Solution

The first example $a = 3\sqrt{3}$, $b = 2\sqrt{2}$ Note that $x^3 – y^3> x^2 – y^2$ for $x > y > 1$. Therefore  $|a^m – b^n| = |(\sqrt{3}) ^3 – (\sqrt{2})^3| >|3^m – 2^n| \geq 1$ $($the last inequality follows from the different parity of the numbers $3^m$ and $2^n$$)$. Hence, the statement of the problem follows.

$\\$ $\\$
The second example $a = 2 + \sqrt{3}$, $b = 3a$. We set $c = 2 – \sqrt{3}$, $d = 3c$. From the binomial formula it follows that the number $A = a^m + c^m = (2 + \sqrt{3})^m + (2-\sqrt{3})^m$ is an integer and not a multiple of $3\,\, ($the terms containing an odd degree $\sqrt {3}$ cancel out and all of the terms except $2^m$ are multiples of $3\,)$.
$\\$ $\\$
At the same time, the number $B = b^n + d^n = 3^n (a^n + c^n)$ is also an integer and divisible by $3$.
But $0 < c < d <1$, therefore $[a^m] = A – 1 ≠ B – 1 = [b^n]$.
$\\$ $\\$ $\\$ $\\$
It’s easy to come up with the right pair of numbers, if you do not require their irrationality. In fact, the requirement of irrationality is not fundamental, it only creates some technical difficulties. The solution presented illustrate two ways to overcome these difficulties.

Answer

Yes, there are.

Hint

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_{2022} \} $.

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_{2022} $ 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 $⌊x / 11⌋= n,$ thus $n + 1 = ⌊x / 10⌋= n + ⌊(n + r) / 10⌋$ that is $10 ≤ n + r< 20,$ $10 – r ≤ n ≤ 19 – r.$ For each r from 0 to 10, we get 10 solutions.

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

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 $a + 1 / a$ 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 $n = 3.$

Answer

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

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 $k^m≤ n.$

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 $h =2021^2/2020.$ Then $h \times 2021^{n-1} = 2021^{n + 1}/2020= m + 1/2020$ and $h \times 2021^n = 2021m + 1 + 1/2020$ $($where m is a natural number$)$. For any positive integer $n,$ the number $m$ is greater than $1,$ so $2021m + 1$ 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

Let $[x] = n$ and $x = n + α,$ where $0 ≤ α < 1.$ Then $x^3 – x + α = 3,$ and it follows from the restrictions on $α$ that $2 <x^3 – x ≤ 3.$ If $x ≥ 2,$ then $x (x^2 – 1) ≥2 \times 3 = 6,$ so the inequality $x^3 – x ≤ 3$ does not hold. If $x < -1,$ then $x (x^2 – 1)< 0,$ therefore the inequality $2 <x^3 – x$ does not hold. Thus, $-1 ≤ x < 2,$ that is, $[x] = -1, 0$ or $1.$ Accordingly, we get the equations $x^3+ 1 = 3,$ $x^3 = 3,$ $x^3 – 1 = 3.$ Their solutions: $x =\sqrt[3]{2},\,\, x =\sqrt[3]{3},\,\, x=\sqrt[3]{4}.$ In this case $[ \sqrt[3]{2}] ≠ -1, \,\,[ \sqrt[3]{3}] ≠ 0,$ and $[ \sqrt[3]{4}] = 1.$

Answer

$x =\sqrt[3]{4}.$

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 all the numbers $[a],\, [2a],\,…,\, [3a]$ are different. But if $a < 1,$ then $1-1/N \leq a \leq 1,$ and $2-2/N \leq 2a < 2$, and so on, $N-1 \leq Na < N,$ 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

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

a) The equation $x^2 – 3x + 3 = 0$ has no roots, since it has a discriminant negative, and the equation $[x^2] – 3x + 3 = 0$ has the roots $x_1 = 4/3, x_2 = 5/3$.

b) Suppose that the function $f (x) = x^2 + 2ax = b = (x + a)^2 + b – a^2$ does not vanish, i.e. it is equal to 0. Then $b – a^2$ is positive, and therefore at least 1 $($a and b are integers$)$. Hence, $f (x) ≥ 1$ for any x. Replacing $x^2$ by an integer reduces the value of the function by less than 1, and therefore leaves it positive.

Marks – 2 + 3.

Answer

a) Yes, there do exist; b) No, there do not.

Hint

Solution

Each time the accountants calculated the integer part of the sum of some two numbers, which is equal to the sum of their integer parts plus, possibly, one. We call this the addition of a unit by chance. It can happen only when both terms were non-integral. Each accountant recorded as a result the sum of the whole parts of the original numbers plus the number of such chance cases that he came across. Since each such case deleted two non-integer numbers, then there was a maximum of 50, and the possible different results were 51.

Example: suppose there were 50 numbers of 0.3 and 50 numbers of 0.7. Let’s show how the accountant can obtain any whole result k from 0 to 50. First, he adds k times 0.3 + 0.7 and gets k units $($at k = 0 he adds 0.3 + 0.3$)$. Next to the already available integer, he adds in turn all the remaining numbers.

Answer

51 accountants.

Hint

–

Solution

Since $a > 1,$ there is a positive integer n such that $n^4 ≤ a < (n + 1)^4$. Therefore, $n^2 ≤ \sqrt{a}
<(n + 1)^2$, and $n ≤\sqrt{4}{a} < n + 1.$ Thus, $⌊\sqrt{4}{a}⌋= n.$
On the other hand, $n^2≤ ⌊\sqrt{a}⌋<(n + 1)^2$, therefore $n ≤ ⌊\sqrt{4}{a}⌋ < n + 1,$ that is, $⌊\sqrt{⌊\sqrt{a}⌋}⌋ = n.$
Hence, the above equality actually takes place.

Answer

Yes, it is valid.

Hint

–

Answer

For example, see the figure below.

Hint

Answer

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

Hint

Solution

Let $a_n$ be the number of ways to return in n jumps to the original vertex, and $b_n$ to get in n jumps to the neighbouring vertex. It is easy to see that $a_{n + 1} = 2b_n, b_{n + 1} = a_n + b_n$. Hence it is not difficult to deduce that $b_{n + 2} = b_{n + 1} + 2b_n, a_{n + 2} = a_{n + 1} + 2a_n$.

From the initial conditions $a_0 = 1, a_1 = 0$, we obtain $a_n = \frac{2^n + 2 \times (-1)^n}{3}$ $($this is easy to prove by induction$)$.

1. Also see problem number 61474.

2. For a general approach to the solution of recurrence equations, see problem number 61458.

Answer

$\frac{2^n + 2 \times (-1)^n}{3}$ ways.

Hint