Solution

The points $a, b, c, d$, divide the segment $[0, 1]$ into no more than 5 interval ranges. At least one of these interval ranges has a length of no less than 0.2. Let $x$ lie at the centre of this interval range. The distance between $x$ and the end of the interval are no less than 0.1 and the distance to the other points $a, b, c, d$, is greater than 0.1. Therefore two of the numbers $\left | x-a \right |, \left | x-b \right |, \left | x-c \right |, \left | x-d \right |$ are no less than 0.1, and the other two are greater than 0.1. Hence their reciprocals are no greater than 10, and two of them are less than 10. Therefore the sum of the reciprocals is less than 40.

Hint

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Solution

First, let’s “glue” the opposite points of the equator. Then we can assume that Aladdin did not move along the equator, but along the circle obtained after gluing the points together. We will also assume that Aladdin did not teleport at any point in time and that he visited all of the planet’s points. Let 2πφ $($t$)$ be the angular coordinate of Aladdin at time t. Since Aladdin did not teleport at any time, the function φ can be chosen to be continuous. Then it suffices to prove the following assertion:

A continuous function $φ (t)$, where ${φ (t)}$ takes all possible values, is given. Prove that $ \max_ {t}φ (t) – \min_ {t} φ (t) ≥ 1$. But this assertion is obvious: if this difference is less than one, then the function φ cannot take values from {$ \max_ {t} φ (t)$} to 1 and from 0 to {$\min_ {t} φ (t)$}.

Answer

See the solution above.

Hint

Solution

Answer: no this line does not exist. Indeed, if the graph of the function y = $2^x$ is symmetric with respect to some line, then for symmetry with respect to this line the horizontal asymptote y = 0 of this the graph should coincide with some asymptote of this graph. But the graph of the function y = $2^x$ has no other asymptotes. Consequently, for this symmetry, the straight line y = 0 goes into itself, and hence the symmetry axis either coincides with the straight line y = 0, or is perpendicular to it. Checking that the straight line y = 0 and the straight lines of the form x = const are not axes of symmetry of the graph of the function y = $2^x$, is left to the reader as an exercise

Hint

Solution

We consider a more general case when we have two gears with
Let $n$ denote the number of removed pairs of teeth $($in our case, $n = 6)$. Then each gear has $n^2 – n + 2$ teeth $($in our case 32$)$. In total, there are $n^2 – n + 1$ such rotations of the top gear relative to the bottom gear, in which all the teeth of both gears are aligned. We call a hole the place on the gear where there is no teeth. Consider an arbitrary hole in the bottom gear. At $n – 1$ positions of the top gear $($except the original one$)$, above this hole is a hole on the upper gear. But there are n holes on the bottom gear, so out of $n^2 – n + 1$ turns of the top gear only for no more than $n(n – 1)$ of them do the holes of both gears coincide. Since $(n^2 – n + 1) – n(n-1)  = 1,$ there is such a rotation of the upper gear, when no holes will coincide. This turn is the desired one.

Hint

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Solution

Assume that in each picture taken by a camera, all 999 other cameras are visible. Therefore the cameras must be located on the vertices of a convex 100-sided polygon, otherwise there will be a camera whose picture does not show all of the other cameras. The sum of the angles in a 100-sided polygon is equal to $180^\circ \times 998 = 179640^\circ > 179^\circ \times 1000$. Therefore one of the vertices must have an angle of greater than $179^\circ$. The camera located on this vertex cannot therefore capture all of the other cameras in its picture. This is a contradiction.

Hint

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Hint

The sections of carpet must lie on top of each other somewhere.

Solution

Since the total length of the rugs is 1000m, ten times greater than the length of the corridor, either the entire length of the corridor is covered by 10 layers of carpet sections or there is a point in the corridor where 11 sections of carpet lie on top of one another. The difference between the number of stretches of corridor covered by sections of carpet, and the number of stretches of corridor not covered by sections of carpet, cannot be greater than 1.

If 11 sections of carpet lie on top of one another then these represent one stretch of corridor covered by sections of carpet. Therefore there can be no more than 9 other stretches of corridor covered by a section of carpet as there are only 9 other sections of carpet left. Therefore can be no more than 10 stretches of corridor covered by sections of carpet – and so no more than 11 sections of corridor not covered by sections of carpet.

One such arrangement is if we take 11 sections of carpet of length 90.5m, and then 9 sections of length 0.5m. In total these have the required length of 1000m. If we place all 11 of the 90.5m sections on top of one another there are $100-90.5=9.5$m of corridor not yet covered by a section of carpet. However the total length of the remaining sections of carpet is just 4.5m. Therefore we can arrange the group of 90.5m sections, and the nine single 0.5m sections, so that there are 11 sections of corridor not covered by any carpet – for example if we left a 0.1m gap before the first section in the corridor, then a 0.1m gap between each of the stretches covered by sections of carpet, then a larger gap at the end.

Hint

The sections of carpet must lie on top of each other somewhere.

Solution

Answer: at the height of $\frac{a}{2 \sqrt2}$ , where a is the side of the square. Let’s consider the following 5 points: the vertices of the square and its centre. If the diameter of the circle is less than $\frac{a}{\sqrt2}$, then it can contain no more than one of these points. Therefore, 4 circles of radius less than $\frac{a}{\sqrt2}$ cannot completely cover a square with side a. It is easy to see that 4 circles, whose diameters are the segments connecting the centre of the square with its vertices, completely cover the square.

Hint

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Solution

Let us define an arbitrary point $A$ which lies in the plane. Draw 7 straight lines through $A$ which are parallel to the lines given. The angles between these lines will be equal to the corresponding angles between the straight lines of the original system. But the 7 lines passing through A have angles between them that sum to $360^{\circ}$, dividing the circle into 14 parts. Therefore there will be at least one angle which is no greater than $360^{\circ} \div 14 = (25 \frac{5}{7})^{\circ} < 26^{\circ}$.

Hint

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Solution

Without loss of generality, we assume that all cyclists travel the track counter-clockwise, the first of them being the fastest and the third being the slowest. Consider the movement of the cyclists with reference to the second cyclist. Then the second cyclist is constantly at some point A, and the first and third move counter clockwise and clockwise, respectively. They periodically meet with each other at regular intervals. Denote by $B_1, B_2, B_3,$ … the consecutive points of their meetings from the beginning of the observation of these athletes $($Figure left$)$. Every two adjacent points $B_n$ and $B_{n + 1}$ $($where n is an arbitrary natural number$)$ are different, since the first cyclist cannot make a full circle counter clockwise, without meeting the third one. We denote by $β_n$ the smaller of the two arcs of the track with endpoints $B_n, B_{n + 1}$. Its length does not exceed 150 m. Since the meetings occur periodically with a shift in one direction, all the arcs βn are equal to each other and the union of several of them covers the whole track. Hence, there is an arc $β_m$ containing the point A. The length of one of the arcs $B_m$A or A$B_{m + 1}$ does not exceed 75 m. Thus, at some point of the meeting of the first and third cyclists they will be no more than 75 m from the second cyclist. Therefore, the photographer will be able to make a successful shot, if d$\geq$75.

Let us give an example of the movement of cyclists, in which they cannot be on any part of the track with a length of less than 75 m at any time. Let the first and third cyclists move with respect to point A with velocities equal in magnitude but different in direction, and in the initial moment they are at the point B, separated from A by a quarter of the circle $($Figure on the right$)$. Then they will meet alternately at the diametrically opposite points B and C $($a distance from A of 75 m$)$, and their positions at each moment in time will be symmetrical with respect to the straight line BC. Hence, at each moment in time the distance from one of them to point A will be no less than 75 m.

Answer

75

Hint

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Solution

a) The first way. After each “move” the grasshopper will not jump to the point where the next grasshopper was before. So, the next one after him will not reach this point in two consecutive jumps. So, continuing in this way, we get that after the 12th jump the first one will not reach its initial position.

The second way. Suppose that one of the grasshoppers $($let’s call it the first one$)$ returned to the starting point A after 12 moves. Then the remaining 11 grasshoppers jumped through point A before the first grasshopper returned to this point. But in one move through point A, no more than one grasshopper could jump over, and on the first move through A no one jumped. This is a contradiction.

b) If at the beginning, one of the grasshoppers is moved clockwise, then the first jump, and the next one after it, will jump further, and the rest will jump as before. For the second jump, three grasshoppers will now jump further, and the rest – as before. Continuing the argument in this way, we see that as a result, for 13 jumps all grasshoppers will jump further. We number the grasshoppers counter-clockwise and move all but the first one to the point O, where the first sits $($first we shift the 2nd, then the 3rd, etc.$)$. At this initial position, it is easy to follow the sequence of jumps: the k-th grasshopper will first leave O on the k-th “go”, jumping by $1/2^k$ from the circumference, and the first after this jump will be at the same distance from O on the other side. As a result, the first grasshopper on his 13th jump will return to point O. Hence, in the initial situation, he will not jump to his starting point.

Marks: 4 + 3

Answer

See the solution above.

Hint

Solution

We arrange the indicated angles in descending order: $α \geq β \geq γ$.

a) α + β + γ 2π, therefore the average value is 2π/3. Obviously, each angle is smaller than π, and $γ \leq 2π/3$. Hence, $D <1/3 (π^2 + π^2 + 2π/3) - (2π/3)^2 = 10π^2/27$.

b) It is clear that $α \geq 2π/3$. Since α $<$ π, then β + γ $>$ π, therefore β $>$ π/2. From these inequalities, we obtain $0 \leq α – 2π/3 < π / 3, - π/6 < β - 2π/3 < π/3$, $-2π/3 < γ - 2π/3 \leq 0$.

Therefore, D = 1/3 $((α-2π/3)^2 + (β-2π/3)^2 + (γ-2π/3)^2) < 1/3 ((π/3)^2 + (π/3)^2 + (2π/3)^2) = 2π^2/9$.

Hint

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Answer

$a_n = C_{2n}^n$.

Hint

Hint

Solution

Nothing to translate.

Hint

Solution

Nothing to translate.

Hint

Solution

No solution given.

Hint

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Solution

Let Q $(x)$ = $(x+1)^n$ – $x^n$ – 1, p $(x)$ = $x^2$ + x + 1, then x + 1 = -$x^2$, $x^3$ = 1 $($ mod P $)$ $($ the comparison of polynomials is analogous to the comparison of numbers $)$
a$)$ Q $(x)$ = $(-1)^n$ $x^{2n}$ – $x^n$ + 1 $($ mod P $)$. Let’s consider all possible cases.

1$)$ n is a multiple of 3. Then $(-1)^n$ $x^{2n}$ – $x^n$ – 1 = $(-1)^n$ – 2 $($ mod P $)$.

2$)$ n =1 $($ mod 6 $)$. Then $(-1)^n$ $x^{2n}$ – $x^n$ – 1 = – $x^2$ -x -1 = 0 $($ mod P $)$.

3$)$ n = 2 $($ mod 6 $)$. Then $(-1)^n$ $x^{2n}$ – $x^n$ – 1 = x – $x^2$ – 1 = 2x $($ mod P $)$.

4$)$ n = 4 $($ mod 6 $)$. Then $(-1)^n$ $x^{2n}$ – $x^n$ – 1 = $x^2$ -x -1 = -2x -2 $($ mod P $)$.

5$)$ n = 5 $($ mod 6 $)$. Then $(-1)^n$ $x^{2n}$ – $x^n$ – 1 = -x – $x^2$ -1 = 0 $($ mod P $)$.

Thus, Q is divisible by P for n ≡ 1, 5 (mod 6). In particular, n is odd.

b) As in Problem 61139, it is sufficient to verify whether Q ‘ is divisible by P. Q’ $(x)$ = n $(x+1^{n-1}$ – $nx^{n-1}$ = n $(x^{2n-2} – x^{n-1})$ $($ mod P $)$.
For n=1 $($ mod 3 $)$ $x^{2n-2}$ – $x^{n-1}$ = 1-1 = 0$($ mod P $)$
For n = 2 $($ mod 3 $)$ $x^{2n-2}$ – $x^{n-1}$ = $x^2$ – x = – 2x – 1 $($ mod P $)$.
With this method, we see that Q divides by $P^2$ for n=1 $($ mod 6 $)$.
c$)$ Q” n$(n-1)$ $(( x+1)^{n-2} – x^{n-2})$ = -n $(n-1)$ $(x^{2n-4} + x^{n-2})$ = -n$(n-1)$ $(x + x^2)$ = n $(n-1)$ $($ mod P $)$. With this method we show that Q divides by $P^3$ only if n=1. In fact the polynomial Q is equal to zero.

Answer

For a$)$ n = 6k ± 1; b$)$ n= 6k + 1; c$)$ n = 1

Hint

Hint

See question number 61013

Answer

a$)$ 1, 3, -2; b$)$ -1, 3

Hint

See question number 61013

Solution

The equation of the tangent to the graph of the polynomial P$(x)$ at the point $(x_0, P (x_0))$ has the form y – P $(x_0)$ = P ‘$(x_0)$ $(x – x_0)$. Let the tangents at the points $(x_k, P (x_k))$ $(k = 0, 1, …, n)$ pass through the point $(a, b)$. Then b – P $(x_k)$ = P ‘$(x_k)$ $(a – x_k)$, that is, the polynomial P $(x)$ – P’ $(x)$ $(x – a)$ – b, whose degree is not greater than n, has n + 1 roots. This gives us a contradiction.

Answer

No, you cannot.

Hint

Solution

a) Using the formula from problem 60537, $σ (320) = σ (23 \times 3² \times 5) = 15 \times 13 \times 6 = 3 \times 390 > 3 \times 320$.

b) Let $p_1, p_2, p_3, …$ be an increasing sequence of all primes.
From problem number 34918, there is an $m$ such that $1 + ½ + … + 1/m > 100.$
Let $k$ and $α$ be such that every number from $2$ to $m$ decomposes into a product of primes not exceeding $p_k$ in powers not exceeding $α.$ We take $n = (p_1 … p_k)^α$. Then  \[\frac{\sigma(n)}{n} = (1+\frac{1}{p_1} + … +\frac{1}{p_1^{\alpha}})…(1 + \frac{1}{p_k} + … + \frac{1}{p_k^{\alpha}}) \geq 1 + 1/2 + … + 1/m>100\] $($all of the numbers 1, ½, …, 1/m and some others are obtained when the parentheses are expanded$)$.

Answer

Yes, it can.

Hint