Problems and solutions

Answer:

$216$

Since $6^3=216$, the side length of the cube is $6\text{m}$. The missing block does not change the surface of the cube, hence the area of the surface is $6\cdot 6^2=216{\text{m}}^2$.

Answer:

$8.75\text{m}$

The base area of one house is half of $35\text{m}\cdot 25\text{m}-300{\text{m}}^2=575{\text{m}}^2$, that is $287.5{\text{m}}^2$. Since one side length of the rectangular house is $10\text{m}$, the other side length must be $28.75\text{m}$. Therefore, we get $b=8.75\text{m}$.

Answer:

$600$

Observe that the option of wearing nothing can be viewed as an additional outfit. Considering the hats, Marcus has to choose between not wearing a hat at all, wearing the first hat, the second one, or the third one, which gives 4 possibilities in total for the hats. Similarly, there are 5 possibilities to wear the sunglasses and 6 possibilities of wearing a T-shirt. Since Marcus has to put on one of the 5 swimming trunks, in total he has $5\cdot 4\cdot 5\cdot 6=600$ possibilities to appear in an appropriate outfit.

Answer:

18 days

Let $v$ be the number of days of Laura’s vacation. Then $v-11$ is the number of days when it did not rain in the morning, and similarly $v-12$ is the number of days when it did not rain in the afternoon. Since there was no day without any rain, we see that

or $v=18$.

Answer:

$2034$

The number $n-Q(n)$ is always divisible by $9$. Since $2016$ is divisible by $9$, also $Q(n)$ and consequently $n$ have to be divisible by $9$. Clearly $n<3000$, so $Q(n)\le 2+9+9+9$, thus $n=2016+2Q(n)\le 2074$. Now the only solution $2034$ can easily be found.

Answer:

$111111111$

If $n$ is a non-zero digit, then there are exactly $10^{n-1}$ numbers starting with $n$ and fulfilling the property from the statement, for these are precisely the integers between $\overline{n0\dots 0}$ and $\overline{n9\dots 9}$. We conclude that there are

such numbers in total.

Answer:

15

A regular $n$-gon is preserved by a rotation precisely if it is by a multiple of the angle between the segments connecting two consecutive vertices with the center. This angle is $360^{\circ }∕n$, so we seek the smallest positive $n$ such that

is an integer. The result is $n=15$.

Answer:

17.06.2345

The month of a happy day either contains a zero or is 12, so either the year does not contain a zero or exceeds 3000. Let us pursue the former case. Since the leading digit of the day is one of 0, 1, 2, 3, we see that the year has to be at least 2145. However, this implies that the day is 30, which collides with the month. The second smallest possible year is 2345. We shall show that there is a happy day in that year. The leading digit of the day has to be 1, so the first possible month is 06. Finally, setting the day to be 17 is enough to complete the date of the happy day.

Answer:

$12$

There are six planes containing the faces of the cuboid, and further for each pair of opposite faces, there are two planes perpendicular to these faces and containing a face diagonal. In total there are $12$ planes.

Answer:

$9$

To get the area of $A$, we have to subtract the area of the circular segment (center $M_2$, radius $r_2$) from the area of the semicircle (center $M_1$, radius $r_1$). For the area of the segment we calculate the area of the quadrant with radius $r_2$ and subtract the area of the isosceles right-angled triangle with leg $r_2$. Using the fact that $r_2^2=2r_1^2$ (the Pythagorean theorem), we infer that the sought area is

Answer:

$6$

Only one of the answers may be correct, so there are either three or four liars among the servants. However, if there were four of them, they would have 28 legs in total, implying that the last servant did not lie–a contradiction. So, the lying servants have 21 legs altogether. If the sole truth-telling servant had eight legs, the total number would be 29, which is not among the answers. We deduce that the truth-teller had six legs (and it was the third one to report the number of legs).

Answer:

25

The price of a single chocolate bar is a common divisor of the amounts in the statement. Should the number of bars be minimal, the price has to be the greatest possible, i.e. the greatest common divisor. Since $\mathrm{GCD}(270,189,216)=27$, we infer that the total number of sold bars is

Answer:

$17$

If Anna ordered $16$ or less pieces of pastry to be taken randomly, trouble among the kids could arise: For example in case of receiving $4$ brownies, $4$ blueberry muffins, $4$ honey scones, and $4$ danish, or less pieces of any of these, there are neither five pieces of the same type nor five pieces different from each other. On the other hand, if she asks for $17$ random pieces, there might be five or more different types of pastry and the children would be happy. Otherwise there are at most four different types; however, in such a case, if there were less than five pieces of each type, there would be at most $4\cdot 4=16$ pieces in total—a contradiction. We conclude that the father suggested Anna to ask the salesgirl for $17$ randomly chosen pieces of pastry.

Answer:

$4:\pi$

Let $r$ be the radius of the circle and $a$ be the side length of the square. Since $2\mathit{\pi r}=4a$, we compute the ratio of areas as

Answer:

22:30

Let $d$ be the duration of the flight and $s$ the difference between the time in Europe and on the Cocos Islands (in hours). The statement may be rewritten as the system of equations

with the solution $d=14$, $s=5.5$. We deduce that Paul returned home at 22:30 of the Cocos Islands Time.

Note: The Cocos Islands indeed use the time zone GMT+6:30.

Answer:

70, 140, 280

Since $n$ divides $14\cdot 20=2^3\cdot 5\cdot 7$, only the primes $2$, $5$, and $7$ may occur in the factorization of $n$, with $5$ and $7$ occurring at most once and $2$ at most three times. Further, from $14\mid 20n$ wee see that $n$ is a multiple of $7$, and similarly, $20\mid 14n$ implies $10\mid n$, so $70\mid n$. It remains to conclude that all of the possible numbers $70$, $140$, $280$ fulfill the conditions from the statement.

Answer:

17

Denote by $R$, $S$ the other two vertices of the rectangle, $B$ the other endpoint of $x$, and $M$ the point on $\mathit{SR}$ such that $\mathit{SM}=\mathit{PA}=10$.

As $\mathit{PQ}=18$ and the area of rectangle $\mathit{PQRS}$ is $180+90=270$, it follows that $\mathit{PS}=\mathit{QR}=270∕18=15$. The formula for the area of trapezoid $T_2$ states that

or $\mathit{BR}=16$. Now $\mathit{BM}=\mathit{BR}-\mathit{MR}=8$ and using the Pythagorean theorem,

Answer:

$(M,F,B,V)=(3,5,6,8)$

Let us denote by $V$ the number of Valentin’s strawberries; obviously, $V\ge 6$. By analysing cases, we will show that Elisabeth cannot distribute less than $22$ strawberries. If $V=6$, there is only one possible distribution, namely $(3,4,5,6)$. In the case $V=7$, each of the possible distributions $(3,4,5,7)$, $(3,4,6,7)$, $(3,5,6,7)$, and $(4,5,6,7)$ uses a different number of strawberries, therefore Valentin, knowing the total number, can determine the distribution. Similarly, for $V=8$ or $V=9$, only the distributions $(3,4,5,8)$, $(3,4,6,8)$, and $(3,4,5,9)$ exist (with less than $22$ strawberries), each one being computable by Valentin.

On the other hand, the distribution $(3,5,6,8)$ satisfies all the conditions: Valentin and Michael cannot distinguish it from $(3,4,7,8)$, whereas Ferdinand and Benedikt can think that the distribution is $(4,5,6,7)$. It remains to show that no other distribution of $22$ strawberries complies with the requirements: From $(3,4,5,10)$, $(3,4,6,9)$, $(3,4,7,8)$, and $(4,5,6,7)$, the third one can be deduced by Benedikt and the remaining three by Valentin.

Answer:

$800$

In the first pass, all the numbers of the form $15k+1$ (for some integer $k$) are marked, starting with $1$ and ending with $991$; the following pass starts with $6$ and ends with $996$, marking all the numbers of the form $15k+6$. Finally, in the third pass one begins with the number $11$ and ends with $986$ (the marked numbers having the form $15k+11$), which is the last number to be marked. Observe that we have marked exactly all the numbers of the form $5k+1$, which comprise precisely one fifth of all the numbers on the circle. We conclude that $4∕5\cdot 1000=800$ numbers remain unmarked.

Answer:

$540^{\circ }$

Denote the tips of the star by $A,B,\dots ,G$ as in the picture; further, let $X$, $Y$ be the intersection of $\mathit{DE}$ with $\mathit{AB}$, $\mathit{AG}$, respectively.

Let $S$ be the sum in question. Since the sum of the internal angles in both quadrilaterals $\mathit{XBCD}$ and $Y\mathit{EFG}$ is $360^{\circ }$, we see that

However, $\mathit{\angle BXY}=180^{\circ }-\mathit{\angle AXY}$ and $\mathit{\angle XY}G=180^{\circ }-\mathit{\angle XY}A$, so

It follows that $S=540^{\circ }.$

Answer:

17/6

One can easily see that Lucy’s procedure is in fact the following: She picks some ordering of the given numbers, say $(a_1,a_2,a_3,a_4,a_5,a_6)$, and computes

Among all these orderings, the highest value of $S$ is achieved for the ascending ordering, since the largest number is divided by the smallest, the second largest by the second smallest etc. Similarly, the smallest value of $S$ is achieved for the descending ordering. Obviously, the largest error will occur for one of these extremal orderings. The arithmetic mean of the given numbers is $35∕6$. If the ordering is chosen ascending, we get $S=67∕8$, yielding the error $61∕24$. If the ordering is descending, we obtain $S=3$ with the error $17∕6$, which is the greater of the two and thus the sought result.

Answer:

$24\text{m}$

The triangles $E{L}_1{L}_2$ and $E{L}_1{L}_5$ are similar, since $\alpha$ and $\angle L_5{L}_{1}E$ occur as interior angles in both of them. Therefore, we get

yielding the desired distance $E{L}_1=24\text{m}$.

Answer:

$10$ and $13$

Denote by $a$ and $b$ the numbers fulfilling the given condition. As the sum of the first $17$ numbers is $153$, we have to solve the equation $153-(a+b)=\mathit{ab}$. Rearranging and adding $1$ on both sides of the equation leads to $154=\mathit{ab}+a+b+1=(a+1)(b+1)$. Since $154=2\cdot 7\cdot 11$ and $2\le (a+1),(b+1)\le 18$, the only possible factorization is $154=11\cdot 14$. Therefore, the sought-after numbers are $10$ and $13$.

Answer:

$\left(\genfrac{}{}{0.0pt}{}{14}{3}\right)=364$

Let us express the $6$-tuple as

where $0\le x,y,z<15$. The condition $a>b>c>d>e>f$ is equivalent to $x>y>z>0$, so the $6$-tuple is uniquely determined by a choice of three positive integers less than 15. It follows that there are $\left(\genfrac{}{}{0.0pt}{}{14}{3}\right)=364$ such $6$-tuples.

Answer:

03:03

The number of minutes is equal to the number of seconds once in each minute, so in 50 minutes this happens 50 times. The event when the number on the display reads the same occurs once in each minute with the units digit not exceeding 5, so this happens 30 times. There are five cases when both these events happen at once: $00:00,11:11,\dots ,44:44$. Therefore the light bulb blinks $50+30-5=75$ times (including at $00:00$) before the bomb explodes; we can disable the bomb when there are only five blinks remaining ($00:00$, $01:01$, $01:10$, $02:02$, $02:20$), i.e. when there is $03:03$ shown on the display.

Answer:

$\frac{1}{3}$

Denote by $M_1$, $M_2$, and $M_3$ the centers of the circles as in the figure and by $r_3$ the radius of the second smallest circle.

Due to the symmetry of the whole figure, $M_1{M}_2\perp M_1{M}_3$ and the Pythagorean theorem helps to find $r_3=\frac{2}{3}$ from the equation

Let $P$ be a point that completes the centers $M_1$, $M_2$, and $M_3$ to a rectangle. Let $A$, $B$, and $C$ be the intersection points of rays $M_{1}P$, $M_{2}P$, and $M_{3}P$ with the respective circles. Since $M_2{M}_1{M}_{3}P$ is a rectangle, we get $\mathit{PB}=\frac{4}{3}-1=\frac{1}{3}$, $\mathit{PC}=1-\frac{2}{3}=\frac{1}{3}$ and $\mathit{PA}=2-M_2{M}_3=2-(1+r_3)=\frac{1}{3}$. Therefore, $P$ has distance $\frac{1}{3}$ to all three points $A$, $B$, and $C$. Due to this fact, these points lie on the circle with center $P$ and radius $\frac{1}{3}$. Since $M_{1}P$, $M_{2}P$, and $M_{3}P$ are straight lines, the points $A$, $B$, and $C$ are the tangent points of the corresponding circles and the circle with center $P$ and radius $\frac{1}{3}$ is the small circle in the picture of the posed problem.

Answer:

$9$

By $n$ denote the number of people at the beginning, when Erich still was playing, and by $x$ denote the average balance of one gambler at that table. From the given statements we obtain the following two equations:

Working out these equations yields

and now we easily obtain $n=9$. So, while Erich was gambling, there were nine people sitting at the table.

Answer:

125

At the beginning, there are $7^3$ unit cubes. By removing one partition we connect two unit cubes and the number of isolated spaces within the cube decreases by one. At the end, we want to have at most $7^3-5^3$ isolated spaces (that is the number of outer unit cubes). It follows that we need to remove at least $5^3=125$ partitions. It is easy to see that $125$ is enough.

Answer:

909

Let $a^2$ be a perfect square ending with $\dots 16$. This means that $a^2-16=(a-4)(a+4)$ is divisible by $100$, so $a=2b$ and $(b-2)(b+2)$ is divisible by $25$. Thus, $b=25n\pm 2$ and consequently $a=50n\pm 4$. As $1404^2<{(1.414\cdot 1000)}^2<{(1000\sqrt{2})}^2=2000000$ and $1454^2>1450^2=2102500>2100000$, the only possibility is $a=1446$, yielding $a^2=2090916$.

Answer:

$18∕5$

Let $D$ be the midpoint of $\mathit{BC}$; then $△\mathit{ABD}$ is a right-angled triangle, so from the Pythagorean theorem, $\mathit{AD}=4$. The part of the triangle not filled with water is therefore a triangle similar to $△\mathit{ABC}$ with ratio of similitude $1∕4$. Since the ratio of the areas (of the non-filled part and the whole triangle) stays the same after rotating the triangle, an analogous similarity has to occur in the new situation, too. Therefore the surface of the water is always in $3∕4$ of the height, so it suffices to compute the height from $\mathit{AB}$. Knowing that the area of $△\mathit{ABC}$ is $\frac{1}{2}\cdot \mathit{AD}\cdot \mathit{BC}=12$, we have $h_{\mathit{AB}}=2\cdot 12∕\mathit{AB}=24∕5$. We conclude that the area filled with water is of height $3∕4\cdot 24∕5=18∕5$.

Answer:

$1,1,3,2,4,6$

Let us trace backwards the possible moves: The final state $(0,0,0,0,0,0)$, i.e. when all the peaches are eaten, can be reached only from $(1,0,0,0,0,0)$, which in turn could have emerged only from $(0,2,0,0,0,0)$ etc.—this way we construct a unique chain of distributions of peaches

ending with $(1,1,3,2,4,6)$, which is the sought-after distribution of seventeen peaches.

Answer:

26

Firstly, the distance between the first and the last up-going chair is thrice the distance of the first and the last down-going chair. Thus, we see that the time between the two moments described in the statement is twice the time from the moment the leading chairs meet till the moment the leading up-going chair meets the last down-going one. We infer that the leading chairs met at 12:13 (exactly in the middle of the lift), hence the time needed to go through the whole lift is 26 minutes.

Answer:

$100^{\circ }$

Since $\mathit{CD}=\mathit{AD}=\mathit{MD}=\mathit{ND}$, the triangles $\mathit{AMD}$ and $\mathit{NCD}$ are isosceles with bases $\mathit{AM}$, $\mathit{NC}$, respectively. Put $𝜃=\mathit{\angle DAB}$; then $\mathit{\angle ABC}=\mathit{\angle ADC}=180^{\circ }-𝜃$. On the other hand, since

we have

and

We conclude that

or $𝜃=80^{\circ }$, therefore $\mathit{\angle ABC}=100^{\circ }$.

Answer:

130

If any column of the table is monochromatic, then the neighboring column(s) must have the other color, so the next column(s) must have the same color as the first one etc., so there are two possibilities to color the table this way, according to which color is used in the first column.

On the other hand, the previous paragraph implies that if there is a column colored with both colors, all the other columns have to use both colors as well. It is easy to see that no matter how we distribute the colors between the upper and lower cells in this case, the resulting coloring will always satisfy the condition from the statement, so there are $2^7=128$ such colorings.

In total there are $2+128=130$ colorings of the table.

Answer:

196

Assume that we have a division in accordance with the problem statement. Let $m$ be the number of diamonds in the smallest pile. If $m\ge 2$, then the smallest pile can be divided into two piles of sizes $1$ and $m-1$ respectively, neither of which has the same size as some other pile; hence $m=1$.

Next, we show that the second smallest pile consists of $3$ diamonds. Since $2$ is excluded, we have to rule out the case when it consists of $n\ge 4$ diamonds. However, this is clearly not possible because of the division $n=2+(n-2)$.

Finally, we prove that if the piles $1,3,\dots ,2k-1$ ($k>1$) are the $k$ smallest piles in the division, then the $(k+1)$-th smallest pile (if it exists) consists of $2k+1$ diamonds. Let $p$ be the size of the $(k+1)$-th smallest pile. Clearly, $p$ is odd, for a pile of even size could be divided into two smaller ones of even size. If $p\ge 2k+3$, the division $p=2+(p-2)$ yields a contradiction. We conclude that the only possibility is $p=2k+1$, which is readily seen to satisfy the conditions from the statement.

It follows by induction that the number of Michael’s diamonds is of the form $1+3+\cdots +(2k-1)=k^2$. The largest square number less than $200$ is $14^2=196$, which is the largest possible number of diamonds in Michael’s possession.

Answer:

$3{!}^3∕9!=1∕1680$

Let us define three sets of three pairs each:

Note that a single game in the tournament ends in a tie if and only if two pairs from the same set out of $D_R$, $D_P$, $D_S$ are played against each other.

Possible outcomes of the whole tournament are all pairs of permutations of the set $D_R\cup D_P\cup D_S$. All the games end in a tie if and only if elements of each set $D_R$, $D_P$, $D_S$ occupy the same three positions in $P_1$’s and $P_2$’s permutation. Consider any permutation and let it denote the order of moves of the player $P_1$ in subsequent games. Number of arrangements of $P_2$’s moves yielding a draw in each game equals $3{!}^3$, no matter what the $P_1$’s permutation was. Therefore the desired probability is

Answer:

$\frac{5}{3}\sqrt{2}$

The described solid can be obtained from a cube in the following way: Each corner of the cube is cut off, the cut going through the centers of the edges adjacent to the removed vertex. The edge length of the cube is $\sqrt{2}$, so its volume is $2\sqrt{2}$. The removed corners are eight (oblique) pyramids, the base of each being an isosceles right-angled triangle with the leg of length $\sqrt{2}∕2$, and the height being $\sqrt{2}∕2$, too. Therefore the volume of one corner is $\frac{1}{3}\cdot \frac{1}{2}\cdot {(\sqrt{2}∕2)}^2\cdot (\sqrt{2}∕2)=\sqrt{2}∕24$ and the volume of the given solid is $2\sqrt{2}-8\cdot \sqrt{2}∕24=5\sqrt{2}∕3$.

Answer:

601

Observe that

and

but only the last factor is greater than $100$. Since we know that a three-digit prime factor exists and $5^4-5^2+1=601$ is clearly divisible by neither of $2$, $3$, $5$, it is a prime, and hence the sought number.

Answer:

20

Let us consider 13 cells shaded in the picture below:

Each three-cell sector contains exactly one shaded cell, so the cell of the little bee must be one of the shaded.

If it is the central cell, then there are exactly two ways to divide the rest of the honeycomb into 12 large bee sectors (the one shown in the picture and the one rotated by 60 degrees). For each of the 6 ‘middle’ shaded cells there is exactly one way to accommodate large bees in the rest of the honeycomb. Finally, for each of the boundary shaded cells there are exactly two ways to divide the remaining cells into three-cells sectors (the one shown and the symmetric one).

This gives a total of $2+6\cdot 1+6\cdot 2=20$ ways to dissect the honeycomb into the sectors as requested.

Answer:

15/8

Since $\mathit{\angle AXB}=\mathit{\angle ACB}=60^{\circ }$ and $\mathit{\angle AXC}=\mathit{\angle ABC}=60^{\circ }$, $\mathit{\angle BXT}=180^{\circ }-\mathit{\angle AXB}-\mathit{\angle AXC}=60^{\circ }$. Let $U$ be a point on $\mathit{AX}$ such that $\mathit{TU}\parallel \mathit{BX}$.

Then $\mathit{TUX}$ is an equilateral triangle and $△\mathit{TUA}\sim △\mathit{BXA}$. Therefore we have

Answer:

$\left(\genfrac{}{}{0.0pt}{}{26}{2}\right)=325$

We see that $\mathit{PA}$, $\mathit{PB}$, $\mathit{PC}$ are among the 27 rays from $P$: If not, we would obtain a quadrilateral, leading to a contradiction. Let us divide the perimeter of $△\mathit{ABC}$ into 27 segments such that each side is divided into segments of equal length; there are altogether $\left(\genfrac{}{}{0.0pt}{}{26}{2}\right)=325$ such divisions, since if we fix $A$ to be the first dividing point, we can freely choose $B$ and $C$ from the remaining 26 points. Finally, observe that each such division corresponds to exactly one shining point and vice versa: Clearly, (the rays from) each shining point gives rise to a division. On the other hand, given the numbers $a$, $b$, $c$ of the segments the respective sides are divided into, we let $P$ be the unique point inside $△\mathit{ABC}$ such that the distances of $P$ to the sides $\mathit{BC}$, $\mathit{CA}$, $\mathit{AB}$ are in the ratio $a:b:c$. A straightforward computation shows that $P$ is indeed the shining point, the rays from which divide $△\mathit{ABC}$ in accordance with the given division.

Answer:

$47$

From the prime factorization $2016=2^5\cdot 3^2\cdot 7$ we obtain $2016^2=2^{10}\cdot 3^4\cdot 7^2$. Therefore, $2016^2$ has $11\cdot 5\cdot 3=165$ positive divisors of which $\frac{1}{2}\cdot (165-1)=82$ are less than $2016$—excluding $2016$, the divisors may be divided into pairs $(x,y)$ such that $x\cdot y=2016^2$ and $x<2016<y$. Note that $2016$ has $6\cdot 3\cdot 2-1=35$ divisors less than $2016$ which naturally are divisors of $2016^2$, too. Hence, the desired number of divisors is $82-35=47$.

Answer:

$\frac{1}{2}(4033\sqrt{4033}-1)$

Observe that for $n\in \mathbb{N}$,

Therefore,

Answer:

423

The number $1$ can be reached only from $3$, which in turn can be obtained from $2$ or $9$. Further, $9$ can arise from $8$ or $27$; $2$ can be reached from $1$ or $6$, but only $6$ is admissible in the light of the problem statement. Let us construct further predecessors: Let $P_n$ be the set of positive integers such that the sequence from the statement reaches $1$ in exactly $n$ steps if and only if $a_0\in P_n$. Clearly, to establish $P_{n+1}$ for $n\ge 3$, we take $3x$ for each $x\in P_n$ and also $x-1$ for each $x\in P_n$ such that $x-1$ is not a multiple of three.

Let $p_n$ be the number of elements of $P_n$, and denote further by $f_n$, $g_n$, $h_n$ the number of elements of $P_n$ of the form $3k$, $3k+1$, $3k+2$, respectively. Observe that for $n\ge 3$, all the elements of $P_n$ are greater than $3$, and so

• $f_{n+1}=p_n$, since for each $x\in P_n$ there is $3x\in P_{n+1}$,
• $g_{n+1}=h_n$, since for each $x\in P_n$ of the form $3k+2$ there is $x-1=3k+1\in P_{n+1}$, and
• $h_{n+1}=f_n$ for similar reasons.

Therefore we have

for $n\ge 4$. The initial calculations show that $p_1=1$, $p_2=2$, and $p_3=3$, and the subsequent terms can be calculated in a straightforward manner using the recurrence above. The sought result is $p_{11}=423$.

Answer:

$74^{\circ }$

Since $\mathit{KJ}$ is a median in triangle $\mathit{BID}$, we have $\mathit{KJ}\parallel \mathit{BI}$, and similarly, $\mathit{KL}\parallel \mathit{CF}$. Thus $\mathit{\angle JKL}=\mathit{\angle IBA}+\mathit{\angle DCF}$. Since $\mathit{\angle AIE}=\mathit{\angle ABE}=90^{\circ }$, the quadrilateral $\mathit{BIAE}$ is cyclic. Consequently,

In the same way we deduce $\mathit{\angle DCF}=37^{\circ }$, therefore $\mathit{\angle JKL}=74^{\circ }$.

Answer:

133

Assume that there are exactly $a$, $b$, $c$ numbers equal to $-1$, $1$, $2$, respectively, among the James’ numbers (those equal to $0$ clearly play no role). The conditions from the statement may then be rewritten as

Our goal is to maximize $-a+b+8c=19+6c$. However, by adding the equalities, we find out that $6c=118-2b$, so $c\le 19$. The value $c=19$ can be obtained with the choice $a=21$, $b=2$, so the sought maximum is $19+6\cdot 19=133$.

Answer:

$876513240$

Denote by $A_k$ the $k$-th digit of the sought number, so the number is $\overline{A_1{A}_2{A}_3{A}_4{A}_5{A}_6{A}_7{A}_8{A}_9}$. There are $10$ possible digits, so exactly one, say $d$, is not used in the decimal representation of the desired number. Let $N_k$ be the number with $k$-th digit crossed out.

We know that $N_2$ is even, so $2\mid A_9$. Furthermore, the number $N_5$ is divisible by $5$, hence so is $A_9$. This means that $A_9=0$.

The number $N_9$ is divisible by $9$, so the digit sum of this number, namely $1+2+3+4+5+6+7+8+9-d=45-d$, is divisible by $9$, which leads to $d=9$.

Numbers $N_8$ and $N_4$ are both divisible by $4$ which means that both digits $A_7$, $A_8$ are even. In addition, the first of these numbers is divisible by $8$, so the 2-digit number $\overline{A_6{A}_7}$ is divisible by $4$. Also numbers $N_3$ and $N_6$ are divisible by $3$ and so is the digit sum of the desired number. From this we get $\{A_3,A_6\}=\{3,6\}$.

As we are looking for the largest possible number, suppose that $A_1=8$, $A_3=6$, $A_6=3$. We then have $\{A_7,A_8\}=\{2,4\}$ but since $4\mid \overline{A_6{A}_7}$, $A_7=2$ and $A_8=4$.

It suffices to check that putting the remaining digits in a decreasing order in the gaps leads to the number $876513240$ which satisfies the remaining condition, i.e. the number $N_7=87651340$ is divisible by $7$.

Answer:

25

Note that a triangle has equal area and perimeter if and only if the incircle has radius $2$. Thus triangles $\mathit{BCP}$ and $\mathit{ADP}$ are congruent; indeed, if $P$ is closer to $\mathit{AD}$ than $\mathit{BC}$, then the inradius of $\mathit{ADP}$ is smaller than the inradius of $\mathit{BCP}$. This means that $P$ lies on one of the axes of symmetry of $\mathit{ABCD}$.

Let $Q$ be the perpendicular projection of $P$ onto $\mathit{BC}$, $M$ the midpoint of segment $\mathit{AB}$ and put $x=\mathit{BQ}$, $y=\mathit{CQ}$. The area of triangle $\mathit{ABP}$ is thus $6x$, and using the Pythagorean theorem in triangle $\mathit{MBP}$, we find $\mathit{BP}=\sqrt{x^2+6^2}$. The equality of area and perimeter of $△\mathit{ABP}$ thus translates into the equation

with the only solution $x=9∕2$.

The value of $y$ can be found similarly: We have $\mathit{BP}=15∕2$ and $\mathit{CP}=\sqrt{y^2+6^2}$, so the condition on triangle $\mathit{BCP}$ implies

with the only positive solution $y=5∕2$.

We conclude that $\mathit{CP}=13∕2$ and the perimeter of triangle $\mathit{CDP}$ equals $25$.

Answer:

145

Denote $c_i$ the number $c$ used in the $i$-th step. After $n$ steps, the number $a$ (i.e. the first coordinate of the pair) will be $a=n{c}_1+(n-1)c_2+\cdots +c_n$. Fix $n$ and let $s=n{𝜀}_1+(n-1){𝜀}_2+\cdots +{𝜀}_n$, where ${𝜀}_i=1$ if $c_i=247$, and ${𝜀}_i=0$ otherwise. Define $t$ in a similar fashion with ${𝜀}_i=1$ if and only if $c_i=-118$. Clearly $a=247s-118t$, so the condition $a=0$ implies $247s=118t$, but as the numbers $247$ and $118$ are coprime, there is an integer $k$ such that $s=118k$ and $t=247k$. It follows that

and since $365=5\cdot 73$, we see that $n$ is at least $2\cdot 73-1=145$.

It remains to show that there are numbers $c_i$ such that $247s=118t$ with $n=145$. Indeed, let $m$ be the smallest positive integer such that

now put $c_i=-118$ for $i\in \{1,\dots ,m\}\setminus \{r\}$ and $c_i=247$ otherwise, where

(In fact, $m=120$ and $r=97$.) This way we get precisely

as desired.

Answer:

416

Every two zigzags can intersect in at most nine points, and for any number of zigzags, we may easily achieve the configuration when every two intersect in exactly nine points (and each point is the intersection of at most two lines). Consider placing the zigzags one by one: The $n$-th added zigzag is divided by the $9(n-1)$ intersection points with the $n-1$ already placed zigzags into $9(n-1)+1$ segments, each dividing an existing region into two. It follows the maximum number of regions definable using $n$ zigzags, $Z_n$, satisfies $Z_1=2$ and $Z_n=Z_{n-1}+9n-8$ for $n\ge 2$. The general result $Z_n=\frac{9}{2}{n}^2-\frac{7}{2}n+1$ then follows easily by induction and in particular, $Z_{10}=416$.