[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 303

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Now that's great:
The level of knowledge in public schools is just as disastrous: половина преподавателей не имеют дипломов (они сами еще учатся, чтобы стать учителями). This nonsense has become ubiquitous because school owners pay a teacher without a diploma far less than a certified teacher.
That's great! The demands in Russia are much higher. Without a teaching degree they won't let you anywhere near a school (in Belokamen). It's true, though, that the crisis and employers' lawlessness are taking the heat off the labour market.
 
Yeah. We don't care if you don't have a degree or three, if you're a schoolteacher, you don't get paid shit.
 

Grade 9.
 
sqrt(2) at sqrt(2) and 1/sqrt(2)
 
Mathemat >>:
Кстати, твое решение как раз приведено в задачнике. 9450 в центре. Но тебе для обоснования нужно намного меньше, чем программа на "пятере". Обрати внимание, что т.е. оба числа снизу и сверху от 9450 делятся соответственно на 11 и 13. Осталось найти способ доказать это без привлечения сложных вычислительных методов. А больше ничего доказывать и не надо :)
So the formula ( (11 * 13) * N + 12) % (2*3*5*7) == 0 is exactly derived from this requirement.
By the way, yawned yesterday ((11 * 13) * N - 12) % (2*3*5*7) == 0 also leads to solution.
Generally, there are only two arithmetic progressions with difference 30030 which form a complete set of solutions.
n = 9440 + 30030*k is the solution to the first equation. And n = 20570 + 30030*k is the solution to the second.
As for "no need to prove anything else" - I guess I can't prove it, because it doesn't work.
But the problem didn't require proving anything, did it? Just find it.
// Prove that a natural chain of more than 21 numbers satisfying the same requirement cannot be constructed!
// But if we add 17 (2*3*5*7*11*13*17) to the set of multipliers, a chain of 25 numbers is possible. (Least solution: n = 217128)
// If we add another 19, we get the maximal length of the chain = 33 // (min(n) = 60044) - oddly enough, the minimal solution is smaller.
// And if we add also 23 - what do you think the maximum chain length will be? // by the way min(n) = 20332472
 
MetaDriver >>:
Но вроде в задачке и не требовалось чего-то доказывать? Только найти.
// А вот докажи, что натуральную цепочку более чем из 21 числа, удовлетворяющую этому же требованию соорудить не удастся!

Well, yes, find and prove that's what it's all about. I had to prove it for my big numbers...

About the addition: we'll see. Maybe it's true.

2 TheXpert: Have you solved it before, Andrei?

 
Mathemat >>:
2 TheXpert: раньше, что ли, решал, Андрей?

I don't get it. The answer is pretty obvious. I've made up my mind too, but I've also decided not to break the game to others... :)

 

I've been thinking about the cube and the box...
how many ways you can paint the cube to look different?

 
Don't be so difficult about it. There's no catch in the problem.
But max( min( x, y + 1/x, 1/y ) ) )... Well, two people have already solved it so quickly, and I'm still thinking.
 
omgwtflol >>:

А я всё над кубиком с коробкой думаю...
сколькими способами можно раскрасить куб чтоб выглядело по-разному?

5*3*2=30

1...296297298299300301302303304305306307308309310...628
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