Problems and solutions

Answer:

$14325$

The sum $16$ is only possible to obtain as $9+7$. Since $R=9$ leads to the contradiction $J=N=3$, we get $S=9$ and $R=7$. Step by step, we deduce the remaining values: $J=5$, $N=1$, $Q=6$, $B=3$, $P=8$, $A=4$ and $O=2$. Therefore, we conclude $\mathit{NABOJ}=14325$, which may be interpreted as the date of Náboj 2025.

Answer:

$14$

Gear $A$ has 14 teeth, gear $B$ has 6, and gear $C$ has 15. Let us first find the smallest number of teeth by which the gear wheels must be turned so that all gear wheels return to their original position; this number must be a multiple of 14, 6, and 15. The least common multiple of these numbers is 210. Gear $C$ must therefore be turned by 210 teeth and consequently must rotate $210∕15=14$ times.

Answer:

$9897989698$

Let us look at the leftmost digit. We set it to 9, the largest possible value, and try to build the desired number by always adding the largest possible digit to the right so that the given condition is satisfied. Next, we cannot add 9, so we use 8 instead. Then we can add 9 again. Now, we cannot add 8 or 9, so the largest possible value is 7. Next, we can again put 9, 8, and 9 but then no digit from $\{9,8,7\}$. After placing 6, we can again add 9 and then, as the tenth digit, 8. This way we get the number $n=9897989698$. We claim that this is the largest ten-digit admissible number. Indeed, let $m$ be another one and look at the leftmost digit where $m$ and $n$ differ. Since our algorithm chose the largest available digit at this position we have $m\le n$ regardless of the remaining digits.

Answer:

$6$

The area of the square must be a multiple of 3, the area of the triomino, and it is easy to see that a $3\times 3$ square cannot be fully covered. However, a $6\times 6$ square can be covered, and one possible arrangement is shown below.

Answer:

$36$

The difference of perimeters is exactly $\mathit{XB}+\mathit{CS}=2\cdot \mathit{CS}=\mathit{CD}$, which further equals $\mathit{BC}$ and $\mathit{XS}$, so the result is $3\cdot (50-38)=36$.

Answer:

$93$

Let $x$, $y$ be the two digits of the number. Observe that the units digit of Elaine’s product is the same as the units digit of the product $x\cdot y$. There are only two ways to multiply two digits and get a number ending with $7$, namely $1\cdot 7=7$ and $3\cdot 9=27$. We can easily rule out the option $17\cdot 71$ because this product is too small, so the correct option is $39\cdot 93=3627$ and the answer is $93$.

Answer:

$37$

If Kurt holds all combinations of three suits and twelve ranks (a total of $3\cdot 12=36$ cards), then it is possible that the card on top of the playing pile has the missing fourth suit and the missing thirteenth rank. In this case, none of Kurt’s cards would be playable, therefore $N$ is at least $37$.

On the other hand, the card on the top of the pile shares a suit with $12$ cards and a rank with $3$ cards. Since the total number of cards is $52$, and at least one of them is in the pile, the maximum number of unplayable cards in Kurt’s hand is $52-1-12-3=36$, so holding $37$ cards, he will certainly be able to play at least one of them.

Answer:

$15^{\circ }$

Since triangle $\mathit{FGS}$ is equilateral, we get $\mathit{FS}=\mathit{GS}$. Therefore, the isosceles right-angled triangles $\mathit{GAS}$ and $\mathit{FSE}$ are congruent, which implies $\mathit{ES}=\mathit{AS}$. Hence, triangle $\mathit{EAS}$ is isosceles. Using $\mathit{\angle ESA}=45^{\circ }+60^{\circ }+45^{\circ }=150^{\circ }$, we obtain $\mathit{\angle SAE}=\frac{1}{2}(180^{\circ }-\mathit{\angle ESA})=15^{\circ }$.

Answer:

$34$

Let us write into each tile how many permissible triangles have this particular tile as their top or bottom corner (depending on the orientation of the tile):

The sought number is simply the sum of all these values.

Answer:

$6075$

Let $N=\overline{\mathit{abcd}}$ be an intriguing number and put $n=\overline{\mathit{cd}}$. Then $N=1000a+100b+n$ and when we remove the hundreds digit, we obtain a number $M=100a+n$. Multiplying the equality $M=\frac{1}{9}N$ by $9$, we get

and further rearranging and dividing by $4$ yields

From this equality we infer that $a+b$ has to be an even number and less than $\frac{2\cdot 100}{25}=8$, since $n<100$. This implies that $a+b$ is at most $6$ and in order to maximize the number $N$, we pick $a=6$ and $b=0$, leading to $n=75$. It is easy to check that $N=6075$ satisfies the condition from the statement.

Answer:

$60$

Since the ship was loaded with $85$ tons on the first journey, it had to carry at least $15$ tons of oil. However, as the amount of oil doubled on the return journey, the ship could carry at most $15$ tons of oil. Therefore, it carried its full capacity of ethanol and mercury. On the return journey, the total load can be calculated as

Answer:

$8$

Let us start covering the shape from one of the inner corners as in the diagram (1); there are two ways to place a domino, but those clearly lead to completely symmetric situations. Therefore, we can select one of them and multiply the result by $2$. Once this domino is fixed, a placement of two more tiles is determined (2). The two ‘squares’ on the left and right can each be covered in two ways (3), and the remainder of the shape can then be covered uniquely (4). Hence, there are $2\cdot 2=4$ ways to proceed with this placement in (1). Accounting for the symmetric option, we obtain $2\cdot 4=8$ ways in total.

Answer:

$30$

The area of each dark gray square is $216∕6=36$, so its side length is $\sqrt{36}=6$. On each face, the uncovered part is twice the size of the light gray part, meaning that each light gray rectangle is half the size of one white square. Therefore we can fit $5$ light gray rectangles with width $6$ along one edge length of the cube, leading to the side length $5\cdot 6=30$.

Answer:

$800\%$

Let $n$ be the price of the notebook, $r$ the price of the ruler, and $p$ the price of the pen. From the given conditions, we have the equations $n=r+p$ and $\frac{3}{2}r=p+n=2p+r$. From the second equation, we deduce that $r=4p$, and substituting this into the first equation gives $n=5p$. Therefore, the price of the pen should increase ninefold, i.e., by $800\%$.

Answer:

$4098600$

Note that $\gcd (x,x-1)=1$ for any positive integer $x$, hence $\gcd (x,\gcd (x-1,a))=\gcd (x,x-1,a)=1$ for any positive integers $a$ and $x$. Therefore, the expression is equal to

Answer:

$12$

The right-angled triangles formed by the regions between the oblique line and the rectangles are similar, with a similarity ratio of 2. Consequently, the width of the gray rectangle is $2\cdot 6=12$, measured in circle diameters.

Answer:

$7.2=\frac{36}{5}$

Let $v$ be Tyler’s original pace (in km/h) and $t$ the time he ran at the fast pace (in hours); then his slow pace is $\frac{3}{4}v$ and the time spent at this pace is $2t$. The total distance is the sum of the two partial distances, hence

therefore

which is the distance covered at the fast pace.

Answer:

$10$

Observe that Laura is totally absent in the constraints, hence she can be placed anywhere. On the other hand, there are only two ways to arrange the other four ladies, so in total, there are 10 options.

Answer:

$220$

For any arrangement of castle walls, the longest possible line segment within the castle is at most as long as the longest segment $\ell$ that connects two points of the castle (note that if some of the interior angles between consecutive walls exceed $180^{\circ }$, $\ell$ might not be entirely contained within the castle). The segment $\ell$ must connect two towers; otherwise, it could be extended by first continuing each of its endpoints in a straight direction until they reach a wall and then further along the walls toward suitable adjacent towers. These two towers divide the walls into two groups, with the sum of each group’s lengths being at least $S$, the length of $\ell$. Thus, we obtain the inequality

and since $S$ has to be a multiple of $10$, we get $S\le 220$. This value can be achieved by splitting the walls as $130+90=220<110+50+70$.

Answer:

$247$

The difference is smallest when the numbers are as close to each other as possible. For this purpose, the thousands digit may only differ by 1. The hundreds digit must be as small as possible for the larger number and as large as possible for the smaller number. Once the hundreds digit is fixed, the same applies to the tens digit and then finally to the units digit. This leads to the numbers 5123 and 4876, whose difference is 247.

Answer:

$62$

If we drop the condition that two neighbouring beds have to be planted with the same type of flowers, then the answer is $2^6=64$. From this result, we subtract the options where the condition is violated, which is the case when the two types of flowers alternate. Those being only two options, this leads to the total of $64-2=62$ valid arrangements.

Answer:

$25\pi -12$

Let $M$ be the center of the circular sheet, and let $A$, $B$, $C$ be the remaining vertices of the rectangle. Denote the radius of the circle as $r$.

We have $\mathit{MA}=r-1$, $\mathit{MB}=r$, and $\mathit{MC}=r-2$, with $\mathit{AB}=\mathit{MC}$. Applying the Pythagorean theorem in the right-angled triangle $\mathit{ABM}$ gives the equation

which can be simplified to

Since $r=1$ leads to an invalid configuration, the only valid solution is $r=5$. The remaining area of the sheet is $r^2\pi -3\cdot 4=25\pi -12$ (in ${\mathrm{dm}}^2$).

Answer:

$23\frac{1}{3}=\frac{70}{3}$

If we mix $x$ units of the former substance and $y$ of the latter, the resulting mixture contains $\frac{4}{5}x+\frac{3}{10}y$ Algebra and $\frac{1}{5}x+\frac{7}{10}y$ Alchemy. To obtain $1:1$ ratio,

must hold, which can be rearranged to $y=\frac{3}{2}x$. In other words, for each $\mathrm{mg}$ Algebry, $1.5\mathrm{mg}$ of Alchema have to be used in the mixture. Hence the maximal amount of Algemy will be produced when Náboicus uses all $14\mathrm{mg}$ of Alchema and $\frac{2}{3}\cdot 14\mathrm{mg}$ of Algebry, producing $\frac{5}{3}\cdot 14=\frac{70}{3}\mathrm{mg}$ of the sought mixture in total.

Answer:

$34$

Let $m^2=M=K+3333$. Since $M$ and $K$ are both $4$-digit perfect squares, we get $32\le n<m\le 99$ and therefore

or

Now

Given the constraints above, the only possible factors are $m+n=101$ and $m-n=33$, yielding the solution $m=67$ and $n=34$. It remains to check that $K=34^2=1156$ indeed has all its digits less than $7$ in this case.

Answer:

$\frac{2\pi \sqrt{3}}{3}=\frac{2\pi }{\sqrt{3}}$

As $X$, $Y$ are centers of the bases of the cylinder, the height of the cylinder equals their distance. Since $X$ and $Y$ are opposite vertices of the cube, they lie on a space diagonal of length $\sqrt{3}$. To determine the radius of the cylinder, pick any other vertex $Z$ of the cube and compute its distance from the diagonal $\mathit{XY}$. The triangle $\mathit{XZY}$ is right-angled at $Z$ ($\mathit{XZ}$ is a face diagonal and $YZ$ is an edge of the cube), and the sought radius is the altitude from $Z$ onto $\mathit{XY}$. By similarity (or area comparisons), this altitude is $\sqrt{2∕3}$. Hence, the volume of the cylinder is

Answer:

$13$

The second assertion cannot be true: If there were at least $31$ liars, they would all respond to the question negatively, making it impossible to have $39$ positive replies. Therefore, there are at most $30$ liars. Since truth-tellers always tell the truth, they must have answered this question negatively, meaning there can be at most $60-39=21$ truth-tellers.

This also implies that the first assertion is false. The $43$ positive responses must have come from all the liars plus some normal people. Since there are at most $30$ liars, at least $43-30=13$ normal people must exist.

This configuration is feasible, as the distribution $17$ truth-tellers, $30$ liars, and $13$ normal people satisfies both conditions. Thus, the minimum number of normal people in the village is $13$.

Answer:

$24$

First, observe that the total number of points awarded is 7, which means that out of the 6 matches, exactly one resulted in the winning team obtaining two points, while the rest resulted in the victors earning only one point each. This implies that the best team defeated all the other teams, with exactly one of those victories being by the larger margin. The matches among the three non-top teams must have resulted in a “cycle”, since each of these teams was awarded only one point, and there are exactly two ways to orient this cycle. In total, there are $4\cdot 3\cdot 2=24$ ways the tournament could have unfolded.

Answer:

$225$

Clearly, the smallest possible value of $M$ is 9, because

If $k\ge 1$, then each substitution $10^k=10\cdot 10^{k-1}$ increases the number of summands by 9. This implies that $M$ has to be a multiple of 9; moreover, the possible values of $M$ form a consecutive set of multiples of 9. The largest possible value of $M$ is 2025, because

Thus, the number of possible values for $M$ is

Answer:

$360$

Let $t_1$ be the time elapsed from the moment the front of the train passes Max and Paul until the rear of the train passes Max. Similarly, let $t_2$ be the time elapsed from the moment the rear of the train passes Max until it passes Paul. Finally, let $\ell$ be the length of the train in meters. Since Max and Paul walk at the same speed, and during $t_1$, Max managed to cover $45$ meters while Paul covered $60-45=15$ meters during $t_2$, the ratio of the time intervals is

Now, consider the motion of the train. During $t_1$, the train advances by $\ell -45$ meters, because initially, its front was at the meeting point, while at the end of this interval, the rear was still 45 meters away from that point. During $t_2$, the rear of the train covers the $105$ meters between Max and Paul. Thus, the speed of the train is

The full length of the train is then given by

Answer:

$33.75^{\circ }={\frac{135}{4}}^{\circ }$

Since triangle $X{C}^{\prime }B$ is isosceles, we have

Furthermore, $\mathit{\angle CXY}=\mathit{\angle Y}X{C}^{\prime }$ because of the folding, hence

Finally, $\mathit{\angle X}{C}^{\prime }Y=\mathit{\angle Y}\mathit{CX}=90^{\circ }$, thus

Answer:

$911$

Observe that, for $k\le n$, the number of all increasing $k$-tuples with entries in the set $\{1,2,\dots ,n\}$ equals $\left(\genfrac{}{}{0.0pt}{}{n}{k}\right)$, since strictly increasing tuples directly correspond to subsets of size $k$. More generally, the number of all increasing $k$-tuples with entries in the set $\{m,m+1,\dots ,n\}$ equals $\left(\genfrac{}{}{0.0pt}{}{n-m+1}{k}\right)$. To determine the position of the tuple, we count the number of preceding increasing 4-tuples, grouping them based on their known entries:

• $(0,\ast ,\ast ,\ast )$: there are $\left(\genfrac{}{}{0.0pt}{}{15}{3}\right)=455$ such tuples;
• $(1,\ast ,\ast ,\ast )$: there are $\left(\genfrac{}{}{0.0pt}{}{14}{3}\right)=364$ such tuples;
• $(2,3,\ast ,\ast )$: there are $\left(\genfrac{}{}{0.0pt}{}{12}{2}\right)=66$ such tuples;
• $(2,4,a,\ast )$ with $a\in \{5,6\}$: there are $\left(\genfrac{}{}{0.0pt}{}{10}{1}\right)+\left(\genfrac{}{}{0.0pt}{}{9}{1}\right)=19$ such tuples;
• $(2,4,7,b)$ with $b\in \{8,9,10,11,12,13\}$: there are 6 such tuples.

Thus, the tuple $(2,4,7,14)$ appears at position $455+364+66+19+6+1=911$.

Answer:

$60$

Let $A$ and $B$ denote the number of apples brought to the market by Adam and Bettina respectively. We know that

Substituting $b=\frac{45}{A}$ and $a=\frac{20}{B}$ into the second equation, we get

Therefore $A=\frac{3B}{2}$, and substituting into the first equation we get $\frac{3B}{2}+B=\frac{5B}{2}=100$, therefore $B=40$ and $A=100-40=60$.

Answer:

$598$

Let $a$, $b$, and $c$ be the hundreds, tens, and units digits of the three-digit number $N$. If $c=9$, then $N>9^3=729$, so $a$ must be $7$ or $8$. However, in both cases, there is no choice of $b$ making the last digit of $729+a+b^2$ equal to $9$.

Next, consider $c=8$. Since $8^3=512$, $a$ must be $5$ or $6$, but

so $a$ can only be $5$. We then need $b$ satisfying

or

whose solutions are $b=1$ and $b=9$. Both yield valid numbers:

(Equivalently, one can note the units digit of $b^2$ must be $1$.)

If $c\le 7$, then

so the largest valid value of $N$ is $598$.

Answer:

$42$

Connecting $P$ to all points that divide the sides of quadrilateral $\mathit{ABCD}$ into thirds creates twelve triangles.

Each set of three triangles along the same side has equal area because they share the same height from point $P$ and have bases of equal length. Let $a$ be the area of triangle $\mathit{PEI}$, $b$ be the area of triangle $\mathit{PJF}$, $c$ be the area of triangle $\mathit{PGK}$ and $d$ be the area of triangle $\mathit{PLH}$. From the given information, we establish the equations

The area of quadrilateral $\mathit{PFBG}$ is given by $b+c$. This can be computed as

Answer:

$237$

Overall there are $\left(\genfrac{}{}{0.0pt}{}{12}{4}\right)=495$ ways of choosing $4$ squares from a $12$-square ring. For each of the four sides, there are eight squares not belonging to that side, giving $\left(\genfrac{}{}{0.0pt}{}{8}{4}\right)=70$ ways of choosing four squares so that this side is left out. Therefore, there would be $495-4\cdot 70=215$ choices so that no side is left out. However, some choices were subtracted twice – namely those where two sides are left out at once. This is possible to do by choosing $4$ out of $5$ squares grouped around a corner ($5$ ways for a corner, $20$ for all $4$ corners) or $4$ out of $4$ opposite middle squares (two ways overall). This gives us $215+22=237$ ways. Because omitting three or four sides is impossible in this situation, this is the final answer.

Answer:

$\frac{6}{5}$

Denote the centres of the circles $X$, $Y$, $Z$, respectively, and let further be $A$, $B$, $C$ the points of tangency as in the diagram below. Triangle $\mathit{XY}Z$ has the side lengths $1+2=3$, $1+3=4$, and $2+3=5$, which is a triple of numbers satisfying the Pythagorean theorem, so it is a right triangle with right angle at $X$. To compute the sought area of triangle $\mathit{ABC}$, we will subtract the areas of isosceles triangles $\mathit{XCB}$, $Y\mathit{AC}$, and $\mathit{ZBA}$ from the area of triangle $\mathit{XY}Z$, which is $\frac{1}{2}(3\cdot 4)=6$.

• Triangle $\mathit{XCB}$ is right-angled, therefore its area is $\frac{1\cdot 1}{2}=\frac{1}{2}$.
• To compute the area of triangle $Y\mathit{AC}$, we first find the length of its height $\mathit{AR}$. Since triangles $\mathit{RY}A$ and $\mathit{XY}Z$ are similar with similarity ratio $YA∕YZ=\frac{2}{5}$, it follows that $\mathit{AR}=\frac{2}{5}\mathit{ZX}=\frac{8}{5}$. Therefore, the area of triangle $Y\mathit{AC}$ is $\frac{1}{2}(\frac{8}{5}\cdot 2)=\frac{8}{5}$.
• Similarly, to get the area of triangle $\mathit{ZBA}$, we utilize the similarity $△\mathit{SAZ}\sim △\mathit{XY}Z$ with ratio $\frac{3}{5}$ to get $\mathit{AS}=\frac{9}{5}$, so the area of triangle $\mathit{ZBA}$ is $\frac{1}{2}(\frac{9}{5}\cdot 3)=\frac{27}{10}$.

Finally, we can calculate the desired area as

Answer:

$300$

It is easy to see that the number of diagonals of an $n$-gon is $\frac{1}{2}n(n-3)$. As 2025 is odd, we can equivalently investigate when the product $P=n(n-3)$ is divisible by 2025. As $2025=3^4\cdot 5^2$, thanks to coprimality, we have to ensure that $P$ is divisible by $3^4=81$ and $5^2=25$. Note that only one of the factors $n$, $n-3$ can be divisible by $5$, so one of them must be divisible by $25$. On the other hand, $n$ is divisible by $3$ if and only if $n-3$ is divisible by $3$, so both of them contribute to the total power of $3$ dividing $P$; however, only one of $n$, $n-3$ can be divisible by $3^k$ for $k\ge 2$. This means that we need one of the factors to be divisible by $3^3=27$.

If one of the factors was divisible by both $25$ and $27$, it would be equal to at least $25\cdot 27=675$. Let us check whether a smaller value can be found by choosing one of the factors (say $m$) to be divisible by $27$ and the other one ($m\pm 3$) to be divisible by $25$ (this way either $n=m$ or $n=m+3$). By the first condition, we have $m=27k$ for some positive integer $k$ and we are interested in the smallest value of $k$ such that $27k\pm 3$ is divisible by 25 (for one of the sign choices). As $25k$ is always divisible by $25$, we can equivalently check $2k\pm 3$, which is first divisible by $25$ for $k=11$ and the positive sign. Therefore $m=27\cdot 11=297$ and $n=m+3=300$.

Answer:

$84$

For each node we compute (1) the number of (oriented) paths from $A$ ending at it, (2) the number of paths to $B$ starting from it. In the diagram below, for example, $3|1$ in the top right node indicates that there are exactly three paths from the starting node that end at this node, and only one path from this node to the final node. For each arrow, the number of paths traversing it in the opposite direction is given by the product of the first number at the endpoint and the second number at the starting point, so we compute this product for every arrow and sum up. The result is 84.

Answer:

$18249$

We can rewrite the computation equivalently as

which further simplifies to

It follows that $N=1$. At this point, we are left with $\overline{\mathit{AAA}}+\overline{\mathit{BB}}+O=2025-1111=914$, implying $A=8$. Further reduction $914-888=26$ yields $B=2$ and finally, $O=4$. The value of $J$ can be arbitrary, but different from the digits already used; hence, the largest possible value of $\mathit{NABOJ}$ is $18249$.

Answer:

$150$

Denote by $YN$ the number of organizers voting for in the first voting and against in the second; similarly define $YY$, $\mathit{NN}$, and $\mathit{NY}$. Finally, let $YX$ be the number of leaving organizers. Then we have the following system of equations:

Substituting $YY=120$ and $\mathit{NN}=70$ and rearranging the terms, we get

Multiplying the second equation by $2$ and summing all equations together yields $3YX=450$, hence $YX=150$. The remaining two variables are equal to $YN=5$ and $\mathit{NY}=155$.

Answer:

$12$

Let $g=\gcd (a,b)$. Then $a=g{a}^{\prime }$, $b=g{b}^{\prime }$ for some positive integers $a^{\prime }$, $b^{\prime }$. Since $g\le a\le b$, the three-term arithmetic sequence is $(g,a,b)$ or its reverse; in either case, $a-g=b-a$ or $b=2a-g$, which becomes $b^{\prime }=2a^{\prime }-1$ after dividing by $g$. From the condition on the sum we obtain

hence $g{a}^{\prime }=675=3^3{5}^2$. This number has $(3+1)\cdot (2+1)=12$ positive divisors and it remains to check that each such divisor yields a valid value of $a^{\prime }$, i.e., one that can be completed to a valid pair $(a,b)$. Indeed, by letting $b^{\prime }=2a^{\prime }-1$, $g=675∕a^{\prime }$, $a=g{a}^{\prime }=675$, $b=g{b}^{\prime }$, we have $a\le b$ since $g{a}^{\prime }\le g(2a^{\prime }-1)$ for any positive integers $a^{\prime }$, $g$, and also

as $a^{\prime }$ and $2a^{\prime }-1$ are coprime for any positive integer $a^{\prime }$.

Answer:

$697$

Note that the set of problems solved by a team is fully determined by the set of unsolved problems and vice versa; this can be also viewed as the set of the problems the team has left on the table at the end of the contest, which can be any set of size at most 3. Therefore there are at most

teams competing.

Answer:

$51$

Using the formula $(x-2)(y-2)=\mathit{xy}-2x-2y+4$, the given conditions can be rearranged to the following equations:

and our goal is to find $(a-2)(d-2)$. Since $b\ne 2$ and $c\ne 2$ because of the second equation, the desired expression can be obtained as

Therefore, $2a+2d-\mathit{ad}=-(-47)+4=51$.

Answer:

$540^{\circ }∕7$

Since $\mathit{ABCDEFG}$ is a regular heptagon, the angle $\mathit{AME}$ is a right angle and $\mathit{AE}$ is the diameter of the larger circle. Instead of measuring the angle between the tangents at $X$, we can equivalently consider the angle between the tangents at $A$, which is the second intersection point of the two circles. This angle is in turn equal to the angle $\mathit{BAE}$ between the corresponding diameters, since those are perpendicular to the tangents. Its size, $\frac{3}{7}\cdot 180^{\circ }$, can be easily determined from the symmetry of the regular heptagon or by recognizing that it is the inscribed angle corresponding to the central angle of $\frac{3}{7}\cdot 360^{\circ }$ on the circumcircle of the heptagon.

Answer:

$\frac{2^{100}\cdot (4^{100}-1)}{3}$

We begin with a more general observation: for any positive integer $n$ and real numbers $x_1,x_2,\dots ,x_n$, if we sum the squared expressions ${(\pm x_1\pm x_2\pm \cdots \pm x_n)}^2$ over all possible choices of signs, the result is always

To see why this holds, consider expanding ${(\pm x_1\pm x_2\pm \cdots \pm x_n)}^2$. Each expansion consists of squared terms $x_i^2$, as well as mixed terms of the form $\pm 2x_i{x}_j$ for $i\ne j$. Each term $x_i^2$ appears in every possible expansion, regardless of the chosen signs. Since there are $2^n$ different sign combinations, these squared terms contribute a total factor of $2^n$. On the other hand, the mixed terms $\pm 2x_i{x}_j$ appear with a positive sign in exactly half of the cases and with a negative sign in the other half, depending on whether $x_i$ and $x_j$ have the same sign or not. Since these contributions perfectly cancel out across all sign choices, they do not affect the final sum. Thus, the total sum simplifies to $2^n$ times the sum of the squared terms, proving the formula.

In our case, we have $x_k=2^{k-1}$, and the desired sum equals

Using the formula for the sum of a geometric series,

we obtain the final result

Answer:

$56$

We define a segment as one side of the square road. Let $m=24$ and $n>m$ be the speeds of the slower and faster cars, respectively. Observe that whether the rubber band ever breaks depends only on the ratio of the speeds, not on their absolute values. Thus, let $m^{\prime }$, $n^{\prime }$ be coprime positive integers with $m^{\prime }:n^{\prime }=m:n$. We claim that the band never breaks precisely when $n^{\prime }-m^{\prime }$ is a multiple of $4$.

First, let us analyze the case when $n^{\prime }-m^{\prime }$ is not a multiple of 4. After the slower car has traveled $m^{\prime }$ segments, it is at a corner. At that same time, the faster car has traveled $n^{\prime }$ segments (due to the ratio of speeds), so it is at a corner, too. Because $n^{\prime }-m^{\prime }$ is not divisible by $4$, these two corners cannot coincide. If they turn out to be opposite corners, the band breaks immediately. If they are neighboring corners, then after traveling another $m^{\prime }$ and $n^{\prime }$ segments, the cars end up at opposite corners, causing the band to break as well.

Let us now assume that $n^{\prime }-m^{\prime }$ is a multiple of 4. First, note that the cars cannot both be at a corner any earlier than when the slower car has traveled $m^{\prime }$ segments. Otherwise, say after some $s<m^{\prime }$ segments of the slower car, the faster car would have traveled $\frac{s}{m^{\prime }}\cdot n^{\prime }$ segments, which cannot be an integer because $m^{\prime }$ and $n^{\prime }$ are coprime. Hence, the first time both cars reach corners simultaneously is when the slower car has traveled $m^{\prime }$ segments and the faster car $n^{\prime }$. By assumption, $n^{\prime }-m^{\prime }$ is a multiple of $4$, so they must land at the same corner. Once they coincide at a corner, the entire situation effectively resets (possibly starting from a different corner), so the band never breaks.

It remains to find the smallest $n>24$ that satisfies the given condition. We must have $n^{\prime }-m^{\prime }$ divisible by $4$. In that case, both $m^{\prime }$ and $n^{\prime }$ must be odd. Since $m^{\prime }$ is a divisor of $m=24$, the only possible values are 1 and 3. If $m^{\prime }=1$, the smallest $n^{\prime }$ that is coprime to $m^{\prime }$ and such that $n^{\prime }-1$ is a multiple of $4$ is $5$. Then $n=\frac{24}{1}\cdot 5=120$. If $m^{\prime }=3$, the smallest $n^{\prime }$ coprime to $m^{\prime }$ with $n^{\prime }-3$ divisible by $4$ is $7$. Here, $n=\frac{24}{3}\cdot 7=56$. Since $56$ is smaller than $120$, we conclude that the desired smallest value of $n$ is $56$.

Answer:

$2^{975}+2025\cdot 2^{2999}=2^{975}\cdot \left(1+2025\cdot 2^{2024}\right)$

Let us denote the players by numbers $1$, $2$, $\dots$, $2025$, and the amount of coins owned by player $p$ after $r$ rounds as $m_{p,r}$. Without loss of generality, assume that player $1$ lost the first round, player $2$ the second one, and so on. Denote $M=2^{3000}$, the number of coins at the end of the game held by all players.

For a player $p$ after their loss at $p$-th round, the amount of their coins is doubling till it reaches $M$ at the end of the game; hence for a round $p\le r$, the amount of coins is

Moreover, since the $r$-th player lost the $r$-th round, they lost the number of coins equal to the amount held by all other players, and since the total amount of coins in the game is $2025M$, it follows that

By rearranging the equation, we get

Plugging in $r=1$, we obtain the desired result

Answer:

$\sqrt{10}$

The fact that the final cone has zero volume means that joining three identical lateral surfaces results in a completely flat shape – a full circle. Consequently, each individual cone surface, when flattened, corresponds to a circular sector with a central angle of $120^{\circ }$. Let $l$ be the slant height of the original cone and $r$ its base radius. From the central angle, we obtain the relation $l=3r$. Observe that the slant height remains unchanged across all three cones, including the degenerate one.

The intermediate cone, formed by joining two lateral surfaces, has a central angle of $240^{\circ }$, giving it a base radius of $2r$. Applying the cone volume formula, we find

From this, we compute the volume of the original cone as