My mother used to wear a t-shirt of his, but I'm not sure she had much of a clue about who he was.

Who is he?

10 points CatGoddess (INTPf): Lawrence Tureaud, aka Mr. T

25 points MoneyJungle (INTPx): Hulk Hogan

Born 1953 as Terry Gene Bollea, starred in Rocky III (1982), famously fought André the Giant.


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ⰋⰆ

According to current estimates, there are on average 100 billion stars per galaxy, and 2 trillion galaxies in the observable universe. Assuming these estimates to be accurate, we can be confident that there are galaxies with precisely the same number of stars.

But according to these figures, at least how many galaxies can we be sure to have the same number of stars?

(Multiverses and star formation and reference frames and dark matter and all that are interesting subjects of course but this is a questions of mathematics, not physics. I think it's more interesting though phrased in terms of realistic estimates for stars and galaxies. Put another way: we have 2 trillion sets with on average 100 billion elements each; what is the least number of sets with the same number of elements?)

edit: We assume that each star (element) is in precisely one galaxy (set).

There's some possible misunderstanding about the question, so without trying to give too much away let's consider an example of 5 galaxies (5 sets) with on average 1 star (1 element) each. Let's name the stars a, b, c, d, e.

We could have galaxies like:

{a,b,c,d,e}, {}, {}, {}, {} - in this case we have four galaxies with the same number of stars (zero)
{a}, {b}, {c}, {d}, {e} - in this case we have five galaxies with the same number of stars (one)
{}, {a,b}, {c}, {d,e}, {} - in this case we have two galaxies with the same number of stars (zero or two)

In fact, no matter how we assign the 5 stars into the 5 galaxies, we see we will always have a group of at least 2 galaxies with the same number of stars: 2 galaxies must have the same number of stars no matter what the assignment. Given 5 galaxies with on average one star per galaxy, the answer would thus be 2.

The question is asking something similar, but for 2 trillion galaxies (sets) with on average 100 billion stars each (elements).

40 points Serac: At least 10 galaxies with precisely the same number of stars

I wrote a long explanation to this one to be thorough, but if tl;dr, jump to "Put more intuitively perhaps ...".

The pigeon hole principle tells us that for natural numbers k and m, if n = km + 1 objects are distributed among m sets, then at least one set will contain (at least) k + 1 objects. So if you have 2 sets and assign 5 objects into them, as a lower bound, one set will have at least 3 elements (it might have 4 or 5).

In the case of galaxies and stars, to make the pigeonhole principle work, we need to turn the problem on its head: sets in this case are rather defined by the number of stars; their elements are galaxies with precisely that number. If we assumed that the max number of stars was 100 billion, then we'd have at most m = 100 billion sets (numbers of stars) and n = 2 trillion objects (galaxies). 2 trillion = k(100 billion) + 1, so k ≈ 20. There would be at least 21 galaxies with the same number of stars.

But the estimate we have for number of stars is an average, not a max. Based on that average, the max is bounded by the case where all the stars are in one galaxy, meaning 2 trillion x 100 billion = 200 sextillion stars in one galaxy, and our previous argument goes out the window. But in this worst case, with all the stars in one galaxy, then 2 trillion (minus one) galaxies would have the same number of stars: zero.

So from 200 sextillion total stars, how many unique numbers of stars can we actually get? Well, we can start with a galaxy with 0 stars, then 1 star, then 2 stars, etc. How far do we get before we use up 200 sextillion? Well the sum of i for (0 > i > n) is n(n-1)/2; for example, the sum of 1, 2, 3, 4, 5 (n=5) is 5(4)/2 = 10. So we can set

n(n-1)/2 = 200 sextillion
n(n-1) = 400 sextillion
n² ≈ 400 sextillion
n ≈ 632 billion

So assigned zero stars to one galaxy, one star to the next galaxy, two stars to the next galaxy, and so on, we can give a unique number of stars to 632 billion galaxies.

But if we number the galaxies up to 632 billion, we'll used up all the stars; we can only have one galaxy with one star, one galaxy with two stars, ..., one galaxy with 632 billion stars, and then with zero stars left, we have 1 trillion 368 million galaxies with zero stars.

What if we divide the stars into two and create two numberings?

n(n-1)/2 = 100 sextillion
n(n-1) = 200 sextillion
n² ≈ 200 sextillion
n ≈ 447 billion (with two numberings, enough to cover 894 billion stars)

What if we divide the stars into y and create y numberings ...

n(n-1) = (400 sextillion)/y
n² ≈ 400 sextillion)/y
n ≈ (632 billion)/sqrt(y)

and then say that these y numberings should cover the 2 trillion stars:

yn = 2 trillion
n = 2 trillion/y

and putting the two together

2 trillion/y ≈ (632 billion)/sqrt(y)
2 trillion ≈ 632 billion × sqrt(y)
3.16455 ≈ sqrt(y)
10.0144 ≈ y

Okay, but y should be a natural number, maybe 10 or 11 since it's close and we've rounded a bit.

Trying 10:

n(n-1)/2 = 20 sextillion
n(n-1) = 40 sextillion
n² ≈ 40 sextillion
n ≈ 200 billion x 10 is 2 trillion!

We better be a little more careful then since we hit the number exactly:

n² - n = 40 sextillion
n² > 40 sextillion (since n is positive)
n > 200 billion
10 n > 2 trillion

So we have enough stars to "number" the galaxies 10 times from 0 to 200 billion.



Put more intuitively perhaps, we can "number" the galaxies using pairs as follows:

(0, 200 billion)
(1, 200 billion - 1)
(2, 200 billion - 2)
(3, 200 billion - 3)
...
(100 billion - 1, 100 billion + 1)

Adding each pair, the mean remains 100 billion. There are 200 billion unique numbers, so we need to cycle these pairs ten times to number 2 trillion galaxies.

Hence we can conclude that according to current estimates for stars and galaxies, that there must be a group of at least 10 galaxies with precisely the same number of stars.

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Name the national dish being prepared here:


20 points MoneyJungle (INTPx): Hákarl (Iceland)

Fermented shark, described to have a flavour blending really really strong cheese and fish.

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Which one-sided battle began over an effort to horde alcohol, leading to over one thousand dead and wounded and, eventually, the force retreating from itself?


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If greed is the ambassador to Britain, what would be the ambassador to France?

25 points lbloom (INTPx): Sloth

According to De Plancy, Hell's ambassadors were as follows:
  • Belfegor, Ambassador of France.
  • Mammon, Ambassador of England.
  • Belial, Ambassador of Italy.
  • Rimmon, Ambassador of Russia.
  • Thamuz, Ambassador of Spain.
  • Hutgin, Ambassador of Turkey.
  • Martinet, Ambassador of Switzerland.
According to Binsfeld's classification of demons we have:
  • Lucifer: pride
  • Mammon: greed
  • Asmodeus: lust
  • Leviathan: envy
  • Beelzebub: gluttony
  • Satan: wrath
  • Belphegor: sloth

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Which divisive issue in the world of software is directly linked to a similarly divisive dining habit in the world of Gulliver's Travels?

20 points Hephaestus (INTPx): Big endian vs. Little endian

In Gulliver's travels, refers to factions in a civil war: big endians think you should eat a boiled egg from the stout end, little endians think you should eat from the pointed end.

In software, refers to the order of bits in a byte, or bytes in the data, where in little endian the least significant bit/byte is sent first, while in big endian the most significant bit/byte is sent first.

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ⰌⰁ

Which EP released in the 90's featured a warning on the cover that two of the three tracks may be illegal to play according to U.K. law? (Special effort had thankfully been made to try to ensure that the third track remained legal.)

40 points Penguinhunter (INTPx): Autechre - Anti EP

It was a protest EP against the Criminal Justice and Public Order Act 1994 in the UK, which tried to indirectly ban raves by specifically banning gatherings where music is played with a succession of regular beats. Autechre thus released a three-song EP with the warning that the first two songs had regular beats. The third song, Flutter, uses unique bars each time on the drum machine.


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ⰌⰂ

Behold the following square:



But as the pattern of filling a quarter of the remaining square with purple continues towards infinity, to what fraction of purple area will the square converge?

20 points lbloom (INTPx): 1/3

We write a geometric series as:

a + ar + ar² + ...

For purple we have:

1/4 + 1/16 + 1/64 + ...

With a = 1/4 and r = 1/4.

Since |r| is less than 1, the sum approaching infinity converges and is a/(1-r) = (1/4)/(3/4) = 1/3.

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ⰌⰃ

Margaret Anne Lake, born July 27, 1942, is a famous practitioner of sun sign astrology (the Western kind we all know and love), which traditionally has assigned a zodiac sign to a person based on the constellation in which the Sun is (most closely) placed at the time of that person's birth. But in which constellation (as defined by the International Astronomical Union) was the Sun placed when Margaret was born?

20 points Penguinhunter (INTPx): Cancer

According to herself, she would be Leo.
She's really cancer ... sorry Cancer.

The issue is due to precession (wobble) in the Earth's axis with a period of 26,000 years. Due to this effect, the star signs have changed a lot in the 2000 years since they were first proposed. The "real" star signs you can find here, for example. While the Sun would have been in Leo on July 27, 1, it was in Cancer on July 27, 1942.

https://www.constellation-guide.com/wp-content/uploads/2011/01/Cancer-constellation-map.gif

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ⰌⰄ

By which popular convention would the following inequalities hold?

45984 < 62396 < 28845 < 47784 < 34949 < 53955 < 76354 < 26262 < 74777

35 points Serac (INTPf): These are poker hands.

9 high < pair 66 < pair 88 < two pair 77,44 < two pair 99,44 < trips 555 < straight < full house < four of a kind

Being super pedantic I guess you could say what if the straight was a straight flush, but you'd have the solution at that stage so I said I'd include it. :)

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ⰌⰅ

Explain the two related aspects that make each of these two images notable.




40 points Rolling Cattle (INTPf): The images are the same backwards as forwards and what they depict are also written the same backwards as forwards.

NURSES RUN and RACECAR are both palindromes

Ava and Bob met on an online dating site last month and naturally enough the conversation turned to the following topic ...


Why might the drink be a bad idea?

10 points Hephaestus (INTPx): Ava is under the legal age for drinking.

40 points if you can explain why, also giving the date on which she was born

40 points C.J.Woolf (INTPx): She was born on 10/02/2001, so she was 17 at the time of the message.

She's not legal age. Both their names are palindromes so the topic comes up. Bob uses the US spelling of "favorite".

  • Ava: Well I was born on one
Must be a palindromic date.

  • Ava: So yeah that's my favourite
U.K. spelling of "favourite" so probably she's thinking DD-MM.

  • Bob: Hence the name I guess?
  • Alice: Yeah :)
Her name is a palindrome.

  • Bob: It's not today by chance?
The date of the message is January 10th (a palindrome in MM-DD: 01-10 or DD-MM 10-01).

  • Alice: No
  • Alice: It's more special than that!
Maybe it's not just the date is a palindrome but also the year. She also says ...

  • Alice: You'd probably disagree
The palindromic dates in DD-MM/MM-DD indicate the same dates (e.g., 03-30 or 30-03 gives March 3rd). In DD-MM-YYYY or MM-DD-YYYY the palindromic dates might differ.

So if she were born in the 20th century, the palindromic date would have to be XX-91-19XX, which isn't possible in any calendar.

She had to be born in the 21st century. The earliest she could have been born was 10-02-2001, or as she sees it 10 February 2001. This still leaves her underage at the time of message (in either the UK or the US). But for Bob he would see that date as 2 October 2001, while 02-10-2001 is not palindromic (hence why he wouldn't agree).

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ⰌⰆ

The Sheikh dies, leaving behind three sons, seventeen camels and the following will:
  • His oldest son shall inherit one in two camels.
  • His middle son shall inherit one in three camels.
  • His youngest son shall inherit one in nine camels.
Now the three don't know how best to respect the will as they are reluctant to kill and chop up a camel. So they ask a wise old friend of the family, who does the following:
  • He brings with him his own camel to donate, so now there are eighteen camels.
  • The oldest gets nine camels.
  • The middle son gets six camels.
  • The youngest son gets two camels.
  • Having shared out the camels to the sons, there is still a camel left: the wise old friend's. He takes it back home with him.
(Just gave you the solution to a classic puzzle there; you're welcome.)

The next day the three sons tell them they have a similar problem with the will, this time with forty-one goats, so the wise old friend brings a goat with him, solves the problem the same way, and again returns home with his own goat. Assuming the oldest son was assigned more goats than the middle son, and that the middle son was assigned more goats than the youngest son, what (whole) fraction of the goats were assigned to each son by the will?

20 points Hephaestus (INTPx): Oldest: 1/2. Middle: 1/3. Youngest: 1/7.

  • He brings with him his own goat, so now there are forty-two goats
  • The oldest gets twenty-one goats.
  • The middle son gets fourteen goats.
  • The youngest son gets six goats.
  • Having shared out the goats to the sons, there is still a goat left: the wise old friend's. He takes it back home with him.
As with the camels, the issue is that 1/2 + 1/3 + 1/7 does not add up to 1, rather to 41/42. Hence adding a goat makes it 42/42 and in the end there will be a goat left.

===========================================================
ⰌⰇ

Name the three bright bodies just below (A), (B) and (C) as well as the constellation above (D) in the mid 2016 image below.


20 points lbloom (INTPx): (A) Saturn, (B) Antares, (C) Mars, (D) Scorpius.

Fun fact: Antares' name means anti-Ares, i.e., Mars' rival, because it has a similar appearance to Mars

===========================================================
ⰌⰈ

We've all seen those puzzles where you say what you see. Here the answer lies in what you don't say.


40 points stigmatica (INTPx): INTP

Based on the silent letters: busIness, autumN, balleT, receiPt

Seen but not heard.

===========================================================

Name the bird species for the following song:


40 points Serac (INTPf): Kauaʻi ʻōʻō

This Hawaiian species has been extinct since 1987. This recording was of the endling: the last surviving member of the species. It is the mating call of a male bird, waiting for a female that would never come.

The end-of-the-line image was intended as a hint.

===========================================================
ⰍⰀ

They say we use base 10 because that's how far we can count up to on our fingers. But if that were really the case, what base should we be using?

25 points Hephaestus (INTPx): 1023


binary > unary

===========================================================
ⰍⰁ

  • A shoe that rebuffs friction
  • A scene may require many
  • The most costly is Japanese
  • A wager to kill the pale
But which English poet am I?

10 points Blorg (INTPx): Keats

40 points if you can explain why


10 points Penguinhunter (INTPx): skate and kates

25 points Robcore (INTPx):

  • A shoe that rebuffs friction: skate
  • A scene may require many: takes
  • The most costly is Japanese: steak
  • A wager to kill the pale: stake
The hint is Kates.

All are anagrams of Keats.

===========================================================
ⰍⰂ

Which Arab has a salad?

10 points Hephaestus (INTPx): An-Nasir Salah ad-Din Yusuf ibn Ayyub, aka Saladin (also suggested but afterwards by CatGoddess and Rolling Cattle on INTPf).

10 points Penguinhunter (INTPx): Al-Maʿarri

10 points CatGoddess (INTPf): Qaboos bin Said al Said. Salad in Arabic translates to "Sulta", which has the double meaning of power/rule/influence. There is only one Sultan left in the Arabic world, and it's Said, who rules Oman.

Rather than change the question to avoid more alternative answers, I will just be more strict: I need the exact answer now.

Had his brother not died, he might have been able to help you see the answer more clearly.

10 points lbloom (INTPx): Assad

Question still open to explain how it answers the original question.

20 points CatGoddess (INTPf): Bashar al Assad is an anagram of Arab has salad.

His brother was to be in power but he died; at the time Assad was a mild-mannered ophthalmologist in the U.K.

===========================================================
ⰍⰃ

The following polygon describes what once-in-a-lifetime trip?


40 points CatGoddess (INTPf): The Seven (Modern) Wonders of the World


===========================================================
ⰍⰄ

Name the painting from which the following sliver is taken:


20 points C.J.Woolf (INTPx): El Greco: View of Toledo [1596-1600]

===========================================================
ⰍⰅ

All and only these popes share a particular distinction; but what is it?

Felix III - Boniface VIII - John XVII - Benedict XI - Gregory IX - Alexander VI​

10 points lbloom (INTPx): They followed antipopes (the question says "All and only" and other popes followed antipopes too; still, close enough for some points)

20 points MoneyJungle (INTPx): Their numeric predecessors were antipopes

There was no:

Felix II - Boniface VII - John XVI - Benedict X - Gregory VIII - Alexander V​

In other words the listed popes skipped a number. More precisely, the above titles had been previously used by antipopes and they decided not to use the same title. In most cases the titles taken by antipopes were later used by popes (for example, Benedict XIII).

===========================================================
ⰍⰆ

The son of a foreigner, this political leader embezzled hundreds of millions of dollars until the scandal broke and he fled to the native country of his father. His daughter later ran for president but narrowly lost.

Who is he?

20 points Madrigal (INTPx): Alberto Fujimori

President of Peru, 1990-2000.

===========================================================
ⰍⰇ

Name the video game:


20 points CatGoddess (INTPf): League of Legends

===========================================================
ⰍⰈ

The protagonist of this novel suffers constant time lapses, which provide glimpses of his time in the war, his later married life, and how he fell in love in an extraterrestrial zoo.

What's the novel?

20 points Penguinhunter (INTPx): Slaughterhouse Five

===========================================================

Though syntactically valid, the following programme code makes a very basic error.

Code:
do you speak'a my language?

he just smiled

gave me a vegemite sandwich

and he said

i come from a land down under

where the beer does flow and the men chunder

can't you hear the thunder?

you better run you better take cover

MEN AT WORK


©
(I've noticed the forum's code tags are not to be trusted; just in case, please see attached file code.txt)

Can you change it to give the correct output?

10 points Rolling Cattle (INTPf): The code needs two space characters removed from after the phrase "and he smiled", and another two space characters removed after the phrase "and he said". This would perfectly set up a binary code of tabs and spaces every other line which when converted to ascii spells out "AUSTRIA".

It is programme code you can run like C or Python or Java, though clearly it is not one of those languages. Rather it is an esoteric language (esolang) like nothing you have ever seen. You need to correct the code to give the right output.

20 points Serac (INTPf)

The code is written in Whitespace, a programming language that ignores non-whitespace characters. You can try it here. The mistake is to print AUSTRIA instead of AUSTRALIA. An example solution:

Code:
do you speak'a my language?

he just smiled   

gave me a vegemite sandwich 

and he said

i come from a land down under

can't you hear the thunder?

where the beer does flow and the men chunder

can't you hear the thunder?

you better run

you better take cover

MEN AT WORK


©
Or removing the redudant characters:

Code:
.

Note that the code tags on here change spaces to tabs so copying and pasting from the post will not work. Check out code.txt if you want to see it in action. You can get the solution on INTPx (the code tags are fine there).

===========================================================
===========================================================[/SPOILER]
 

Attachments

Utisz

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Final Scores

[TABLE="class: grid"]
[tr]
[th]Member[/th]
[th]Forum[/th]
[th]Score[/th]
[/tr]
[tr]
[td]lbloom[/td]
[td]INTPx[/td]
[td]245[/td]
[/tr]
[tr]
[td]Serac[/td]
[td]INTPf[/td]
[td]240[/td]
[/tr]
[tr]
[td]Penguinhunter[/td]
[td]INTPx[/td]
[td]185[/td]
[/tr]
[tr]
[td]CatGoddess[/td]
[td]INTPf[/td]
[td]165[/td]
[/tr]
[tr]
[td]Hephaestus[/td]
[td]INTPx[/td]
[td]145[/td]
[/tr]
[tr]
[td]Rolling Cattle[/td]
[td]INTPf[/td]
[td]130[/td]
[/tr]
[tr]
[td]C.J.Woolf[/td]
[td]INTPx[/td]
[td]80[/td]
[/tr]
[tr]
[td]MoneyJungle[/td]
[td]INTPx[/td]
[td]65[/td]
[/tr]
[tr]
[td]stigmatica[/td]
[td]INTPx[/td]
[td]40[/td]
[/tr]
[tr]
[td]Limes[/td]
[td]INTPx[/td]
[td]35[/td]
[/tr]
[tr]
[td]Blorg
jawdropper
Madrigal
MarkovChain
Robcore[/td]
[td]INTPx
INTPf
INTPx
INTPx
INTPx[/td]
[td]25[/td]
[/tr]
[tr]
[td]scarydoor[/td]
[td]INTPx[/td]
[td]10[/td]
[/tr]
[/TABLE]
[TABLE]
[TR]
[TD][/TD]

[TD][/TD]

[TD][/TD]
[/TR]
[/TABLE]
 

Serac

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Didnt fully understand the rules. Only the first person counts? So if I'm first and get 1 question right, I win?
 

CatGoddess

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3rd to last is League of Legends. Second to last is Slaughterhouse Five. I might go for more but I'm on my phone right now.

@Serac The rules mean that the first person to answer a question correctly gets the points for that specific question.
 

Utisz

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Didnt fully understand the rules. Only the first person counts? So if I'm first and get 1 question right, I win?
As CatGoddess mentioned, the first person to get the answer (on either forum) gets the points. The overall winner will be the points total.

3rd to last is League of Legends.
Correct! 20 points.

Second to last is Slaughterhouse Five. I might go for more but I'm on my phone right now.
Correct! But you were just beaten to the punch on the other forum.
 

Rolling Cattle

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ⰃⰊ

Billy is a 2d being with perfect 1d eyesight who lives on an infinite 2d plane (in the positive x and y directions). Currently Billy is at the origin (0,0) of a plane with a point at (a,b) for all a and b whole positive numbers. The points on the plane are mathematical points, meaning that they have no width or height, but if Billy looks right at them, he can see blue. On the other hand, if he is not looking at a point (in the arbitrary distance), he sees white. (If he looks along the x or y axis or away from the positive quadrant he sees nothing.)

Here we see Billy looking at the point (1,1), seeing blue.



Can Billy ever see white from his current position?

===========================================================
ⰄⰊ

It's Billy time again!

If Billy sees a point (1,1), he cannot see the points directly in line behind it like (2,2), (3,3). We will call points like (1,1) visible and points like (2,2) hidden. Assume that the plane is actually finite extending to the point (n,n) (again for n a positive whole number). As n approaches infinity, the fraction of visible points converges to a particular value. What is that value?
Billy i) The chances of billy looking in a direction where there is no points is infinitely more than than if he happens to look in a more rational point direction (like 1, 1). So very probably yes billy will see white, almost certainly.

Billy ii) As n approaches infinity, the fraction of visible points converges to 6 / π² or 0.607927102
 

Serac

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Billy on a 2D plane one:

yes, he can see white: if y = bx represents his line of sight, then he can pick b as an irrational number between 0 and 1, e.g. 1/pi. Then there is no integer x that y = bx is an integer, i.e. such a line doesn't intersect any of the points (x, y) where x, y are integers.
 

Utisz

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Billy i) The chances of billy looking in a direction where there is no points is infinitely more than than if he happens to look in a more rational point direction (like 1, 1). So very probably yes billy will see white, almost certainly.
Correct! 20 points

Billy ii) As n approaches infinity, the fraction of visible points converges to 6 / π² or 0.607927102
Correct! 20 points

Billy on a 2D plane one:

yes, he can see white: if y = bx represents his line of sight, then he can pick b as an irrational number between 0 and 1, e.g. 1/pi. Then there is no integer x that y = bx is an integer, i.e. such a line doesn't intersect any of the points (x, y) where x, y are integers.
Correct! But just beaten to it here by Rolling Cattle.
 

CatGoddess

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Did I not get the triathlon question?

Also,
utisz said:
Which one-sided battle began over an effort to horde alcohol, leading to over one thousand dead and wounded and, eventually, the force retreating from itself?
Was it the Battle of Karánsebes?
 

CatGoddess

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utisz said:
He was born in the 50's, but most of us now know him by another name. Probably the best-known movie he acted in was in 1982, but that's not what he's most famous for. He's a big fella, but not as big as some of the bigger fellas he had supposed run-ins with. My mother used to wear a t-shirt of his, but I'm not sure she had much of a clue about who he was.

Who is he?
Lawrence Tureaud, aka Mr. T?

Also, ambassador to France: Mission 1 million?

Britain's version of the American game show Greed is also called Greed, but in France the name's Mission 1 million.
 

Utisz

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Lawrence Tureaud, aka Mr. T?
Not the one I had in mind, but it fits pretty well (my mother doesn't have his t-shirt but you're not to know that I guess). I'm thinking about giving 10 points as an alternative, but I'm not sure about this part: who did he have a run in with who was bigger than him?

Also, ambassador to France: Mission 1 million?

Britain's version of the American game show Greed is also called Greed, but in France the name's Mission 1 million.
Not the one, no. The ambassador part is important.
 

Utisz

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Just to note, per the rules, editing posts is discouraged. :)
 

The Grey Man

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Ugh, deleted
 

Serac

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The raft one:

according to my calculations, yes, it's possible (blue points are the daughters):

4096


calculated center of gravity to be origin (0, 0) as follows:

if vector of weights is w = [0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1], then

sum( sin( k* pi/6) * w[k])
for k = 1, 2, ... 12, is the center of gravity along x-dimension, and
sum( cos( k* pi/6) * w[k])) along y-dimension
 

CatGoddess

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Can you provide any more info on the program code (last question)? I don't get why it's code, or what it's supposed to do...? If it's supposed to display the lyrics, there should either be an actual command statement that says to write the words, or the lyrics should be commented out with //.
 

Utisz

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The raft one:

according to my calculations, yes, it's possible (blue points are the daughters):

View attachment 4096

calculated center of gravity to be origin (0, 0) as follows:

if vector of weights is w = [0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1], then

sum( sin( k* pi/6) * w[k])
for k = 1, 2, ... 12, is the center of gravity along x-dimension, and
sum( cos( k* pi/6) * w[k])) along y-dimension
Looks like a solution yes! 20 points

Xzibit is happy because of KFC?

Not what I'm looking for!
I guess there must be a lot of memes like that. :)

I'm looking for something more specific.

Can you provide any more info on the program code (last question)? I don't get why it's code, or what it's supposed to do...? If it's supposed to display the lyrics, there should either be an actual command statement that says to write the words, or the lyrics should be commented out with //.
There's not much more I can say for now.

I think you'll have to trust me that the question is fine; I'm sure you'll find the answer satisfying.
 

Serac

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What is the smallest number (if any) of the form 1[0]ᵐ1[0]ⁿ1 that is prime, where [0]ᵐ denotes m 0's and [0]ⁿ denotes n 0's (m ≥ 0, n ≥ 0)?

For example, 111, 1011, 10101, 10001001, 10000000011, 10100000001, and so on, are all numbers of this form.
there is no such prime number, because these numbers can be generated as multiples of 3, in particular

367 * 3 = 1101
3667 * 3 = 11001
33667 * 3 = 101001

etc

very cool questions btw, thank you for putting them together
 

Utisz

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there is no such prime number, because these numbers can be generated as multiples of 3, in particular

367 * 3 = 1101
3667 * 3 = 11001
33667 * 3 = 101001

etc
Could you provide an argument as to why all of them have to be multiples of 3?
 

Utisz

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there is no such prime number, because these numbers can be generated as multiples of 3, in particular

367 * 3 = 1101
3667 * 3 = 11001
33667 * 3 = 101001

etc
Looking in more detail, the pattern is clear enough to award the points. 20 points

I'll post a different explanation in the answer.
 

Serac

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there is no such prime number, because these numbers can be generated as multiples of 3, in particular

367 * 3 = 1101
3667 * 3 = 11001
33667 * 3 = 101001

etc
Could you provide an argument as to why all of them have to be multiples of 3?
the numbers can be expressed as e.g. 1001001 = 1000000 + 1000 + 1. Dividing each term by 3 leaves 1/3 as remainder for each, which sums up to 1.
 

Utisz

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very cool questions btw, thank you for putting them together
Thanks! And you're welcome!

the numbers can be expressed as e.g. 1001001 = 1000000 + 1000 + 1. Dividing each term by 3 leaves 1/3 as remainder for each, which sums up to 1.
Nice argument! And no worries; the pattern you posted is indeed pretty definitive. See the answer now posted below the question for another argument.
 

Serac

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triathlon: 11.069

time on land: sqrt(9 + x^2)/10
time in water: sqrt(9 + (8-x)^2) / 2

where x is point along x-axis where you enter the water. add them, differentiate wrt x and solve differential = 0.
 

Serac

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===========================================================

There is a letter in the English alphabet that if you write it down, then flip it 180° and write it down, then flip it 180° again and write it down, you arrive at a three letter word for something that roughly half of us could never officially be while staying in one piece. What's the word?
nun?
 

Utisz

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triathlon: 11.069

time on land: sqrt(9 + x^2)/10
time in water: sqrt(9 + (8-x)^2) / 2

where x is point along x-axis where you enter the water. add them, differentiate wrt x and solve differential = 0.
I think I may have messed up my own calculation (I used a different method, but yours seems sound).

The problem is that lbloom on the other forum already posted this answer (so he would get the points), but it also makes me more certain the problem lies with me; apologies if so. I will double-check now.
 

Utisz

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triathlon: 11.069

time on land: sqrt(9 + x^2)/10
time in water: sqrt(9 + (8-x)^2) / 2

where x is point along x-axis where you enter the water. add them, differentiate wrt x and solve differential = 0.
My apologies; I made a bit of a mess of this question. You are correct and your method is sound, but lbloom got there first. I had marked his answer wrong because my own answer is wrong. However I'll award you 10 points since you provided your own method.

The answer can also be gotten from Snell's law. I posted this in the answer under the question.
 

Serac

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lbloom is a beast.

in the stars & galaxies question, does "sets with the same number of elements" mean sets with non-unique element count? E.g. if you have a collection of sets with cardinalities 2, 2, 4, 4, 6, 6, then the number of non-unique counts would be 0
 

jawdropper

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ojalá by silvio rodríguez Uhg. bye again.
 

Serac

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chess one:
bishop at 2F to C8
black rook at E1 to B1
white pawn at D3 to C3
 

Serac

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taking a wild stab at the sheet music: Liszt – Liebestraume no. 3 in A-Flat major
 

Utisz

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lbloom is a beast.

in the stars & galaxies question, does "sets with the same number of elements" mean sets with non-unique element count? E.g. if you have a collection of sets with cardinalities 2, 2, 4, 4, 6, 6, then the number of non-unique counts would be 0
Right! In this case the question claims that based on the figures given, there can be no sets with unique counts. The question then is, without knowing how the elements are assigned into the sets, how many sets can we be sure to have the same number of elements?

Hmm, I get that it might be hard to wrap one's head around. It's hard to give more explanation without giving away part of the puzzle, but I might provide an example later.

ojalá by silvio rodríguez Uhg. bye again.
Correct! 25 points
Bye! :)

chess one:
bishop at 2F to C8
black rook at E1 to B1
white pawn at D3 to C3
Not the answer.

taking a wild stab at the sheet music: Liszt – Liebestraume no. 3 in A-Flat major
Not so wild that stab!
Correct! 25 points
 

CatGoddess

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Uhmmm... The program won't display the different lines, so you need /n?
 

Utisz

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I had to change the chess question slightly; I had overlooked something that means there would be more than one solution if removing pieces were allowed. Now you cannot get rid of pieces; you can only move them from their current square to a vacant square. Apologies for that (I think it was more of an issue on the other forum).

Uhmmm... The program won't display the different lines, so you need /n?
Not the answer. :)
 

Rolling Cattle

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The Code one:

do you speak'a my language?

he just smiled

gave me a vegemite sandwich

and he said

i come from a land down under

where the beer does flow and the men chunder

Can't you hear, can't you hear the thunder? <---Correction

you better run you better take cover

MEN AT WORK


©
 

CatGoddess

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Oh, wait, is the Arab one a pun...? Saladin/An-Nasir Salah ad-Din Yusuf ibn Ayyub?
 

Serac

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chess: remove rook at e1, move black king to e8

edit: nvm, saw the change in question now
 

Rolling Cattle

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ⰂⰊ

Name the three people who have contributed to this "face":
I know Mike Tyson for sure.
I'm going to guess Lena Headey (Cersei Lannister from GOT)??
Prince Philip ??
 

Utisz

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The Code one:

do you speak'a my language?

he just smiled

gave me a vegemite sandwich

and he said

i come from a land down under

where the beer does flow and the men chunder

Can't you hear, can't you hear the thunder? <---Correction

you better run you better take cover

MEN AT WORK


©
I don't think so? Maybe you could explain your thinking a little?

Oh, wait, is the Arab one a pun...? Saladin/An-Nasir Salah ad-Din Yusuf ibn Ayyub?
Not a pun no.

ⰍⰂ

Which Arab has a salad?
Saladin Ahmed?
Not the one!

Hephaestus on the other forum guessed the same.

It's something more ... specific.

chess: remove rook at e1, move black king to e8

edit: nvm, saw the change in question now
This wouldn't have been enough.



Why is Xzibit so happy in this photo?


from https://knowyourmeme.com/memes/xzibit-yo-dawg :

"The original photograph of Xzibit comes from a set of studio portraits that were originally used to promote the 2006 sports drama film Gridiron Gang, in which the rapper plays the role of a minor character named Malcolm Moore."
There's something else going on in the question. :)

chess:

rook at e1 to a5, black king to e8
Not the one!

ⰂⰊ

Name the three people who have contributed to this "face":
I know Mike Tyson for sure.
I'm going to guess Lena Headey (Cersei Lannister from GOT)??
Prince Philip ??
Not the three.
 

Utisz

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Just to mention, it looks like the chess one has been solved on the other forum, but I need to squint at it for a bit to make sure. It's the solution I had in mind anyways.
 

Serac

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galaxy one: 84% of the 2 trillion

I'm sure there's some clever trick for that one, but too lazy to think so just wrote a numerical solution for a smaller universe of 1000 galaxies
 

Utisz

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Just to mention, it looks like the chess one has been solved on the other forum, but I need to squint at it for a bit to make sure. It's the solution I had in mind anyways.
I'm closing this. I'm relatively satisfied my original solution is okay.

galaxy one: 84% of the 2 trillion

I'm sure there's some clever trick for that one, but too lazy to think so just wrote a numerical solution for a smaller universe of 1000 galaxies
Not sure I understand?

Ah, you mean at most 16% of galaxies can have a unique cardinality? Uff, that's not the question I had in mind but I see how you could interpret the question that way. I would have to think about this and will give you partial points if you're correct.

What I'm looking for is a number, like we can be certain that there are at least 2, or 3, or n galaxies with the same number of stars. I will add that example soon I think. Apologies for any misunderstanding.
 

Serac

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@Utisz that was my interpretation yes.

But as long as they are not all of unique cardinality (which they can't be, since the minimum average for that case would be (n+1)/2 = 1 trillion + 0.5 ), then clearly at least 2 must have the same cardinality

unless you're looking for the maximum multiplicity of cardinalities, which would be a slightly more challenging problem :phear:
 
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