Problems and solutions

Answer:

$46$

When the game ends, the winner has $10$ points, while all the other players have at most $9$ points, which sums up to $10+4\cdot 9=46$.

Answer:

$21$

Let us consider two cases:

I. $B$ contains one card and $A$ contains three. Let us consider the possible values of $B$:

• $B=1$: $A$ is any permutation of $2,3,6\to 6$ options.
• $B=2$: $A$ must end with $6\to 2$ options.
• $B=3$: $A$ is any permutation, since the sum of digits is divisible by $3\to 6$ options.
• $B=6$: $A$ must end with $2\to 2$ options.

II. $A$ and $B$ both contain two cards. Then the ratio $A∕B$ will be less than 6, so it can be $1$, $2$, $3$, $4$, or $5$. Let us work out these possibilities:

1:

This is not possible, as it would imply $A=B$.

2:

$A$ must end with $2$ or $6$. If $A$ ends with $2$, then $B$ must end with $6$ or $1$. In the former case, we get $32$ and $16$, in the latter, $62$ and $31\to 2$ options. If $A$ ends with $6$, then $B$ must end with $3$, we get $26$ and $13\to 1$ option.

3:

$A$ must start with $6$ or $3$. In the former case, $B$ must start with $2$, we get $63$ and $21$. In the latter case, $B$ must start with $1$, we get $36$ and $12\to 2$ options.

4:

$A$ must start with $6$ and end with $2$, since it is even. So $A=62$, and it is not divisible by $4$.

5:

$A$ must end with $5$ or $0$, but this is not possible.

Summing everything up, we get $21$ numbers.

Answer:

$18$

The squares have side lengths in the ratio of $3:2:1$. Hence, the area of the largest square is ${\left(\frac{3}{2}\right)}^2\cdot 8=18$.

Answer:

$124$

We denote the number of magpies, mathematicians, and centaurs by $g$, $m$ and $c$. Then the condition on the number of tails implies $g+c=15$, while the condition on the number of hands translates to $2m+2c=94$. The number of legs is $2g+2m+4c$, which is $2(g+c)+(2m+2c)=30+94=124$.

Answer:

$64$

There will be twelve channels in the sequence. The channels in odd positions must be the $2$nd for sure, and each of the channels in even positions could be either the $1$st or $3$rd. There are six such positions, so we get $2^6=64$.

Answer:

$27^{\circ }$

Let us label the points as on the picture below. The common diagonal $AB$ bisects the angle $\angle FBD$, hence

Moreover, since the dashed line $EF$ is parallel to $AB$, we get for the sum of angles $\angle EFB+\angle FBA=180^{\circ }$. Subtracting the known right angle $\angle AFB$, we get

$$$$

Answer:

$2016$

We take the value of interest $x^3-14x+2024$ and subtract $x(x^2-4x+2)=0$ in order to cancel $x^3$, obtaining $4x^2-16x+2024$. Now we want to cancel $4x^2$, so we subtract $4(x^2-4x+2)=0$ and obtain $2016$ which is the answer.

Answer:

$123$

The sought number cannot be a single-digit number because its digit is either even or odd, and hence it is counted in one of those categories. Similarly, a two-digit number is not feasible as it would need to end with the digit $2$, which is even. Assume that the number of digits is $3$ therefore, the sum of even and odd digits is also $3$. Hence, the possible numbers are $123$, $213$, and $303$. After applying the transformation from the problem statement, all of these numbers end up as $123$, including $123$. Hence, $123$ is the sought after solution.

Answer:

$16$

Let us extend the pentagon with a parallelogram to an equilateral triangle of side length $11=4+a=2+b$ as in the picture.

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We conclude that $a=7$, $b=9$, and $a+b=16$.

Answer:

$12$

The time needed to dig a well is assumed to be proportional to its volume. Because it is a right circular cylinder, its volume is given by the formula $V=\frac{\pi }{4}\cdot D\cdot W^2$, where $D$ stands for depth and $W$ stands for width. Now we can reason that the girl needs $7$ days to dig out $\frac{\pi }{4}\cdot \frac{D}{3}\cdot W^2=\frac{1}{3}V$ of the earth, thus she needs $21$ days to dig out the whole volume $V$ of earth. Similarly, Janko needs $7$ days to dig out $\frac{\pi }{4}\cdot D\cdot {\left(\frac{W}{2}\right)}^2=\frac{1}{4}V$ of the earth, thus he needs $28$ days to dig out the whole volume $V$ of earth. Therefore, the girl can perform $\frac{1}{21}$ of the well in a day, and her companion can perform $\frac{1}{28}$ of the well in a day. Together, they can perform $\frac{1}{21}+\frac{1}{28}=\frac{1}{12}$ of the well in a day. Finally, we can conclude that they need $12$ days to build a well together.

Answer:

$360$

First we make the observation that every $2\times 1$ rectangle contains exactly $2$ diagonals of length $\sqrt{5}$. So our goal is to enumerate the number of $2\times 1$ rectangles. Suppose that the rectangle is oriented vertically. Then we have $10$ possibilities to place it in a column and $9$ possibilities for a row. So we have $90$ possibilities to place this rectangle in a $10\times 10$ grid. We can also orient the rectangle horizontally with the same number of possible placements. In conclusion, we have two diagonals in each of $90+90=180$ rectangles, so we have $360$ diagonals in total.

Answer:

$5963$

Since $\mathit{MATH}$ and $\mathit{HAHA}$ have the same hundreds digit and the corresponding digit in $2024$ is $0$, we see that there is no carry when adding the tens digits on the left-hand side. Therefore, $\mathit{TH}=\mathit{HA}+24$ and $M=H+2$. However, adding $A$ and $4$ must produce a carry, for otherwise we would have $T=H+2=M$, and the digits are assumed to be distinct. This implies that $H=A+4-10=A-6$, hence $M=A-4$ and $T=A-3$. From this, we easily get that $\mathit{MATH}$ is one of $3741$, $4852$, or $5963$, the last value being the greatest.

Answer:

$48$

Since the altitudes from the two vertices to the diagonal have equal lengths for all triangles involved, the shaded area is exactly $\frac{3}{7}$ of the total area. Hence the solution is $\frac{3}{7}\cdot 8\cdot 14=48$.

Answer:

$116$

Let us denote the number of girls by $g$ and the number of boys by $b$. The statement gives us the following system of equations:

Plugging in $b=\frac{5}{4}g+\frac{5}{4}$ from the first equation into the second equation, we get

which solves for $g=51$, hence $b=\frac{13}{10}\cdot 50=65$. Therefore, the answer is $g+b=51+65=116$.

Answer:

$50$

There are $7$ squares of side length $1$ and two squares of side length $2$. To reduce the number of squares to $5$, one has to break exactly $4$ squares. To remove a square bounded by matchsticks $3,7,6,10$, one needs to remove at least three of them. Hence, it is one of the squares that will survive. It is important to note that matchstick $6$ cannot be removed as it would break this square and we cannot collect the remaining matchsticks from this square.

Removing matchsticks $11$ or $13$ breaks both large squares simultaneously. If that happens, then one square needs to be broken with two removals. Consider a case where a matchstick of $11$ is drawn; then there is a possibility of taking pair off matchsticks $12$, and $13$ or the pair of $18$, $20$. On the other hand, if matchstick $13$ was taken, then it could be either the pair of $1$, $4$, the pair of $11$, $12$ or the pair of $16$, $19$. Out of which, the removal of matchsticks $11$, $18$, and $20$ has the highest sum of $49$. If both large squares are intact, that means that the only matchsticks that can be removed are $5$, $12$, and $17$ and that does not result in a valid configuration.

Since matchstick $6$ cannot be taken and one large square needs to be broken, it is safe to assume that both matchsticks from one of the following pairs are removed: $18$, $20$ or the pair $16$, $19$, or the pair $1$, $4$. Such removal takes two squares at the same time, one large and one small. After that, there needs to be one matchstick that breaks exactly two squares. In the case of the pair $18$, $20$, it can be either matchstick $11$ breaking one small and the large one or $12$ breaking two small circles, leaving a total sum of $50$. Matchstick $13$ cannot be taken, as matchstick $15$ would not be part of any square. In a similar manner, it can be assumed that breaking pair $16,19$ alongside matchstick $13$ results in a total sum of $48$. Hence $50$ is the sought answer.

Answer:

$65$

Let $r$ denote the base radius. From the first configuration, the volume of water in the bottle is $13\pi r^2$. Similarly, from the second configuration, the volume of air in the bottle is $(21-14)\pi r^2=7\pi r^2$. Hence, the bottle has a total volume of $(13+7)\pi r^2=20\pi r^2$ and the desired percentage is

Answer:

$28$

There are four paths from $A$ to $B$ that do not pass through $C$. One of them allows two ways to get to $C$, one path has exactly one possibility to get to $C$ and for the other two paths, it is not possible to get to $C$ without walking a street twice. Altogether, there are three valid paths to the cinema via his girlfriend. Two of them have lengths of $10$, the other one has a length of $8$, hence the result is $28$.

Answer:

$44$

Let guard $A$ be the one who needs $14$ minutes (rectangular path) and $B$ the one who needs $12$ minutes (square path) to complete one round. There are two possible meeting points for the guards. If they meet at the left one after $a$ full rounds made by $A$ and $b$ full rounds made by $B$, then the condition

must hold. This equation simplifies to $7a=6b+3$, implying $7\mid 6b+3$. By trying $b\in \{0,1,2,\dots \}$ we see that $b=3$ and $a=3$ is the smallest solution. Similarly, the guards meet at the right point if

is satisfied. Hence, we get $7a=6b-1$. So we have $7\mid 6b-1$, which forces $b\ge 6$. Hence, they meet for the first time after $14\cdot 3+2=44$ minutes.

Answer:

$26$

Square both equations and compute their sum to obtain $2b^2=392$. Since $b$ has to be positive, $b=14$ is the only solution. Plugging this value into the square of the first equation, we obtain $a^2+24=c^2$ or $c^2-a^2=24$. Put $d=c-a$; then $d$ is an even number, since $c$ and $a$ are either both even or both odd.

The value of $c^2-a^2=(c-a)(c+a)$ can be bounded from below by $d(d+2)$, which for $d\ge 6$ is at least $48$, a number larger than $24$. Hence, the only options for $d$ are $d=2$ and $d=4$. In the former case, we obtain $a+c=12$ with the solution $a=5$ and $c=7$. In the latter case, $a+c=6$ with the solution $a=1$ and $c=5$. Therefore, the largest possible value of $a+b+c$ is $5+14+7=26$.

Answer:

$\frac{3}{4}$

Subdividing into congruent triangles as in the figure and counting leads to the answer $\frac{18}{24}=\frac{3}{4}$.

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Alternative solution. Let $r$ be the radius of the circle. Both hexagons can be subdivided into $6$ equilateral triangles; for the inscribed hexagon, the height of each triangle is $\frac{1}{2}\sqrt{3}r$, while for the circumscribed one it is $r$. Therefore, the scaling factor for lengths is $k=\frac{1}{2}\sqrt{3}$, and consequently, the scaling factor for areas is $k^2=\frac{3}{4}$.

Answer:

$20$

Any square birthday number needs to be composed of one or multiple factors of the form $p^k$ where $k>1$. All such factors less or equal to $196$ are $S=\{4,8,9,16,25,27,32,36,49,64,81,100,121,125,128,144,169,196\}$. We can see that the product of at least two numbers from $S$ greater than $27$ either belongs to $S$ or is larger than $196$. From the numbers smaller or equal to $27$, only $8=2^3$ and $27=3^3$ are not perfect squares. Since a product of perfect squares is a perfect square, it either means that the result would either belong to $S$ or be larger than $196$. Finally, the products using $8$ or $27$ and some other numbers from $S$ that are smaller than $196$ and not already in $S$ are $27\cdot 4=108$ and $8\cdot 9=72$. It follows that there are $18+2=20$ such numbers.

Answer:

$3^8\cdot 2=13122$

The organizers have three questions to choose from in the contest nr. $1$. They can choose one of the $4-1=3$ possible questions from the second contest, as one question number is already occupied. It is easy to see that this pattern stays, namely that in the $k$-th contest there are already $k-1$ questions unavailable due to (some) previous choices leaving three options, until contest nr. $9$ which contains a question number $11$ which is too large, thus leaving only two questions to choose. Finally, from the test nr. $10$, there are questions nr. $11$, and nr. $12$, which cannot be chosen, leaving only one suitable question. Altogether, there are $3^8\cdot 2=13122$ such tests.

Answer:

$142857$

Knowing that the last digit of the number is $1$, one can try to reconstruct the number backwards as follows:

Indeed, $142857\cdot 3=428571$ holds.

Alternative solution. Any positive integer starting with the digit $1$ and at least two digits can be written as $10^k+a$ for some $k\ge 1$ and some $k$-digit number $a$. After relocating the digit $1$ from the beginning to the end, the number changes to $10a+1$. Hence, we want to solve the equation

for $k$ and $a$; let us simplify it to

The number on the left-hand side is nothing else than a $2$ followed by $k$ nines. Take as many nines as needed and divide the number $2999\dots$ by $7$ until there is no remainder. This process leads to $a=42857$ and thus to the solution $142857$.

Answer:

$58$

Denoting $x$ the sidelength of the smaller squares and using the Pythagorean theorem for the shaded right triangle in the picture, we get

This equation simplifies to $x^2=2x$, hence we get $x=2$. The answer is then $2(2^2+5^2)=58$.

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Answer:

$18$

Let us denote the length of the unstretched rope by $l$ and the height of the wall by $h$. When the climber meets the mark, one half of the rope (stretched) covers twice the distance of the climber from the top, so we have

When the climber touches the ground, we similarly obtain

which, after substituting $l=\frac{20h}{9}$ from the first equation, gets easily solved and yields $h=18$.

Answer:

$10$

Let $b$ denote the number of black socks. Then the probability that both socks are black is $\frac{b}{n}\cdot \frac{b-1}{n-1}$. Since this expression is equal to $\frac{2}{15}$, the following equation must hold:

Since both $3$ and $5$ divide the left-hand side and both are coprime to $2$, they must divide $n\cdot (n-1)$ on the right-hand side. Now start with small multiples of $3$ and $5$ for $n$ to discover that $n=6$ leads to $15\cdot b\cdot (b-1)=2\cdot 6\cdot 5=60$. However, $b\cdot (b-1)=4$ cannot be fulfilled by any integer, therefore $n=6$ is not a solution. If $n=10$, then $b\cdot (b-1)=12$ can be achieved by $b=4$. Therefore, the answer for $n$ is $10$.

Answer:

$9876504$

We will use the condition of divisibility by $11$: a number is divisible by $11$ if and only if the difference between its sum of digits on odd positions and its sum of digits on even positions is divisible by $11$.

Among all numbers with a specified number of digits, seven in this case, the largest are the ones that start with the largest digits. Thus, let us start looking for a solution by choosing the digits from $9$ downwards. After writing $98765$, we see that the sum of the “odd” group is $9+7+5=21$ and the sum of the “even” group is $8+6=14$. The difference is $7$ and we have to make it divisible by $11$, using just two digits from the set $\{0,1,2,3,4\}$. The only way is to add $0$ to the “even” group and $4$ to the “odd” group, which produces the solution $9876504$. Because all the other solutions would have to begin with a five-digit sequence smaller than $98765$, this is indeed the largest one.

Alternative solution. Start with the largest number, $9876543$, consisting of seven distinct digits. Observe either using the rule about division by $11$ or doing long division that this number is not divisible by $11$, but $9876537$ is, and this is the largest multiple of $11$ smaller than the number we started with. Since its digits are not distinct, we subtract $11$ and check the digits of the newly obtained number with respect to distinctness. After some steps

we obtain the sought-after number, $9876504$.

Answer:

$13$

Let us create an equilateral triangle $BCE$ with $E$ on the segment $CD$. Then $AED$ is a triangle where $AE=8$, $ED=7$, and $\angle DEA=120^{\circ }$. Therefore, by the Law of Cosines, $A{D}^2=8^2+7^2-2\cdot 7\cdot 8\cdot \cos 120^{\circ }=169$, so $AD=13$.

$$$$

Alternative solution without using the Law of Cosines. If the triangle $AED$ is extended by half of an equilateral triangle of side length $7$, we can use the Pythagorean theorem to get $A{D}^2={(5+3+3.5)}^2+{(3.5\cdot \sqrt{3})}^2=169$.

Answer:

$18$

First of all we need the prime factorization of $2024$, which is

Since $a$, $b$, $c$, and $d$ are positive integers, we have $2+a\ge 3$, $2+c\ge 3$, and $4+d\ge 5$. A factor $1$ or a factor $2$ on the right hand side may occur only once in $(0+b)$ and a factor $4$ may appear only either in $(2+a)$ or $(2+c)$.

Since the product on the right-hand side of the given equation consists of four factors, there are four possible factorizations of $2024$, with at most one factor being less than 4, namely

For the first factorization, we have $b=1$, and the remaining factors can be assigned to $a+2$, $c+2$, and $d+4$ in any order, yielding $6$ solution. In the second factorization, we have $b=1$, then either $a+2=4$ or $c+2=4$. And in each of these cases, the remaining factors can be assigned in two ways, so there are in total $4$ solutions for the second factorization. Analogously, there are $4$ solutions for each of the third and fourth factorizations. Therefore, in total, there are $18$ different solution quadruples:

Answer:

$3540$

Considering $2^x\cdot 3^y=24^k$, we have

Then, $2^x\cdot 3^y={(2^3\cdot 3^1)}^{15\cdot 59}$, which means that $x=3\cdot 15\cdot 59=45\cdot 59$ and $y=15\cdot 59$. Therefore, $x+y=60\cdot 59=3540$.

Answer:

$21$

Let us focus just on the first and last dozen for the moment. If the middle semi-dozen of apples (i.e., apples $7$–$12$) contains just green apples, both the first and last dozen are missing only one green apple, which is easily fixed by placing a green apple into both the first and last semi-dozens, so eight green apples are enough. If some of the middle-dozen-apples are red, one would need more green apples in total, as for each green apple removed from the middle semi-dozen one needs to add two apples (one into the first semi-dozen and the other one into the last semi-dozen). Therefore, $8$ is the minimal amount of green apples needed, with six of them being placed in the middle, one in the beginning, and one towards the end.

However, the first and last green apples cannot be placed into their semi-dozens arbitrarily. In order to meet the condition for every dozen consecutive apples, the distance between the two green apples cannot exceed $12$, e.g. if the first green apple is placed at position $2$, the last can be placed either at position $13$ or at position $14$. Depending on the position of the first apple, the last apple thus has $1$ to $6$ possible positions. Adding these up gives $1+2+3+4+5+6=21$ possible sequences.

Answer:

$12$

Let $a$, $b$, $c$ be the number of $1\mathrm{ct}$, $2\mathrm{ct}$, and $5\mathrm{ct}$ coins, respectively, in the sister’s pile. Then, we have the equations

and

By subtracting the first equation from the second, we obtain $b+4c=115$. This equation has multiple solutions of the form $b=115-4c$ for any given $c$. However, the constraints $0<115-4c<106$ for $b$ imply $c\in \{3,4,\dots ,28\}$. But not all solutions have a valid amount of $1\mathrm{ct}$ coins. Therefore, we have to add the condition $0\le 85-(115-4c+c)=-30+3c\le 33$ for $c$. We can see that only values of $c$ from the set $\{10,11,\dots 21\}$ lead to a valid solution. Hence, there are 12 possible ways.

Answer:

$36^{\circ }$

Let us recall that, given a circle $\omega$, the angle under which a segment $XY$ with endpoints on $\omega$ is visible from another point $Z$ on $\omega$ is given only by on which of the two arcs (in which $X$ and $Y$ split the $\omega$) it lies and the sum of these two values equals $180^{\circ }$.

$$$$

In our situation, labelling some of the points as in the picture above and using the well-known measures of internal angles of regular pentagon and triangle, we have

Similarly, we have

Finally, since $ABC$ is an isosceles triangle, we have

and hence, because of how these three angles at $E$ overlap, we compute

Answer:

$11296$

Let us count the number of configurations where at least one rook is not attacked by any other rook. Such a rook needs to be alone in the row and the column at the same time, which means that there is at most one such rook. It can be placed at any square in $4\cdot 4=16$ ways. Removing the corresponding row and column leaves nine squares where the remaining eight rooks must fit. The empty square can be chosen in $9$ ways. Altogether, there are $16\cdot 9=144$ ways of doing that. The total number of ways to choose nine squares from sixteen squares equals $\left(\genfrac{}{}{0.0pt}{}{16}{9}\right)=11440$, and hence the desired result is $11440-144=11296$.

Answer:

$9409$

Since $N$ is not a prime number, it is either $1$ or there is a prime number $p<N$ that divides $N$. The requirement $p<100$ leads to

Observe that $N=97^2=9409$ satisfies the conditions.

Now suppose there is a number $N^{\prime }>9409$ that also meets the given conditions. If $p\le 97$ is a prime number that divides $N^{\prime }$, the quotient $\frac{N^{\prime }}{p}$ is an integer greater than $97$. This yields $\frac{N^{\prime }}{p}\in \{98,99\}$, since any divisor of $N^{\prime }$ has to be smaller than $100$. But then $N^{\prime }$ is divisible by $k\in \{2,3\}$, and the given condition leads to

which is a contradiction.

Therefore, $N=9409$ is the sought number.

Answer:

$845$

Denote a bus stop $s_0$ if all three bus lines stop, and denote $s_k$ the $k$-th bus stop after $s_0$ by the bus lane $A$. Now calculate in how many ways Danko can reach the bus stop $s_6$ from the bus stop $s_0$.

1.

Danko can reach the bus stop $s_1$ in exactly one way, and that is by the bus lane $A$.

2.

There are two ways to get to the bus stop $s_2$, and that is either by bus lane $A$ from the bus stop $s_1$ or by bus lane $B$ from the bus stop $s_0$.

3.

To reach the bus stop $s_3$, Danko either took the bus lane $C$ from the bus stop $s_0$ or the bus lane $A$ from the bus stop $s_2$, which can be reached by $2$ ways; hence, there are $3$ possible ways.

4.

To reach the bus stop $s_4$, it is either lane $A$ from stop $s_3$ or lane $B$ from stop $s_2$ hence, there are $3+2=5$ ways.

5.

To reach the bus stop $s_5$, there is only one way, and that is by lane $A$ from stop $s_4$ hence, there are $5$ ways.

6.

Finally, to reach the bus stop $s_6$, it is either by the bus lane $A$ from the bus stop $s_5$, or by the bus lane $B$ from the bus stop $s_4$, or by the bus lane $C$ from the bus stop $s_3$ hence, there are $3+5+5=13$ ways.

Bus stops where all lines stop are the central station $c$, the bus stop nr. $6$ and the bus stop nr. $12$. Since any bus stop where all bus lines stop can act as the bus stop $s_0$ there are $13$ ways for Danko to reach the bus stop nr. $6$ from the central station $c$ and the bus stop nr. $12$ from the bus stop nr. $6$. It can be deduced that to reach the $17$-th stop from stop nr. $12$ is the same as to reach the bus stop $s_5$ from the $s_0$ and hence there are $5$ ways. Together, there are $5\cdot 13\cdot 13=845$ ways for Danko to reach the stop nr. $17$.

Alternative solution. Let us refer to the stops by their numbers, setting $c=0$. Every stop $s$ is reachable via line $A$, so every journey to the preceding stop $s-1$ can be extended to stop $s$ via line $A$. If line $B$ stops at $s$, then the journeys can be extended from stop $s-2$ to $s$ via line $B$. A similar fact holds for stop $s$ where line $C$ stops. Therefore, if we denote by $J(s)$ the number of ways Danko can reach stop $s$, then for $s\ge 1$ we obtain

Since the central stop is “reachable” in only one way, we have $J(0)=1$ and we can determine $J(17)$ using the given recurrence; the arrows under the table show which values are added to give the number in each cell.

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Answer:

$3034+\frac{\sqrt{3}+1}{2}=3035+\frac{\sqrt{3}-1}{2}$

Note that $a_1$ has the decimal part $a_1-\lfloor a_1\rfloor =\sqrt{3}-1$. Hence, it can be written as $a_1=1+\sqrt{3}-1$. We calculate the first few terms

Observe that terms $a_1$ and $a_3$ have the same decimal part $\sqrt{3}-1$ and the difference $a_3-a_1=3$. The same reasoning applies to the terms $a_2$ and $a_4$, which have the same decimal part $\frac{\sqrt{3}-1}{2}$ and difference $a_4-a_2=3$. This leads us to the hypothesis $a_{2k+1}=3k+1+\sqrt{3}-1$ and $a_{2k+2}=3k+2+\frac{\sqrt{3}-1}{2}$, where $k=0,1,\dots$. The validity for all $k$ can be proven by induction; it is clear for $k=1$ and $k=2$. For the rest, it suffices to plug the formulas into the definition $a_{n+1}=\lfloor a_n\rfloor +\frac{1}{a_n-\lfloor a_n\rfloor }$. Observe that

Hence, $a_{2024}=3034+\frac{\sqrt{3}+1}{2}=3035+\frac{\sqrt{3}-1}{2}$.

Answer:

$2520$

Let us first note that $A=[2,4]$ and $B=[6,3]$ with respect to the lower left corner of the square. Reflect $A$ and $B$ across the sides of the square and label everything according to the picture below.

$$$$

Consider the trajectory from $A$ to $B$ that bounces off $KN$ and then $MN$ and reflect its parts adjacent to the points $A$ (resp. $B$) across $KN$ (resp. $MN$). Due to the reflection angle rule, we obtain precisely the segment $A_1{B}_2$. (In the sequel, we will say that the trajectory was straightened into $A_1{B}_2$). Note that this is the only admissible trajectory that bounces off precisely these two sides of the square. Indeed, straightening a possible trajectory $A\to MN\to KN\to B$ results in the segment $A_2{B}_1$, which does not intersect the square $KLMN$. This is due to the properties of reflection implying that $N$ is the midpoint of both $A_1{A}_2$ and $B_1{B}_2$. Therefore, such a trajectory is not possible.

Similarly, all the desired trajectories bouncing off two adjacent sides of the square $KLMN$ straighten into one side of the quadrilateral $A_1{B}_2{A}_3{B}_4$ or its shifted copy $B_1{A}_2{B}_3{A}_4$. From the corresponding congruent sides, exactly one is used because the other side is in the invalid configuration. It follows that these trajectories contribute to the sum of lengths squared by the sum of squares of the lengths of sides of $A_1{B}_2{A}_3{B}_4$. Using Pythagoras’ theorem and the fact that its diagonals are perpendicular and intersect at point $C=[6,4]$ (see the picture below), we calculate that

$$$$

Now it remains to consider the trajectories bouncing off two opposite sides of the square $KLMN$, such as the two shown below.

$$$$

In this case, both orders are possible, resulting in a contribution equal to the sum of squares of the diagonals of parallelograms $A_2{B}_2{B}_4{A}_4$ and $A_1{A}_3{B}_3{B}_1$. Using the fact that the sum of squares of diagonals in a parallelogram equals the sum of squares of its sidelengths, we obtain

for both of the parallelograms. Hence, the total sum is

Answer:

$94590$

Note that since all numbers $a$, $b$, and $c$ are distinct, and from the divisibility condition, it must hold that $2\cdot a\le b$ and $2\cdot b\le c$. Let $s(k)$ denote the number of digits of $k$. From the divisibility conditions, it follows that digit counts $s(a)$, $s(b)$ and $s(c)$ must be non-decreasing. Therefore, there are only two cases:

1.

Digit counts satisfy $s(a)=1$, $s(b)=1$, and $s(c)=3$. Then the number $a$ is at most $4<\frac{9}{2}$. This leads to the result $a=4$, $b=8$, and $c=992$.

2.

Digit counts satisfy $s(a)=1$, $s(b)=2$, and $s(c)=2$. Then the number $b$ is at most $49<\frac{99}{2}$. To maximize the number $a\parallel b\parallel c$, assume that the first digit is $9$. Then the maximal possible $b$ is $b=45$ and hence $c=90$. Considering any smaller $a$ leads to a smaller result.

Therefore, the maximal possible number is $94590$.

Answer:

$471$

We look for $x$ such that the fraction $\frac{2x^3+2x+4}{x+5}$ is an integer. Since

it is sufficient for $x+5$ to divide $256=2^8$. Since $x>5$, we search for divisors that are at least $10$. All such divisors are $16$, $32$, $64$, $128$, and $256$. Therefore, the sought-after solution is the sum

Answer:

$7∕20$

Let us do straightforward combinatorics: The probability of the first event is

while for the second event we obtain