Solution

Create the numbers $ b_1=\{a_1\}, \ b_2=\{a_1+a_2\}, \ b_3=\{a_1+a_2+a_3\}, \ …, \ b_{1998}=\{a_1, a_2, a_3,…,a_{1996} \} $.

Consider only the decimal part of each $b$ number. Divide the range $[0,1]$ into 1001 line segments of length $\frac{1}{1001}$. Some two of the numbers $ b_1,…,b_{1996} $ will end up in the same line segment, let these be $b_i$ and $b_j , i < j$.

This means that $a_i+1+…+a_j$ differs from a whole number by no more than $\frac{1}{1001} < 0.001$.

Hint

–

Solution

Let x = 11n + r, where n $≥$ 0, 0 $≤$ r $≤$ 10. Then, [x / 11] = n, n + 1 = [x / 10] = n + [$($n + r$)$ / 10], that is, 10 $≤$ n + r $<$ 20, 10 - r $≤$ n $≤$ 19 - r. For each r from 0 to 10, we get 10 solutions.

Answer

If x $≥$ 220, then x / 10 $≥$ 2 + x / 11, therefore the integer parts of the numbers x / 10 and x / 11 differ by at least 2. If x $<$ 220, then these integer parts differ by no more than 2. Note that if [x / 10] = [x / 11] + a, then [$($x + 110$)$/10] = [$($x + 110$)$/11] + a + 1. This implies that exactly one number from the pair {x, x + 110} $($x $<$110$)$ is the solution of our equation. Thus, it has exactly 220/2 = 110 solutions.

Hint

–

Solution

We denote the nth word by $W_n$, and the indicated operation by f: $W_n + 1 = f (W_n)$. Note that $f (UV) = f (U) f (V)$. We can assume that
$W_0 = B$. Since $W_2 = W_1W_1W_0, W_{n + 1} = W_nW_nW_{n-1}. (*)$
$\\$ Let $W_n$ contain $a_n$ letters A and $b_n$ letters B. It follows from $(*)$ that $a_{n + 1} = 2a_n + a_{n-1}. (**)$
$\\$ By hypothesis $b_{n + 1} = a_n$.
$\\$ a) We can calculate successively: $a_0 = 0, a_1 = 1, a_2 = 2, a_3 = 5, a_4 = 12, a_5 = 29, a_6 = 70, a_7 = 169, a_8 = 408, a_9 = 985$. Consequently, the 1000th the letter will already appear in the word $W_10 = W_9W_9W_8 = W_9W_5 … = W_9W_4W_4 … = W_9W_4W_3W_3 … = W_9W_4W_2W_2 …$
$\\$ The word $W_9$ contains 985 + 408 = 1393 letters, the word $W_4$ – 12 + 5 = 17 letters, $W_2W_2$ = AAB(A)AB $($the marked letter is the 1000th letter A$)$. Thus, the 1000th letter A is on the 1414th place $(1393 + 17 + 4 = 1414)$.
$\\$ b) Suppose that there exists Then Then, passing to the limit in the equality obtained from $(**)$ , we see that l is the root of the equation l = 2 + 1/l, that is, irrational.
$\\$ Suppose that our sequence is periodic, the pre-period contains c letters, the period contains a letters A, b letters B and fits $k_n$ times in the word $W_n$. Then $k_na ≤ a_n $<$c + (k_n + 1) a, k_nb ≤ b_n $<$c + (k_n + 1) b$; hence, . It follows that $a_n/b_n$ tends to a rational number a/b. This is a contradiction.

Answer

a) At the 1414th place.

Hint

Solution

Solution 1:$\\$
a$)$ Let’s denote a as a = 2 + $\sqrt{3}$. Since $(2 + \sqrt{3}) (2 – \sqrt{3}) = 1, {a} + {1/a} = (\sqrt{3} – 1) + (2 – \sqrt{3}) = 1$. $\\$
b$)$ Assume that the number a is rational, and we represent it in the form of an irreducible fraction: a = m / n. If {a} + {1 / a} = 1, then, obviously, a + 1 / a is an integer; we denote it by k. So, m / n + n / m = k, that is, $m^2$ + $n^2$ = kmn. It follows that $m^2$ is divisible by n, and $n^2$ is divisible by m, and since m and n are relatively prime, this is possible only for | m | = | n | = 1, that is, for a = ± 1. But in this case {a} + {1 / a} = 0. Which is a contradiction.

Solution 2:$\\$
It follows from the condition that a + 1 / a is an integer, but for an irreducible fraction a = p / q this is impossible: the fraction is also irreducible. Thus, equality holds if and only if a + 1 / a is an integer n, and a is not an integer. Since the equation $a^2$ – na + 1 = 0 for n $>$ 2 can not have rational roots $($see problem 61013$)$, the answer in part b is negative. As an example $($ part a $)$, it is only necessary to solve this equation, for example, for n = 3.

Remarks: $\\$

points: 3 + 5

Answer

a$)$ For example, a = 2 + $\sqrt{3}$. b$)$ It can not.

Hint

Solution

We will plug the “holes” in the graph with vertical segments $($by connecting the limit on the left and the limit on the right at the break point$)$. If we now interchange the coordinate axes, we obtain a graph of a continuous non-decreasing function. Therefore, we can assume that the original function is continuous. For the same reasons, we can assume that its graph contains points $($0, 0$)$ and $($1, 1$)$.

$\\$ Let $A_k (k = 0, 1, …, N)$ be the point of intersection of the graph with the line x + y = 2k/N. The section of the graph between $A_k$ and $A_{k + 1}$ is covered by a rectangle with sides parallel to the axes of coordinates, $($ which is non-degenerate with diagonal $A_kA_{k + 1}$ or degenerate is the segment $A_kA_{k + 1}$$)$ of the perimeter 4/N, that is, an area that is not larger than $1/N^2$. Increasing each rectangle to the area $1/N^2$, we obtain the required covering.

Remarks:$\\$
1. 9 points $($the preparatory version, where the continuity of the function f was also assumed, is worth 8 points$)$.
$\\$ 2. Compare with problem number M884 from the “Quant” problems.

Answer

See the solution above.

Hint

Solution

Add n to both parts. Let us prove that in the resulting equality both sides are equal to the number of pairs $(k, m)$ of natural numbers satisfying the inequality $k^m$ $≤$ n. Indeed, the integer part of the number a $≥$ 1 is equal to the number of natural numbers less than or equal to a. therefore$\\$

[$n^{1 / m}$] is the number of pairs $(k, m)$ for which $k^m$ $≤$ n,$\\$

[$log_{k}$n] is the number of pairs $(k, m)$ for which $k^m$ $≤$ n. $\\$

Points: 15.

Hint

Solution

We will draw an arrow to each female cat from the male cat sitting next to her, which is fatter than she is. Then, the number of arrows is 19: the same as the number of cats. But, on the other hand, no more than two arrows can leave from each male cat since the arrows point to neighbouring cats, and yet, they do not point to all of the female cats, but only to the thinner ones. Consequently, the arrows lead from all male cats, as required.

Answer

See the solution above.

Hint

Solution

See the problem 87972 b) which says the following:$\\$

The solution to the problem rests on the obvious fact that the sum of an odd number of odd numbers is odd.

Answer

See the solution above.

Hint

Solution

See problem number 87972 b) which says the following:

The solution to the problem rests on the obvious fact that the sum of an odd number of odd numbers is odd.

Answer

See the solution above.

Hint

Solution

The solution is similar to the solution of the problem 86118.

Hint

Solution

By hypothesis, the function y = sin x + a – b x vanishes exactly at two points $x-1$ and $x_2$, $x_14$ $<$ $x_2$. These points divide the numerical axis into 3 intervals $($-∞, $x_1$], $(x_1, x_2], (x_2, + ∞)$, Since b ≠ 0 and | sin x | $≤$ 1, then on the intervals $(-∞, x_1)$ and $(x_2, + ∞)$ the function has different signs, therefore on some two adjacent intervals $(-∞, x_1)$, $(x_1, x_2)$ or $(x_1, x_2)$, $(x_2, + ∞)$ the function has the same sign, and then either the point $x_1$ or the point $x_2$ is an extremum point and the derivative on it is y '= $($sin x + a – bx$)$' = cos x - b vanishes.

Hint

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

First, we notice that seven socks are enough. Indeed, this means that all of the pairs of socks are taken, except one. Consequently, in the suitcase there were two pairs of different colours and sizes, namely the pair, which differs from the one only in colour, and a pair which differs from the one only in size.

If you leave one sock from each of the two pairs of the same colour in the drawer, then in the suitcase there will not be two pairs of different colours. Consequently, six socks may not be enough.

Answer

7 socks.

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 propose h =$1969^2$/1968. Then $h \times 1969^{n-1}$ = $1969^{n + 1}$/1968= m + 1/1968 and $h \times 1969^n$ = 1969m + 1 + 1/1968 $($where m is a natural number$)$. For any positive integer n, the number m is greater than 1, so 1969m + 1 is not divisible by m.

Answer

Yes, there does exist.

Hint

Solution

Let’s single out one city and connect it with an airline with each of the other 49 cities. This will require 49 airlines.

A smaller number of airlines cannot do. Indeed, according to problem number 31098 d) there are at least 49 edges in a connected graph with 50 vertices.

Answer

See the solution above.

Hint

Solution

This is possible, for example, if all points are connected in a chain.

Answer

Yes, it is.

Hint

Solution

See problem number 31098 b) which says the following:

A tree has a pendant vertex $($see problem number 30786$)$. We delete it along with the edge that emerges from it. The remaining graph is also a tree. Therefore, it has a pendant vertex, which we also remove along with the edge that emerges from it. Having done this operation n – 1 times, we get a graph consisting of one vertex $($in which, of course, there are no edges$)$. Since each time exactly one edge was removed, there n -1 of them at first n – 1.

Answer

See the solution above.

Hint

Solution

Answer: {N-1}/N $\leq$ $|a|$ $\leq$ M/{M-1}

The numbers [x] and [y] are different only when the numbers [-x] and [-y] are also different. Therefore, it is sufficient to examine the case when a $>$ 0. If a $<$ {N-1}/N, then out of the numbers [a], [2a], ..., [Na] there must be some number that are the same, since these N numbers are made up of only N-1 differs numbers: 0, 1, ..., N-2. Therefore a $\geq$ {N-1}/N. By considering similar principles for the numbers of the form 1/a, we can see that 1/a $\geq$ {M-1}/M, i.e. a $\leq$ M/{M-1}. We can show that, if {N-1}/N $\leq$ $|a|$ $\leq$ M/{M-1}, then all the numbers [a], [2a], ..., [Na] are different. Indeed, if a $\geq$ 1, then even all the numbers [a], [2a],..., [3a] are different. But if a $<$ 1, then 1-1/N $\leq$ a $\leq$ 1, 2-2/N $\leq$ 2a $<$ 2,...., N-1 $\leq$ Na $<$ N, and so [ka] = k-1 for k = 1, 2,..., N. For the numbers [1/a], [2/a], ..., [M/a] the reasoning is analogous.

Hint

–

Solution

When five segments are drawn from the end of the segment, 5 new free ends appear and one old one disappears. As a result, the number of free ends increases by 4. Therefore, if five of the segments are drawn k times, then the number of free ends is 4k + 1. For k = 250, we get the required number of free ends.

Answer

Yes.

Hint