Solution

If the first ladybug told the truth, then the second and third should also have spoken the truth, since they should have as many points on their backs as the first. But the second and third statements contradict each other, so at least one of them is lying. Therefore, the first ladybug is also lying. $\\$ $\\$
Let each of the first three ladybirds lie, then all the rest also lied, since none of these three told the truth. So all the ladybugs are liars. But in this case, the first ladybug still told the truth, which can not be the case. Therefore, the first three bugs can not all lie at the same time, so either the second or the third told the truth, and the other two were the liars. Thus, each of the other ladybirds told the truth. $\\$ $\\$
Consequently, there are two ladybugs with 4 spots on their backs and a few ladybirds with six dots on their backs and a total of either 26 or 30 dots on the backs of all the ladybirds. But $30 – 2 \times 4 = 22$, which is not divisible by 6, so there are 26 spots in total, and there were $(26 – 2 \times 4) /6 = 3$ four spot ladybirds. So, 5 ladybirds were gathered in the clearing.

Answer

5 ladybirds

Hint

Solution

We put all the diamonds from the first pile on one scales, and on the other 17 diamonds from the second $($ or from the third $)$. If the balance is in balance, then in the first pile all the stones are real. If there is no balance, then the pile, whose stones do not participate in the weighing, contains only real diamonds.
$\\$ $\\$
Other solutions are possible.

Hint

Solution

In his first move Winnie the Pooh can take four pine cones, and then take one pine cone with each subsequent turn. In this case, after every move of the donkey, under the tree there will be an odd number of cones. Since the number of cones under the tree will gradually decrease, then inevitably there will come a time when under the fir tree there will be only one pine cone. Taking it, Winnie the Pooh will win.

Answer

Winnie the Pooh.

Hint

Solution

Solution 1
$\\$ $\\$ $\\$
Suppose that P $(x)$ has no real roots. Then P $(x)$ has an even power of at least 2. Indeed, any polynomial of odd degree has at least one real root, and if P $(x)$ = const, then from the condition we get that P $(x)$ ≡ 0.
Since P $(x)$ does not have real roots, it takes values of the same sign. We assume that P $(x)$ $>$ 0 $($ otherwise we multiply it by -1 $)$. Since P $(x)$ has an even power, there is a point $t_0$ in which a $($ global non-strict $)$ minimum P $(x)$ is attained, that is,
P $(x)$ $≥$ P $(t_0)$ = A $>$ 0. Consider an $x_0$ such that $t_0$ = $a_3x_0$ + $b_3$. Then P $(a_1x_0 + b_1)$ + P $(a_2x_0 + b_2)$ $≥$ 2A $>$ A = P $(t_0)$ = P $(a_3x_0 + b_3)$. And so we have a contradiction.
$\\$ $\\$ $\\$
Solution 2
$\\$ $\\$ $\\$
If $a_1$ ≠ $a_3$, then there is an $x_0$ such that $a_1x_0$ + $b_1$ = $a_3x_0$ + $b_3$. Substituting x = $x_0$ in the given equation, we obtain after reduction P $(a_2x_0 + b_2)$ = 0, that is, P $(x)$ has a root. The case $a_2$ ≠ $a_3$ is treated similarly.
There remains only the case $a_1$ = $a_2$ = $a_3$ = a ≠ 0. Let $p_0$ be the leading coefficient of the polynomial P $(x)$, and let n be its degree. Then the leading coefficients of the polynomials on the left and right sides of the given equality are $2p_0a^n$ and $p_0a^n$, therefore $p_0$ = 0. This means that P $(x)$ ≡ 0.

Hint

Solution

The first method. 2012 = 503 + 503 + 503 + 503. Replacing the zero with five, we will increase the total by 200; for compensation, we replace one of the fives in the hundreds digit with a three: 2012 = 353 + 553 + 553 + 553. $\\$
The second method. Let’s try to find two numbers that are written with two digits and in their sum they give 2012/ 2 = 1006. Of the tens digit, there must be a digit carried forward, so the sum of the digits in the hundreds digit without this transfer should be 9. Hence, in this digit there are two unknown numbers which have a sum of 9. On the other hand, the sum of the digits in the units column ends with a 6. Therefore, either this sum is 6 and is obtained by 3 + 3 with a second digit of 6 $($ but then we can not get the sum of 10 in the tens column $)$ or it is equal to 16 and is obtained by 8 + 8 with the second digit being a 1. Hence we obtain the representation
$1006 = 118 + 888 = 188 + 818; 2012 = 118 + 888 + 118 + 888 = 188 + 818 + 188 + 818 = 118 + 888 + 188 + 818$.
$\\$The complete solution. Let the numbers a and b be used in the record of our numbers, and a $<$ b. We subtract from each of our numbers the number $111 \times a$. The resulting numbers will also consist of only two digits – 0 and c = b – a. Thus, they can be represented as a product of c by a number written only with the digits 0 and 1. The sum of the resulting numbers is 2012 – $444 \times a$; dividing this sum by c, we see that each digit of the quotient $\\$

does not exceed 4. Now the solution can be found by a search: only the pairs a = 3, c = 2 $($ q = 340, b = 5 $)$ and a = 1, c = 7 $($ q = 224, b = 8 $)$ fit. By various ways of arranging the digits found in the given number q, we arrive at the four possibilities indicated in the answer.

Answer

2012 = 353 + 553 + 553 + 553 = 118 + 118 + 888 + 888 = 118 + 188 + 818 + 888 = 188 + 188 + 818 + 818.

Hint

Solution

So that TICK is as small as possible, you must first make the number representing the letter T as small as possible, then – the number I, then – the number of C, and then – K.
$\\$Let’s try to take T = 1. Then S is 2 or 3. Then if S = 3, then I is no less than 5, and if S = 2, then we can take I = 3. We take C = 4, K = 5 $($ in this case, TICK has the minimum possible value $)$ and try to pick up the remaining numbers. From the equality 13ME + 1345 = 2P31, we obtain that
$E = 6, M = 8, P = 7.$

Answer

1386 + 1345 = 2731.

Hint

Solution

All those sitting at the table break up into pairs of friends; hence, there is an equal number of knights and liars. Consider any couple of friends. If they are sitting next to each other, then the knight will answer “Yes” to the question, and the liar says “No”. If they do not sit side by side, their answers will be opposite. In any case, exactly one of a pair of friends will answer “Yes.” Hence, all the remaining 15 answers will be “no”.

Answer

0.

Hint

Solution

Since the order of reading the answers to the questions to the spectator is unknown, he must make an unmistakable choice, knowing only the number of answers which are “no.” If N notes are sent, then the number of “no” answers can take any integer values from 0 to N, that is, N + 1 variants are possible. This number must determine the number of the prize box, so its value should be no less than the number of boxes, that is, N $≥$ 99. $\\$
Here is an example of a set of 99 notes that satisfies the condition. Let the viewer give the presenter notes with the same phrase: “Is it true that the box number with the prize does not exceed the number K?” for all K from 1 to 99. Suppose that the prize is in the box with the number m. Then the viewer will hear m – 1 “no” answers. Adding 1 to the number of “no” answers, he will determine the number of the prize box.

Answer

99 notes.

Hint

Solution

We divide the weights into pairs with a difference of 20 g: $(1, 21)$, $(2, 22)$, …, $(20, 40)$. If there is exactly one weight from each pair on the scales, then $($ regardless of the choice of weights in the pairs $)$ the weight of one on the odd cup is comparable to zero modulo 20, and the weight of one on the even one is equal to 10 modulo 20. This gives us a contradiction.

Hint

Solution

Let us consider the greatest possible values of the number DE. The numbers 89 and 79 are prime, so do not fit. Next, consider the number 78, we will go over its single-digit divisors and get the answer. Further search can be made to make sure that this answer is the only one.

Answer

$13 \times 6$ = 78.

Hint

Solution

Let Kristen record the specified set of numbers on the walls of the cells. Then the first two prisoners will change places every night, that is, each of them will return to their original place every two days. The remaining three prisoners will return to their places once every three days. Since the numbers 2 and 3 are mutually prime, we get six different locations of prisoners in the cells.

Answer

For example: 2, 1, 4, 5, 3 $($in the order of the cell numbers$)$.

Hint

–

Solution

A comment. Note that on Hannah’s calculator any number can be increased by 1: $($ $x \times 3$ + 3$)$ / 3 = x + 1. Therefore, in principle, from the number 1 on the calculator one can obtain any natural number.

Answer

For example, $((1 \times 3 \times 3 \times 3$)$ + 3 + 3$)$ / 3 = 11 or (1 \times 3 $ + 3)/ 3 + 3 + 3 + 3 = 11.

Hint

–

Solution

It follows from the condition that the quadratic equation f $(y)$ – y – f $(x)$ = 0 is solvable with respect to y for any x. Substituting x = – a / 2, we obtain the equation $y^2$ + $(a-1)$ y + $a^2$ / 4, whose discriminant is equal to $(a – 1)^2$ – $a^2$ = 1 – 2a, whence a $≤$ 1/2. On the other hand, if a = 1/2, then for any x we can put y = -x: then f $(y)$ = $x^2$ – 1/2 x + b = $(x^2 + 1 x + b)$ – x = f $(x)$ + y , as required.
$\\$ $\\$
1. Also suitable is y = x + ½.
$\\$
2. It is easy to verify that, for x = -a / 2, the minimum of the discriminant of the trinomial f $(y)$ – y – f $(x)$ is attained. Therefore, if it is nonnegative for x = -a / 2, then it is always nonnegative.

Answer

a = ½.

Hint

Solution

Arrange the boxes according to the increasing order in the number of pencils. Note that in the n-th box there are at least n pencils $($ of no fewer than n colors $)$. From the first box, take any pencil, from the second – a pencil of a different color, from the third – a pencil of the third color $($ different from the first two $)$, etc.

Hint

Solution

We number the vertices of a regular pentagon with the numbers from 1 to 5. We name the sides and diagonals using the numbers of these vertices. A pair of cards hidden by the viewer corresponds to one of the segments. Among the three cards, the assistant has a pair corresponding to the parallel segment. He names that segment to the magician.

Hint

Solution

Let A and B be factions. Then = is also a faction and, hence, = A B – is also a faction.

Hint

Solution

We denote our operations by $f (x) =\frac{1+x}{x}, g (x) =\frac{1-x}{x}$. We first prove that by successively applying them, we can obtain operations inverse to f and g. We have $f (g (f (x))) = -x$ and $f (g (f (g(f (x))))) = \frac{1}{1+x}$ for $x \neq 1$. Therefore, $f (g (f (f (g(f(x)))))) = x$ for x -1 and $f (g (f (g (f(g (x)))))) = x$ for $x \neq 1$. Further, from 2 one can get -2 $($and vice versa$)$. In addition, $f (-2) = ½, g (½) = 1$. Hence, one can obtain 1 of 2. Since $g (-1) = -2$, and from -2 it is possible to obtain 2, one can obtain 2 from -1. Since $g (2) = – ½$ and $f (- ½ ) = -1$, so -1 can be obtained from -2. Thus, the invertibility is proved. Hence, it suffices to prove that from the number 1, one can obtain all of the positive rational numbers. We prove this by induction on the sum of the numerator and the denominator of a fraction in its simplest form representing our rational number m/n. The induction base for m + n = 2 is obvious. Suppose that for m + n $<$k the number m/n can be obtained from 1. Let us prove that then for m + n = k the number m/n is obtained from the number 1. If m $>$ n, then $\frac{m}{n} = 1 + \frac{\frac{1}{n}}{m-n}$ and n + (m-n) = m $<$ m + n, therefore, the number $\frac{n}{m-n}$ can be obtained by the induction hypothesis. If m $<$ n, then $\frac{m}{n} = \frac{1}{1+\frac{n-m}{m}}$ and (n-m) + m = n $<$ m + n, therefore, the number $\frac{n-m}{m}$ can be obtained by the induction hypothesis. Hence, in both cases the number m/n can be obtained from the number 1, and so the induction transition is proved.

Answer

See the solution above.

Hint

Solution

See problem number 77989.

Hint

Solution

Similar to the previous problem, let’s look at the first two numbers. The first number is single-digit, and the second is two-digit. Therefore, their difference is less than 100. Therefore, the digit in the hundred digit each time increases by no more than 1, from where D = 1, C = 2, E = 3. Therefore we get the record: A, B2, 13F, 2GH, 2B3, 3KG. Similar to the previous problem, 3d = 2B3 – B2 = 201, d = 67. From here it is easy to restore the inscription: 5, 72, 139, 206, 273, 340.

Answer

5, 72, 139, 206, 273, 340.

Hint

Solution

Since the integral of a polynomial over a segment is equal to the increment of its antiderivative, our problem is a particular case of problem number 98268.

Answer

For example, a$)$ the segments [-1, -1/2], [1/2, 1] are black, [-1/2, 1/2] are white.
b$)$ the segments [-1, – 3/], [- 1/4, 0], [1/4, 3/4] are white, [- 3/4, – 1/4], [0, 1/4], [3/4, 1] are black.

Hint