Problems and solutions

Answer:

$48$

If Peter pulls out all the green socks and seven red socks, he cannot form eight pairs as desired, so $47$ is not sufficient. But if he takes out $48$ socks, there are at least $\frac{48}{3}=16$ socks of one colour and at least $48-40=8$ ones not having this colour among them, which enables him to form eight two-coloured pairs of socks.

Answer:

30

Using the given equality $1234=28^2+2\cdot 15^2$, we get

Since we know that there is only one such pair, we obtain $x=30$.

Answer:

$450$

There are two hours during which the digit five is displayed all the time: $5$ and $15$. That makes $120$ minutes. In the rest of the day, five can be seen during the last ten minutes of each hour ($22\cdot 10=220$ minutes) and then five times during the remaining fifty minutes ($22\cdot 5=110$ minutes). To sum up, five is shown exactly $450$ minutes.

Answer:

$1120$

Since the side lengths of the small rectangles are in ratio $2:5$, let us denote them by $2x$ and $5x$. The side lengths of the big rectangle are $10x$ and $7x$, hence $2\cdot (10x+7x)=34x$ is the perimeter of the big rectangle. This yields $x=4\text{cm}$, thus the area is $10\cdot 7\cdot 4^2{\text{cm}}^2=1120{\text{cm}}^2$.

Answer:

$10$

Let $a$ be the area of a small chocolate, i.e. the one with side length $1\text{cm}$. Then the area of a big chocolate is $4a$, the total area of all chocolates is $\mathit{na}+4\mathit{na}=5\mathit{na}$, and the area of the box is $s^{2}a$ as the shape of the box is just the shape of a small chocolate stretched by factor $s$ in all directions. It follows that $5n=s^2$, so $s$ is a multiple of $5$.

It can be proven that it is impossible to fit five big chocolates in a box of side length $5\text{cm}$, so $s\ne 5$. However, one can easily find a composition of $20$ small and $20$ big chocolates inside a box of side length $10\text{cm}$.

Answer:

$21$

On one hand, every number of socks chosen is the sum of an even number of socks, which are used to build pairs of socks, and a certain number of unpaired socks, which can be at most six due to the number of colours. If Peter takes $21$ socks out of his wardrobe, there cannot be six unpaired socks among them because $21-6=15$ is not even. Thus there are at most five single socks and the remaining socks form at least $\frac{21-5}{2}=8$ pairs. On the other hand, it does not suffice to take $20$ socks because Peter could have taken seven pairs of white socks and six single socks, one of each colour. It follows that Peter has to take at least $21$ socks out of his wardrobe.

Answer:

$153^{\circ }$

Denote the vertices as in the picture. Then the polygon $A{B}_1{B}_2{C}_1{C}_2{D}_1{D}_2$ is the intersection of the two given polygons.

Since the figure is symmetric with respect to the line $\mathit{AC}$, it suffices to check the interior angles at $A$, $B_1$, $B_2$, and $C_1$. Clearly, the first one is $90^{\circ }$ and the last one is $135^{\circ }$. As $B_2{D}_1$ is parallel to $\mathit{XY}$, we have $\angle D_1{B}_2{B}_1=\mathit{\angle Y}\mathit{XW}=108^{\circ }$, and since $\angle C_1{B}_2{D}_1=\mathit{\angle CBD}=45^{\circ }$ ($B_2{D}_1\parallel \mathit{BD}$), the angle at $B_2$ is $153^{\circ }$. Finally, triangle $B_{1}B{B}_2$ is right-angled, from which we easily obtain that the angle at $B_1$ is $117^{\circ }$. Thus the largest angle is $153^{\circ }$.

Answer:

$20\text{mm}$

Having counted the numbers of teeth, one can observe that one full rotation of wheel 1 forces $\frac{20}{15}=\frac{4}{3}$ of a full circle of the double-wheel. Similarly, when the double-wheel performs one full rotation, wheel 2 covers $\frac{18}{10}=\frac{9}{5}$ of a full circle. Hence whenever wheel 1 rotates one full circle, wheel 2 rotates $\frac{4}{3}\cdot \frac{9}{5}=\frac{12}{5}$ of the full circle. The circumference of circle 2 must therefore be $\frac{5}{12}$ times the circumference of circle 1. The ratio of circumferences equals the ratio of diameters, which allows us to compute the diameter of circle 2 as $\frac{5}{12}\cdot 48\text{mm}=20\text{mm}$.

Answer:

31

Observe that if Robert was telling the truth on one day, then he must have been doing so on all the previous days. If he was telling the truth only for $k<31$ days, it would contradict the fact that he was lying for $62-k>31$ days. Similarly, telling the truth for more than 31 days would result in lying too little. We infer that Robert lied on exactly 31 days.

Answer:

$16$

The decomposition of the $9\times 9$ cell grid in figure 1 shows that $16$ shots are necessary since every $5\times 1$ rectangle must be hit by at least one shot.

But $16$ shots are sufficient as well, as we see from figure 2.

Therefore $16$ is the correct answer.

Answer:

$155$

Since $5\cdot 13\cdot 31=2015=a+b+c=\gcd (a,b,c)\cdot (a_0+b_0+c_0)$ for some distinct positive integers $a_0$, $b_0$, $c_0$, we obtain $a_0+b_0+c_0\ge 6$. Hence $\gcd (a,b,c)$ can be at most $5\cdot 31=155$. We can reach the maximum by taking any distinct positive integers that sum up to $13$. For instance, $a_0=1$, $b_0=5$, $c_0=7$ yield $a=1\cdot 155$, $b=5\cdot 155$, $c=7\cdot 155$.

Answer:

31

There are four possibilities where the photographed I-C-C carriage sequence can be, and for each of them there are $2^3$ ways to complete the train. However, this results in counting the train I-C-C-I-C-C twice. Thus the number of different trains is $4\cdot 8-1=31$.

Answer:

17

The structure can obviously fit in a box that is two cubes wide, five cubes high and five cubes long. Let us divide it in half and analyse its two $1\times 5\times 5$ parts separately. If we look at the structure from the front, we can see mirror ‘1’. Hence in the right part, there is one square visible from the front and five squares visible from above. The only way to achieve that is to put five cubes in one row. Similarly for the left part, there are five squares visible from the front and three squares visible from above, so we get the maximum number of cubes if we place five cubes in each of the three columns.

The resulting structure looks like the one in the following picture. If we look at it from the right, we can see 17 cubes.

Answer:

$7999$

Denote by $Q(r)$ the sum of digits of $r$. If the tens digit of $n$ differs from $9$, then we have $Q(n+10)=Q(n)+1$. Hence the tens digit of $n$ has to be $9$. If the hundreds digit differs from $9$, we have $Q(n+10)=Q(n)-8$, but if the hundreds digit is $9$ and the thousands digit is not, we get $Q(n+10)=Q(n)-17$, so we can make both $Q(n)$ and $Q(n+10)$ divisible by $17$. To keep $n$ as small as possible, suppose that this is the case and $Q(n)=2\cdot 17=34$. The sum of all digits but the hundreds and tens one is therefore $34-2\cdot 9=16<2\cdot 9$, which means that two more digits are enough. Now it is easy to see that $n=7999$ is the desired result.

Answer:

20, 110

Assume first that Lisa has the most expensive ticket (i.e. the one from $A$ to $D$) in her collection; thus its price is $70$. As this price is the sum of the tickets for the sections $\mathit{AB}$, $\mathit{BC}$, and $\mathit{CD}$, at least two of which are already in Lisa’s possession, we see that the only possibility for the prices of these three tickets is $10$, $20$, and $40$, so the missing price has to be $20$. It is easy to see that these prices can be assigned to the sections so that they agree with the rest of the tickets.

If the most expensive ticket is missing, then the ticket costing $70$ has to be for a journey with one stop; the only way to decompose this price to a sum of two which Lisa already has is $10+60$. We infer that the ticket for the remaining section costs $40$ and the longest journey costs $10+40+60=110$. It is again easy to check that these values are satisfactory.

Answer:

$\sqrt{3}\text{cm}$

Denote by $P_x$ the point on the boundary of the box where the hour hand points at $x$ o’clock, and let $C$ be the center of the box. Since $\angle P_{2}C{P}_1=\angle P_{3}C{P}_2=30^{\circ }$, we see that the point $P_2$ is the incenter and centroid of the equilateral triangle $C{C}^{\prime }{P}_1$, where $C^{\prime }$ is $C$ reflected through $P_3$. Now, the distance $P_1{P}_2$, which we are looking for, is $2∕3$ of the height of an equilateral triangle of side length $3\text{cm}$, i.e. $P_1{P}_2=\frac{2}{3}\cdot \frac{1}{2}\cdot \sqrt{3}\cdot 3\text{cm}=\sqrt{3}\text{cm}$.

Answer:

728163549

Let $x,y\in \{1,2,\dots ,9\}$ be distinct digits. A pair $\mathit{xy}$ will be called valid whenever there exist $k,l\in \{1,2,\dots ,9\}$ such that $10x+y=\mathit{kl}$ holds. Since the only valid pair involving $9$ is $49$, the block $49$ must appear at the end of the desired nine-digit number $z$. There are two valid pairs involving $7$, namely $27$ and $72$. They cannot occur simultaneously, and the end of $z$ is already occupied. Therefore $72$ must be put at the beginning of $z$. Since valid pairs involving $8$ are $18$, $28$, $48$ and $81$, and $4$ has been already used to form $49$, the only option is to form a block $281$. Hence $z=7281\dots 49$. Now we have to handle the remaining digits $3$, $5$ and $6$. Since neither $13$ nor $34$ is a valid pair, and the only valid pair $\mathit{xy}$ such that $x\in \{5,6\}$ and $y=3$ is $63$, we find $z=728163549$. Clearly, $z$ satisfies all imposed conditions.

Answer:

89

Instead of looking for the largest prime $p$, let us look for the smallest composite number $n$ such that $210-n$ is a prime. We have $210=2\cdot 3\cdot 5\cdot 7$. Consequently, if $n$ was divisible by 2, 3, 5, or 7, so would be $210-n$, i.e. $210-n$ would not be prime. The next smallest composite candidate for $n$ is $11^2$ which gives $p=210-121=89$, which is a prime.

Answer:

20

Denote by $T$ the total number of pencils and by $a$, $b$, $c$ the numbers of pupils in the corresponding classes. From the statement we know that $T=9(a+b+c)$, $T=30a$, and $T=36b$. We are looking for $T∕c$. Substituting $a=T∕30$ and $b=T∕36$ in the first equation yields

Answer:

1681

Since for $k\ge 50$ the inequality ${(k+1)}^2-k^2>100$ holds, $50^2=2500$ is the only square starting with $25$ (similarly for $3600$, $4900$, $6400$, and $8100$). Hence the first two digits have to be $16$. The only square greater than $1600$ and less than $1700$ is $41^2=1681$, which clearly satisfies the problem conditions.

Answer:

4/7

Let $x$ be the fraction of time spent in the sections under repair. Then $1-x$ is the fraction of time spent on the rest of the highway. Thus we have

which yields $x=4∕7$.

Answer:

90

By summing the areas of the squares, the area of the rectangle is $464=2^4\cdot 29$. The lengths of both sides of the rectangle have to be at least $9$ because of the square with side length $9$. Hence the only possible factorization is $16\cdot 29$, which yields the perimeter $90$.

Such a decomposition exists, as one can see in the figure:

Answer:

$36\text{cm}$

By checking angles, we see that all triangles in the statement picture are right-angled and similar to each other. If we denote the side lengths of the triangle in the lower right corner by $6x$, $8x$, and due to the Pythagorean theorem $10x$, then by “unfolding” we see that the side length of the square is $18x$. This implies that the lower left triangle has one side of length $18x-6x=12x$, as can be seen in the following picture. The other sides of this triangle are of length $9x$ and $15x$ due to the scaling factor $3∕2$ of similarity. Now we can see that the side length of the given square is also $15x+6\text{cm}$, which yields $x=2\text{cm}$. Therefore the side length of the paper is $36\text{cm}$.

Answer:

96

When Simon returns to the back of the ship, he has made $300$ steps and the ship has moved on $240-60=180$ steps. Hence during the time in which Simon makes $60$ steps, the ship moves $180:5=36$ steps forward. However, Simon reaches the back of the ship after $60$ steps, so the length of the ship must be $60+36=96$ steps.

Answer:

$992016=996^2$

Let ${(1000-n)}^2$ be the desired number ($n\ge 1$). We can calculate the squares for $n=1,2,3,4,\dots$ by using the formula ${(1000-n)}^2=1000\cdot (1000-2n)+n^2$ to get $999^2=998001$, $998^2=996004$, $997^2=994009$, $996^2=992016$, $\dots$ It remains to see that the digits $2$, $0$, $1$ are pairwise different and that $99\cancel{2}\cancel{0}\cancel{1}6=996$ is the square root of $992016$. Therefore $992016$ is the desired number.

Answer:

596

Notice that if we turn a digit upside-down, then either it does not change (0, 2, 5, 8), or it changes ($6\leftrightarrow 9$), or we do not get a digit at all (1, 3, 4, 7). Let us denote by $x$ the Linda’s number and by $x^{\prime }$ the same number upside-down. Since the number 369 ends with 9, there are three candidates for the units digit of $x$: 0, 6, and 9 (all other admissible numbers lead to a non-admissible units digit of $x^{\prime }$).

If $x$ ended with 0, then $x^{\prime }$ would begin with 0, which is impossible. Furthermore, $x$ ending with 9 would result in $x^{\prime }$ ending with 8, i.e. $x$ beginning with 8 and $x^{\prime }$ beginning with 6, which is a contradiction since $x>x^{\prime }$. Finally, if 6 is the units digit of $x$, we have $x=\overline{5a6}$ and $x^{\prime }=\overline{9a^{\prime }5}$ for some admissible digit $a$ and its upside-down version $a^{\prime }$. However, $a=a^{\prime }$ is not possible in the light of $x^{\prime }-x=369$, so $a$ has to be either 6 or 9. Direct computation shows that $x=596$ is the sought value.

Answer:

80

Denote the families $A$, $B$, and $C$. If the sons from $A$ marry two sisters from some other family (w.l.o.g. $B$), then the daughters from $A$ have to marry the sons from $C$ (otherwise at least one son from $C$ would marry his sister). It follows that the sons from $B$ marry the daughters from $C$. This yields $2\cdot 2^3=16$ ways—there are two ways to match the families, and then each pair of daughters has two options of marrying the sons from the matched family.

If, on the other hand, the sons from $A$ decide to have wives from different families, their sisters must do the same in order to prevent inter-family marriages. The two sons together have eight choices of wives and the same holds for the daughters. After this matching is done, in each family $B$ and $C$ there is exactly one son and one daughter left unmarried, which gives the one and only way to complete the marriage scheme. Therefore this scenario gives $8\cdot 8=64$ ways.

We conclude that there are $16+64=80$ ways to marry the people.

Answer:

$68$, $136$

Let us number the slices by the number of olives put on them. Note that the cutting line cannot separate the slices $1$ and $50$ (or the slices $25$ and $26$) because clearly

Hence we can assume that $n$, $n+1$, $n+25$, $n+26$ are the numbers of the slices adjacent to the cutting line, where $1\le n\le 24$. The sum of these numbers is $4n+52$. Now we compute $(n+1)+(n+2)+\cdots +(n+25)=25n+\frac{1}{2}\cdot 25\cdot 26=25(n+13)$ and $1+2+\cdots +50=\frac{1}{2}\cdot 50\cdot 51=25\cdot 51$. Then we have two possibilities:

The first possibility yields $n=4$ and the desired sum is $4n+52=68$. From the second possibility we obtain $n=21$ and $4n+52=136$. In total, there are two solutions, namely $68$ and $136$.

Answer:

$421$

If $p$ is a prime fulfilling the given condition, then there exists an integer $k>2$ with $k^3=19p+1$. Therefore we get the equation

Due to $k>2$, both factors on the right-hand side are proper divisors of $19p$. Since $19p$ is a product of two primes, we have either $k-1=19$ or $k^2+k+1=19$. The first case leads to $k=20$ and $p=400+20+1=421$, which is a prime. In the second case, the quadratic equation $k^2+k-18=0$ has no integer solution. Therefore $p=421$ is the only solution.

Answer:

$\frac{7}{30}$

In the following, we shall denote the area by square brackets. Since the triangles $\mathit{EHD}$ and $\mathit{CHB}$ are similar, the equation

follows. Due to the similarity of the triangles $\mathit{FBG}$ and $\mathit{CDG}$, we get $\frac{\mathit{DG}}{\mathit{GB}}=\frac{3}{2}$. Therefore $\mathit{DH}=\mathit{BG}=\frac{2}{5}\cdot \mathit{DB}$ and $\mathit{HG}=\frac{1}{5}\cdot \mathit{DB}$. Since

we get

Answer:

$85595$

Let $\overline{\mathit{abcde}}$ be a five-digit number such that $\overline{\mathit{abc}}=9\cdot \overline{\mathit{de}}$ and $\overline{\mathit{cde}}=7\cdot \overline{\mathit{ab}}$. Then we have

so $\overline{\mathit{de}}=\frac{1007c}{53}=19c$. Analogously we get $\overline{\mathit{ab}}=17c$. If $c\ge 6$, then the numbers $17c$ and $19c$ are greater than $100$. It follows that the maximum possible value of $c$ is $5$, i.e. the maximum possible value of $17119c$ is $85595$.

Answer:

$42$

The first man to raise his hand was the first one to see all three special cards. This man must have received one special card as his fifth card because otherwise he could have raised his hand earlier. Moreover, he must have received one special card at the very beginning because otherwise his left-hand neighbour would have seen all special cards after the previous round. Denote $C_1$ the first card, $C_3$ the fifth card, and $C_2$ the remaining special card that the man got in the meantime. After the subsequent looks at their cards, exactly the persons who encountered $C_2$ and at least one other card know or can deduce the positions of all three special cards—the positions of $C_1$ and $C_3$ are known to everyone, but their labels are not. It is easy to see that there are six such men after having seen six cards and seven after having seen seven cards, so the answer is 42.

Answer:

$\pi \frac{3-2\sqrt{2}}{4}=\pi \frac{{(1-\sqrt{2})}^2}{4}=\pi \frac{1}{4{(1+\sqrt{2})}^2}=\pi \frac{1}{4(3+2\sqrt{2})}$

If an isosceles right-angled triangle is cut into two equal triangles, then they are similar to the original one with the ratio $1:\sqrt{2}$, and so are the radii of the incircles. Thus each new incircle has half the area of the original one. In other words, the total area of the incircles is not changed when cutting the big triangle. Hence it is sufficient to determine the area of the incircle of the big triangle, radius of which we compute using equal tangents. Since the length of the legs of the big triangle equals $\frac{\sqrt{2}}{2}$ by Pythagorean theorem and the perpendiculars from the incenter to the legs form a square, we get that the radius is $\frac{\sqrt{2}}{2}-\frac{1}{2}$, so the area equals

Answer:

$7$, $11$, $19$, $79$

As $p$ is a prime, it is either equal to $11$ (which clearly satisfies the given condition) or coprime to $p+11$. In the latter case, the product in question is divisible by $p+11$ if and only if $(p+1)(p+2)$ is. This reduced product modulo $p+11$ equals $(-10)\cdot (-9)$, thus $p+11\mid 90$. This is satisfied for $p\in \{7,19,79\}$.

Answer:

$\frac{28!}{2^{14}}$

A way to dress Jim’s legs can be represented by a $28$-tuple consisting of $14$ socks and $14$ shoes in which the sock belonging to a specific leg is in a position previous to that of the shoe belonging to the same leg. For the first pair of a sock and a shoe, there are $\left(\genfrac{}{}{0.0pt}{}{28}{2}\right)$ possibilities. For the second pair, there are $28-2=26$ places, hence $\left(\genfrac{}{}{0.0pt}{}{26}{2}\right)$ possibilities. Continuing this way, there remains only $\left(\genfrac{}{}{0.0pt}{}{2}{2}\right)=1$ possibility for the last pair. Therefore the result is

Answer:

128

It suffices to plug in $x^3=8-4x$ into the given expression as follows:

Answer:

$36^{\circ }$

Let $F$ be the point on $\mathit{BC}$ such that $\mathit{AE}\parallel \mathit{DF}$. Now $\mathit{DF}=\frac{1}{2}\mathit{AE}=\mathit{CD}$, so the triangle $\mathit{FCD}$ is isosceles with base $\mathit{FC}$. Denote $\phi =\mathit{\angle BAC}$. Then

Finally, since $\mathit{\angle CAE}=\frac{1}{2}\phi$, we have (using $△\mathit{AEC}$)

so $\phi =36^{\circ }$.

Answer:

$\frac{1}{1008}$

By subtracting the recursive formulas for $n$ and $n-1$, we get $a_n=n^2\cdot a_n-{(n-1)}^2\cdot a_{n-1}$, which can be simplified to $a_n=\frac{n-1}{n+1}{a}_{n-1}$. Thus

and therefore $a_{2015}=\frac{2\cdot 2015}{2015\cdot 2016}=\frac{1}{1008}$.

Answer:

1, 9

Denote by $c_1$, $m_1$, $y_1$, $c_2$, $m_2$, $y_2$ the numbers of faces of respective colours on the dice. We know that $c_1+m_1+y_1=12$, $c_2+m_2+y_2=12$, and $y_1=4$. Moreover, from the total of $12^2=144$ possible outcomes of throwing the dice, exactly one half results in same colours, so

Expressing the left-hand side in terms of $c_1$, $c_2$, and $m_2$ using the relations above, we obtain

or

Using $-3\le c_1-4\le 3$ and $-9\le c_2-m_2\le 9$, we deduce that $c_2-m_2$ is either $8$ or $-8$, which together with $0<y_2=12-c_2-m_2$ yields that $m_2$ is either $1$ or $9$. A straightforward check shows that both these values are possible.

Answer:

32

Let us pick a woman, then the one $24$ seats to her right, then the next one $24$ seats to the right etc. After a certain number of such steps, say $s$, we get back to the original woman. This $s$ is readily seen to be the smallest positive integer such that $24s$ is a multiple of $n$. Hence $s=n∕d$, where $d$ is the greatest common divisor of $n$ and $24$.

Observe that the truthfulness of the women has to alternate, i.e. every woman is truthful if and only if the one $24$ seats to her right is a liar. If $s$ were odd, this would lead to a contradiction. Therefore $n$ has to be divisible by a higher power of $2$ than $24$ is. The smallest candidate for $n$ is $32$, and it is easy to check that this number satisfies the given conditions.

Answer:

$15^{\circ }$

Denote by $a$, $b$ the legs and by $c$ the hypotenuse of the remaining triangles, and assume $a\le b$. A straightforward computation shows that the area of the small square equals two thirds of the area of the big square. Hence the area of one of the triangles is one eighth of the area of the small square, that is

Let $\alpha$ be the smallest internal angle in the triangle. Then $a=c\sin \alpha$ and $b=c\cos \alpha$, so

or

Since $\alpha \le 45^{\circ }$, it follows that $2\alpha =30^{\circ }$, and finally, $\alpha =15^{\circ }$.

Answer:

$\frac{4}{3}\sqrt{14}$

Denote $O_1$, $O_2$, $O_3$ the centers of ${\omega }_1$, ${\omega }_2$, ${\omega }_3$, respectively, and let $T_1$, $T_2$, $T_3$ be the feet of perpendiculars from $O_1$, $O_2$, $O_3$ to the segment $\mathit{AB}$ (hence $T_1$ and $T_2$ are the points of tangency of $\mathit{AB}$ to ${\omega }_1$, ${\omega }_2$). Since $O_1{T}_1\parallel O_2{T}_2\parallel O_3{T}_3$, $O_1{T}_1=1$, $O_2{T}_2=2$, and $O_1{O}_3=2O_2{O}_3$, where $O_1$, $O_2$, $O_3$ are collinear, we easily get $O_3{T}_3=\frac{5}{3}$ (consider similar triangles). Applying the Pythagorean theorem to the triangle $A{O}_3{T}_3$, we conclude that

Answer:

84

There are clearly three two-digit numbers satisfying both given answers about divisibility by $a$ and $c$ (two corresponding to the Sphinx’ first answer and one after the correction). Moreover, both these answers have to be positive since a negative answer would result in more than three eligible two-digit numbers. Hence these numbers are multiples of $\mathrm{lcm}(a,c)=m$. It follows that $25\le m\le 33$, but only two numbers within this range are the least common multiples of two digits, namely $28=\mathrm{lcm}(4,7)$ and $30=\mathrm{lcm}(5,6)$. The latter is impossible because of $c-a\ge 2$, so we have $a=4$ and $c=7$. Now if $b$ were $5$, then there would be no two-digit number being divisible by $a$, $b$, and $c$ simultaneously. In the case $b=6$, the negative answer about $b$ corresponds to the numbers $28$ and $56$, while the positive answer yields $S=84$.

Answer:

3:20 p.m.

Since the time in question does not depend on the chosen reference frame, we may assume that the car is not moving at all. Under this assumption, the motorcycle needed one hour from where it met the car to its meeting point with the Vespa, whereas the Vespa needed five hours for the same distance, thus the motorcycle was five times faster. Similarly, one may deduce that the motorcyclist was twice as fast as the bicyclist, hence the ratio of the speeds of the Vespa and the bicycle was $2:5$.

If the Vespa needed $t$ hours to get from the car to its meeting point with the bicyclist, then the bicyclist needed $t-2$ hours. The ratio of these required times equals the inverse of the ratio of the speeds, so

or $t=10∕3$. Finally, using the fact that the Vespa met the car at 12 noon, we infer that it met the bicyclist at 3:20 p.m.

Answer:

20

If the robovac does not start in one of the corner $3\times 3$ ‘multisquares’, there always remains a non-cleaned part of the carpet—when the robovac leaves the edge of the carpet (i.e. no later than in its seventh move), it divides the remaining non-cleaned squares into two separate regions. We can routinely verify that the squares with coordinates $(1,2)$, $(2,1)$, $(2,3)$, $(3,2)$ and their symmetric counterparts in the other corners do not satisfy the given conditions for similar reasons. However, those with coordinates $(1,1)$, $(2,2)$, $(3,3)$, $(3,1)$, $(1,3)$ do. Thus there are $4\cdot 5=20$ possible starting squares in total.

Answer:

$22^{\circ }$

The inscribed angle theorem implies that $\mathit{\angle CEB}=\mathit{\angle CDB}$, so let us compute $\mathit{\angle CDB}$. Since the quadrilateral $\mathit{BCEF}$ is cyclic, we have

Moreover, $\mathit{\angle DGF}=180^{\circ }-\mathit{\angle DGB}=56^{\circ }$, so $\mathit{AD}$ and $\mathit{BC}$ are parallel. This implies $\mathit{\angle ADB}=\mathit{\angle DBC}$ and using inscribed angles we further infer $\mathit{\angle DBC}=\mathit{\angle DEC}=34^{\circ }$. Since $\mathit{\angle ABD}=90^{\circ }$ (Thales’ theorem) and trapezoid $\mathit{ABCD}$ is cyclic, we obtain

and finally

Answer:

14

Denote the couples $(a_1,a_2)$, $(b_1,b_2)$, $(c_1,c_2)$, $(d_1,d_2)$, $(e_1,e_2)$. We may w.l.o.g. assume that $a_1$ is the person who attended only two events and that the first of these events was also attended by $b_1$, $c_1$, $d_1$, and $e_1$, whereas the second one by their counterparts. Every other event could have been attended by at most two people from $b_1,b_2,\dots ,e_1,e_2$ simultaneously, so there had to be at least $12$ other events. If further $a_2$ takes part in any disjoint four of these events, we obtain the situation described in the statement, thus the smallest value of $E$ is 14.

Answer:

$678$

Since $0<a<b$, we have $b\ge 2$ and $6\cdot \overline{\mathit{bac}}>1200$, so the result obtained by Alex must be a four-digit number, say $6\cdot \overline{\mathit{bac}}=\overline{\mathit{defg}}$. On the other hand, we know that $6\cdot \overline{\mathit{abc}}=\overline{\mathit{dfeg}}$. Subtracting these two equalities, we get

Thus $540(b-a)=90(e-f)$, or $6(b-a)=e-f$. As $e$, $f$ are digits, we have $e-f\le 9$. It follows that $e-f=6$ and $b-a=1$ ($b-a>0$ because $b>a$). Substituting $b=a+1$ in $6\cdot \overline{\mathit{bac}}=\overline{\mathit{defg}}$ gives

This means that $\overline{\mathit{defg}}$ must differ from some multiple of $660$ by a non-zero multiple of 6 not exceeding $42$ ($c\ge 3$). Remembering that $e-f=6$, and considering successively $a=1,2,\dots ,7$, we determine all candidates for the number $\overline{\mathit{defg}}$: $1938$, $2604$, $3930$, $3936$, $4602$, $4608$. Only for $\overline{\mathit{defg}}=4608$ the number $\overline{\mathit{bac}}=\overline{\mathit{defg}}∕6=768$ turns out to have distinct non-zero digits. Now we can directly check that indeed $6\cdot \overline{\mathit{abc}}=6\cdot 678=4068=\overline{\mathit{dfeg}}$.

Answer:

$70$

Observe that the animals may meet only in the middle row. Furthermore, the path of each animal is completely determined by the visited squares in the middle row. It is easy to see that the animals do not meet if these middle parts of their paths do not intersect, so we are looking for the number of pairs of disjoint segments of the middle row.

Let us first handle the case when there is at least one unused square between the segments. Then the number of pairs may be computed as $2\cdot \left(\genfrac{}{}{0.0pt}{}{6}{4}\right)$, since one chooses four out of six edges of squares in the middle row and then chooses the animal which uses the segment between the first two of the chosen edges (the second animal uses the segment between the second two edges). If there is no gap between the segments, we obtain $2\cdot \left(\genfrac{}{}{0.0pt}{}{6}{3}\right)$ pairs since the segments are described by three edges.

In total, there are $2\cdot \left(\genfrac{}{}{0.0pt}{}{6}{4}\right)+2\cdot \left(\genfrac{}{}{0.0pt}{}{6}{3}\right)=70$ possible ways.

Answer:

172

Observe that $\lfloor \sqrt[3]{n}\rfloor =k$ if and only if $k^3\le n\le {(k+1)}^3-1$. Out of the $3k^2+3k+1$ numbers in this range, every $k$-th is divisible by $k$ starting with $k^3$, thus there are $3k+4$ such numbers. It remains to sum this expression for all numbers $k$ such that ${(k+1)}^3-1\le 1000$ (i.e. $k\le 9$) and add one for the number $1000$, which satisfies the given condition as well. Hence the desired result is

Answer:

281/2

Let $a$, $b$, $c$ be the roots of the given equation, which are also side lengths of a right-angled triangle. Assume w.l.o.g. that $0<a,b<c$, hence by Pythagoras’ theorem, the equation $a^2+b^2=c^2$ holds. By Vieta’s formulas (or working out $(x-a)(x-b)(x-c)$ and then comparing coefficients), we get

Squaring $15\sqrt{2}-c=a+b$ leads to $450-30\sqrt{2}c=2\mathit{ab}$. After multiplying by $c$ and substituting $\mathit{abc}=195\sqrt{2}$, we get the quadratic equation

having roots $c_1=\sqrt{2}$ and $c_2=13\sqrt{2}∕2$. Since the constraints $0<a,b<c$ and $\mathit{abc}=195\sqrt{2}$ allow only $c=13\sqrt{2}∕2$, the desired number $m$ can be calculated via

Answer:

$4$ green and $21$ red

Let $g$ be the number of green apples and $r$ the number of red apples. The statement may be rewritten as

equivalently

The latter can be viewed as a quadratic equation with variable $r$ and parameter $g$ . The discriminant

must be a square of an integer, otherwise the roots would be irrational. Due to $g\ge 2$, we obtain the inequalities

Therefore $169g^2-170g+1$ has to be equal to ${(13g-7)}^2$, which implies $12g=48$, or $g=4$. Then the roots of the quadratic equation are $-24$ and $21$, but since $r>0$, we get $r=21$ as the only solution. Hence there are $21$ red and $4$ green apples in the basket.

Answer:

$10$

We solve the problem in a more general way. Let $\mathit{ABCD}$ be a square of side length $1$ with corner $A$ in the origin of a coordinate system. Point $P$ is on $\mathit{AB}$ and has coordinates $(b,0)$ with $0<b<\frac{1}{2}$. If focus $F_1$ has coordinates $(f,f)$, then focus $F_2$ has coordinates $(1-f,1-f)$ (both foci have to lie on the diagonal $\mathit{AC}$ symmetrically with respect to the other diagonal).