6 ms·

Ridiculous Math Problems

219 points by martinlaz 6y ago

praptak 6y ago
My math teacher used to give us silly problems like how long will it take the raising water level to reach the top step of a ladder hanging down from a ship. It trained us not to apply formulas blindly.
- em-bee 6y ago
  it depends on the size of the hole at the bottom of the ship
dan-robertson 6y ago
A ridiculous problem I remember seeing was approximately this one: a/(b+c) + b/(a+c) + c/(a+b)=896. a,b,c positive integers I think it probably used random symbols instead of letters and didn’t have the integer requirement. Indeed if you allow real numbers or 0 then it is easy to find a solution. I think it didn’t go particularly viral because it was too hard. I think there’s a slightly easier version if you replace 896 by 16.
- contravariant 6y ago
  The version I first saw used '4' instead of '896'. Which frankly made the answer large enough that no reasonable amount of brute force could find it. A good write up can be found here: https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y%2Bz-%2B-frac-y-z%2Bx-%2B-frac-z-x%2By-4/answer/Alon-Amit https://www.quora.com/How-do-you-find-the-positive-integer-s...
  - kens 6y ago
    That is an amazing, detailed explanation. (Don't be fooled by the quora location.) Spoilers: the solution to the equation is 80-digit numbers, which can be obtained by using elliptic curves, which the link explains in detail.
  - IanCal 6y ago
    That is fascinating thank you, and a great lead into some more advanced topics.
foolfoolz 6y ago
the math equivalent of a dumb programming interview question
- adrianmonk 6y ago
  The similarity extends to the part where the real motivation for asking is to make the person asking look smart.
deleted 6y ago
[deleted]
mike00632 6y ago
Reminds me of the cute banana, apple, pineapple math challenge that is actually an elliptic curve. https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4 https://www.quora.com/How-do-you-find-the-positive-integer-s...
murgindrag 6y ago
There's a difference between a ridiculous math programs (which are often amusing) and incorrect ones (for example, adding different units, or absolute units rather than relative ones). The article mixes the two. One type is funny and clever. The other type reinforces misconceptions and is damaging to students.
arkitaip 6y ago
This reminds me of the riddle by the good soldier Svejk: ----- "Would you know how to calculate the diameter of the globe?" "No, I'm afraid I wouldn't," answered Svejk, "but I'd like to ask you a riddle myself, gentlemen. Take a three-storied house, with eight windows on each floor. On the roof there are two dormer windows and two chimneys. On every floor there are two tenants. And now, tell me, gentlemen, in which year the house-porter's grandmother died?" ----- https://english.radio.cz/good-soldier-svejk-a-literary-character-a-legend-8072464 https://english.radio.cz/good-soldier-svejk-a-literary-chara...
- gweinberg 6y ago
  I'm guessing it's a trick question and she's still alive.
- pgtan 6y ago
  Which reminds me of the riddles in Yerofeyev's Moscow—Petushki (sorry, only in russian, my russian and my english are too bad to translate) «Знаменитый ударник Алексей Стаханов два раза в день ходил по малой нужде и один раз в два дня – по большой. Когда же с ним случался запой, он четыре раза в день ходил по малой нужде и ни разу – по большой. Подсчитай, сколько раз в год ударник Алексей Стаханов сходил по малой нужде и сколько по большой нужде, если учесть, что у него триста двенадцать дней в году был запой». «Когда корабли Седьмого американского флота пришвартовались к станции Петушки, партийных девиц там не было, но если комсомолок называть партийными, то каждая третья из них была блондинкой. По отбытии кораблей Седьмого американского флота обнаружилось следующее: каждая третья комсомолка была изнасилована; каждая четвертая изнасилованная оказалась комсомолкой; каждая пятая изнасилованная комсомолка оказалась блондинкой; каждая девятая изнасилованная блондинка оказалась комсомолкой. Если всех девиц в Петушках 428 – определи, сколько среди них осталось нетронутых беспартийных брюнеток?» And my favorite: «Как известно, в Петушках нет пунктов А. Пунктов Ц тем более нет. Есть одни только пункты Б. Так вот: Папанин, желая спасти Водопьянова, вышел из пункта Б1 в сторону пункта Б2. В то же мгновенье Водопьянов, желая спасти Папанина, вышел из пункта Б2 в пункт Б1. Неизвестно почему оба они оказались в пункте Б3, отстоящем от пункта Б1 на расстоянии 12-ти водопьяновских плевков, а от пункта Б2 – на расстоянии 16-ти плевков Папанина. Если учесть, что Папанин плевал на три метра семьдесят два сантиметра, а Водопьянов совсем не умел плевать, выходил ли Папанин спасать Водопьянова?»
Quekid5 6y ago
My personal favorite along these lines are the Busy Beaver numbers. It's not just huge... it's actually impossible to calculate (beyond very low inputs).
- saagarjha 6y ago
  Impossible to compute, only feasible to calculate for the first few.
  - Quekid5 6y ago
    Good correction, thanks! :)
godelski 6y ago
For more ridiculous math problems I want to submit: The Jewish Problems[0]. These were used when Jewish people were trying to get into grad school in Russia in the '70's. Basically designed so Jewish people wouldn't get in. [0] https://arxiv.org/abs/1110.1556 https://arxiv.org/abs/1110.1556
- saagarjha 6y ago
  These seemed like fairly decent problems to put on a mathematics competition…
  - messe 6y ago
    Yeah, but absolute hell in an oral exam when applying to a school.
- IncRnd 6y ago
  The interesting part of thost problems is that once you learn how to answer one, you can answer almost all of them. Of course, that was the issue, the answer wouldn't be accepted. It was a shame these problems were used in that manner.
  - saagarjha 6y ago
    > The interesting part of thost problems is that once you learn how to answer one, you can answer almost all of them. I can't see how that would be true. How would you get the solutions to the others by knowing how to do one of them?
    - ddxxdd 6y ago
      I think the parent comment meant to say that once you see the answer to one of the questions, the question seems easy and simple in retrospect (i.e. the questions are NP hard, in a certain respect). That's the fact that is usually touted in regards to these questions.
      - 6y ago
raister 6y ago
"If a ship has 26 sheep and 10 goats on board, how old is the ship's captain?" oh boy
- cyptus 6y ago
  42
  - raister 6y ago
    Unfortunately, the actual answer is "there is not enough information on the question to provide a response". It's an exercise as to the limits of problem formulation, from what I understood.
    - cyptus 6y ago
      are you sure? https://www.google.com/search?q=answer+to+the+ultimate+question+of+life%2C+the+universe%2C+and+everything https://www.google.com/search?q=answer+to+the+ultimate+quest...
nephanth 6y ago
Not about algebra, but one good example of ridiculous problems leading to great discoveris in math is how, according to the legend, Cauchy invented complex analysis (at least the theorems on holomorphic and meromorphic functions) just to solve some random integral somebody had given him
kazinator 6y ago
> Well this is a bizarre 4th degree polynomial equation, \(x^4-2x^2 -400x=9999\) and he applies one approach to solving it, then comes to a dead end and says "Hence ingenuity is called for" and finally finds that his unknown number is 11. I, personally, would not have had a clue how to solve it. Here is what you can do. If there is an integer solution, we can keep the constant term 9999 on the opposite side and factor the formula. As in: x^4 - 2x^ - 400x = 9999 x(x^3 - 2x - 400) = 9999 Ok, so no we have x f(x) = 9999. From this we know that if there is an integer solution for x, x itself must be a factor of 9999: either a prime factor, or a product of factors. So we just factor 9999: 3^2 11 101 = 9999 9999 has 3 as a degree 2 factor, and then also 11, and 101. Looking at just the factorization breakdown alone, x could be the product of these combinations of factors: { (3), (11), (101), (3, 3), (3, 11), (3, 101), (11, 101), (3, 3, 11), (3, 3, 101), (3, 11, 101) }. If we substitute these factors for x, in order from least to greatest, we will soon hit upon the solution. Intuitively we can guess that x is small, because the opposite factor is cubing it. So for instance, it can't be the case that x is the product of 3, 11, and 101, such that x^3 - 2x - 400 then works out to the remaining factor of 3. It's obvious that x can't be too small, like 3, because x^3 is only 27; that's not going to leave us a positive factor when we subtract 400 alone, let alone -2x.

219 points by martinlaz 6y ago

My math teacher used to give us silly problems like how long will it take the raising water level to reach the top step of a ladder hanging down from a ship. It trained us not to apply formulas blindly.

em-bee 6y ago
it depends on the size of the hole at the bottom of the ship

dan-robertson 6y ago

A ridiculous problem I remember seeing was approximately this one: a/(b+c) + b/(a+c) + c/(a+b)=896. a,b,c positive integers I think it probably used random symbols instead of letters and didn’t have the integer requirement. Indeed if you allow real numbers or 0 then it is easy to find a solution. I think it didn’t go particularly viral because it was too hard. I think there’s a slightly easier version if you replace 896 by 16.

contravariant 6y ago
The version I first saw used '4' instead of '896'. Which frankly made the answer large enough that no reasonable amount of brute force could find it. A good write up can be found here: https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y%2Bz-%2B-frac-y-z%2Bx-%2B-frac-z-x%2By-4/answer/Alon-Amit https://www.quora.com/How-do-you-find-the-positive-integer-s...
- kens 6y ago
  That is an amazing, detailed explanation. (Don't be fooled by the quora location.) Spoilers: the solution to the equation is 80-digit numbers, which can be obtained by using elliptic curves, which the link explains in detail.
- IanCal 6y ago
  That is fascinating thank you, and a great lead into some more advanced topics.

foolfoolz 6y ago

the math equivalent of a dumb programming interview question

adrianmonk 6y ago
The similarity extends to the part where the real motivation for asking is to make the person asking look smart.

deleted 6y ago

[deleted]

mike00632 6y ago

Reminds me of the cute banana, apple, pineapple math challenge that is actually an elliptic curve. https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4 https://www.quora.com/How-do-you-find-the-positive-integer-s...

murgindrag 6y ago

There's a difference between a ridiculous math programs (which are often amusing) and incorrect ones (for example, adding different units, or absolute units rather than relative ones). The article mixes the two. One type is funny and clever. The other type reinforces misconceptions and is damaging to students.

arkitaip 6y ago

This reminds me of the riddle by the good soldier Svejk: ----- "Would you know how to calculate the diameter of the globe?" "No, I'm afraid I wouldn't," answered Svejk, "but I'd like to ask you a riddle myself, gentlemen. Take a three-storied house, with eight windows on each floor. On the roof there are two dormer windows and two chimneys. On every floor there are two tenants. And now, tell me, gentlemen, in which year the house-porter's grandmother died?" ----- https://english.radio.cz/good-soldier-svejk-a-literary-character-a-legend-8072464 https://english.radio.cz/good-soldier-svejk-a-literary-chara...

gweinberg 6y ago
I'm guessing it's a trick question and she's still alive.
pgtan 6y ago
Which reminds me of the riddles in Yerofeyev's Moscow—Petushki (sorry, only in russian, my russian and my english are too bad to translate) «Знаменитый ударник Алексей Стаханов два раза в день ходил по малой нужде и один раз в два дня – по большой. Когда же с ним случался запой, он четыре раза в день ходил по малой нужде и ни разу – по большой. Подсчитай, сколько раз в год ударник Алексей Стаханов сходил по малой нужде и сколько по большой нужде, если учесть, что у него триста двенадцать дней в году был запой». «Когда корабли Седьмого американского флота пришвартовались к станции Петушки, партийных девиц там не было, но если комсомолок называть партийными, то каждая третья из них была блондинкой. По отбытии кораблей Седьмого американского флота обнаружилось следующее: каждая третья комсомолка была изнасилована; каждая четвертая изнасилованная оказалась комсомолкой; каждая пятая изнасилованная комсомолка оказалась блондинкой; каждая девятая изнасилованная блондинка оказалась комсомолкой. Если всех девиц в Петушках 428 – определи, сколько среди них осталось нетронутых беспартийных брюнеток?» And my favorite: «Как известно, в Петушках нет пунктов А. Пунктов Ц тем более нет. Есть одни только пункты Б. Так вот: Папанин, желая спасти Водопьянова, вышел из пункта Б1 в сторону пункта Б2. В то же мгновенье Водопьянов, желая спасти Папанина, вышел из пункта Б2 в пункт Б1. Неизвестно почему оба они оказались в пункте Б3, отстоящем от пункта Б1 на расстоянии 12-ти водопьяновских плевков, а от пункта Б2 – на расстоянии 16-ти плевков Папанина. Если учесть, что Папанин плевал на три метра семьдесят два сантиметра, а Водопьянов совсем не умел плевать, выходил ли Папанин спасать Водопьянова?»

Quekid5 6y ago

My personal favorite along these lines are the Busy Beaver numbers. It's not just huge... it's actually impossible to calculate (beyond very low inputs).

saagarjha 6y ago
Impossible to compute, only feasible to calculate for the first few.
- Quekid5 6y ago
  Good correction, thanks! :)

godelski 6y ago

For more ridiculous math problems I want to submit: The Jewish Problems[0]. These were used when Jewish people were trying to get into grad school in Russia in the '70's. Basically designed so Jewish people wouldn't get in. [0] https://arxiv.org/abs/1110.1556 https://arxiv.org/abs/1110.1556

saagarjha 6y ago
These seemed like fairly decent problems to put on a mathematics competition…
- messe 6y ago
  Yeah, but absolute hell in an oral exam when applying to a school.
IncRnd 6y ago
The interesting part of thost problems is that once you learn how to answer one, you can answer almost all of them. Of course, that was the issue, the answer wouldn't be accepted. It was a shame these problems were used in that manner.
- saagarjha 6y ago
  > The interesting part of thost problems is that once you learn how to answer one, you can answer almost all of them. I can't see how that would be true. How would you get the solutions to the others by knowing how to do one of them?
  - ddxxdd 6y ago
    I think the parent comment meant to say that once you see the answer to one of the questions, the question seems easy and simple in retrospect (i.e. the questions are NP hard, in a certain respect). That's the fact that is usually touted in regards to these questions.
    - 6y ago

raister 6y ago

"If a ship has 26 sheep and 10 goats on board, how old is the ship's captain?" oh boy

cyptus 6y ago
42
- raister 6y ago
  Unfortunately, the actual answer is "there is not enough information on the question to provide a response". It's an exercise as to the limits of problem formulation, from what I understood.
  - cyptus 6y ago
    are you sure? https://www.google.com/search?q=answer+to+the+ultimate+question+of+life%2C+the+universe%2C+and+everything https://www.google.com/search?q=answer+to+the+ultimate+quest...

nephanth 6y ago

Not about algebra, but one good example of ridiculous problems leading to great discoveris in math is how, according to the legend, Cauchy invented complex analysis (at least the theorems on holomorphic and meromorphic functions) just to solve some random integral somebody had given him

kazinator 6y ago

> Well this is a bizarre 4th degree polynomial equation, \(x^4-2x^2 -400x=9999\) and he applies one approach to solving it, then comes to a dead end and says "Hence ingenuity is called for" and finally finds that his unknown number is 11. I, personally, would not have had a clue how to solve it. Here is what you can do. If there is an integer solution, we can keep the constant term 9999 on the opposite side and factor the formula. As in: x^4 - 2x^ - 400x = 9999 x(x^3 - 2x - 400) = 9999 Ok, so no we have x f(x) = 9999. From this we know that if there is an integer solution for x, x itself must be a factor of 9999: either a prime factor, or a product of factors. So we just factor 9999: 3^2 11 101 = 9999 9999 has 3 as a degree 2 factor, and then also 11, and 101. Looking at just the factorization breakdown alone, x could be the product of these combinations of factors: { (3), (11), (101), (3, 3), (3, 11), (3, 101), (11, 101), (3, 3, 11), (3, 3, 101), (3, 11, 101) }. If we substitute these factors for x, in order from least to greatest, we will soon hit upon the solution. Intuitively we can guess that x is small, because the opposite factor is cubing it. So for instance, it can't be the case that x is the product of 3, 11, and 101, such that x^3 - 2x - 400 then works out to the remaining factor of 3. It's obvious that x can't be too small, like 3, because x^3 is only 27; that's not going to leave us a positive factor when we subtract 400 alone, let alone -2x.