10 interesting tasks

The speed of convergence of trains is 100 km / h, the distance between them is 100 km, which means they will meet in 1 hour. The fly will fly during this hour at a speed of 200 km / h and fly a total of 200 km .

In the triangle MBC, the CM legs are smaller than the hypotenuse BM. MC = AM. Hence, MB> AM. Consider the triangle ABM, in it the side MB is greater than the side AM. And on the opposite side of the larger side lies a larger angle. This means ∠CAB> ∠ABM. CTD.

A trap task. The rule “there is exactly one boy in a family” does not affect the probability of having a boy as a whole in the country. The correct answer is 50% .

Suppose that the entire task consists of 28 tasks. Half of the task (14 tasks) will be solved by the first one with Reshebnik for 14/7 = 2 hours, and the second one during this time will solve 2 * 4 = 8 tasks. Armed with Reshebnik, the second will solve the remaining 28-8 = 20 tasks in 20/7 <3 hours, and the first during this time will solve only 3 * 4 = 12 tasks, and all she has left is 14 tasks. Consequently, the second girl will cope with the task earlier.

We recall the law of Archimedes: on a body immersed in a fluid, a pushing force acts, equal to the force of gravity of the fluid displaced by this body. It should be noted that the body must be completely surrounded by liquid or intersect with the surface of the liquid. For example, Archimedes' law cannot be applied to scrap, which lies at the bottom of the basin. Thus, while ice and scrap float, the mass of scrap and ice is equal to the mass of water displaced. When scrap reaches the bottom, the volume of water displaced by it is equal to the volume of scrap. The mass of ice is equal to the mass of water, which it turned into when it melted, so the contribution of ice can be ignored. (By the way, for the same reason, if the scrap did not appear in the original task, the answer would be that the water level would not change.) Before melting, scrap crowded out more water than after melting, since water has a lower density than scrap. Answer: the water level in the pool will go down . (If we assume that the ice and scrap floated in the water column, and not on the surface, the justification is quite simple: the ice volume decreased when the ice melted, and the scrap volume did not change.)

Recall Newton's bin .

In the last expression, each binomial coefficient is divided by p, since p is simple and is in the numerator in the formula of the binomial coefficient , and in the denominator - the product of numbers less than p. CTD.

Denote by F (N) the number of rows of length N, not containing two units in a row. Denote by F0 (N) the number of rows of length N, which do not contain two units in a row and end with 0. Denote F1 (N) the number of rows of length N, which do not contain two units in a row and end with 1.

We derive the recurrence formulas for F0 and F1.
F1 (N) = F0 (N-1). Since two units cannot go in a row, the number of lines ending in 1 is equal to the number of lines 1 shorter, ending in 0.
F0 (N) = F0 (N-1) + F1 (N-1). Zero can be added to any valid line, without violating the conditions. This means that the number of lines ending in 0 is equal to the number of lines in 1 shorter.

F0 (N) = F0 (N-1) + F1 (N-1) = F0 (N-1) + F0 (N-2). So, F0 is a Fibonacci series with a certain shift. F1 (N) = F0 (N-1), then F1 also consists of Fibonacci numbers, with a “lag” of 1 relative to F0.

F (N) = F0 (N) + F1 (N). The F (N) term is obtained when a Fibonacci number F0 is added to a Fibonacci number F1, which is the previous number for it. And the sum of the Fibonacci number with the previous Fibonacci number is also the Fibonacci number (according to the formula of Fibonacci numbers). So, we have proven that the number of rows that do not include two units in a row will be the Fibonacci number . It remains to determine the value of F (N) for the first two N. F (1) = 2 (two suitable sequences: "0" and "1"), F (2) = 3 (three suitable sequences: "00", "01" , "ten").

The key condition is the fact that at the beginning no shareholder sits behind his sign. Let the number of shareholders be equal to N. The number of table states is also N. On one of the table states (initial) no shareholder sits at his plate, therefore the table conditions at which comparisons of shareholders with tablets are possible are N-1. Let's sort through these N-1 states. During this time, all N shareholders must "go" for their own signs. According to the Dirichlet principle , at least one of the enumerated N-1 states must have at least two of N comparisons. In other words, there is a state in which at least two shareholders are sitting at their own signs. CTD.

Denote the white cap unit, black cap zero. The last sage adds the colors of the caps in front of the standing sages modulo 2 (x them). Calls the color corresponding to the resulting amount. The further fate of the last sage is not so interesting: perhaps the named color coincided with the color of its cap, perhaps not. The next sage, having calculated the sum of the caps in front of him, adds it to the color called the last sage, and receives the color of his cap, which he calls. The next one adds three numbers already: the number of white caps standing in front and the numbers said by the two previous sages. Subsequent sages operate in the same way. It turns out at least N-1 correct answers.