Solution

F $(x + 1)$ $(F (x) + 1)$ = -1, hence, F does not vanish anywhere. For each x, either
$\\$
a$)$ F $(x + 1)$ $>$ 0, F $(x)$ + 1 $<$ 0, or $\\$ b$)$ F $(x + 1)$ $<$ 0, F $(x)$ + 1 $>$ 0.
$\\$
If a$)$ holds for at least one x, the function is discontinuous: F $(x + 1)$ and F $(x)$ have different signs.
$\\$
If b$)$ holds on the whole axis, then at every point -1 $<$ F $(x)$ $<$0. Then | F $(x + 1)$ | $<$ 1, | F $(x)$ + 1 | $<$ 1, and therefore the product of these numbers is not equal to 1. Hence we have a contradiction.

Hint

Solution

Robin Hood, being nothing but a fool, prepared for the duel: he drank beforehand from the first source $($ it could also have been from any other, except the tenth $)$. And so the poison from the source N 10 that the dark wizard gave to him acted on him as an antidote, and not as a poison. Furthermore Robin Hood gave the dark wizard ordinary water. After that, the dark wizard came to his cave, drank the poison from the source N 10 and was poisoned and died.

Hint

Solution

Let the Olympiad consist of 6 tasks. Then we will evaluate each task according to the 6-point system $($ the number of points varies from 0 to 5 $)$. Each participant is assigned an 8-digit number, the first two digits of which express the sum of the points scored by the student, and each of the remaining six digits are equal to the number of points for the corresponding task. If there are more tasks, the design is similar.

Hint

Solution

First, let the first guard come running to the top of B, and the second to C. Let the monkey find itself on one side of the triangle ABC. If the monkey is on BC, the guards move towards each other along the track BC. If the monkey is on AB or AC, the guard catches the monkey, as it is moving to the top corner at A. Let this not happen. Without loss of generality, we can assume that the monkey is in the right half of the zoo, that is, on one side of the triangle CEF. Next, the first guard runs to the corner E, and the second is at the vertex C. When the first guard reaches the corner of E, the monkey either remains in the right half of the zoo, or is on the DE segment. In the first case, the guard is caught by the monkey, acting in the same way as described above for the triangle ABC. In the second case, the second guard runs to the corner B. When he runs, the monkey is on one of the segments BD or DE, and the guard catches it when moving to D.

Compare with problem number 78751.

Answer

Yes they can.

Hint

Solution

Let the first player, starting with their second turn, make moves symmetric to the moves of the second player relative to the center of the field. Then, as soon as the second player takes one of the corner cells, the first one can occupy the opposite corner cell.

Hint

Solution

Suppose that Callum is a truth-teller. Then nine of the ten players of the truth-tellers team scored on an odd number of goals $($ one, three or five $)$, and Callum scored an even number of points. But then the team of truth-tellers scored an odd number of goals, which contradicts the condition. Consequently, our assumption is incorrect and Callum is a liar.

Answer

He’s lying.

Hint

Solution

Put two coins on the scales and weigh them. We consider two cases.

1$)$ One of the bowls outweighs the other. Therefore on the heavier bowl lie the coins of 4 g and 2 g or of 4 g and 3 g, and on the lighter bowl – the coins which weigh 1 g and 3 g or those which weigh 1 g and 2 g. For the next two weighings, let us compare the coins on each bowl with each other and determine which coins weigh 1 g and 4 g. With the fourth weighing we compare the remaining two coins, determining which one weighs 2 g, and which one weighs 3 g.

2$)$ The scales are in balance. Therefore on one bowl there are the coins which weigh 2 g and 3 g, and on the other the coins which weigh 1 g and 4 g. In the next two weighings we compare the coins on each bowl with each other, and then compare the two coins that turned out to be heavier, determining which one weighs 4 g, and which one weighs 3 g. By comparing the two coins, we automatically find out which one turned out to be lighter using the results of the second and third weighings.

Note that four weighings are not sufficient to perform the standard sorting algorithm for four objects.

Answer

Yes you can.

Hint

Solution

Suppose that the council consisted of seven members, and the votes were distributed, as shown in first the table.

The points were distributed as follows: A – 13, B – 11, C – 6, D – 12. The candidate with the most points was candidate A, the one with the least – Candidate B.
When B withdrew his candidacy, the preferences table changed $($ second table $)$. The scores also changed: A – 6, C – 7, D – 8.

1. Strangely, the candidate with the lowest rating changed the results of the elections, by simply ceasing to participate in them, and the leader then became an outsider. This and other paradoxes of elections and the famous Arrow theorem are described in detail in the book by R.E. Klima, J.K. Hodge. “Mathematics of elections”. Moscow. MCNMO. 2007.

2. 2 points.

Answer

Yes it could.

Hint

Solution

Evaluation. Suppose that among the rational numbers there is a 0 and it is written on the top row of the table. Suppose that x irrational and 50 – x rational numbers are written along the left side of the table. Then along the upper side there are written 50 – x irrational and x rational numbers. We note that the product of a nonzero rational number and an irrational number is irrational. Hence, in the table there are at least x $(x – 1)$ + $(50 – x)^2$ = $2x^2$ – 101x + $50^2$ irrational numbers. The parabola’s vertex y = $2x^2 – 101x + 50^2$ is at the point 25,25, therefore the minimum value of this function at an integer point is reached at x = 25 and is equal to $25 \times 24 + 25^2 = 1225$.
If 0 is replaced by a non-zero rational number, then the number of irrational numbers can only increase. Therefore in the table in any case there are no more than 2500 – 1225 = 1275 rational numbers.

Example, when the rational number is 1275. Along the left side, we put the numbers 1, 2, …, 24, 25, $\sqrt{2}, 2\sqrt{2}, …, 25 \sqrt{2}$, and along the upper side we place the numbers 0, 26, 27, …, 49, $26\sqrt{2}, 27\sqrt{2}, …, 50 \sqrt{2}$. Then only $25 \times 24$ + $25^2$ will be irrational = 1225 products of nonzero rational and irrational numbers.

Answer

1275 products.

Hint

Solution

Suitable weightings could be carried out with the weights of 49.5 g, 50.5 g and 51.5 g.

1. Let us prove that this example is unique.
None of the weightings involved exactly one weight: otherwise its weight would be greater than 90 g. If all three weights were involved in some weighing, then this could only be the third weighing, that is, the sum of the weights of all the weights is 102 g. Then, so that the pharmacist can get 100 g and 101 g, one of the weights must weigh 1 g, and the other is 2 g. Hence, the remaining weight weighs 99 g, which does not suit us.
Therefore, in each weighing, exactly two weights were used, each time different ones. Hence, the doubled weight of all the weights is equal to 100 + 101 + 102 = 303 g, that is, the sum of the weights of three weights is 151.5 g. Therefore, the weight of the lightest weight is 151.5 – 102 = 49.5 g, the weight of the next –
151.5 – 101 = 50.5 g, and the weight of the heaviest – 151.5 – 100 = 51.5 g.

2. 4 points.

Answer

Yes it could.

Hint

Solution

There are many other examples in which one of the terms is equal to zero, and also only one example in which all terms are nonzero:
1/ 3/ 6 + 5/ 8/ 9 + 7/ 2/ 4 = 1.

Answer

For example, 2/ 1/ 4 + 9/ 3/ 6 + 0/ 5/ 7 = 1 or 5/ 4/ 3 + 7/ 6/ 2 + 0/ 8/ 9 = 1.

Hint

Solution

First, move all the flashlights into the right box, without touching the switches. Next, move from the right box to the left box any hundred flashlights, while switching each, and the goal will be achieved. Let us prove this.
Let’s switch exactly k lights on. So, k left lights were turned off and 100 – k were switched on. And in the right box there are 100 – k switched on and k switched off.

Hint

Solution

Let a be the number of excellent students, b – the number of slackers, c – the number of mediocre students, who made a mistake in answering the first question, answered the second correctly and were mistaken in the answer to the third question.
By the condition a + b + c = 19, b + c = 12, c = 9. Hence b = 3, a = 7. Hence, the number of players is 30 – 7 – 3 = 20.

Answer

20 mediocre students.

Hint

Solution

Let’s paint the doors in black and white colors, alternating them. For the first attempt Hannah Montana checks all the white doors. If after that she did not manage to leave the room, then a black door was the one that was unlocked. Michelle will lock that one and unlock a neighboring white one. In this case, Hannah Montana just needs to once again check all the white doors.

Compare with the task number 65893.

Hint

Solution

Note that at each step, if both participating cubs had at least one berry, then at least one berry remains with each of them. Therefore at the end each bear will have at least one berry.
Let us prove that the fox can leave each bear cub with exactly one berry, that is, move from the position $(1, 2, …, 2^{99})$ to the position $(1, 1, …, 1)$. To do this, we prove by induction that in the case where there are n bear cubs, the fox from the position $(1, 2, …, 2^{n-1})$ can go to the position $(1, 1, …, 1, 2^{n-1})$, and from the last – to the position $(1, 1, …, 1)$.
Base. For n = 1, all three positions coincide.
Inductive step. From the position $(1, 2, …, 2^{n-1}, 2^n)$, forgetting about the last bear cub, we can proceed by induction first to $(1, 1, …, 1, 2^{n-1}, 2^n)$, and then to $(1, 1, …, 1, 2^n)$. Using the last two cubs, the fox moves to the position $(1, 1, …, 1, 2^{n-1}, 2^{n-1})$. Now, again forgetting about the last bear cub, you can go to the position $(1, 1, …, 1, 2^{n-1})$, and from it, forgetting about the first one, to the position $(1, 1, …, 1)$ . For the case where there are 100 cubs, this means that the fox has eaten $2^{99}$ + $2^{98}$ + … + 2 + 1 – 100 berries which equals $2^{100}$ – 101.

5 points

Answer

$2^{100}$ – 101.

Hint

Solution

Note that at each step, if both participating cubs had at least one berry, then at least one berry remains with each of them. Therefore at the end each bear will have at least one berry.
Let us prove that the fox can leave each bear cub with exactly one berry, that is, move from the position $(1, 2, …, 2^{99})$ to the position $(1, 1, …, 1)$. To do this, we prove by induction that in the case where there are n bear cubs, the fox from the position $(1, 2, …, 2^{n-1})$ can go to the position $(1, 1, …, 1, 2^{n-1})$, and from the last – to the position $(1, 1, …, 1)$.
Base. For n = 1, all three positions coincide.
Inductive step. From the position $(1, 2, …, 2^{n-1}, 2^n)$, forgetting about the last bear cub, we can proceed by induction first to $(1, 1, …, 1, 2^{n-1}, 2^n)$, and then to $(1, 1, …, 1, 2^n)$. Using the last two cubs, the fox moves to the position $(1, 1, …, 1, 2^{n-1}, 2^{n-1})$. Now, again forgetting about the last bear cub, you can go to the position $(1, 1, …, 1, 2^{n-1})$, and from it, forgetting about the first one, to the position $(1, 1, …, 1)$ .

5 points

Answer

100 berries.

Hint

Solution

We give examples of how this could happen.
The first example. We will arrange the pupils in a circle. Suppose that everyone on their birthday went up to all the classmates, except the one following him clockwise. Then every two pupils A and B met at all of the parties, except for two: the one that A did not attend, and the one that B did not attend. Hence, each pair of pupils met 21 times.
The second example. Let’s select from the class two pupils A and B. Let on A’s birthday all classmates came to the celebration, except for B, and to the birthday party of B only A came, and to the remaining birthdays came only B. Then every couple in which there is no B met only on the birthday of A, and all the pairs containing B, met exactly once in the other celebrations. Therefore in total, each pair met exactly once.

Answer

It could happen

Hint

Solution

Clearly, not all those present are knights and not all are liars: in these cases, none of them could say the first phrase.
If we take one knight, one of the liars next to him can say the first phrase, and the remaining liars can say the second. This gives an example of the case when there is a single knight.
If we take two knights on opposite places around the table, then all the liars will be able to say the second phrase. This gives an example of the case when there are two knights.
The fact that there are no more than two knights can be proved in different ways.  
The first method. Suppose there are two knights sitting next to each other. Let’s move from them in a circle until we reach the first liar. Then we find a combination of KKL; But in this case the knight in the middle cannot pronounce any of the phrases. This gives a contradiction. Therefore, every knight is surrounded by liars. But in such conditions each knight will utter the first phrase. Therefore, there are no more than two knights.
The second method. Suppose a knight said the second phrase. Then both his neighbors are knights. Consider his neighbor on the right. He is a knight, and to the left of him is a knight. He cannot lie, he speaks the first phrase, and he must say that both his neighbors are knights. Then consider his neighbor on the right; and so on. It turns out that all those present at the table are knights, which can not be the case. Hence, all the knights present are obliged to speak the first phrase, and there are only two such phrases. Therefore, there are no more than two knights.

Answer

Either one or two.

Hint

Solution

Note that when dividing by 7, the number of matches taken by Anna gives a remainder of 5, and the number of matches taken by Natasha gives a remainder of 2. The number 2016 is a multiple of 7, so after each turn of the player who went second, the number of remaining matches will also be a multiple of 7.
Since there were two matches left on the table, for the last move the number of matches which gives a remainder of 5 on dividing by 7 was taken from the table. This means that the last move was made by the Anna and she won.
All moves, except for the last, are divided into pairs, in which the total number of matches taken is a multiple of 7, so the first move was also made by Anna.

Answer

The first was made by Anna, and she won.

Hint

Hint

2016000 = $2^8 \times 3^2 \times 5^3 \times 7$.

Solution

For example: $10 \times 24 \times 10 \times 28 \times 30$ = 2016000 or $10 \times 28 \times 10 \times 24 \times 30$ = 2016000.

Answer

Yes there is.

Hint

2016000 = $2^8 \times 3^2 \times 5^3 \times 7$.