Problems and solutions

Answer:

$105^{\circ }$

Note that the extra segment is a base of an isosceles triangle with angles $150^{\circ }$, $15^{\circ }$, $15^{\circ }$. The sought angle is $180^{\circ }-60^{\circ }-15^{\circ }=105^{\circ }$.

Answer:

$6$

Starting at the rightmost number $72$ and moving left, the next number always equals the former one divided by a positive integer $n>1$. Since $72=2^3\cdot 3^2$ and with each step exponent of some prime decreases, hence we cannot have more than $1+3+2=6$ athletes. This is attained e.g. by the sequence $1,2,4,8,24,72$.

Answer:

$44$

Let $n$ be the sought number of group members. When we add the counts $23$ and $32$, we obtain the total number of members, but with counting twice each member who can play both the instruments. Since there are $n∕4$ such members, we infer that

hence $n=44$.

Answer:

$21$

Assume that $n$ is Cecil’s first number, then $C=n(n+1)(n+2)(n+3)(n+4)$. David starts with $n+1$, so $D=(n+1)(n+2)(n+3)(n+4)(n+5)$. We have

Solving this equation for $n$ leads to $n=20$. Since we seek David’s smallest number, the answer is $21$.

Answer:

$168$

There are $18$ boundary triangles with two neighbours and $3$ triangles with just one neighbour. The remaining $64-18-3=43$ triangles have three neighbours and hence the result is $3\cdot 43+2\cdot 18+3=168$.

Answer:

$4$

Since for any integer $n$ the number $n(n-5)$ is a product of odd and even numbers, we infer that $n^2-5n+6=n(n-5)+6$ is even. Since the only even prime number is $2$, we get the equation $n^2-5n+6=2$, the solutions of which are $1$ and $4$. The sought largest integer is therefore $4$.

Another possible solution is to note directly that $n^2-5n+6=(n-2)(n-3)$ can be a prime number only if one of the brackets equals $\pm 1$. Largest such number is $4$ and the quadratic expression evaluated at $4$ indeed yields the prime number $2$.

Answer:

$2\sqrt{3}\doteq 3.46410$

The dashed segment is located $1$ unit above the horizontal diameter of the big circle. The centre of the big circle, one of the endpoints of the dashed segment and the point “on the top" of the big circle thus form an isosceles triangle with vertical base and one other side of length $2$. The result is thus twice the length of an altitude in the equilateral triangle of side length $2$ and can be computed e.g. by the Pythagorean theorem.

Answer:

$3∕5$

Let $t_a$ be the time for which Daniel was absent and $t_p$ the time for which he was present. Adam and Daniel ate the same amount of pretzels, so

Answer:

$58$

We have to solve the equation $11.8=0.4\cdot x_1+0.5\cdot x_2+1\cdot x_3+2\cdot x_4+4.1\cdot x_5+9.9\cdot x_6$ where all $x_i$ must be in the set $\{0,1\}$. Since $0.4+0.5+1+2+4.1<11.8$, we know that $x_6$ must be $1$. Therefore, the equation reads $1.9=0.4\cdot x_1+0.5\cdot x_2+1\cdot x_3+2\cdot x_4+4.1\cdot x_5$. Since by choosing $x_4=1$ or $x_5=1$ the amount spent will exceed 11.8 euros, we need to set $x_4=x_5=0$. Because of $0.4+0.5+1=1.9$, the only solution is $x_1=x_2=x_3=x_6=1$ and $x_5=x_4=0$. Therefore, Mark will get $1+2+5+50=58$ Czech crowns.

Answer:

$27$

Seven types of rectangles can be found in the shown part of the grid:

There are $7$ unit squares, $2$ squares of size $2\times 2$, $4$ squares like the one with dotted lines, and $2$ like the one with dashed lines. In addition, there are $8$ rectangles of size $1\times 2$, $2$ rectangles of size $1\times 3$, and $2$ rectangles like the one with the dashed lines.

Altogether, we count $7+2+4+2+8+2+2=27$ rectangles.

Answer:

$39$

We wish to solve the equation $\mathrm{lcm}(60,n)=777+\gcd (60,n)$. The greatest common divisor must be a divisor of $60$, i.e. it must be one of the numbers $1,2,3,4,5,6,10,12,15,20,30,$ or $60$. Of those divisors, only the number $3$ can be added to $777$ to obtain a multiple of $60$. Thus $\gcd (60,n)=3$ and $\mathrm{lcm}(60,n)=780$. Then $60\cdot n=3\cdot 780$, so $n=39$.

Answer:

$\sqrt{7∕12}\doteq 0.76376$

Let us “unwrap” the surface of the tetrahedron as shown by the figure. The ant sits at the point $A$ and wants to get to $B$. Since the net is planar, the shortest way has to be a line segment (dashed line). We get $|\mathit{PD}|$ from the Pythagorean theorem as $|\mathit{PD}{|}^2=|\mathit{CD}{|}^2-|\mathit{CP}{|}^2=1^2-{(1∕2)}^2=3∕4$, i.e., $|\mathit{PD}|=\sqrt{3∕4}$. Thus

Since $|\mathit{DB}|=1∕2$, we can apply the Pythagorean theorem once again and compute

which gives the answer

Answer:

$27$

If Doina got $3$, she would lie. Therefore, exactly one of the others would have to lie, which is not possible, since two ladies would have to have drawn the same number. If Doina got $9$ or $12$, all others would lie, since it is not possible to draw any multiple thereof, thus Doina got $6$. The only one of the others who could tell the truth is Agnieszka. Thus she got $12$. Therefore, the two liars are Brunhilda and Cecilia, who got $3$ and $9$ in some order. The product of their numbers is $27$.

Answer:

$420$

Recall that for any positive integer $n$ with factorization $n={p_1}^{a_1}\cdots {p_k}^{a_k}$ such that the $p_i$ are distinct primes, the number of all positive divisors of $n$ is equal to $(a_1+1)\cdots (a_k+1)$.

Obviously, the number of odd divisors can be found by just omitting all factors $2$ in the prime decomposition of $n$. Because $24=8\cdot 3$ and with the fact that the number of odd divisors has to be 8, we get that the power of $2$ must be $2$. Since $2\cdot 2\cdot 2=4\cdot 2=8$, we have to consider either first powers of the three smallest prime numbers or the third power of the smallest one and the first power of the second smallest one or the 7-th power of the smallest one. But $3\cdot 5\cdot 7<3^3\cdot 5<3^7$, so the smallest number $n$ is equal to $2^2\cdot 3\cdot 5\cdot 7=420$.

Answer:

$\sqrt{\pi }\doteq 1.77246$

Since the grey regions have the same area, the circle and the square have the same area, because this area equals the area of the overlap plus the quadruple of the area of a single grey region. If $a$ denotes the side length of the square and $r$ the radius of the circle, we get $a^2=r^2\pi$ and therefore $a:r=\sqrt{\pi }$.

Answer:

$5$

Note that one will be positive, since minuses were put only in between the numbers.

If all the signs are plus, the result is $210$. She got $18$ less, so she had to place a minus sign in front of numbers with combined sum $9$. These are one triplet (the smallest one) $(2,3,4)$, three pairs $(2,7)$, $(3,6)$, $(4,5)$ and the single number $9$. Thus we are left with five possibilities.

Answer:

$1∕81$

The key observation is that the length of the side of the yellow square is one third of the base of the triangle; this can be easily seen by extending the “outer” triangles to squares.

From this we conclude that the area ratio of the yellow square to the triangle is

Using the observation again, the side length of the cyan square is one sixth of that of the yellow square, hence their area ratio is $1∕36$. We get the desired ratio by multiplying these two, obtaining $1∕81$.

Answer:

$26$

We can easily see that we cannot get any of $1,2,3,4,6,10$, since we have $a,b\ge 1$ and $\gcd (a,b)=1$. On the other hand take the coprime couples $(a,b)$ of the form $(a,1)$ to write all odd numbers $n\ge 5$, take $(2k-3,2)$ to express all $n=4k\ge 8$, and $(2k-5,4)$ for numbers $n=4k+2\ge 14$. Therefore the desired sum is $1+2+3+4+6+10=26$.

Answer:

$4$

Since the polynomial has two roots, it can be represented as $c(x-a)(x-b)$, where $a,b$ are the integer roots and $c$ is the leading coefficient, thus being integer as well. The sum of the coefficients is easily computed as $c(a-1)(b-1)$. Since this equals $2020=2\cdot 2\cdot 5\cdot 101$, and since $(a-1)$ and $(b-1)$ differ by one, we are left with four possibilities, namely $2020=1010\cdot 1\cdot 2$, $2020=1010\cdot (-1)\cdot (-2)$, $2020=101\cdot 4\cdot 5$, and $2020=101\cdot (-4)\cdot (-5)$.

Answer:

$19$

Note that there are altogether $40\cdot 41∕2=820$ passengers on the train. When the train consists of $22$ carriages, its last $20$ carriages have to be full, for otherwise there would have been only at most $1+2+39+19\cdot 40=802$ passengers. This leaves $20$ passengers to sit in the first two carriages and since the carriage $2$ cannot be full, carriage $1$ holds still only $1$ passenger. It follows that there are $19$ passengers in carriage $2$.

Answer:

$18$

Let the side lengths of the envelope be $a$ (the width) and $b$ (the height), and the diagonal of the square be equal to $d$. Looking at the triangles formed by the vertical folding lines (they are right angled and isosceles, so the length of the horizontal altitude equals half of the vertical hypotenuse), we observe that $d=b+a+2$. In order to fold properly, the inequality $\frac{d-b}{2}\le b$, i.e. $b\ge \frac{d}{3}$ must hold. Together it gives $a\le d-\frac{d}{3}-2=\frac{2}{3}d-2=18$.

Answer:

$55$

Let $a+b=d^2$, $b+c=e^2$, $c+a=f^2$ for (necessarily distinct) positive integers $d$, $e$, $f$. WLOG $d<e<f$ and since Martha’s numbers are positive,

Moreover, the sought value equals $(d^2+e^2+f^2)∕2$. Hence we are looking for a triple of squares satisfying $(♡)$ and having its sum even and minimal at the same time. Such a triple is $(d^2,e^2,f^2)=(25,36,49)$ and the answer is $55$.

Answer:

$162=2\cdot 3^4$

The path is completely determined by the following choices: First go either up or down (2 options) and then for each of the 4 vertical pairs of nodes: either do not use any vertical arrow, or use one of them, or use both (3 options each). Hence the answer is $2\cdot 3^4=2\cdot 81=162$.

Answer:

$29$

We code the numbers in the following way: we write U if the right digit is bigger than the left one and D if the right digit is smaller than the left one, e.g. 1474 is encoded as UUD and 1414 UDU. There are eight possible codes in total and for each of them the corresponding number is determined by the first digit:

In total, there are $3+6+4+4+7+4+1=29$ sought numbers.

Answer:

$\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\doteq 0.41421$

Let $x$ be the side length of the small square. The radii $r$ of the circles are $\frac{1-x}{2}$. The diagonal of the unit square is divided by the centres of the circles and their point of tangency into four segments. Two of them are the radii and two of them are diagonals in squares of sides $r$. Altogether, we have

and solving this equation gives $x=1-\frac{\sqrt{2}}{1+\sqrt{2}}=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$.

Answer:

$2020∕2019$

If we sum all the given equations, we obtain

so the sought result is $2020∕2019$.

Answer:

$620$

Let us call winning the numbers for which there exists a strategy guaranteeing victory, and the rest of the numbers losing. Observe that a number $n$ is winning if and only if there is a losing number in the interval $[\frac{n}{3},\frac{n}{2}]$; on the other hand, a number $n$ is losing if and only if every number from the interval $[\frac{n}{3},\frac{n}{2}]$ is winning (which includes the option that there is actually no integer in that interval).

Clearly, $1$ is a losing number, so according to the rules above, $2$ and $3$ are winning. This in turn implies that the numbers $4,\dots ,7$ are losing, since for each of these numbers we have $[\frac{n}{3},\frac{n}{2}]\cap \mathbb{N}\subseteq \{2,3\}$. Next we get that $8,\dots ,21$ are winning etc. Continuing in the fashion, we obtain additional winning numbers $44,\dots ,129$, $260,\dots ,777$. Altogether there are $620$ winning numbers.

Answer:

$60$

Thales theorem yields two right angles at the point $D$ so the points $B$, $C$ and $D$ lie on one line. From the Pythagorean theorem we compute $B{D}^2=17^2-(7^2-4^2)=16^2$ and thus $\mathit{BC}=16\pm 4$ (indeed, the right angles can be either identical or form a straight line). Connecting the centres of the circles we obtain a similar (with ratio $\frac{1}{2}$) triangle with $\mathit{ABC}$ and thus the desired distance equals either $20∕2=10$ or $12∕2=6$ and the desired product is $60$.

Answer:

$20$

The trolleybuses must not be in a distance of two stations. Obviously odd and even stations are independent and also obviously, both can accommodate at most 4 trolleybuses. Thus one has to accommodate 3 and the other 4, so the solution is $2\cdot N$, where $N$ is the number of ways in which 3 stones could be placed at seven places in a row such that they are separated by a free spaces. We first place the four free spaces in a line and then we place the three stones in between the free spaces. The first stone has 5 possibilities, the second 4 and the third 3, since the stones are indistinguishable the result reads $N=(5\cdot 4\cdot 3)∕(3\cdot 2\cdot 1)=10$. The final answer is therefore $2\cdot 10=20$

Answer:

$4251$

First, let us calculate how many digits will remain the same. It is easy to see that for every one or two digit number the operation $+11$ will not leave any digit intact. So every digit that will be intact has to be on position of hundreds or thousands. If the last two digits of a page are labelled from $00$ to $88$, the first two digits will be intact. This is the case for the hundreds digit for $89$ numbers in each range of hundred. In addition, for every number from $1000$ to $1988$ we have one more stable digit—the thousand digit. So, considering all numbers up to $1999$, we have $19\cdot 89+989=2680$ digits that remain the same. From $2000$ to $2020$, there are $21$ numbers that will not change their hundred or thousand digit which gives $2\cdot 21=42$ stable digits. Hence $2680+42=2722$ digits remain the same.

Now, how many digits are there in total? From $1$ to $9$ there are $9$ digits. From $10$ to $99$ there are $90\cdot 2$ digits. From $100$ to $999$ there are $900\cdot 3$ digits and from $1000$ to $2020$ there are $1021\cdot 4$ digits. So there are $6973$ digits overall. Therefore, $6973-2722=4251$ digits have to be changed.

Answer:

$65$

We label each intersection point with the number of paths the puck can take from that point on. We start from the bottom of the box where there is only 1 path, and work our way up. Each number is the sum of the number or numbers immediately below it.

Answer:

$68$

Notice that the king and the three knights to his left were all served chicken. So after the third pass, the first vegetarian meal to the king’s left was passed to a meat-eater, and the first vegetarian to the king’s right received meat. These are the two knights who had to trade places.

Everyone else was served the same dish as the person 3 places to their left, so the seating arrangement was

Then every vegetarian and every meat-eater to the right of a vegetarian had the wrong meal after the first pass. So there are $64∕2=32$ vegetarians. The remaining $100-32=68$ knights (as well as the king) ate chicken.

Answer:

$150$

Let us denote the inner angles of triangle $\mathit{ABC}$ by $\alpha$, $\beta$ and $\gamma$, respectively, and note that the other three triangles have to have the same inner angles. Obviously $|\sphericalangle \mathit{EAC}|=\beta$ and $|\sphericalangle \mathit{AEC}|=\alpha$.

Since the sizes of angles $\mathit{EDA}$ and $\mathit{EDB}$ add up to $180^{\circ }$, which is also the sum of $\alpha$, $\beta$ and $\gamma$, they have to be the same, that is, they are both right. Moreover, they cannot equal $\beta$ as that would mean that triangle $\mathit{BDE}$ would have two right inner angles, so they are equal either to $\alpha$ or to $\gamma$. Straightforward angle chasing gives a unique solution to both possibilities, see figure below. The first one trivially means $\alpha =90^{\circ }$, while the other implies $3\alpha =180^{\circ }$, i.e. $\alpha =60^{\circ }$. The sum of all possibilities thus equals $90^{\circ }+60^{\circ }=150^{\circ }$.

Answer:

$240$

Let us draw the workers as nodes in a diagram, where a segment between two nodes represent the shift of the two workers. We are looking for the number of ways to draw all the $5\cdot 4∕2=10$ segments so that no two subsequent segments have a common node. The first segment can be chosen in $10$ ways, the second in $3$ ways and in the next two steps we always have two choices. However, no matter what choices we do, we get essentially the same situation, i.e. we can get any sequence of choices from any other by renumbering the nodes. The fifth choice may lead to two different outcomes: Either we obtain a cycle of five segments, or a cycle of four segments with a “tail”. The former option leads to $1$, $2$, $1$, $1$, and $1$ choices in the following steps, whereas the latter option cannot be completed in a valid way, since the final two segments would have to contain the “end of the tail”. We conclude that there are

possible time schedules.

Answer:

27721/200

Let $a<b<c$ be the side lengths of Lucy’s triangle. The situation described in the statement can happen only when lengths satisfy $a:b=b:c$. Now there are three options: (1) $a=32$, $b=50$, which leads to $x=c=50\cdot \frac{50}{32}$; (2) $a=32$, $c=50$, which leads to $x=b=\sqrt{32\cdot 50}=40$; (3) $b=32$, $c=50$, leading to $x=a=32\cdot \frac{32}{50}$. It is easy to check that none of these situations violate the triangle inequality. The result is the sum of the three values.

Answer:

$902$

Let us observe that the runner does exactly $\frac{40}{\mathrm{GCD}(40,a)}$ steps (runs) of size $a$ edges where $\mathrm{GCD}(x,y)$ stands for the greatest common divisor of positive integers $x$, $y$. For all possible numbers $d=\mathrm{GCD}(40,a)$ (divisors of $40$) we list the possible values of $a$:

• $d=1$ for $a\in \{1,3,7,9,11,13,17,19,21,23,27,29,31,33,37,39\}$,
• $d=2$ for $a\in \{2,6,14,18,26,34,38\}$,
• $d=4$ for $a\in \{4,12,28,36\}$,
• $d=5$ for $a\in \{5,15,25,35\}$,
• $d=8$ for $a\in \{8,16,24,32\}$,
• $d=10$ for $a\in \{10,30\}$,
• $d=20$ for $a\in \{20\}$,
• $d=40$ for $a\in \{40\}$.

The total number of steps can now be obtained by summing the products of $\frac{40}{d}$ and the size of the set on the respective row¹ above. We obtain $40\cdot 16+20\cdot 8+10\cdot 4+8\cdot 4+5\cdot 4+4\cdot 2+2\cdot 1+1\cdot 1=903$. It means that the runner rested $902$ times.

Answer:

$2∕9$

Let us denote the initial vertex (i.e. the home university) by $A$, its neighbours by $B_1$, $B_2$ and $B_3$, their neighbours (different from $A$) by $C_1$, $C_2$ and $C_3$ and finally the last vertex by $D$. It is easy to see that after an odd number of steps (i.e. trips) the only possible positions are $B_i$ or $D$ and after an even number only $C_i$ or $A$ (which is not allowed before the $2020$-th step) are possible. Let us denote $P_n(V)$ the probability of reaching vertex $V$ after the $n$-th step. Due to symmetry, we have $P_n(B_1)=P_n(B_2)=P_n(B_3)=:P_n(B)$ for every $n$ and analogously for vertices $C_i$. The only non-zero probabilities in the first steps are easy to compute by the given rules:

• $P_0(A)=1$
• $P_1(B)=\frac{1}{3}$
• $P_2(C)=\frac{1}{3}$
• $P_3(D)=\frac{1}{3}$, $P_3(B)=\frac{2}{9}$
• $P_4(C)=\frac{1}{3}$.

Observing that after $4$ steps the situation is identical to the one after $2$ steps, we infer that the process is periodic and $P_{2019}(B)=\frac{2}{9}$, hence (now returning to $A$ is again possible)

Answer:

$27$

Noticing that $x\star 3=3$ for any real $x$ we easily compute that $a=(3\star 2)\star 1=5\star 1=27$ is the only solution.

Answer:

$2052$

Let us label the courses $1$, $2$, $3$ and $4$ while the friends $A$, $B$, $C$ and $D$. Let us assume that course $1$ is the one which is selected by more than one of them (after calculating the result with this assumption, it will only need to be multiplied by $4$). Therefore we have only five ways of assigning course $2$ (now it can be chosen by $A$, $B$, $C$ and $D$ or nobody). Similarly, we have five ways of assigning courses $3$ and $4$. On summary, there are $5^3=125$ ways of assigning these three courses. In one of these possibilities, no friends got no courses; in $4\cdot 3\cdot 2=24$ possibilities, three of the friends got one course each.

Now, we have to calculate how many possibilities are there for exactly one student to get some courses. There are $4\times 3$ possibilities where one of them got exactly one course and the remaining two courses were null (chosen by nobody); $4\times 3$ possibilities where one of them chose exactly two courses and one course was null; and $4$ possibilities where one of them got all three considered courses. In total, this gives $4\cdot 3+4\cdot 3+4=28$ possibilities. It also means that there are $125-1-24-28=72$ possibilities for exactly two students to get some courses.

Having discovered that, we can get back to assigning course $1$, the last of them, which is also the only multiple one. We know that every student without previous choices has to get this one, while every student with previous choices can get it or not, provided that there are at least two students with this course. It means that there is exactly 1 way of assigning course $1$ if no students have previous courses; exactly $2$ ways if one student has previous courses; exactly $4$ ways if two students have previous courses; exactly $7$ ways if three students have previous courses (seven because at least one of these three students has to choose course $1$ nevertheless). This renders the final calculation: $4(1\cdot 1+28\cdot 2+72\cdot 4+24\cdot 7)=2052$.

Answer:

$259$

Let $n$ be a number consisting only of digits $1$, $2$, and $4$. If we replace all its digits $1$ by $2$, all $2$ by $4$, and all $4$ by $1$, we obtain another number having only these digits, and applying this operation two more times produces the original number $n$. Let us group all the summed numbers into triples, in which the numbers can be obtained from each other using this cyclic substitution.The sum of every such triplet is equal to the number $B$ consisting of $1000^{1000}$ digits $7$, because each digit is the sum of $1$, $2$, and $4$ in some order. The total number of summed numbers is $3^{1000^{1000}}$, hence the number of triplets is $3^{1000^{1000}-1}$ and the sum in question is thus equal to

Since we are interested only in last three digits, let us compute remainders modulo $1000$. Clearly, $B\equiv 777(\mathit{mod}1000)$. Furthermore, as $3$ and $1000$ are coprime we may use Euler’s Theorem to handle the large power of $3$: We have $\phi (1000)=400$, which is clearly a divisor of $1000^{1000}$, hence

where the negative exponent stands for modular inverse. Since

we conclude that the modular inverse of $3$ is $-333\equiv 667(\mathit{mod}1000)$. Therefore, our sought number is

Answer:

$120$

Denoting the lengths of sides $a$, $b$ and $c$ in a usual way and $\beta$ the angle by $B$. Form the sine law we get

which gives the value of $\cos \beta =\frac{a}{2b}$. Again from the sine law, trigonometric equalities and calculated value of $\cos \beta$ we get now

The smallest $a$ for which there exist $b$ and $c$ satisfying this equation and being the side lengths of a triangle is $a=6$, with $b=4$ and $c=5$. Thus, the product of the side lengths is $120$.

Answer:

$1860497$

Let $A(n)$ be the number of such choices for a table with $n$ seats; for convenience, we will also include the choice of zero seats in this number. We will prove the recurrence relation

for $n\ge 5$. Let us call a subset $M$ of $\{1,\dots ,n\}$ cyclically sparse if it contains no consecutive numbers and does not contain $1$ and $n$ at the same time. Clearly, $A(n)$ is equal to the number of cyclically sparse subsets of $\{1,\dots ,n\}$.

Let $B(n)$ be the number of sparse subsets of $\{1,\dots ,n\}$, i.e. those not containing consecutive numbers (there is no condition on $1$ and $n$). Then $B(n)$ satisfies the relation $B(n)=B(n-1)+B(n-2)$ — indeed there are $B(n-1)$ such subsets not containing $n$ and $B(n-2)$ such subsets containing $n$.

Let us perform a similar analysis on $A(n)$: The number of cyclically sparse subsets containing $n$ is equal to $B(n-3)$, because such a subset does not contain $1$ and $n-1$, while the rest of the elements form any sparse subset of $\{2,\dots ,n-2\}$. Further, if a cyclically sparse subset does not contain $n$, then the remaining elements may form any sparse subset of $\{1,\dots ,n-1\}$. Therefore

a relation which holds for any integer $n\ge 3$. Since the (shifted) sequences $B(n-1)$ and $B(n-3)$ satisfy the desired relation, so does their sum $A(n)$, hence we have proved (R) for $n\ge 5$.

It is easy to see that $A(3)=4$ and $A(4)=7$. Using (R), we can now compute all further values, in particular $A(30)=1860498$. Since the statement excludes the possibility of no seat chosen, the result is one less.

Answer:

$5\sqrt{\frac{7}{2}}\doteq 9.354134$

Let us draw a line parallel to $\mathit{DF}$ through point $A$ as in the figure and denote its intersection with the line $\mathit{CD}$ by $E$. Denoting $x$ the desired length $\mathit{CD}$, we conclude from similar triangles $\mathit{CDF}\sim \mathit{CEA}$ that $\mathit{ED}=\frac{4}{10}x=\frac{2x}{5}$ and $\mathit{EA}=\frac{10+4}{10}\cdot 5=7$. Since $\mathit{\angle AEC}=\mathit{\angle FDC}=\mathit{\angle DBC}$, the quadrilateral $\mathit{AEBC}$ is cyclic. It follows in turn that $\mathit{\angle EAB}=\mathit{\angle ECB}=\mathit{\angle ACD}$. Hence $\mathit{EDA}\sim \mathit{EAC}$ and thus

Answer:

$1011$

We claim that $f(n)\le n+1$ for all $n\in \mathbb{N}$. Since $f(n+1)\ge f(n)+f(f(1))-1\ge f(n)$ for any $n\in \mathbb{N}$, $f$ is non-decreasing. Assume that

in other words $f(m)=m+c$ for some $c\in \mathbb{N}$, $c\ge 2$. Then

and by applying this argument inductively we get $f(2^{r}m)\ge 2^{r}m+2^r(c-1)+1$. Combining this inequality, the one from statement and the fact that $f$ is non-decreasing, we obtain

so again by the monotonicity $f(2^{r}m+1)=f(2^{r}m+2)=\cdots =f(2^{r}m+2^r(c-1)+1).$ For all $k\in \mathbb{N}$ choose $r_k\in \mathbb{N}$ such that $2^{r_k}(c-1)>k$. Then

and hence $f(f(k))\le 1$ meaning $f(f(k))=1$ for all $k\in \mathbb{N}$. Hence also $1=f(f(m))=f(m+c)\ge f(m)$ and hence $f(m)=1$ contradicting the assumption $\mathrm{\text{(}}\text{1}\text{)}$. Hence indeed $f(n)\le n+1$ for all $n\in \mathbb{N}$.

In fact, for a given positive integer $N>1$, the value $f(N)$ can be any element of the set $\{1,2,\dots ,N+1\}$. To see this, let first $A<N$. Define $f_1(n)=1$ if $n\le A$ and $f_1(n)=A$ if $n>A$. The function $f_1$ satisfies the condition and $f_1(N)=A$. Secondly, the function $f_2(n)=n$ also satisfies the condition and $f_2(N)=N$. Lastly, the function $f_3(n)=N\lfloor \frac{n}{N}\rfloor +1$ gives $f_3(N)=N+1$ and it also satisfies the condition. Indeed, we have

where we used that $\lfloor \frac{n}{N}\rfloor +\frac{1}{N}<\lfloor \frac{n}{N}\rfloor +1$ for $N>1$ and that $\lfloor x\rfloor +\lfloor y\rfloor \le \lfloor x+y\rfloor$ for any real numbers $x,y\in (0,\infty )$.

Since $2020>1$, the result is simply the average of all the positive integers from $1$ to $2021$, i.e. $\frac{2022}{2}=1011$.

Answer:

$88$

First of all, we show that the triangles $△\mathit{BEC}$, $△\mathit{AED}$, $△\mathit{DCF}$ and $△\mathit{ABF}$ have an identical excircle lying in the angle $\mathit{EAF}$.

Let $G$ denote the tangent point of the $B$-excircle of triangle $△\mathit{BEC}$ lying on $\mathit{AB}$. The distance between $B$ and $G$ is half the perimeter of the triangle, i.e. $\mathit{BG}=\frac{1}{2}(\mathit{CB}+\mathit{BE}+\mathit{EC})$. Now let $G^{\prime }$ be the tangent point of the $A$-excircle of triangle $△\mathit{AED}$ lying on $\mathit{AB}$. Using $a+b=60=c+d$ we get

and hence $G=G^{\prime }$. Therefore, the tangent point $O$ on side $\mathit{CE}$ is uniquely determined for both triangles and it follows that the triangles $△\mathit{BEC}$ and $△\mathit{AED}$ have an identical excircle. By analogy, we can show that the triangles $△\mathit{DCF}$ and $△\mathit{ABF}$ have also the same excircle. It is even identical for all four triangles under consideration since the center point of the excircle has to lie on the angular bisector of $\mathit{\angle EAF}$ and on the angular bisector of $\mathit{\angle ECF}$. Since the tangents to a circle have equal lengths, from $\mathit{AG}=\mathit{AH}$ we get that the triangles $△\mathit{AED}$ and $△\mathit{ABF}$ have perimeters of equal lengths. Hence we obtain

as the length of $\mathit{CF}+\mathit{FD}$.

Answer:

$1568$

Let us write the equation in the form

Note that if the last bracket is nonzero, then it equals at least two and the numbers $p,q,r$ cannot be all equal. Hence $k\ge (1+1+2)\cdot \frac{1}{2}\cdot 2\ge 4$. The triple $(p,q,r)=(n,n,n+1)$ is a solution for $k=3n+1$ for any $n\ge 1$. Similarly, the triple $(p,q,r)=(n,n+1,n+1)$ is a solution for $k=3n+2$ for any $n\ge 1$. If $3\mid k$ then by rewriting the equation as

we see that also $3\mid p+q+r$ and thus necessarily $9\mid k$. On the other hand, the triple $(p,q,r)=(n-1,n,n+1)$ solves the equation with $k=9n$ for any $n\ge 2$. It only remains to investigate the case $k=9$ If the positive integers $p,q,r$ are pairwise different, we have $k=\frac{1}{2}(p+q+r)\left({(p-q)}^2+{(q-r)}^2+{(r-p)}^2\right)\ge (1+2+3)\frac{1}{2}(1+1+4)=18$ and if $p=q\ne r$ we have $9=(2p+r){(p-r)}^2$ and since $2p+r>1$ we must have $2p+r=9$ and $|p-r|=1$ which is impossible as then $r=p\pm 1$ and $9\ne 3p\pm 1$. In conclusion, all the admissible $k\in \{1,\dots ,2020\}$ can be obtained by removing from the set the numbers smaller than $4$ and multiples of $3$ and then adding the multiples of $9$ larger than $9$. Hence the result is $2020-3-672+223=1568$.