Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 59
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Therefore it was sufficient to prove the coincidence of three consecutive values of the series for the series to coincide
Yes, but it may not have been Fibs.
And I didn't actually solve the system, I just noticed a literal coincidence with them, which eliminated the need to solve it.
And can you explain what the buckeyes are?
The coordinates of MM with the dog -- (x1, y1);
The coordinates of MM with the hat -- (x2, y2);
So, there is a MM with coordinates -- (x1, y2); (X).
What can you say about X? It is not higher than the MM with the dog as it is in the same longitudinal row as it is and not lower than the MM with the hat as it is in the same transverse row as it is.
The coordinates of MM with the dog -- (x1, y1);
The coordinates of MM with the hat -- (x2, y2);
So, there is a MM with coordinates -- (x1, y2); (X)
What can be said about X? It is not higher than the MM with the dog, since it is in the same longitudinal row with it, and not lower than the MM with the hat, since it is in the same transverse row with it.
Two armies of mega-brains come out to fight: pointed and blunt-tipped. Each army has 2*N men. Each megabrain has a gun, which can kill no more than one enemy when fired. Megabrains follow the rules of combat: first shoot the sharp-tipped ones, then shoot the blunt-tipped ones and then shoot the sharp-tipped ones again. After these three volleys the battle ends. Question: what is the maximum number of mega-brains that could have died in this battle? Justify that this number is the maximum.
3*N apparently (i.e. N will remain). Scenario -- N -- N
Consider 2 cases:
1. In the first salvo less than N people are killed (K). Then the minimum number is 4N - K - (2N - K) - K = 2N - K > N
2. In the first salvo more than N people are killed (L). Then the minimum number is 4N - L - (2N - L) - (2N - L) = L > N
Very brief, the chain is not very clear. I had a more authentic one.
I.e. in the first salvo the sharp-tips kill K people. The blunt-tipped ones have 2N - K people, the sharp-tipped ones still have everyone alive, i.e. 2N.
In the second one, they shoot 2N-K blunt-edged men and kill... how many?
In short, it's not clear where minimality comes from. There is only one parameter, not two.
The first salvo kills K MM, the second L. Obviously L <= 2N - K. I.e., the first two salvos killed S MM, which is no more than
S = K + L <= 2N. (1)
After two salvos 4N - S MM is left. With the last salvo no more than
floor( (4N - S) /2 ), and total killed is not more than S + floor( 2N - S/2 ), where floor() is the nearest integer from below.
S + floor( 2N - S/2 ) monotonically increases along with growth of S, and, taking into account (1) does not exceed 3N
My rationale (credited):
RATIONALE:
Suppose the first volley of sharp-edged men kills X blunt-edged men with 2*N-X left alive. X is killed.
Then 2N-X blunt-pointed men kill Y pointed men, leaving 2N-Y. Another Y is killed.
Finally 2N-Y pointy-tails kill Z pointy-tails, which leaves 2N-X-Z. Another Z is killed.
In total X+Y+Z are killed, and this value must be maximized. There are restrictions:
0<=X<=2N
0<=Y<=2N-X
0<=Z<=2N-Y
0<=2N-X-Z
X>=0, Y>=0, Z>=0
X<=2N, Y<=2N, Z<=2N
Rewrite the problem:
X+Y+Z -> max (0)
0<=X+Y<=2N (2)
0<=Y+Z<=2N (3)
0<=X+Z<=2N (4)
X>=0, Y>=0, Z>=0 (5)
X<=2N, Y<=2N, Z<=2N (6)
Obviously, (5) and (6) restrict a part of the space inside the cube in the positive octant with vertex at zero coordinates and side 2*N. In fact the domain (6) is redundant for the problem. The really important constraints are (2)-(5) and the maximization condition (0).
(2) defines a region of three-dimensional space bounded by a "vertical" plane X+Y=2N with the origin "inside".
Similarly, (3) and (4) are two more similar regions, only oriented differently.
On the other hand, the plane X+Y+Z = const is also easily visualized: it carves an equilateral triangle in the cross section of the positive octant of space. It remains, by moving the plane from the origin of coordinates, to find its maximal distance from zero coordinates at which conditions (2)-(4) hold.
Due to complete symmetry of all variables, the required maximum is reached when X=Y=Z=N. The number of killed is 3*N. In each salvo the army kills exactly half of the opposite one.
I have another solution, it came a little later... Let's keep your X, Y, Z
Obviously Y <= 2N - X; Z <= 2N - Y, i.e.
X + Y <= 2N (1)
Y + Z <= 2N (2)
On the other hand, the total number of killed is no more than 2N + Y - all blunt-endings are killed
X + Y + Z <= 2N + Y, or
X + Z <= 2N (3) //I just saw that the previous two lines are redundant. The number of dead ends killed is at most 2N.
Add all three inequalities and divide by 2, we get
X + Y + Z <= 3N
Yes, short and to the point. Thank you both!
(4), not scored
It is snowing (falling vertically). With very little friction, two identical carts roll with inertia. On each one sits a megabrain. One constantly cleans the cart from the snow (shovels it to the side perpendicular to the movement trajectory), the other does not. The trolleys gradually but slowly slow down from friction. The snow does not melt. The mega-brains are wearing tuluk and valenki, which don't allow any heat to penetrate. Which cart will go the furthest?
(3), not yet scored, but confident in his own solution:
Which is greater: sin(cos(x)) or cos(sin(x))?
It is snowing (falling vertically). With very little friction, two identical trolleys roll with inertia. A mega-brain sits on each one. One constantly cleans the trolley of snow (shovels it to the side perpendicular to the trajectory of movement), the other does not. The trolleys gradually but slowly slow down from friction. The snow does not melt. The mega-brains are wearing tuluk and valenki, which don't allow any heat to penetrate. Which trolley will go farther?