When the impossible goal is to solve a 10x hard problem in 1x time live during an oral exam, no. But it did have a filter effect, so that the Jewish students who did pass the entrance test were much higher ability than their average peers.
I doubt it. Those problems are close to impossible to solve for ordinary students, therefore many talented but not necessarily genius students lost the opportunity to get good education. I'm a firm believer that students need to be pushed to stay in their discomfort zone, but Jewish Problems can easily push most students into panic zone.
1. They're an entrance exam. They were explicitly designed to discriminiate against otherwise worthy candidates.
2. The key feature of the problems is that they are not difficult because they are an average example of a difficult branch of math. Instead they are easy problems, but they are only easy if you can figure out the correct substitution. Otherwise, they are extremely difficult. In other words, getting good at the problems won't make you a better mathematician, it'll just make you better at passing that particular test.
I was only agreeing with the parent that students need to pushed out of their comfort zone, but just a little. Not a lot like the Jewish Problems were doing.
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.
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.
Which is funny because we could say the same about competitions like Putnam, but that doesn't make Putnam any easier. But that's the problems with solutions, they are elegant and obvious, but only after the fact. Competition problems are explicitly created to be like that. These problems similarly, but because they could show a simple solution you can "justify" the racial exclusion.
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.
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.
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.
This reminds me of the riddle by the good soldier Svejk:
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"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?"
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?»
Using Google Translate doesn't produce anything understandable. But searching for the text of the "favorite" part it can be seen that is is from this work:
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?"
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.
> 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.
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.
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
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).
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