Ridiculous Math Problems

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?»

I'll take a shot at translating those in a somewhat intelligible way while trying to preserve hilarity, with my commentary in parenthesis when appropriate.

"A famous udarnik (a shock worker, someone who was a lot more productive than others) Aleksei Stakhanov went number one 2 times a day and number two 1 time every two days. However, when in zapoy (drinking continuosly) he went number one 4 times a day and didn't go number two at all. How many times he went number one and number two in a year, considering that he was in zapoy 312 days a year?"

"When American Seventh fleet ships moored to Petushki station, there was no party member girls (Communist Party obviously, there was no other party) there, but if you consider Komsomol (kind of a youth division of the Party) girls belonging to the party, a third of them were blonde. When American Seventh fleet ships left, the following was discovered: every third Komsomol girl was raped, every forth raped girl was a Komsomol member, every fifth raped Komsomol girl was a blonde, every ninth raped blonde was a Komsomol member. Considering that there was a total of 428 girls in Petushki, how many non-party brunettes were left untouched?"

"As we know, there is no point A's in Petushki. All the more, there is no point C's. The only points present are B. So, Papanin (a last name), going to save Vodopyanov (another last name), started moving from point B1 to point B2. At the same moment Vodopyanov, going to save Papanin, started moving from point B2 to point B1. It is unknown why, but they both ended up at point B3, which is 12 Vodopyanov's spits away from point B1 and 16 Papanin's spits away from point B2. Considering that Papanin's spits land 3m 72cm away, and Vodopyanov can't spit at all, was Papanin actually going to save Vodopyanov?"

UPD: typo

> 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.

it depends on the size of the hole at the bottom of the ship

Good correction, thanks! :)