Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 219
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The average probability would probably be the integral of the function divided by the length. Taking as length not 101 landing points, but just one
The answer would be Mathemat.
The answer is correct, except where did you get it from? There are inaccuracies in the reasoning, and *** appears like hell out of a box :) However, the logic of reasoning is correct.
About the integral: yes, integral (at least I have it that way). But you can also sum the degrees of integers (squares) and then take the number of points to infinity. Then everything will be without integrals.
P.S. Well, yes, I didn't take into account that you may not have forgotten integrals yet...
There are no probabilities here.
You need a logical sequence that, in no more than 10 steps , is guaranteed to lead the megamosk to the opening of the doors. No special knowledge is required, and the task is completely honest, with no pitfalls.
Sure, he's lucky, but his luck is 100% due to his brain, not his luck.
The "Cube" problem.
/There was a correct solution - MathematAre the hands put in simultaneously or is it possible to put one first and then the other?
Only at the same time. All this is strictly monitored by a security guard.
The Cube problem /There was a correct solution - Mathemat/.
The first point is clear. But the second one is not clear. How do we choose two "-s"?
Only at the same time. All this is strictly monitored by a security guard.
Only at the same time. There's a security guard keeping a close eye on all this.
The first point is clear. It's the second that's not clear. How do we get two "-s"?
A regular N-gon is inscribed to a circle of unit radius. Find the product of the lengths of all its diagonals drawn from one vertex (counting the adjacent sides).
The task is here. The weight is 5.
The answer is intuitively clear, if you calculate it for the first small values of N. The main thing is the reasoning.
Moderators of the resource state that there is a school solution, but it is not nice. And there is a non-school solution, short and beautiful (I argue that), and I have already got it.
So it goes like this. Solve through the chord equation:
then the general formula for diagonals (except the largest one - it is equal to two radii of the circumcircle) will be
Two people play the following game. An even number of number cards are laid out in a row on the table. Players take turns picking one of the cards from either end of the row. Whoever wins has to get the higher amount, otherwise it's a draw. Who does not lose in this game? What is a no-lose strategy?
You have to go first, then the last card is guaranteed to be at least as big as your opponent's.
When making a move, you have to calculate if the cards are in the form: L1, L2, L3, ... P3, P2, P1
MAX (L1 - MAX( L2 - L3, P1 - P2), P1 - MAX( L1 - L2, P2 - P3))
if the right number is larger, pull from the right side, if the left one is larger, pull from the left side.
Okay, that's pretty good. All that's left is to multiply them. What's the big deal, some product of N sines... The problem is a two-time spit and scratch :)
Two people play the following game. An even number of number cards are laid out in a row on the table. Players take turns picking one of the cards from either end of the row. Whoever wins has to get the higher amount, otherwise it's a draw. Who does not lose in this game? What is a no-lose strategy?
You have to go first then the last card is guaranteed to be at least as big as your opponent's.
...
...if the higher number is right, we draw from the right, if the higher number is left, we draw from the left.
It's not about one last card, it's about the total. Try to take a broader view. Of course you have to do the calculation, but it is easier than you think.