Hint

Appropriate fractions $P_k$ / $Q_k$ to the indicated infinite continued fraction are simultaneously suitable fractions to finite continued fractions of the form
[$a_0$; $a_1$, …, $a_n$] for n $>$ k. Therefore, the conditions of problem 60602 are fulfilled for them. The existence of the limit α follows from parts b$)$ and d$)$ of this problem. It remains to verify that $a_k$ = [$α_k$] when expanding according to the algorithm from problem number 60606.

Hint

Appropriate fractions $P_k$ / $Q_k$ to the indicated infinite continued fraction are simultaneously suitable fractions to finite continued fractions of the form
[$a_0$; $a_1$, …, $a_n$] for n $>$ k. Therefore, the conditions of problem 60602 are fulfilled for them. The existence of the limit α follows from parts b$)$ and d$)$ of this problem. It remains to verify that $a_k$ = [$α_k$] when expanding according to the algorithm from problem number 60606.

Solution

a$)$ See the problem 35111.

b$)$ First we turn all the coins except the first, then all but the second, etc. As a result, each coin will be turned 19 times.

Answer

a$)$ It is impossible; b$)$ it is possible.

Hint

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Hint

Substantially use the condition that not one of the labels indicates the contents correctly.

Solution

We take out the ball from the urn with the WB sign. If we removed a white ball, then in this urn there are two white balls $($otherwise the table would indicate the contents correctly$)$. Next, we determine that in an urn with a BB sign there can be only one white and one black ball $($otherwise the table would indicate the contents correctly$)$. There was only one possibility left – in the urn with a sign of WW there are two black balls. If we removed a black ball from the urn with a black ball, we similarly define that there are two black balls in it, in the urn with a sign of WW there is one white and one black ball, and in an urn with a sign of BB there are two white balls.

Answer

One extraction is enough!

Hint

Substantially use the condition that not one of the labels indicates the contents correctly.

Hint

Note that as a result of the “entanglement” of the wires, the number of units in the representation of the transmitted letter does not change – this will greatly reduce the number of variants of the transmitted word.

Solution

Let’s write out under each letter of the received text ЭАВЩОЩИ all letters in which the representation has the same quantity of units, as the initial letter. Next, we try to “read” the meaningful text, choosing from each column of the resulting table one letter. The only option is the word STEAM.

Hint

Note that as a result of the “entanglement” of the wires, the number of units in the representation of the transmitted letter does not change – this will greatly reduce the number of variants of the transmitted word.

Hint

Begin with division.

Solution

For division: $(a*b)*1=1-(a*b)/1=1-(1-a/b)=a/b$.

For multiplication: $a \times b = a/(1/b) = (a*((1*b)*1))*1$.

For subtraction: $a – b = (b*a) \times a = ((b*a)*((1*a)*1))*1$.

For addition: $a+b = a – (0-b) = a – ((b*0)*((1*0)*1))*1$.

Hint

Begin with division.

Hint

8 is a power of 2. First solve this problem for 4 glasses and then for 8.

Solution

Let us separate the glasses into pairs A-A’, B-B’, C-C’, D-D’, and then equalise the volume of water in each pair of glasses. Then we have two identical sets of four glasses: A, B, C, D and A’, B’, C’, D’. We will then equalise the volume of water in the first set of four glasses and aftwerwards use the same method for the second set of four glasses. Thus we have made the problem into one with four glasses. Similarly, we separate the four glasses into pairs, equalise the volume of water in each pair, and thus transform the problem into one with two glasses. However, we can equalise the water volume in two glasses as stated in the problem.

Hint

8 is a power of 2. First solve this problem for 4 glasses and then for 8.

Hint

The occurrence of some combination may have the meaning of “repeating the coin toss”.

Solution

One possible way is as follows. First, we throw a coin two times. If both times we get tails, then the first goes, if we get tails the first time and heads the second time, then the second one goes to get the juice, if we get heads the first time and tails the second, then the third one goes, if we get heads both times then we carry out a second round of the coin toss, similar to the first, i.e. again we throw a coin twice, and so on. It is clear that in each round the probability each of the three friends losing is 1/4, and with a probability of 1/4 they will have to carry out a new coin toss. Theoretically, it is possible that we get heads all of the time which means that the process will not end. But the probability of this event is 0. Indeed, the probability that after n rounds the coin toss is not finished is $1/2^{2n}$ $($we get heads in all 2n tosses$)$; but $1/2^{2n}$ tends to zero as n tends to infinity.

Hint

The occurrence of some combination may have the meaning of “repeating the coin toss”.

Hint

Those who have not had their salary added a pound to lose a pound relative to the other employees.

Solution

Those who have not had a pound added to their salary lose this pound relative to the other employees. We will not pay the employee with the highest salary until his salary is equal to the one that was the smallest $($if the there are multiple employees with the highest salary then choose any of them$)$. Thus, at least two employees will have the smallest salary. Then, again, we select the employee with the highest salary and we will not pay them until their salary is equal to the one that was the smallest, and we get at least three employees with the same salary. Having done this operation no more than nine times, Hannah will be able to pay all the salaries.

Hint

Those who have not had their salary added a pound to lose a pound relative to the other employees.

Hint

It is necessary to ask questions so that each subsequent question reduces the number of remaining options by approximately two times.

Solution

In the beginning, when we do not know anything about the phone number, there are $10^5$ = 100,000 possible options for the phone number. We ask the question: “Is it true that the number is greater than 50,000?”. Regardless of the answer, we halve the number of remaining options. So, after the first question there are no more than 50,000 possible options for the phone number. If the answer to the first question is “yes”, then we ask the question: “Is it true that the number is greater than 75,000?”. After answering this question, there will be no more than 2500 variants. And so on – if there are k possible variants left after the n-th question, then it is possible to ask the $($n + 1$)$th question in such a way that after the answer to it leaves no more than [k / 2] +1 possible variants. Thus, we get the following dependence of the number of remaining possible options relative to the number of questions asked: After the first question, 50,000 options, after the second question – 25,000, 3 – 12,500, 4 – 6250, 5 – 3125, 6 – 1563, 7 – 782, 8 – 391, 9 – 196, 10 – 98, 11 – 59, 12 – 30, 13 – 15, 14 – 8, 15 – 4, 16 – 2, 17 – 1. So, after 17 questions there will be one option, that is you can guess the phone number. On the other hand, you can not guess the number from 16 questions. Indeed, after 16 questions there are $2^16$ $<$ 100000 sequences of answers to the questions. Hence, one of the sequences of answers corresponds to at least two variants of the telephone number. Thus, if the desired number is either of these two numbers, then the sequence of answers to the questions will be the same, i.e. Uniquely determining the number will not be possible.

Answer

17 questions.

Hint

It is necessary to ask questions so that each subsequent question reduces the number of remaining options by approximately two times.

Hint

It is not necessary to fry the cutlet on one side immediately followed by the other side.

Solution

In the first 5 minutes, let’s fry the first two cutlets on one side. Then turn the first cutlet over, remove the second from the pan and put in the third. Fry for another 5 minutes. Now the first cutlet is ready, and the second and third are done on one side. Within the next five minutes, fry the second and third cutlets on the other, not-cooked side.

Answer

15 minutes.

Hint

It is not necessary to fry the cutlet on one side immediately followed by the other side.

Hint

Can I say anything about members of the company A and B, if A knows B $($if A does not know B$)$?

Solution

In order to directly determine that this person is Z, you have to ask 2 * $($N-1$)$ questions: N-1 to him $($does he know every member of the company$)$ and N-1 to the other members of the company $($do they know him$)$. Clearly, such a path is hardly optimal. Consider the method of elimination. Obviously, there can be only one Z in the company. Let the journalist ask A if he knows B. Then if A does not know B, then A is not Z, if A knows B, then B is not Z. Thus, with every question, regardless of the answer, the journalist reduces the search by one person. It is clear that he will be able to do this with N-1 questions, since it is sufficient to check N-1 people, and leaving just the person Z.

Answer

N-1.

Hint

Can I say anything about members of the company A and B, if A knows B $($if A does not know B$)$?

Hint

The hourglasses can be turned over even if there is still some sand in it.

Solution

Let the first hourglass be the 7 minute one, and the second the 11 minute one. Let’s turn both of these hourglasses over at the same time.

The first way. After 7 minutes, let’s turn over the first hourglass, and after 4 minutes, that is, when there is no sand in the second hourglass, let’s turn the hourglass over again. At this moment, in the first hourglass, there will be sand for 4 minutes, and at the top – for 3 minutes. So, if we turn over the first hourglass, we can measure out another 4 minutes.

The second way. After 7 minutes, that is, when all of the sand has reached the bottom in the first hourglass, the second hour will have 4 minutes of sand at the top. From this moment, we start counting. As soon as all of the sand has gone down in the second hourglass, we turn it over and measure out another 11 minutes.

Hint

The hourglasses can be turned over even if there is still some sand in it.

Hint

If more than five guests remain then the guest, who has the smallest shoe size among the remaining guests, can put on the largest remaining pair of shoes.

Solution

We number the guests and their pairs of shoes with numbers from 1 to 10 in order of increasing shoe size. Suppose that there are 6 guests left $($and accordingly 6 pairs of shoes$)$. Then the smallest number of the remaining guest is not more than 5, and the largest number of the remaining pairs of shoes is not less than 6, so the guest with the lowest number will be able to put on the shoes with the largest number. This is a contradiction. On the other hand, if guests with numbers 1, 2, 3, 4, 5 left in order and put on shoes with numbers 10, 9, 8, 7, 6, then none of the remaining five guests could put on any of the remaining shoes.

Answer

5

Hint

If more than five guests remain then the guest, who has the smallest shoe size among the remaining guests, can put on the largest remaining pair of shoes.

Hint

If you recolour each of the rows, you get a table, all the rows of which are of the same colour.

Solution

First, apply the repainting described in the problem to each row of the table. After that, each of the rows consists of cells of the same colour. This, in particular, means that each column has the same number of cells in each colour. Now it is enough to repaint each of the columns – after which all of the columns will be of one colour, and all of them will be painted in the same colour.

Answer

Yes, it is always possible.

Hint

If you recolour each of the rows, you get a table, all the rows of which are of the same colour.

Hint

Use differentiation and induction.

Solution

We prove the assertion by induction on n.

Base case. For n = 1 the assertion is obvious.

Inductive step. Consider the product Q $(x)$ = $(x + 1)^ n$ P $(x)$, where P $(x)$ is a nonzero polynomial. Without loss of generality, we can assume that the free term of the polynomial P $(x)$ is non-zero $($ multiplication by x does not change the number of nonzero coefficients $)$, then the free term of the polynomial Q $(x)$ is also nonzero, and the derivative Q ‘ $(x)$ by one non-zero coefficient is less than that of Q $(x)$. But Q ‘ $(x)$ = $(x + 1)^n$ P’ $(x)$ + n $(x + 1)^{n-1}$ P $(x)$ = $(x + 1)^{n-1}$ $((x + 1) P ‘(x) + nP (x))$, that is, Q ‘ $(x)$ is obtained by multiplying the polynomial $(x + 1)^{n-1}$ by a non-zero $($ as it is easy to verify $)$ polynomial. By the induction hypothesis, the polynomial Q ‘ $(x)$ has at least n non-zero coefficients, hence the polynomial Q $(x)$ has at least n + 1 non-zero coefficients.

Hint

Use differentiation and induction.

Solution

By induction on n, we prove a stronger assertion:
if a polynomial of degree n takes rational values at a rational point n + 1, then its coefficients are rational.

Base case. For any polynomial of degree zero $($ constant $)$ the assertion is obvious.

Inductive step. Let the polynomial P $(x)$ of degree n take rational values at rational points $x_0$, $x_1$, …, $x_n$. By Bezout’s theorem $($ see Problem 60965 $)$
P $(x)$ – P $(x_0)$ = $(x – x_0)$ Q $(x)$, $(*)$
where Q $(x)$ is a polynomial of degree n – 1. The values at the points $x_1$, …, $x_n$ are rational:

$Q(x_i) = \frac{P(x_i) – P(x_0)}{x_i – x_0}$. By the induction hypothesis, all the coefficients of the polynomial Q $(x)$ are rational. Expanding the brackets in $(*)$, we obtain that all the coefficients of the polynomial P $(x)$ are rational.

Using one of the formulas for the interpolation polynomial $($ see problem 61448 or task 61052 $)$, we can recover the polynomial P $(x)$ of degree n from its values at the points 0, 1, …, n. It is easy to see that the coefficients in the expressions obtained are rational.

Hint

Solution

Nothing to translate.

Hint

–

Solution

Nothing to translate.

Hint

–

Solution

The numbers J and X are even, because the numbers EX and AJ are divisible by 4. The numbers OJ and OX are smaller than 25, otherwise, when multiplied by 4, they will cease to be two-digit numbers. Hence, 0 is a two or a one. Let us consider the first case. The number 20 as OX or OJ is not suitable, since 80 ends with the same digit. The number 22 also does not fit. This leaves only the number 24, but the numbers OD and OX must be different. If an even number is added to an even number, which is 4 times as large, then a multiple of 10 is obtained. Consequently, J + X = 10. Then
OJ + OX = 10 + 10 + 10 = 30, and the sum of all four numbers is five times larger, that is, 150.
$\\$
$\\$1. A small search can be used to find all the numbers: OX and OJ are 12 and 18 $($ in any order $)$, and AJ and EX are 48 and 72 respectively.
$\\$2. 4 points.$\\$

Answer

150.

Hint

Solution

Consider any girl. The colors of the dresses of her neighbors on the left and on the right could be: blue – blue, blue – red, red – blue, red – red. The girl answers “yes” only in the first two cases; that is exactly in those cases when her neighbor on the left is in a blue dress.
$\\$But the neighbor on the left was in a blue dress for exactly 17 girls, so the answer “yes” was said 17 times.

Answer

17 girls.

Hint