[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 289
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Richie, it seems you have your own program on Vasik for calculations with crazy accuracy, you boasted about it once. Try to calculate a number, the square of which is what the problem requires.
I had a lot of fun in my fourth year at university. We had a good computer science teacher, a graduate of Moscow State University, who used to set such interesting problems. Then all these modules were not useful to me, and were eliminated for lack of use, as well as numerous copybooks. Now the accuracy of more than 5 characters is not used.
In general, your tasks are interesting. I do not even know how to approach it :)
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When I have a lot of free time I will try to restore what I once did. I remember I struggled with those zeros back then.
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Well, the number: 3,16...................e+99
It's obvious. How many signs there are in the ellipsis, who knows? Of course, it's not proof.
OK, let's listen to those who try to solve it...
Consider the difference between two adjacent squares, n^2 and (n+1)^2. It is 2*n+1.
Now look at our 199-digit number. If it must be the square of some number k, then k < 3.2*10^99. Consequently, the difference between the adjacent squares of integers around k can never be more than 2*3.2*10^99 + 1 < 6.4*10^99 + 1 < 10^100 -1.
On the other hand, 100 digits assigned to the original 99 are in any case a number no smaller than 0, but no larger than 10^100-1. I.e. there is bound to be some kind of square placed in that range. That's it.
ОК, вот решение задачки про 99 девяток.
Рассмотрим разность между двумя соседними квадратами, n^2 и (n+1)^2. Она равна 2*n+1.
Теперь - про наше 199-разрядное число. Если оно и должно быть квадратом некоторого числа k, то k < 3.2*10^99. Следовательно, разница между соседними квадратами целых в районе k никак не может быть больше 2*3.2*10^99 + 1 < 6.4*10^99.
С другой стороны, 100 цифр, приписанных к исходным 99, в любом случае составляют число не меньше 0, но не больше 10^100-1. Т.е. в этом диапазоне обязательно разместится некий квадрат. Всё.
Super. Bravo!
I've seen such a nice line of reasoning somewhere, but it came in handy now (I only remember the beginning, related to construction of number alpha). I think it came up in the theory of transcendental numbers.
Proof.
Let alpha = (sqrt(2))^sqrt(2). Then, obviously, alpha^sqrt(2) = 2. We don't know what the freak number alpha is, so let's reason.
Let's assume that alpha is irrational. Then the last equality solves the problem.
Now suppose alpha is rational. Obviously, it is not equal to 1. Then there exists a natural n such that alpha^(1/n) is irrational. Hence, (alpha^(1/n))^(n*sqrt(2)) = alpha^sqrt(2) = 2. We have again found a pair of irrationals satisfying the problem: alpha^(1/n) and n*sqrt(2). Proved.
P.S. The proof is "not really constructive". Those wishing to construct an explicit example, try it yourself. By the way, a simpler number, alpha = 2^sqrt(2), is also suitable for the proof.
1) Maximum number of dice rolled = 25 (number of prime numbers in range from 1 to 89 + 1).
// minimum number of dice to get the maximum number = 15
2) Average of final sums = 7.449704470311508;
How I solved the second item. Very simple - I made a script in mql5. :) :)
I've found a very brilliant algorithm, because it's simple. The simplicity is that you don't have to build a decision tree, everything is solved in one go.
The script and a text file with the results in the trailer. If you have any questions about the algorithm, ask, I'll answer them.
Доказать, что существуют иррациональные a, b такие, что a^b рационально. 20_
Где-то такое чудесное рассуждение видел, но вот сейчас пригодилось (помню только начало, связанное с конструированием числа alpha). Кажись, встретилось мне в теории трансцендентных чисел.
Доказательство.
Пусть alpha = (sqrt(2))^sqrt(2). Тогда, очевидно, alpha^sqrt(2) = 2. Мы не знаем, что это за уродец такой, число alpha, поэтому давайте рассуждать.
Допустим, что alpha иррационально. Тогда последнее равенство решает задачу.
Теперь допустим, что alpha рационально. Очевидно, оно не равно 1. Тогда существует такое натуральное n, что alpha^(1/n) - иррационально. Следовательно, (alpha^(1/n))^(n*sqrt(2)) = alpha^sqrt(2) = 2. Мы снова нашли пару иррациональных, удовлетворяющих задаче: alpha^(1/n) и n*sqrt(2). Доказано.
P.S. Доказательство "не совсем конструктивно". Желающие построить явный пример - попробуйте сами. Кстати, число попроще, alpha = 2^sqrt(2), тоже подходит для доказательства.
Well done. On close reading, I found a simpler one. I reproduce the whole thing (I copy the beginning from the board, add my own in green):
Proof.
Let alpha = (sqrt(2))^sqrt(2). Then, obviously, alpha^sqrt(2) = 2. We don't know what the freak number alpha is, so let's reason.
Let's assume that alpha is irrational. Then the last equality solves the problem.
Now suppose alpha is rational. Then the solution is alpha = (sqrt(2))^sqrt(2);
That's it. :))
Теперь допустим, что alpha рационально. Тогда решением является alpha = (sqrt(2))^sqrt(2);
Oh, right, yup :) Damn, sometimes I don't see the obvious.
And there's something suspicious with your script. Let's see.