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Rules:
 20 points awarded for each question to start with.
 Points increase with time: 25 points after 12 hours, 30 points after 24 hours, 35 points after 36 hours, 40 points after 48 hours.
 You may use Google.
 For answers that are "guessable" (i.e., have limited possible answers), you need to explain your answer to get the points to show you're not just guessing.
 First to post the full answer (on either forum) gets the points.
 If you edit your post, you may not be awarded the points.
 I believe that all questions have a uniquely satisfying answer. Otherwise I might award partial points for good alternative answers.
 Moaning about some question or other is allowed but I reserve the right to ignore you if you don't have a valid point.
 I'll post hints after a week and offer partial points for remaining question.
 A tie for winner will be broken by number of questions answered, if still equal, by last to start answering.
Questions remaining: 1/50
Link to INTPx thread.
Hope you enjoy. Good luck!
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Open (with hints): 1
To get full points, I will need a full explanation.
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ⰌⰅ
Explain the two related aspects that make each of these two images notable.
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?
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 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.
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Closed: 49
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Ⰰ
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:
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  dp 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.
The first and by far the most obscure one is William Basinski  dp 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
25 points CatGoddess (INTPf): Second song is Brian Eno  Thursday Afternoon.
So:
Fun fact: even if you never heard of Brian Eno before, you'll probably recognise this work of his.
So:
Fun fact: even if you never heard of Brian Eno before, you'll probably recognise this work of his.
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Ⰳ
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).
The star is regular, but not convex (plus the points are not its vertices but rather travel along it).
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Ⰴ
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).
▀▀▀▀▀ 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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Ⰵ
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 straightline 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 + (8y)^2) km, the time in water is sqrt(9 + (8y)^2)/2 h. Hence the total time is t = sqrt(9 + y^2)/10 + sqrt(9 + (8y)^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) Originally I had thought to 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).
Unfortunately working it out is horrible (when I worked it out before posting the quiz, I made a dumb mistake of taking a constant hypotenuse; fixing the error makes things 10 times more difficult)
3) Code it out I guess and bruteforce the minimum time
Fun fact: if you were light, you would automatically find the fastest route through the two media; this is Fermat's principle. Light finds the fastest way.
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 + (8y)^2) km, the time in water is sqrt(9 + (8y)^2)/2 h. Hence the total time is t = sqrt(9 + y^2)/10 + sqrt(9 + (8y)^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) Originally I had thought to 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).
Unfortunately working it out is horrible (when I worked it out before posting the quiz, I made a dumb mistake of taking a constant hypotenuse; fixing the error makes things 10 times more difficult)
sin(θa)/sin(θb) = 5
sin(θa) = y/sqrt(y²+9)
sin(θb) = (8y)/sqrt((8y)²+9)
(y/sqrt(y²+9)) / ((8y)/sqrt((8y)²+9)) = 5
(y sqrt((8y)²+9)) / ((8y) sqrt(y²+9)) = 5
(y²((8y)²+9)) / (8y)²(y²+9) = 25
y²(y²16y+73) / (y²16y+64)(y²+9) = 25
(y⁴16y³+73y²) / (y⁴16y³+64y²+9y²144y+576) = 25
(y⁴16y³+73y²) / (y⁴16y³+73y²144y+576) = 25
y = 7.434 (don't ask how I solved this, haha)
land distance = srqt(y²+3²) = 8.0165
water distance = sqrt((8y)²+3²) = 3.0529
total distance = 11.069 km
sin(θa) = y/sqrt(y²+9)
sin(θb) = (8y)/sqrt((8y)²+9)
(y/sqrt(y²+9)) / ((8y)/sqrt((8y)²+9)) = 5
(y sqrt((8y)²+9)) / ((8y) sqrt(y²+9)) = 5
(y²((8y)²+9)) / (8y)²(y²+9) = 25
y²(y²16y+73) / (y²16y+64)(y²+9) = 25
(y⁴16y³+73y²) / (y⁴16y³+64y²+9y²144y+576) = 25
(y⁴16y³+73y²) / (y⁴16y³+73y²144y+576) = 25
y = 7.434 (don't ask how I solved this, haha)
land distance = srqt(y²+3²) = 8.0165
water distance = sqrt((8y)²+3²) = 3.0529
total distance = 11.069 km
3) Code it out I guess and bruteforce the minimum time
Fun fact: if you were light, you would automatically find the fastest route through the two media; this is Fermat's principle. Light finds the fastest way.
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Ⰷ
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.
20 points CatGoddess (INTPf): Accoding to the metadata in the photo he's on an island in a lake ... on an island in a lake ... on an island.
The Rocco Castoro hint is a reference to the fact that he was involved in uploading a photo of John McAffee (yes that McAffee; also a fugitive at the time) to Twitter giving away his location, helping authorities to apprehend him a few days later..
The Rocco Castoro hint is a reference to the fact that he was involved in uploading a photo of John McAffee (yes that McAffee; also a fugitive at the time) to Twitter giving away his location, helping authorities to apprehend him a few days later..
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Ⰸ
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.
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
He coined the paradox of tolerance.
Doctoral supervisor was Karl Bühler.It seems to me certain that more people are killed out of righteous stupidity than out of wickedness.
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:
Dey coming o'er here, takin' our Academy Awards for Best Director
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
Dey coming o'er here, takin' our Academy Awards for Best Director
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The altright 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 spacefilling fractal/curve.
It's a spacefilling fractal/curve.
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Name the three people who have contributed to this "face":
35 points Limes (INTPx):
 Bernie Sanders (first proposed by lbloom 5 points)
 Helena Bonham Carter (first proposed by Blorg 5 points)
 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)
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.
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.
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.
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 crosssum (the sum of all its individual digits) is divisible by 3. For example, the crosssum 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 crosssum 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.
As another explanation: any number (base 10) is divisible by 3 if and only if its crosssum (the sum of all its individual digits) is divisible by 3. For example, the crosssum 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 crosssum 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.
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:
Re1 to g8: Solves (2) and (5)
Kc4 to e8: Solves (1) and (3)
a3 pawn to b3: Solves (4)
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:
 Both kings in check at the same time
 Double check on white king not possible (no discovered check possible)
 Double check on black king not possible (no discovered check possible)
 White has two pairs of doubled pawns indicating two captures but black is only missing one piece
 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 darksquared bishop or the rook could have been promoted)
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 lightsquared 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.
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 lightsquared 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.
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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ⰋⰅ
He was born in the 50's, but most of us now know him by another name. Probably the bestknown 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 runins with. My mother used to wear a tshirt 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.
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).
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(n1)/2; for example, the sum of 1, 2, 3, 4, 5 (n=5) is 5(4)/2 = 10. So we can set
n(n1)/2 = 200 sextillion
n(n1) = 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(n1)/2 = 100 sextillion
n(n1) = 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(n1) = (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(n1)/2 = 20 sextillion
n(n1) = 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.
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(n1)/2; for example, the sum of 1, 2, 3, 4, 5 (n=5) is 5(4)/2 = 10. So we can set
n(n1)/2 = 200 sextillion
n(n1) = 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(n1)/2 = 100 sextillion
n(n1) = 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(n1) = (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(n1)/2 = 20 sextillion
n(n1) = 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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ⰋⰇ
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.
Fermented shark, described to have a flavour blending really really strong cheese and fish.
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ⰋⰈ
Which onesided battle began over an effort to horde alcohol, leading to over one thousand dead and wounded and, eventually, the force retreating from itself?
20 points CatGoddess (INTPf): Battle of Karánsebes
Probably one of the most bizzare friendly fire incidents you're likely to read about.
Probably one of the most bizzare friendly fire incidents you're likely to read about.
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Ⰼ
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:
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.
 Lucifer: pride
 Mammon: greed
 Asmodeus: lust
 Leviathan: envy
 Beelzebub: gluttony
 Satan: wrath
 Belphegor: sloth
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ⰌⰀ
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.
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 threesong 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.
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 threesong 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/(1r) = (1/4)/(3/4) = 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/(1r) = (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.constellationguide.com/wpcontent/uploads/2011/01/Cancerconstellationmap.gif
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.constellationguide.com/wpcontent/uploads/2011/01/Cancerconstellationmap.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.
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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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.
 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.
The next day the three sons tell them they have a similar problem with the will, this time with fortyone 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 fortytwo goats
 The oldest gets twentyone 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.
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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 antiAres, i.e., Mars' rival, because it has a similar appearance to Mars
Fun fact: Antares' name means antiAres, i.e., Mars' rival, because it has a similar appearance to Mars
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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.
Based on the silent letters: busIness, autumN, balleT, receiPt
Seen but not heard.
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Ⰽ
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 endoftheline image was intended as a hint.
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 endoftheline image was intended as a hint.
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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
binary > unary
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ⰍⰁ
 A shoe that rebuffs friction
 A scene may require many
 The most costly is Japanese
 A wager to kill the pale
10 points Blorg (INTPx): Keats
40 points if you can explain why
40 points if you can explain why
10 points Penguinhunter (INTPx): skate and kates
25 points Robcore (INTPx):
All are anagrams of Keats.
 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
All are anagrams of Keats.
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ⰍⰂ
Which Arab has a salad?
10 points Hephaestus (INTPx): AnNasir Salah adDin Yusuf ibn Ayyub, aka Saladin (also suggested but afterwards by CatGoddess and Rolling Cattle on INTPf).
10 points Penguinhunter (INTPx): AlMaʿarri
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.
10 points Penguinhunter (INTPx): AlMaʿarri
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.
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 mildmannered ophthalmologist in the U.K.
His brother was to be in power but he died; at the time Assad was a mildmannered ophthalmologist in the U.K.
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ⰍⰃ
The following polygon describes what onceinalifetime trip?
40 points CatGoddess (INTPf): The Seven (Modern) Wonders of the World
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ⰍⰄ
Name the painting from which the following sliver is taken:
20 points C.J.Woolf (INTPx): El Greco: View of Toledo [15961600]
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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:
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).
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).
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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, 19902000. His father was from Japan. He embezzled hundreds of millions of dollars and threw in a few human right's violations for good measure. When the corruption scandal broke, he fled to Japan and tried to resign by fax; instead he was empeached. In 2005 he flew to Chile in the hope of reviving his political career. Instead he was extradited (the first democratic president to be extradited to the country of his presidency) and tried and sentenced first for six years, and later for a lot more years based on other counts. His daughter Keiko ran for presidency of Peru in 2011 and came in second place.
President of Peru, 19902000. His father was from Japan. He embezzled hundreds of millions of dollars and threw in a few human right's violations for good measure. When the corruption scandal broke, he fled to Japan and tried to resign by fax; instead he was empeached. In 2005 he flew to Chile in the hope of reviving his political career. Instead he was extradited (the first democratic president to be extradited to the country of his presidency) and tried and sentenced first for six years, and later for a lot more years based on other counts. His daughter Keiko ran for presidency of Peru in 2011 and came in second place.
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ⰍⰇ
Name the video game:
20 points CatGoddess (INTPf): League of Legends
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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
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Ⰾ
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
©
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 nonwhitespace characters. You can try it here. The mistake is to print AUSTRIA instead of AUSTRALIA. An example solution:
Or removing the redudant characters:
The code is written in Whitespace, a programming language that ignores nonwhitespace 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
©
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).
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