Problems and solutions

Answer:

$6$

If $x$ denotes the current age of Punto, then the current age of Punto’s mum is $3(x-3)+3$, which is $3x-6$. According to the second statement, Punto’s dad is currently $3x$ years old. Therefore, the difference of the parents’ ages is $6$ years.

Answer:

$30$

All the triangles in the picture have the topmost point as a vertex. Further, one side of all the triangles lies completely on one of the two parallel horizontal line segments. Therefore, each triangle is determined by a choice of one of the two horizontal segments and two distinct vertices on that segment. Since there are six vertices on each segment, the number of triangles is

Answer:

$15$

Since in an equilateral triangle each internal angle is $60^{\circ }$, we get $\mathit{\angle CBE}=90^{\circ }-\mathit{\angle EBA}=30^{\circ }$. Due to $\mathit{EB}=\mathit{AB}=\mathit{BC}$, the triangle $\mathit{BCE}$ is isosceles and we obtain

Finally, the size of the angle sought after is $\mathit{\angle DCE}=90^{\circ }-\mathit{\angle ECB}=15^{\circ }$.

Answer:

$136$

By chance Sandi might have taken all coins with labels less than 10 and nine coins each of those with labels from 10 to 19. In total, this amount of coins is $(1+2+\cdots +9)+9\cdot 10=135$. Therefore, taking out $135$ coins does not guarantee getting ten equally labelled ones.

However, if Sandi takes out $136$ coins, at least $91$ coins show labels greater than 9. By the pigeonhole principle, one of the ten types of coins with labels 10 to 19 has been taken out at least ten times. Therefore, the minimum number of coins is $136$.

Answer:

$42$

If $n$ is the number of candies bought originally, then we can rewrite the distribution of the candies via the equation

Solving for $n$ yields $n=42$.

Answer:

$24+4\sqrt{6}$

By the Pythagorean theorem, we get $\mathit{BD}=\sqrt{6^2+8^2}=10$ and using this result we obtain

Hence, the area of the quadrilateral $\mathit{ABCD}$ is

Answer:

$27$

We know that Kate wants to print on nine sheets of paper. When using the duplex capability of the printer, printing takes $9\cdot 9=81$ seconds. On the other hand, when printing the sides separately, every sheet gets printed twice so that the printing itself takes $2\cdot 3\cdot 9=54$ seconds. Hence, the manual part of the job has to take $81-54=27$ seconds.

Answer:

$784913526$

Let us arrange the digits $1,\dots ,9$ into a diagram, adding an arrow from $x$ to $y$ if the two-digit number $\overline{\mathit{xy}}$ is divisible by $7$ or $13$.

(The solid arrows stand for divisibility by $7$, the dashed ones for $13$, and the dash-dotted one for divisibility by both numbers.) It is clear from the diagram that the number in question has to begin with digits $784$. If the next digit is $9$, then we have to go on with $1$, $3$, and $5$, and the remaining two digits have to be in the order $2$, $6$. This way we obtain the solution $784913526$.

If, on the other hand, we carry on as $7842$, there are two cases two consider. Firstly, continuing to $1$ and subsequently $3$, there is no way for the number to contain both $5$ and $9$. Secondly, we run into the same issue when carrying on as $784263$, so the solution above is the only one.

Answer:

$54$

Since the triangles adjacent to the sides of square $B$ are isosceles, the side of $B$ lying on the diagonal is precisely the middle third of the diagonal. Therefore, if $s$ is the side length of the large square, the side length of $B$ is $\frac{1}{3}\cdot \sqrt{2}\cdot s$ and the one of $A$ is $\frac{1}{2}\cdot s$. Therefore, the area ratio of the inscribed squares is

so the area of square $A$ is $48\cdot \frac{9}{8}=54$.

Answer:

$0$

Fiona has $9^3=729$ white and $10^3=1000$ black unit cubes. In order to build up the surface of a cube of side length $12$ she needs $12^3-10^3=1728-1000=728$ unit cubes. As a consequence, she can compose a cube of side length $12$ with all six faces being completely white. Therefore, the answer is $0$.

Answer:

$29$

In order to find out the size of the class precisely, we would also need the information about how many students lack two of the skills for each pair of the skills. However, it is easy to see that we obtain the largest number if we assume that everyone who lacks at least two skills in fact lacks all three of them. In that scenario, the total number of pupils is

We have to subtract the number of totally unskilled pupils twice, since when adding the sizes of the three other groups, we count these pupils three times.

Answer:

$44$

Let $\mathit{ABC}$ be a triangle with $\mathit{\angle BAC}=90^{\circ }$. Connect $A$ with the midpoint $M$ of $\mathit{BC}$ to get the median, draw the altitude from $A$, and denote its intersection with $\mathit{BC}$ by $H$.

By the theorem of Thales, the vertices of triangle $\mathit{ABC}$ lie on a circle with center $M$. W.l.o.g. let $\mathit{\angle CBA}=23^{\circ }$. Since triangle $\mathit{ABM}$ is isosceles, we also have $\mathit{\angle BAM}=23^{\circ }$. Furthermore, triangle $\mathit{AHC}$ is similar to triangle $\mathit{ABC}$. This yields $\mathit{\angle HAC}=23^{\circ }$. Finally, we get $\mathit{\angle MAH}=90^{\circ }-2\cdot 23^{\circ }=44^{\circ }$ for the measure of the desired angle.

Answer:

$376$

Clearly, we have $a,b\le 20$. Adding $b$ to both sides of the given equation we get $20(a+b)=365+b$. Now the left-hand side is divisible by 20, hence the right-hand side must equal 380. This yields $b=15$, hence $a=4$ and we just need to compute $20b+19a=380-a=376$.

Answer:

$2017$

The regularity of the polygons involved plays no role and we may assume that the $2019$-gon is made from the $2018$-gon by dividing one of the edges by a new vertex. This new vertex is connected by $2016$ new diagonals to $2016$ non-neighbor vertices and there is also one new diagonal connecting its two neighbors, so in total, the numbers of diagonals increases by $2017$.

Answer:

$2$, $5$, $-6$

Since $x^2-4x+5={(x-2)}^2+1\ge 1$, the base is always a positive real number. Hence the only possibilities to fulfill the equation are the base equals 1 or the exponent equals 0. In the first case, $x^2-4x+5=1$ is equivalent to ${(x-2)}^2=0$ which yields $x=2$. In the second case, $x^2+x-30=(x-5)(x+6)=0$ has the two solutions $x=5$ and $x=-6$. All in all, there are three solutions.

Answer:

$4$

Suppose $1$ is the starting number. In order to comply with the condition about increasing, the permutation has to be $(1,4,3,2)$. But since erasing $1$ yields a decreasing sequence of numbers, this permutation does not fulfill the given conditions. Therefore, $1$ cannot be the starting number, and by symmetry it cannot be the ending number, too. Similarly, $4$ cannot be neither the first nor the last number. Hence $1$ and $4$ must be in the middle. Both orders $(1,4)$ and $(4,1)$ yield two valid permutations:

Therefore there are four permutations which satisfy the conditions.

Answer:

$\frac{15}{2}$

By the Pythagorean theorem, we get $\mathit{AC}=\sqrt{A{B}^2+B{C}^2}=\sqrt{8^2+6^2}=10$. Let $S$ be the intersection of $\mathit{AC}$ and its perpendicular bisector. Clearly, $S$ is the midpoint of the diagonal, i.e. we have $\mathit{AS}=5$.

Since $\mathit{\angle CAB}=\mathit{\angle SAX}$ and $\mathit{\angle XSA}=\mathit{\angle CBA}=90^{\circ }$, triangles $\mathit{ASX}$ and $\mathit{ABC}$ are similar, yielding $\mathit{SX}:\mathit{AS}=\mathit{BC}:\mathit{AB}$ or

Finally, the length we are looking for is $\mathit{XY}=2\cdot \mathit{SX}=\frac{15}{2}$.

Answer:

$3435$

By looking at the units digits, we immediately get $R=0$. Since $\mathit{FIV}E$ is divisible by $5$ and $R=0$, we must have $E=5$. From the hundreds digits and knowing that the digit $0$ already is taken, we conclude $O=9$ and get a carry of $1$ from the tens digits and one into the thousands digits. Therefore, $U+V$ must be greater than $10$ and $N$ has to be an odd digit greater than $1$. On the other hand, $U+V\le 7+8=15$ because the digit $9$ is already taken. This limits the possibilities for $N$ to $N=3$. Consequently, $U=6$ (because of the divisibility by $4$), $V=7$ and $F=1$.

Since $\mathit{NINE}$ is divisible by $3$, its digit sum $N+I+N+E=3+I+3+5=11+I$ must be divisible by $3$, too. Out of the three digits $1$, $4$ and $7$ which would be suitable for $I$, only $I=4$ is still available. Therefore, the answer is $\mathit{NINE}=3435$ and the cryptarithm translates into $1960+1475=3435$.

Answer:

$3+\sqrt{3}$

Denote further by $G$ the intersection of $\mathit{AC}$ and $\mathit{DE}$. Note that the two right-angled triangles having interior angles $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$, i.e. $\mathit{BEF}$ and $\mathit{GCD}$ are congruent, since they both have one side of length $1$ of the square in common. Such a right triangle is half of an equilateral triangle with its side length being the length of the hypotenuse of the triangle. Therefore $\mathit{BE}=2\mathit{BF}$ and by the Pythagorean theorem, $B{E}^2=E{F}^2+B{F}^2$. Together with the fact $\mathit{EF}=1$ we obtain $\mathit{BF}=\sqrt{3}∕3$. Therefore, the side length of triangle $\mathit{ABC}$ is $1+\sqrt{3}∕3$ and its perimeter is $3+\sqrt{3}$.

Answer:

$42$, $-42$

If $r$ is the triple root, then

Comparing the coefficients yields $3r^2=588$ or $r=\pm 14$. Hence we get $a=\pm 42$.

Answer:

$120$

Note that the square island and the rightmost island in the middle are two ’special’ islands: All bridges start in one of them, and from each ’usual’ island there are bridges to both of the special islands. So, we always go from one of the special islands to the other through one of the usual islands. Therefore, all we need to determine the journey is to order the usual islands, which can be done in $5!=120$ ways.

Answer:

$63$

Let us distinguish two cases: Firstly, assume that none of the numbers equals $2000=2^4\cdot 5^3$. Then one of the number is equal to $2^4\cdot 5^k$ for any $k\in \{0,1,2\}$ and the other one to $2^l\cdot 5^3$ for any $l\in \{0,1,2,3\}$, hence there are $24$ such pairs (counting both possible orderings). Secondly, if one of the numbers is $2000$, then the other number can be any divisor of $2000$ and there are $(4+1)\cdot (3+1)=20$ of them, so when taking the ordering into account, we obtain $2\cdot 20-1=39$ pairs; we subtract one for the pair $(2000,2000)$, which is counted twice. In total, there are $24+39=63$ pairs.

Answer:

$98\sqrt{2}$

Let $M$ be the center of the circumcircle of the given octagon. Since $\mathit{\angle AMC}=\frac{2}{8}\cdot 360^{\circ }=90^{\circ }$, the radius of the circumcircle has to be $7$ and the diameter is $14$.

Due to the area rearrangement shown in the picture we get the area by multiplying $\mathit{AC}$ with the diameter. Therefore the area sought is $14\cdot 7\sqrt{2}=98\sqrt{2}$.

Answer:

$232$

Firstly note that a triple of languages can be assigned to a single person in exactly four ways—we can equivalently choose the language the person is not going to learn. Hence there are $4^4=256$ ways of assigning the triples of languages to four people when disregarding the extra condition.

Let us now count the assignments causing that the four people have no course in common. The only possibility is that for each of the four languages, there is a person not attending the course. Hence there are $4!=24$ assignments.

We conclude that the number of assignments satisfying the given condition is $256-24=232$.

Answer:

$24,42$

Clearly, $n$ has at least two digits. If $n$ has precisely two non-zero digits $a$ and $b$, then the given statement translates into

or

Since the right hand side has to be positive, we get $\mathit{ab}<10$. Therefore we have to work out only a couple of cases obtaining that $a$, $b$ are $2$ and $4$ in some order.

Finally, if $n$ has $k\ge 3$ digits, then the left-hand side of the equality is at least ${(10^{k-1})}^2$, whereas the right-hand side is less than $1000+10^k$. As a consequence, $n$ cannot have more than two digits.

Answer:

$110$

Let $S$ be a point on $\mathit{BC}$ such that $\mathit{TS}$ is parallel to $\mathit{AB}$. Then $\mathit{\angle ETS}=40^{\circ }$ and

Since $\mathit{CT}$ is the bisector of $\mathit{\angle DCB}$, triangle $\mathit{CTS}$ is isosceles, $\mathit{\angle DCT}=\mathit{\angle TCS}$ and $\mathit{\angle DCB}=2\cdot 35^{\circ }=70^{\circ }$. Hence $\mathit{\angle CDA}=180^{\circ }-\mathit{\angle DCB}=110^{\circ }$.

Answer:

$10$

Let $m_1$, $m_2$, $w_1$, $w_2$ be the number of men and women in the houses, respectively. Since $m_1{m}_2=85=5\cdot 17$, we see that w.l.o.g. $m_1=5$ and $m_2=17$ (the possibility with one of the numbers being $1$ is easily excluded). Furthermore, there were $85+162=247$ greetings altogether, and from

we obtain $m_1+w_1=13$, $m_2+w_2=19$ (the other factorizations are again easily excluded), hence $w_1=8$, $w_2=2$ and the answer is $8+2=10$.

Answer:

$120∕17$

The Pythagorean theorem shows that the triangle is a right triangle. The segment connecting the vertex of the right angle with the centre of the circle splits the original triangle into two triangles. The radii to the tangent points are perpendicular to the shorter sides of the triangle, hence they are altitudes in the smaller triangles. Let $r$ be the radius of circle $c$. We calculate the area of the original triangle in two ways, using the known dimension of the original triangle and the smaller triangles:

which implies $r=120∕17$.

Answer:

$1-\sqrt[19]{0.5}$

Note that Margarita can play the machine unsuccessfully at most $19$ times, since after spending her last €$1$ she will have only €$0.5$ left, which is less than the machine accepts. The probability of each loss is $1-p$, therefore the probability of not winning the jackpot is ${(1-p)}^{19}$. In order to have at least $50\%$ chance of winning the jackpot, ${(1-p)}^{19}\le 0.5$ has to hold, which is equivalent to $p\ge 1-\sqrt[19]{0.5}$ and $1-\sqrt[19]{0.5}$ is the sought smallest value of $p$.

Answer:

$3435$

As $6^6\ge 10000$, none of the digits may exceed $5$. On the other hand, if all the digits were $4$, the equation would not hold, and if there were at most three $4$’s, the number would be at most $3\cdot 4^4+3^3<1000$. Therefore one of the digits has to be $5$. Since $5^5=3125$, we see that the digit $5$ has to appear exactly once, for if not, the first digit of the number would have been greater than $5$. Now $3000<5^5<5^5+3\cdot 4^4<4000$, hence the first digit of the number is $3$.

At this moment we know that the number in question is at least $5^5+3^3+2\cdot 1^1=3154$, which does not work and neither does $3155$. The next number not containing zeros and digits exceeding $5$ is $3211>5^5+3\cdot 3^3$, therefore one of the digits has to be $4$. There cannot be another $4$ since that would make the second digit exceed $5$ and working out the three remaining cases, we conclude that the only solution is $3435$.

Answer:

$16$

All the five numbers have to end with an even digit. To satisfy the non-divisibility by three, numbers ending with $0$ or $6$ have to start with one of $1$, $5$, $7$, those ending with $2$ or $8$ have to start with one of $3$, $5$, $9$, and the one ending with $4$ has to start with one of $1$, $3$, $7$, $9$. If we pick the number ending with $4$ to be $14$, then there are only two choices for the tens digit for $0$ and $6$ (either $50$, $76$, or $56$, $70$), and consequently, there are only two choices for the tens digit for $2$ and $8$ (either $32$, $98$, or $38$, $92$), hence four options in total. A similar argument applies for whatever choice of the tens digit for $4$, and since there are four digits to choose from, the total number of quintuples is $4\cdot 4=16$.

Answer:

$5439$

Let

Clearly, $f$ is a non-decreasing function of $n$. Further, $f(n)-f(n-1)=1$ if $n$ is divisible by exactly one of the numbers $5$ or $7$, and $f(n)-f(n-1)=3$ if $n$ is divisible by $35$; in all the other cases $f(n)=f(n-1)$ holds. Since

we obtain an estimate for the solution

and since $n$ is an integer, $n\ge 5436$. Now $f(5436)=2018$; the closest larger number divisible by $5$ is $5440$ and the one divisible by $7$ is $5439$, hence by the above, $f(5439)=2019$, $f(5440)=2020$, and $5439$ is the only solution.

Answer:

$6$

Since for any positive integer $a$, $S(a)$ and $a$ leave the same remainder after division by $9$, we infer from the statement that the numbers $n$, $x$, $y$, and $x+y$ leave the same remainder after division by $9$. This implies that $x$ and consequently $n$ have to be a multiple of $9$. If $n=9m$ for a positive integer $m$, then for the choice $x=y=10^m-1$ we obtain the equality from the statement. The largest digit sum for numbers at most one million is $54$, so there are six such $n$: $9$, $18$, $27$, $36$, $45$, and $54$.

Answer:

$\sqrt{5}$

Let $c$ and $d$ be the length of the longer and shorter projection of a side of a square onto a side of the box, respectively.

Then

therefore $c=2$ and $d=1$. Using the Pythagorean theorem we deduce that

Answer:

$15∕64$

Assume that the chocolate has three columns and five rows. We may view the process as follows: Paul picks a sequence of C’s and R’s of total length $(5-1)+(3-1)=6$, and based on this sequence, he eats the columns or rows of the chocolate. There are two possibilities: Either the sequence contains exactly two C’s (and four R’s), in which case only the sweeter square remains after the process, or the number of C’s or R’s is at least the number of columns or rows, respectively, which leads to the whole chocolate being eaten. In the latter case, the last step cannot consist of eating solely the sweeter square, since there are not enough columns or rows eaten to reduce the last row or column to a single square.

There are $2^6$ sequences of C’s and R’s (of length 6) in total, out of which $\left(\genfrac{}{}{0.0pt}{}{6}{2}\right)$ contain exactly two C’s, hence the sought probability is

Answer:

$477$

No prime numbers except $2$ and $5$ may end with a $5$ or an even digit, therefore each of the digits

has to appear at least on the tens position of some number in Greg’s set. Further, each of the remaining digits

has to appear at least on the units position of some number in Greg’s set. Therefore the sum is at least

On the other hand, this sum is achievable by forming the prime numbers as follows:

Answer:

1, 5

Factoring the left-hand side of the equation $T(t)-J(j)=0$ produces $(t+j-3)(t-j+5)=0$ and therefore either $\{t,j\}=\{1,2\}$ or $|t-j|=5$.

Answer:

$\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Define $E$ as refection of $B$ with respect to the midpoint $M$ of $\mathit{AC}$ and $F$ the perpendicular projection of $E$ on line $\mathit{BC}$. Then it follows from the first given condition that $\sin (\mathit{\angle EBC})=\frac{\mathit{EF}}{\mathit{BE}}=\frac{1}{2}$ and thus $\mathit{\angle EBC}=30^{\circ }$ (it cannot be obtuse because of the second given condition). Hence $\mathit{\angle ABM}=\mathit{\angle ABE}=75^{\circ }-30^{\circ }=45^{\circ }$.

From the law of sines in triangles $\mathit{ABM}$ and $\mathit{CBM}$ we compute

and

By dividing these equalities (note that $\mathit{AM}=\mathit{CM}$) and using the law of sines in triangle $\mathit{ABC}$ we finally get

Answer:

$32$

We consider four separate cases according to the symbol in the middle square and the symbol which appears five times in the table. If there is a cross in the middle square and there are five crosses in the table (we call this the case “Cross-Cross”) we must put four more crosses into the table. We can not use any whole diagonal and any whole “axis”. Consequently we end up with a pattern (Figure 1) which is not symmetric with respect to any rotation or axial symmetry and therefore this case contributes with $8$ different draw endings.

In the case “Cross-Nought” we have to place five noughts in the table with cross in the middle. We can not use all four corners since the fifth would make noughts win. On the other hand we have to put at least one nought on both diagonals, otherwise the crosses win. Hence we have to put either exactly three noughts or exactly two noughts in the corner squares and in both these subcases we have exactly one possibility to complete the table to a draw ending. Both of them are one of the $4$ rotations (both have an axis of symmetry) of one of the patterns (Figure 2 and Figure 3).

The cases “Nought-Nought” and “Nought-Cross” are analogous to the two cases discussed above and thus the total number of draw endings is $2(8+4+4)=32$.

Answer:

$32$

Taking reciprocals we obtain

The interval between $0.72$ and $0.75$ has length $0.03>1∕34$, hence for all $a\ge 34$ there is a solution. For $a=33$ there is a solution $b=24$ (giving $b∕a=0.7272\dots$). For $a=32$ there is no solution as $24∕32=0.75$ and $1∕32>0.03$.

Answer:

$68$, $76$, $91$

Let $A$, $B$, and $C$ be the three distances of the peaks from Passau measured in $\mathrm{km}$. These distances have to satisfy $2261\mid \mathit{ABC}$ and $2261\mid (110-A)(110-B)(110-C)$. Due to $2261=7\cdot 323=7\cdot 17\cdot 19$ and $7\cdot 17=119>110$ none of the distances can obtain more than one prime factor of $2261$.

W.l.o.g. $7\mid A$, $17\mid B$, and $19\mid C$. Since the same considerations apply to the distances $110-A$, $110-B$ and $110-C$, we get the two possibilities $7\mid (110-B)$ and $7\mid (110-C)$ due to $7\nmid (110-A)$. In the case $7\mid (110-B)$ we get $19\mid (110-A)$ from $19\nmid (110-C)$, and finally $17\mid (110-C)$. In the case $7\mid (110-C)$ we get $17\mid (110-A)$ and $19\mid (110-B)$ in a similar way.

Since $\mathrm{GCD}(7,19)=1$, the only way to decompose $110$ as $a\cdot 7+b\cdot 19$ with $a$, $b$ non-negative integers is $110=13\cdot 7+1\cdot 19$ (all the decompositions are of the form $110=(13+19k)\cdot 7+(1-7k)\cdot 19$ for $k\in \mathbb{Z}$ and the coefficients are non-negative only for $k=0$). Similarly, we get the decompositions $110=4\cdot 17+6\cdot 7$ and $110=4\cdot 19+2\cdot 17$. This leads to the two solutions

and

Heiko’s remark during the second stop indicates that the third peak is at least $80\mathrm{km}$ away from Passau. As a consequence, the sought distances are $68$, $76$, and $91$.

Answer:

$3\sqrt{3}∕8$

Firstly, note that $\mathit{\angle EKA}=90^{\circ }$; in fact, triangles $\mathit{AEK}$ and $\mathit{ABC}$ are congruent since $\mathit{AK}=\mathit{AC}$, $\mathit{AE}=\mathit{AB}$, and $\mathit{\angle EAK}=\mathit{\angle BAC}$. Further, the centre $S$ of square $\mathit{ABDE}$ lies on the circumcircle of triangle $\mathit{ABC}$ because $\mathit{ASB}$ and $\mathit{ACB}$ are right angles. Since $\mathit{AS}=\mathit{BS}$, the corresponding inscribed angles $\mathit{ACS}$ and $\mathit{BCS}$ are equal as well. Therefore $S$ lies on the angle bisector $\mathit{CJ}$. Let us reflect the triangle $\mathit{JKE}$ across $S$; then $E$ is mapped to $B$, $J$ is mapped to $H$, which is the intersection of $\mathit{AB}$ and $\mathit{CJ}$, and $K$ is mapped to $I$, which lies on $\mathit{CJ}$ and satisfies $\mathit{\angle IBC}=90^{\circ }$.

Now $\mathit{IBC}$ is an isosceles right-angled triangle with right angle at $B$, so its area is ${(\sqrt{3})}^2∕2=3∕2$. Denoting by square brackets the area of triangle, we have

Furthermore, triangles $\mathit{ACH}$ and $\mathit{BIH}$ are clearly similar, the coefficient of similarity being

We conclude that

Answer:

$5∕9$

Let us denote $c$ the colour the first prisoner drew and $o$ the other colour. Then the probability that he drew from box with $c,c,o$ is $2∕3$ and the probability that he drew from the box with $c,o,o$ is $1∕3$. If the second prisoner chooses the same box as the first one, he will draw colour $c$ with probability

and thus colour $o$ with probability

On the other hand if he chooses the other box, he will draw colour $c$ with probability

and colour $o$ with probability

In case $c$ is white, the second prisoner will survive if he draws $c$. Since $\frac{4}{9}>\frac{1}{3}$, it is better for him to choose the other box, and he will survive with probability $\frac{4}{9}$. In case $c$ is black, the second prisoner will survive if he draws $o$. Since $\frac{2}{3}>\frac{5}{9}$, it is better for him to choose the same box, and he will survive with probability $\frac{2}{3}$. Clearly $c$ is white with probability $\frac{1}{2}$ and black with probability $\frac{1}{2}$, hence the second prisoner will survive with probability

Answer:

$18$

For $n<18$ take the first $n$ elements of the Fibonacci sequence given by $a_1=a_2=1$, $a_{k+2}=a_{k+1}+a_k$:

Clearly the greatest number of any triple formed from these numbers is greater or equal to the sum of the two others and hence no such triple can form a nondegenerate triangle. For $n=18$, let $x_1\le \cdots \le x_{18}$ be the chosen numbers. If no three of them form a nondegenerate triangle, we get

obtaining in each step a term of the mentioned Fibonacci sequence and thus we end with $x_{18}\ge 987+1597>2019$, which is impossible. The desired number is therefore $n=18$.

Answer:

$432=2^4\cdot 3^3$

Recall that if

is the prime factorization of $n$, then

The condition that the highest common factor is not a power of two can be equivalently restated that there is an odd prime number $q$ dividing both $\sigma (n)$ and $\sigma (n^3)$. Since

this number is never divisible by $3$, therefore the smallest possible value of $q$ is $5$. Further note that $q$ cannot divide ${\alpha }_i+1$ and $3{\alpha }_i+1$ at the same time, for otherwise it would have to divide

Therefore there are distinct $i,j\in \{1,\dots ,t\}$ such that $q\mid {\alpha }_i+1$ and $q\mid 3{\alpha }_j+1$. Since we are looking for the smallest number, we may assume that $t=2$, $i=1$, and $j=2$.

If $q=5$, the smallest possible values for ${\alpha }_1$, ${\alpha }_2$ are ${\alpha }_1=4$, ${\alpha }_2=3$, and taking the smallest possible primes, i.e. $p_1=2$, $p_2=3$, we obtain $n=2^4\cdot 3^3=432$.

If $q\ge 7$, then ${\alpha }_1\ge 6$ and ${\alpha }_2\ge 2$, yielding

showing that $432$ is indeed the smallest possible value of $n$.

Answer:

$3\sqrt{5}$

Using the formulas for the segments of the hypotenuse, we compute

The radius of circle $t$ (which is equal to $\mathit{DT}$) can be computed using the well-known formula by dividing the area of the triangle $\mathit{ACD}$ by the half of its perimeter: We get that $\mathit{DT}=54∕(36∕2)=3$. Let the circle inscribed in the triangle $\mathit{BCD}$, denoted by $\omega$, touch the line $\mathit{CD}$ at $S$. Similarly, its radius is equal to $\mathit{DS}=4$. The homothety with the center at $C$ and ratio $\mathit{CT}∕\mathit{CS}=9∕8$ sends the circle $\omega$ to the circle $c$. Therefore, the radius of the circle $c$ is $4\cdot 9∕8=9∕2$. Let $M$ be the midpoint of the segment $\mathit{XY}$ and $O$ the center of the circle $c$. We know that $\mathit{XO}=9∕2$ and $\mathit{OM}=\mathit{DT}=3$. Henceforth, using the Pythagorean theorem we conclude that

Answer:

$1128$

Let us make two observation. Firstly, since the colours of any two consecutive squares are clearly different then if there are more than two colours present in a row then there are three consecutive cells coloured by three different colours in that row. Secondly, such a triple completely determines the colouring of the respective three columns, e.g. triple $1-2-3$ forces the vertically neighbouring triples to be $3-4-1$ and these two triples have to alternately fill the respective three columns making them all contain just two colours. The same holds for columns instead of rows as well. Hence there cannot be more than two colours used in some row and more than two colours used in some column at the same time.

Assume now that the table has $6$ rows and $7$ columns. Let us compute the number of colourings using only two colours in every row: We choose the two colours for the first row (then the pair of colours for any other row is also determined) and then choose the starting colour in each of the $6$ rows. This yields $\left(\genfrac{}{}{0.0pt}{}{4}{2}\right)\cdot 2^6=6\cdot 2^6$ colourings. Analogously the number of colourings using only two colours in every column is $6\cdot 2^7$. Finally we have to subtract the number of colourings using only two colours in every row and every column. Since such a colouring is determined by the upper left $2\times 2$ square there are $4\cdot 3\cdot 2=24$ of them.

The result is $6\cdot 2^6+6\cdot 2^7-24=1128$.

Answer:

$597+3^{-99}$

All weights are in grams. After the first step, the second child has $8+4∕3$. In the second step, it gives one third of their chocolate to the third child which has $12+8∕3+4∕3^2$ then. After the third step, the fourth child has $16+12∕3+8∕3^2+4∕3^3$. It is easy to observe that in the end, the $100$th child has

Let us denote

Then we have

Using the sum of geometric series,

we get

Finally we obtain

Answer:

$3,4,5,6,8,14$

We would like $n$ to satisfy ${(n-2)}^2\mid {(n-1)}^{n-1}-n^2+2019\cdot (n-1)$. This is not influenced by adding ${(n-2)}^2$ to the right-hand side and it allows us to get rid of the term $n^2$, since $n^2-4n+4-n^2=-4(n-1)$. We obtain an equivalent condition