Problems and solutions

Answer:

4444

By swapping the “inner” corners into “outer” corners, the decagon can be transformed into a rectangle of dimensions $2018$ and $70+134=204$. Therefore the perimeter is $2\cdot (2018+204)=4444$.

Answer:

$34$

Since the hour hand passes $360^{\circ }:12=30^{\circ }$ in one hour, it takes 2 minutes to pass $1^{\circ }$. Therefore, it passed $137^{\circ }-4\cdot 30^{\circ }=17^{\circ }$ after four o’clock, which took $17\cdot 2=34$ minutes.

Answer:

18

One can see that if pampam stands for greater, then Kevin has the largest score and Liam has the smallest score, and if pampam means smaller, it is just the other way round. In any case, Madison and Natalie always have the middle two scores, namely 6 and 12. Therefore the sought sum is 18.

Answer:

17

Since Jack’s $4$ is John’s $16$, there is a segment of houses where John’s numbering is larger by $12$. However, this segment has to end before Jack reaches $12$, since otherwise John would get $12+12=24$. Since the number always drops by the total number of houses when the end is reached, we see that there are $24-7=17$ houses on the square.

Answer:

$19$

In the original recipe, the vinegar makes $10\%$ of one of $5$ parts. The same concentration is achieved if the vinegar makes $40\%=4\cdot 10\%$ of one of $4\cdot 5=20$ parts. Hence we need $19$ extra parts of water.

Answer:

$110^{\circ }$

We can add points $D$, $E$ lying on $h$, $g$, respectively, so that the known angles and the angle in question become the internal angles of pentagon $\mathit{ABCDE}$. Since the sum of the newly added angles is $180^{\circ }$ (we can make them right as in the figure below, but that is not necessary) and the sum of internal angles of any pentagon is $540^{\circ }$, we infer that the sought value is $540^{\circ }-180^{\circ }-105^{\circ }-145^{\circ }=110^{\circ }$.

Answer:

$67.5^{\circ }$

Let $X$, $Y$ be the two points with angle $𝜀$. Then $\mathit{\angle AXY}=\mathit{\angle AY}X=𝜀$. Further, since $\mathit{\angle XAY}=\mathit{\angle CAB}=45^{\circ }$, the interior angles of triangle $\mathit{XY}A$ satisfy

or $𝜀=67.5^{\circ }$.

Answer:

$7$

Let Cederic be $c$ years old and his mother $m$ years old, $c$ being equal to $m$ reversed. The numbers $c$ and $m$ have the same number of digits (with $c$ possibly having a leading zero, if $m$ ends with a zero), which is at least $2$. Let $a$ and $b$ be the units digit of $c$ and $m$, respectively. As Cederic’s mother is $27$ years older, we see that either $a+7=b$ or $a+7=10+b$. If the mother was at least $100$ years old, the difference of the first digits of their ages could be at most $1$, which is not possible, as those are precisely the digits $b$ and $a$. Thus, both numbers $c$ and $m$ have $2$ digits.

Therefore, we want to find all numbers $\overline{\mathit{ab}}$ such that

We know that $a>b$, so the condition $a+7=b$ cannot hold. Hence we consider the condition $a+7=10+b$ or $a=b+3$. Since $a\le 9$, we have $b\le 6$. For every digit $b\in \{0,1,\dots ,6\}$, we get digit $a$ as $a=b+3$. It is easy to see that for these digits the equality $\overline{(b+3)b}=\overline{b(b+3)}+27$ holds. So the desired situation can happen $7$ times: When Cederic has $3$, $14$, $25$, $36$, $47$, $58$, and $69$ years.

Answer:

$3∕4$

If a unit cube is placed in a corner of the big cube, then three of its faces are visible, if it is on one of the edges, two faces are visible, otherwise only one face is shown. There are eight corners and each of the twelve edges of the big cube consists of two unit cubes—altogether $32$ places, and it is clear that the highest proportion of white area is achieved if the white cubes are placed precisely to these places. In this placement every face of the big cube looks the same, showing twelve white and four black faces. Hence the proportion for the whole surface is equal to $12∕16=3∕4$.

Answer:

3

Since nobody failed in both the psychological and the experience test, all the participants failing in at least two tests must have failed because of their health. This gives us $(100-26)-60=14$ people failing only the health check. Together with $83$ people failing in psychology or experience, there were $97$ participants dismissed, hence only $3$ astronauts from the whole group were selected.

Answer:

$5:4$

Denote by $s$ the side length of square $B$. Observe that by symmetry, the midpoint of the circle divides the side of square $B$ which lies on the diameter in two equal parts of length $s∕2$. Then the Pythagorean theorem yields

and therefore the ratio is $5:4$.

Answer:

$10$

Since there is $2\cdot 5$ in the product, the units digit will be $0$. We obtain the tens digit as the last digit of the product $3\cdot 7\cdot 11\cdot \dots \cdot 37$. It is sufficient to consider only the units digits of multiplicands. Moreover, we can ignore ones. So we have to determine the last digit of the product

As $3\cdot 7=21$ has the units digit $1$, we can remove pairs of $3$ and $7$. After this, there is only $9\cdot 9$ left, whose units digit is again $1$. Therefore, the last two digits of the given product are $10$.

Answer:

3

Let $n$ be the number of friends of the last suspect. The suspect with five friends is a friend of everyone else, so omitting him from the group simply decreases everyone’s number of friends by one. Then we can also omit the suspect with one friend only, since his number of friends dropped to zero and he is no longer significant in any way. This way we obtain a group of four suspects, about which we know that they have $1$, $2$, $3$, and $n-1$ friends among themselves, respectively. Repeating the two steps produces yet smaller group with values $1$ and $n-2$; it is clear then that $n-2=1$ or $n=3$.

Note: This solution can be turned to a way of producing such a group of suspects. The result is shown in the following diagram:

Answer:

40

Let $P$ be the product of numbers on opposite faces. Clearly, the higher $P$, the higher the total sum. Since $P$ has to be divisible by all the three shown numbers, its smallest value of $P$ is their least common multiplier, $P=36$. Under these circumstances, the numbers not shown are $3$, $4$, and $6$ and the sought sum is $6+9+12+6+4+3=40$.

Answer:

$99^{\circ }$

Let us denote the points as in the picture.

Angle $\mathit{CBD}$ is the difference of interior angles of octagon and pentagon, hence $\mathit{\angle CBD}=135^{\circ }-108^{\circ }=27^{\circ }$. We also easily obtain that $\mathit{\angle ABD}=135^{\circ }$. Since both triangles $\mathit{ABD}$ and $\mathit{CBD}$ are isosceles,

Therefore

Answer:

50

The time the minister gained by getting up earlier (1 hour) splits into the unknown length $t$ of the walk and the time which would remain for the car to get from the meeting point to the minister’s house, which is obviously half of the total saved time, therefore

or $t=50$.

Answer:

725

Let $d$ be the first digit of the number, $k$ the number obtained after erasing the first digit, and $n$ the number of digits of $k$. Then the original number equals $10^{n}d+k$ and the assertion can be rewritten as

or

Since $28=2^2\cdot 7$, in order for the right-hand side to be divisible by $28$, $d=7$ and $n\ge 2$ has to hold. Finally, the choice $n=2$ (implying $k=25$) gives the smallest possible number $725$.

Answer:

44

The minute hand does 24 revolutions in 24 hours, the hour hand does 2 revolutions in 24 hours. Therefore the minute hand laps the hour hand 22 times in 24 hours. In each of this 22 times the minute hand and the hour hand are perpendicular 2 times, therefore the answer is 44.

Answer:

1111, 1221

Let $\overline{\mathit{abba}}$ be such a palindrome. Since it is a sum of two three-digit numbers, it does not exceed $1998$, so $a=1$. Let $\overline{1\mathit{bb}1}$ be equal to $\overline{\mathit{cdc}}+\overline{\mathit{xyx}}$, then

Since the left-hand side ends with $1$, $c+x$ also ends with one. As both $c$ and $x$ are at least one and at most $9$, $c+x=11$. Plugging in and simplifying produces

Since $d$ and $y$ are digits, the right-hand side does not exceed $18$, so $b-1$ is either $0$ or $1$. Both options are possible: $1111=505+606$, $1221=565+656$.

Answer:

$31∕49$

The area of each of the three small triangles is

of the large equilateral triangle, because the height is $1∕7$ and the base $6∕7$ of the corresponding lines in the large equilateral triangle. Therefore the ratio of the area of the smaller equilateral triangle to the area of the larger equilateral triangle is

Answer:

$(2,3,2,1)$, $(3,1,3,1)$

No number can appear in the table more than five times; however, the number five cannot appear, since it would take a position used by the number appearing five times. Hence only the numbers $1$, $2$, $3$, and $4$ can be filled in.

Let us show that $d=1$: If $d=2$, then one of $a$, $b$, $c$ has to be $4$, and because there are only two free slots, it has to be $b$. However, $(2,4,2,2)$ is clearly not a valid quadruple. The options $d=3$ and $d=4$ lead even faster to a contradiction.

We now know that $a\in \{2,3\}$. Assuming that $a=2$, we get $b,c\in \{2,3\}$ (there cannot be any other ones and fours), but $b=2$ would contradict the condition on $b$ (there would be three twos), hence $b=3$, and $c=2$ follows directly. If $a=3$, there has to be one more one in the table, and that cannot be $c$ (there is already another three), so $b=1$, and again $c=3$ is immediate.

Answer:

2030

Firstly, consider the case when the subwords ‘math’ and ‘drama’ do not overlap. Then there are two ways to order them: ‘dramamath’ and ‘mathdrama’.

If they do overlap, there is only one possible way to arrange them: ‘dramath’. There are three ways to choose places for the two remaining letters: ‘$\text{∗∗}$dramath’, ‘$\ast$dramath$\ast$’, ‘dramath$\text{∗∗}$’. In every case, there are $26^2=676$ ways to choose these two letters. Hence, in these cases, there are $676\cdot 3=2028$ possible passwords.

In total, there are $2028+2=2030$ possible passwords.

Answer:

$42$

There are $n-1$ pairs of consecutive numbers in the set $\{1,2,3,\dots ,n-1,n\}$ and there are $\frac{1}{2}n(n-1)$ possibilities to choose two arbitrary different numbers. Therefore we have

yielding $n=42$.

Answer:

20

Observe that whenever Charlie wins a round, it has no impact on the number of rounds won by Arthur or Ben, neither has it any impact on the number of rounds when Charlie does not play. Hence we may assume that Charlie always loses. In other words, every victory of Arthur over Ben increases Arthur’s overall score by two (unless it happened in the last round), and vice versa. Since the number of Arthur’s victories is odd, we see that the last round had to be Arthur vs. Ben, won by Arthur. Therefore, if we add one more round (Arthur vs. Charlie won by Arthur), the total number of rounds when Charlie did not play would be one half of the sum of final scores of Arthur and Ben, i.e. $(18+22)∕2=20$.

Answer:

52

Denote $k$ the original number of ratings and $x$ their sum. Further denote $a$, $b$ the two this week’s ratings. Then

or

Equation $(1)$ implies that $k$ is a multiple of $50$. Moreover, after subtracting $(1)$ from $(2)$, we get

As $a,b\le 5$, the left-hand side is a positive integer not exceeding $3$, hence $k\le 75$. We conclude that $k=50$ and after adding the two customers rating this week, we see that $52$ people have rated so far.

Answer:

9

For the sake of brevity, we will use numbers 1, 2, 3, 4 instead of the names of the days. Observe that each day is assigned three distinct numbers: The actual number of the day and the two numbers of the socks being worn. Therefore, we may equivalently describe the assignment of socks by a single number for each day—the only number out of the four not appearing in the aforementioned triple. We infer that the valid assignments of socks correspond to the rearrangements of $(1,2,3,4)$, i.e. permutations which leave none of the numbers at its original position.

The number of rearrangements can be computed as follows: There are three options of placing 1, let $n\ne 1$ be its position. Now $n$ has also three options where to be put. It is easy to see that the remaining two numbers are now assigned in an unique way, hence there are $3\cdot 3=9$ rearrangements and the same number of choices of Juliette’s socks.

Answer:

21

In total, $26\cdot 5=130$ votes were distributed among the films. On one hand, if a film received $20$ or less votes, the remaining $110$ votes can easily be distributed so that there are five films getting $21$ votes each. On the other hand, if the film received at least $21$ votes, it not being nominated would imply at least five films getting at least $22$ votes, resulting in at least $21+5\cdot 22=131$ votes cast, a contradiction.

Answer:

$4∕7$

From the equation $f(-2)-2f(3)=-2$ we see that it is equivalent to determine $f(3)$ instead. Since $f(3)+3f(-2)=3$, we have two linear equations for the unknown values $f(-2)$ and $f(3)$. Multiplying the second equation by $2$ and adding both equations we get $f(-2)=4∕7$.

Answer:

$471$

Firstly, let us factorize $4420=2\cdot 2\cdot 5\cdot 13\cdot 17$. Since $13$ and $17$ are primes, one of the numbers $n$, $a$, $b$, $o$, $j$ has to be divisible by $13$ and one by $17$. As the smallest common multiple of $13$ and $17$ is $221$, there is no two-digit number divisible by both of them. Without loss on generality, we can assume that $n$ is divisible by $17$ and $a$ is divisible by $13$. This means that $n\le 85=5\cdot 17$ and $a\le 91=7\cdot 13$.

Suppose now that $n=85$ and $a=91$. We see that $n=85$ is divisible by $5$, so we only need to guarantee the divisibility by $4$. As $n$ and $a$ are odd, $4$ has to divide $\mathit{boj}$. Therefore, one of the numbers $b$, $o$, $j$ is divisible by $4$, or there are two numbers divisible by $2$. We get the greater sum in the second case, when $b=o=98$ and $j=99$. We have found the sum $n+a+b+o+j=85+91+98+98+99=471$.

Finally, let us check the possibility that $n<85$ or $a<91$. Since the numbers $n$ and $a$ have to be divisible by $17$ and $13$, respectively, this would mean $n\le 68=85-17$ or $a\le 78=91-13$. Then the sum $n+a+b+o+j$ could be at most $68+91+3\cdot 99=456$ (in the former case) or $85+78+3\cdot 99=460$ (in the latter case) and that is less than we have previously achieved.

Answer:

$10(1+\sqrt{3})$

The space diagonal of the box passes through the centers of the handball and two tennis balls, and also through the points of tangency of these three balls. The only area where the diagonal is not inside any ball are the segments between a tennis ball and corner of the box; the distance from the center of a tennis ball to the corner is one half of the space diagonal of the circumscribed cube of the ball. So the length of the diagonal of the box is the sum of

• ($2\times$) $1∕2$ of the diagonal of a cube circumscribed to a tennis ball,
• ($2\times$) the radius of a tennis ball,
• the diameter of the handball.

Hence the length of the diagonal is equal to

and the length of the edge is equal to

Answer:

$4$

On one hand, the power $2^{29}$ can be computed by hand with reasonable effort: For example, use $2^{10}=1024$, compute $1024^2$ and $1024^2\cdot 1024$. Finally, divide the result by $2$ to get $2^{29}=536870912$.

On the other hand, you can use the fact that an integer and its digital sum have the same residue class modulo $9$. Moreover, the residue class of $2^n$ modulo $9$ is periodic with period of length $6$. Since the sum of all digits is $45$, we end up having

where $x$ denotes the missing digit in the decimal representation of $2^{29}$. This leads to $x\equiv 4(\mathit{mod}9)$. Therefore the missing digit is $4$.

Answer:

2501

Denote by $\mathrm{ftd}(n)$ the first two digits after the decimal point of $\sqrt{n}$. It is clear that as $n$ increases from one square number to the next one, $\mathrm{ftd}(n)$ increases as well; since we are looking for the smallest integer $n$, this has to be of the form $k^2+1$ for some positive integer $k$.

When $\sqrt{k^2+1}$ is rounded down to its integer part, the result is $k$, therefore $\sqrt{k^2+1}-k$ is a number strictly between $0$ and $1$. The assertion that $\mathrm{ftd}(k^2+1)=0$ can thus be equivalently restated as

Adding $k$ to both sides, squaring (both sides are positive) and rearranging yields

The right-hand side is a number strictly between $49$ and $50$; since $k$ is an integer, $k\ge 50$ follows. We conclude that $n=50^2+1=2501$ is the sought smallest number.

Answer:

$9^{8068}$

Any filling of the 8068 cells forming the two bottom rows and the two leftmost columns defines the whole filling explicitly as we can uniquely fill in the remaining numbers on consecutive diagonals—see the example in the picture. On the other hand, obviously, each correct filling induces some filling of the 8068 cells. Therefore the number of arbitrary fillings of those cells is the same as the sought number of correct fillings of the table.

Answer:

$(3,18)$, $(6,66)$

The given equation is equivalent to $m^2-{(2^n)}^2=260$ and factorizing the left-hand side leads to $(m-2^n)(m+2^n)=260$. The prime factorization of $260=2^2\cdot 5\cdot 13$ leads to the following possible decompositions:

taking into account that $(m-2^n)<(m+2^n)$. Due to $(m+2^n)-(m-2^n)=2^{n+1}$ we get the two possibilities $26-10=2^4$ and $130-2=2^7$ leading to the two pairs $(3,18)$ and $(6,66)$ fulfilling the the given equation.

Answer:

4033

Instead of reflecting the light ray, we shall let it proceed straight and subsequently reflect the triangle along the side with which the ray is incident. Let us show that one row of these reflected triangles is sufficient for the light ray to reach a vertex of one of the triangles.

Let $E$ be the intersection of the ray $\mathit{BD}$ with line through $A$ parallel to $\mathit{BC}$, and $F$ a point on $\mathit{BC}$ such that $\mathit{EF}\parallel \mathit{AC}$. Then the triangles $\mathit{BCD}$ and $\mathit{BFE}$ are similar and $\mathit{BF}=2018\mathit{BC}$. This implies that the point $E$ may be obtained by reflecting the triangle $\mathit{ABC}$ and it is clearly first such point on $\mathit{BD}$.

It is easy to see that the segment $\mathit{BE}$ intersects $2\cdot 2017-1=4033$ segments in the row of triangles, which equals the number of reflections of the light ray in the original problem. The picture illustrates the solution with the initial condition changed to $\mathit{DC}:\mathit{AC}=1:5$.

Answer:

$9∕4$

Let us denote the various intersection points as in the figure.

All the three triangles $\mathit{KM}{D}^{\prime }$, $A^{\prime }\mathit{MC}$, and $L{A}^{\prime }B$ are right and similar to each other, as straightforward angle chasing shows, hence we seek the ratio of similitude between the two triangles in question. Taking into account that $\mathit{KD}=K{D}^{\prime }$ and $\mathit{LA}=L{A}^{\prime }$, we have

and

Since $A^{\prime }L$ corresponds to $\mathit{KM}$ and $\mathit{LB}$ corresponds to $K{D}^{\prime }$ in the aforementioned similarity, this shows that the ratio is $3∕2$. As we are interested in areas, the result is ${(3∕2)}^2=9∕4$.

Answer:

$185$

The polynomial can be rewritten as

Now, since

we see that all the brackets on the right-hand side of the equality are divisible by $x^2-1$. As the degree of $6x+1$ is smaller than the degree of $x^2-1$, we infer that $6x+1$ is the sought remainder. Now solving $1111=6x+1$ produces the answer $x=185$.

Answer:

$30$

Observe that when assigning the lines on the “outer pentagon”, two of the companies (denoted by $X$ and $Y$) have to get two lines and one company ($Z$) obtains one line. The companies also have to be in order $\mathit{XY}\mathit{XY}Z$ starting from some town. It is easy to show that when we have assigned these lines, the rest of the network is assigned in a unique way: The lines of the “inner pentagon” are assigned to the same companies as their outer counterparts and the “connecting” lines must each go to the sole unused company.

This means that the number of correct assignments of the whole network is equal to the number of assignments of the outer pentagon. There are six ways to assign the three companies to $X$, $Y$, and $Z$, and five possibilities for the town where the assignment $\mathit{XY}\mathit{XY}Z$ starts from. This gives us $5\cdot 6=30$ ways in total.

Answer:

$25-13\sqrt{2}$

Consider the triangle whose vertices are centers of the three circles in the corner. This is a right triangle with legs of length $14$ and $34$, respectively; hence, by the Pythagorean theorem, the length of the hypotenuse is

Further, its inradius can be computed as $(14+34-26\sqrt{2})∕2=24-13\sqrt{2}$. Since the sides of the original triangle are parallel to the new ones and their distance is $1$, the incenters of the two triangles coincide and the sought inradius is just the one we have computed plus the distance, i.e. $25-13\sqrt{2}$.

Answer:

17

For convenience, let us number the positions in the list starting with zero instead. In that case, an easy induction argument shows that the position of a number written in the base $4$ system describes what operations (and in what order) were applied to obtain the number on the given position. For example, this is what Marge’s list looks like after two iterations of marging:

(The numbers under the entries are their positions in base $4$.) Since $2017$ written in base $4$ is $133201$, the number on position $2017$ is

Answer:

25

The equation can be also viewed as a quadratic equation in $n$ with solution

(the other solution would lead to negative $n$ since $\sqrt{x^2+y^2}>x$), or

Let $d=\mathrm{GCD}(x,y)$ and let $x=x_{0}d$, $y=y_{0}d$. Plugging in and simplifying yields

Now since $x_0$ and $y_0$ are coprime, also $y_0^2$ and $x_0^2+y_0^2$ are coprime and $x_0^2+y_0^2\mid n$ follows. Further, the presence of $\sqrt{x_0^2+y_0^2}$ forces $x_0^2+y_0^2$ to be a square. It is well known that $5^2=25$ is the smallest square which is the sum of two squares, $3^2+4^2$. Hence $n\ge 25$ and plugging in $x=4$, $y=3$ shows that $n=25$ indeed gives a solution.

Answer:

8

Let us examine the spider’s possible paths in the net of the cuboid. It is clear that the shortest path becomes a straight line segment when the net is put into a plane. The white circle represents the fly, whereas the black circle stands for the spider, its location in the plane dependent on the way we decompose the net of the cuboid.

There are (up to a symmetry) three possible ways for the spider to reach the fly, crossing one, two, or three of the long faces of the cuboid; the paths are marked by $A$, $B$, and $C$ in the picture. Clearly, a path using four of these faces could be reduced to a shorter one. The length of path $A$ is $8.4m$ and using the Pythagorean theorem, we obtain that the length of path $B$ in meters is $\sqrt{66.32}$ and for path $C$ it is $8$. Therefore path $C$ is the shortest one and the answer is $8m.$

The image below shows the shortest path in three dimensions:

Answer:

49

Let us set $V(x,y)={(6+2\cos (x)-\cos (y))}^2+{(8+2\sin (x)-\sin (y))}^2$. Recall that the circle with center $(C_1,C_2)$ and radius $R>0$ can be parametrized (i.e. coordinates of all points lying on it can be expressed) by angle $\alpha$ as $(x_1,x_2)=(C_1+R\cos (\alpha ),C_2+R\sin (\alpha ))$. Let us consider circles $k_1$, $k_2$ with centers $(0,0)$, $(6,8)$ and radii $1$, $2$, respectively. Then it follows from the Pythagorean theorem that $V(x,y)=A{B}^2$ where $A\in k_1$ with angle $x$ and $B\in k_2$ with angle $y$. It follows that the minimum of $V(x,y)$ is the square of the distance of the closest pair of points on $k_1$ and $k_2$ and we can compute it using the distance of the centers and the radii of $k_1$ and $k_2$: $\sqrt{6^2+8^2}-1-2=7$. Thus the minimum of $V(x,y)$ equals $7^2=49$.

Answer:

$105263157894736842$

Let us call $N$ the number in question. When we remove the units digit, we should get all the digits of $2N$ except for the leftmost one. Therefore, since $N$ ends with $2$, $2N$ has to end with $4$, hence the tens digit of $N$ is $4$. Let $d_i$ be the $i$-th digit of $N$, this time counting from right to left (i.e. $d_1$ is the units digit). Taking into account how multiplying by two works digit-wise, we see that the digits of $N$ have to satisfy

for all $i>2$. This way we can directly write down the digits of $N$. We stop when we obtain the digit $1$ and in the next step the digit $2$: The number starting with this $1$ is $N$, because multiplying by $2$ precisely removes the last digit, but puts $2$ in front of the number. The result is

Answer:

$19∕2$

We use the notation as in the following picture.

Due to the area ratios, $S$ divides $\mathit{QB}$ in the ratio $1:2$ and $\mathit{PC}$ in the ratio $2:3$. Introducing the straight line $\mathit{AS}$ and denoting the area of triangle $\mathit{ASQ}$ by $b$ and the area of triangle $\mathit{APS}$ by $a$ leads to the following two equations:

These are equivalent to

which gives the solutions $a=5$ and $b=\frac{9}{2}$. Therefore the area of quadrangle $\mathit{APSQ}$ is $\frac{19}{2}$.

Answer:

1152

Since each brother’s wealth is a multiple of the wealth of the poorest one, their sum, $2018$, has to be divisible by this number as well. However, prime factorization $2018=2\cdot 1009$ gives only three options for the poorest one: $1$, $2$, or $1009$. Clearly, $1009$ is impossible, as this would be larger than any of the remaining three amounts. Further, if it was just $1$, the rest would be left with $2017$ euros, which is a prime number, hence the second poorest brother would have had only $1$ as well—a contradiction. Therefore the poorest one has $2$ euros and the remaining three have $2016$ altogether.

Let $a<b<c$ be the fortunes of the three brothers; these are subject to the conditions $a\mid b\mid c$ and $a+b+c=2016$. The divisibility together with strict inequality implies that $2a\le b$ and $2b\le c$; if we could achieve equalities, we would clearly get a solution with the smallest value of $c$. Luckily, $1+2+4=7$ divides $2016$, therefore we can indeed divide this sum as

and the answer is $\frac{4}{7}\cdot 2016=1152$.

Answer:

1365

Let $A_n$ be the sequence on the blackboard after Andrew carried out the replacement procedure for the $n$-th time (with $A_0=(♣)$) and $h_n$ the number of pairs $♡♡$ in $A_n$. Since each pair $♡♡$ in $A_n$ arises only from the pair $♡♣$ in $A_{n-1}$, which, on the other hand, comes either from $♡♡$ or $♣$ in $A_{n-2}$, we see that $h_n=h_{n-2}+2^{n-3}$ for $n\ge 3$, since there are exactly $2^{n-3}$ symbols $♣$ in $A_{n-2}$. Therefore, for odd $n$, we have

since $h_1=0$. The desired result is $h_{13}=1365$.

Answer:

$10^{\circ }$

We will prove that the quadrilateral $\mathit{OCNP}$ is cyclic; since $\mathit{\angle ONC}=90^{\circ }$, this is equivalent to $\mathit{\angle OPC}=90^{\circ }$. This can be seen as follows: As both $C$ and $M$ lie on $\varrho$, $\mathit{OC}$ = $\mathit{OM}$. Easy computation also shows that $\mathit{\angle MOC}=60^{\circ }$, so $△\mathit{OCM}$ is equilateral. Now $P$, being the midpoint of $\mathit{OM}$, satisfies $\mathit{\angle OPC}=90^{\circ }$.

Another easy calculation reveals that $\mathit{\angle OCN}=70^{\circ }$, hence $\mathit{\angle OPN}=180^{\circ }-\mathit{\angle OCN}=110^{\circ }$. Using the triangle $\mathit{OQP}$, we get that

Answer:

$(3,2,2013)$

Xena’s statement means just that $x$ is odd; if it was even, $y$ and $z$ could have been the same.

Assume now that $y$ is odd; that means that Yena knew from the beginning that $x$ and $z$ are different. If, moreover, $y\ge 1009$, Yena would have already known that $x$ and $z$ were different from $y$ and thus would not need Xena’s statement. On the other hand, if $y\le 1007$, then despite Xena’s statement Yena still could not tell if her number was different from $x$. We infer that $y$ is even and consequently, $z$ is odd.

If $y$ was a multiple of $4$, then $x+z=2018-y\equiv 2(\mathit{mod}4)$, i.e. it could be a double of an odd number; in such a case Yena could not deduce that $x$ and $z$ are different. However, if $y\equiv 2(\mathit{mod}4)$, then $x$ and $z$ have to give distinct remainders modulo $4$ and Yena’s statement is justified.

Finally, let us examine Zena’s statement. Firstly, $y=2$, for otherwise $y$ could be decreased by $4$ and $x$ increased by $4$ and Zena could not tell the difference. For similar reasons, $x\le 4$, hence either $x=1$ or $x=3$. However, in the former case, knowing $2018-z=x+y=3$ Zena could have determined $x$ and $y$ without Yena’s statement (using just what Xena had reported). We conclude that $x=3$ and $z=2013$.

Answer:

$45∕128$

The exact amount of HP is not important—it clearly suffices to know that Arithmetix loses after being hit once and Combinatorica after being hit twice. Suppose that we are in the state of the duel when Combinatorica has been already hit once and it is Arithmetix’s turn to cast. Denote the probability that Arithmetix wins by $q$. In this state, Arithmetix can win either if his attack succeeds, which happens with probability $0.9$, or if he misses and so does Combinatorica in her turn—that happens with probability $0.1\cdot 0.4$, and subsequently, Arithmetix wins again with probability $q$. Therefore, we obtain the equation

which yields $q=15∕16$.

Let us now compute the probability $p$ that Arithmetix wins the duel. If Arithmetix hits and Combinatorica misses (probability $0.9\cdot 0.4$), the duel gets to the situation of the previous paragraph and Arithmetix wins with probability $q=15∕16$. On the other hand, if Arithmetix misses and Combinatorica misses, too (probability $0.1\cdot 0.4$), then Arithmetix can again win with probability $p$. So we can write the equation

Solving it shows that Arithmetix wins the battle with probability $p=45∕128$.

Answer:

3867

If $a(1)=1$ we also have $a(a(1))=1\ne 3\cdot 1$ which is impossible. Since the sequence is increasing it follows that $1<a(1)<a(a(1))=3$ and thus $a(1)=2$. From the equation we deduce $a(3n)=a(a(a(n)))=3a(n)$ for all $n$. We easily prove by induction (starting with $a(1)=2$) that $a(3^m)=2\cdot 3^m$ for every $m$. Using this we also obtain $a(2\cdot 3^m)=a(a(3^m))=3^{m+1}$.

There are $3^n-1$ integers $i$ such that $3^n<i<2\cdot 3^n$ and there are $3^n-1$ integers $j$ such that $a(3^n)=2\cdot 3^n<j<3^{n+1}=a(2\cdot 3^n)$. Since $a(n)$ is increasing there is no other option than $a(3^n+b)=2\cdot 3^n+b$ for all $0<b<3^n$. Therefore $a(2\cdot 3^n+b)=a(a(3^n+b))=3^{n+1}+3b$ for all $0<b<3^n$. Since $2018=2\cdot 3^6+560$ we have $a(2018)=3^7+3\cdot 560=3867$.

Answer:

1346

Each vertex in the grid can be assigned its distances to the three sides of $T$ (taking as the unit the height of a small triangle); it is easy to see that for each vertex, these three integers add up to $2018$. On the other hand, given a triple of non-negative integers with sum $2018$, there is a unique vertex of the grid with these numbers being its distances from the sides, thus we may equivalently consider such triples instead of the vertices. We will refer to the three numbers as coordinates.

The independence condition translates to the assertion that no two triples in the set have equal the first, the second, or the third coordinate. Let

be an independent set. Since the numbers $x_1,\dots ,x_k$ are distinct non-negative integers, their sum is at least

The same holds for the sums $y_1+\cdots +y_k$ and $z_1+\cdots +z_k$. On the other hand, we have $x_i+y_i+z_i=2018$ for each $i=1,\dots ,k$, and so

It follows that

or $k\le 1346$.

The following two sequences of points form together an independent set of size $1346$:

We conclude that the sought maximal number of elements is

.

The following picture illustrates the construction of an independent set for a triangle of side length $11$ instead of $2018$: