Problems and solutions

Answer:

$130$

It is just the area of the sliding window overlapping the fixed part, which is a rectangle $10\text{cm}\times 13\text{cm}$.

Answer:

$42$

Looking at the picture we notice that the perimeter of the large rectangle equals the sum of the perimeters of the four outer small rectangles with given perimeter minus the perimeter of the middle one. Therefore, the answer is

Answer:

$6301$

Since the last digit of a (four-digit) prime number cannot be even, the prime number was of the form $630\ast$. Moreover, the last digit cannot be $5$, because the number would be divisible by $5$. So the options $1$, $3$, $7$ and $9$ are left. But since $630$ is divisible by $3$, the digits $3$ and $9$ are impossible. Similarly, $630$ is divisible by $7$ and so would be $6307$. Therefore, the number was $6301$.

Answer:

$\frac{41}{70}$

Since we are only interested in the fraction of areas, we can use $5\text{m}$ as a unit. By summing up the areas of the three right-angled triangles, we obtain the result

Answer:

$4$

It is necessary to remove four dots, because the four $1\times 1$-squares in the corners of the given grid do not share any dots in common. Removing two opposite corners of the grid and two centre dots along the other diagonal provides an example to show that this number is sufficient.

Answer:

$5$

The digits in the unit position of square numbers show up periodically having period 10. We have

so the last digit is $5$. Consequently, the last digit of $1^2+2^2+\cdots +2010^2$ is $5$ due to $201\cdot 5=1005$. Furthermore, the last digit of the sum $2011^2+2012^2+\cdots +2017^2$ is $0$. Altogether, the desired digit is $5$.

Answer:

$\frac{11}{12}$

The given quotient can be written in the following way:

Answer:

$163$

We denote the number of trains between Linz and Waldkirchen by $r$, the one between Linz and Regensburg by $w$, and the one between Waldkirchen and Regensburg by $l$. Anna counts all the trains between Linz and the other two cities. Therefore, we obtain the equation $r+w=190$. We get the equations $l+w=208$ and $l+r=72$ similarly. Adding the first two equations and subtracting the third one gives $2w=190+208-72$ which leads to $w=\frac{1}{2}\cdot 326=163$.

Answer:

$256$

Assume that $x$ is equal to $a^4$ for some even integer $a$ and that it can be transformed to $b^4$ for some odd integer $b$.

Since $10000=10^4$, both $a$ and $b$ are less than $10$. By squaring even one-digit integers and then squaring their last digits, we find out that $a^4$ always ends with $6$. Considering the possible values of $b$, we see that only $5^4=625$ and $9^4=6561$ contain the digit $6$. However, any permutation of digits of $9^4$ would be divisible by $3$ (digit sum does not change), so the only possible result would be $6^4=1296$, which cannot be permuted to $9^4$. For $b^4=625$ we easily find $x=a^4=256$ as the only result.

Answer:

$1:7$

The triangles $\mathit{EBF}$ and $\mathit{CDF}$ are similar with a scale factor $\mathit{EB}:\mathit{DC}=1:6$. As a consequence $\mathit{BF}:\mathit{FD}=1:6$ as well, hence $\mathit{BF}:\mathit{BD}=1:7$.

Answer:

$200$

Since the maximum number of people in a flat is three, exactly three people live in each flat from No. 51 to No. 100. Therefore, there are $56-2\cdot 3=50$ people living in the first up to the $50$th flat in total. This leads to a total number of $50+150=200$ inhabitants for the entire house.

Answer:

$3\cdot 2^{1008}$

It is easy to see that in each stage, the red number is greater than the green one. Assuming that in the $n$-th stage the numbers $R_n$, $G_n$ are written with red and green, respectively, then in the stage $n+1$, we have $R_{n+1}=R_n+G_n$ and $G_{n+1}=R_n-G_n$ and in the stage $n+2$,

Therefore, both the numbers are doubled each two stages. Between the 1st and the 2017th stage this happens 1008 times, therefore the result is $3\cdot 2^{1008}$.

Answer:

$124$

By the Pythagorean theorem, the length of the diagonal of the rectangle is $\sqrt{60^2+80^2}=100$. For example by the cathetus theorem, one can determine the shorter segment cut by the altitude to the diagonal as $60^2∕100=36$. Now we obtain $100-36=64$ for the longer segment of the diagonal. Using the altitude theorem, the length of the altitude is $\sqrt{36\cdot 64}=48$. Altogether, the zigzag path is of length $48+(64-36)+48=124$.

Answer:

$8$

Since all digits of $2017$ are different, all digits $a$, $b$, $c$, $d$ are distinct and we should erase each letter exactly once. Observe that after erasing, either the first or the last digit of $2017$ will be equal to $a$. In the former case we have $a=2$ and arrive at an analogous problem for a six-digit palindrome with digits $0$, $1$, $7$. In the latter case we have $a=7$ and arrive at an analogous problem for a six-digit palindrome with digits $2$, $0$, $1$. Again, in both cases we have two choices for the first digit, so we obtain four problems for a four-digit palindrome. Again, each of these problems splits into two problems for a two-digit palindrome, which has a unique solution. Hence, overall we have $2^3=8$ options.

Answer:

$9$

For a positive integer $n$ consisting of only one digit, we always have $S(n)+P(n)=2n>n$. Now we consider positive integers $n$ having more than one digit. Let $m\ge 1$ and $n=a_{m}10^m+\cdots +a_0$ be an integer with $0\le a_k\le 9$ for $0\le k\le m$ and $a_m\ne 0$. Then we have

and equality can hold only for $m=1$. Therefore, for $n$ to satisfy the given condition we must have

which is equivalent to $a_1(9-a_0)=0$, that is to $a_0=9$. We conclude that exactly the nine numbers $19$, $29$, $39$, $49$, $59$, $69$, $79$, $89$, and $99$ have the desired property.

Answer:

$19$

Denote by $x$, $y$, $z$ the number of work team leaders, labourers, and part-time workers, respectively. The conditions from the statement can be rewritten into a system of equations

in positive integers. Expressing $z$ from $(2)$ yields $z=2000-20y-100x$ with the right-hand side divisible by $20$, so $z$ can be written as $z=20k$ for some positive integer $k$. Therefore, $(2)$ can be simplified to $5x+y+k=100$. Subtracting $(1)$, wherein $z=20k$ is used as well, leads to $4x=19k$. As $4$ and $19$ are coprime, it follows that $x$ is a multiple of $19$. Since $x$, $y$, and $z$ are positive, $(2)$ implies $x<20$. Consequently, $x=19$ (with $y=1$ and $z=80$) is the only solution.

Answer:

$51$

Firstly, consider the dark grey triangle being rotated by $30^{\circ }$ about the center of the circle such that its vertices coincide with vertices of the hexagon. Then it is clear that it covers half of the hexagon.

Secondly, observe that a striped triangle is equilateral having the same side length as the hexagon. Therefore, the area of a striped triangle is $\frac{1}{6}$ of the area of the hexagon. Altogether, the area of the dark grey triangle is $3\cdot 17=51$.

Answer:

$15$

By starting to place the tiles along the narrow strips, it is obvious that the way of tiling is determined up to the $3\times 5$-area:

There are three options for the next tile: Either it can be put in the same direction as the last one (case (1) below), which produces two separated $3\times 2$-areas, or it can be put perpendicularly (cases (2) and (3)). In these cases we may carry on tiling in a unique way until all but one $3\times 2$-area remains.

Each $3\times 2$-area can be tiled in three ways as in the following picture:

Therefore, there are $3\cdot 3=9$ ways to finish the tiling in the case (1) and $3$ ways in both cases (2) and (3). Altogether, we get $9+3+3=15$ possibilities.

Answer:

$2021$, $10085$

For the chosen divisor $d$ of $n$, we have $n=\mathit{kd}$ for some integer $k$. Now the equation reads

Since $2017$ is a prime number, we obtain either $d=1$ or $d=2017$. In the first case, we get $n=k=2017+4=2021$. In the second case, we obtain $k=5$ and hence $n=2017\cdot 5=10085$.

Answer:

$60$

Once Grisha has taken one of the five places, there are only $2$ positions left to seat Vanechka in order not to be next to Grisha. This yields $5\cdot 2$ possibilities. For each of them, the three remaining students can be seated in $3!$ ways, which leads to $5\cdot 2\cdot 3!=60$ ways in total.

Answer:

$73^{\circ }$

Denote the point of intersection of the lines $\mathit{AC}$ and $\mathit{DM}$ by $S$ and the point of intersection of the lines $\mathit{AC}$ and $\mathit{BD}$ by $T$.

From $\mathit{\angle MAC}=56^{\circ }$ and the given right angle at $S$, we get $\mathit{\angle SMA}=34^{\circ }$ using the sum of angles of triangle $\mathit{AMS}$. Further, $\mathit{\angle BMD}=180^{\circ }-\mathit{\angle SMA}=146^{\circ }$. As both $\mathit{MD}$ and $\mathit{MB}$ are radii of the circle, $△\mathit{BDM}$ is isosceles ($\mathit{\angle MBD}=\mathit{\angle MDB}$) and using the sum of angles again results in $\mathit{\angle MDB}=17^{\circ }$. Our goal is to find $\mathit{\angle CTB}$, which equals $\mathit{\angle STD}$; however, this can be computed using the sum of angles of the right triangle $\mathit{DST}$ with the knowledge of $\mathit{\angle SDT}=\mathit{\angle MDB}$. The result is $\mathit{\angle CTB}=73^{\circ }$.

Answer:

$488$

Let $f_n$ denote the length of the thick boundary in the $n$-th stage. Note that in each stage three congruent figures from the previous stage are glued together along two unit segments which are no longer parts of the boundary. This observation yields $f_{n+1}=3\cdot f_n-2\cdot 2$ for all $n\ge 1$ which combined with $f_1=4$ gives $f_6=488$ after direct computation.

Answer:

$1345$

Firstly, observe that there are no villains sitting next to each other: if that was the case, another villain would have to sit on either side, next to this one another one etc., implying that no superheroes are present, which is not true because the statement says that there is at least one. Assuming first that all superheros tell the truth, every superhero sits between another superhero and a villain, so the whole party at the table can be divided into segments of the form superhero-superhero-villain. The lying superhero can now be either seated between two superheroes, or inserted together with one more villain between a superhero and a villain. The remainder of the division of the total number of people by three would be $1$ in the former case and $2$ in the latter case. Since $2017$ gives remainder $1$ when divided by $3$, the former case applies and we conclude that there are $\frac{2}{3}\cdot 2016+1=1345$ superheroes at the table.

Answer:

$\sqrt[2016]{2017}$

Since $x$ is positive, we may raise both sides to the power $1∕x$ and obtain $x^{2017}=2017x$ or $x^{2016}=2017$. The solution is $x=\sqrt[2016]{2017}$.

Answer:

$10$

A regular dodecahedron has 20 vertices and at every vertex 3 edges meet. Now consider only the graph model consisting of nodes (vertices) and edges of the dodecahedron. After each step of colouring a path of connected edges with one colour, remove those coloured edges from the model. If a closed path is taken out, at each node either 0 or 2 edges are removed. If the beginning node $b$ and the ending node $e$ of the removed path are different, exactly one edge is taken out at the nodes $b$ and $e$ whereas at the remaining nodes 0 or 2 edges are eliminated. Therefore, each node has to be a starting/ending node of at least one path and so there must be at least 10 paths (colours).

The following picture shows that a colouring using 10 colours exists:

Answer:

$24$

The area of the path can be subdivided into three trapezoids having height $2$ and midline $a$, $b$, and $c$, respectively.

Since the area of a trapezoid can also be computed by multiplying its height by its midline, we get

as the result for the area of the path in ${\text{m}}^2$. The amount of gravel to be ordered can now be computed as $600{\text{m}}^2\cdot 0.04\text{m}=24{\text{m}}^3$. So the answer is $24$.

Answer:

$7744$

Let $N$ be the number in question and denote by $x$ and $y$ its first and last digit, respectively. Then we have

so $N$ is divisible by $11$. But $N$ being a square number, it also has to be divisible by $11^2$. Thus we get $N=11^2{k}^2$ for some positive integer $k$ and $100x+y=11k^2$. Since the left-hand side of this equality is a three-digit integer $\overline{x0y}$ with the tens digit being zero, $k^2$ has to be a two-digit square number whose digits sum up to $10$. The only number of this kind is $8^2=64$. Therefore, we have $100x+y=11\cdot 8^2$ and obtain $N=11^2\cdot 8^2=88^2=7744$.

Answer:

$\frac{51}{58}$

If Johanna puts exactly one white ball each in three of the four boxes, she wins in any case if she chooses one of these boxes. Given the distribution $(1,0),(1,0),(1,0),(15,14)$ for $(w,b)$, she wins with probability $\frac{1}{4}(1+1+1+\frac{15}{29})=\frac{51}{58}$.

It is easy to see that every other distribution leads to a smaller probability for a win: the probability of a loss is the sum of probabilities of choosing each single black ball and for a single ball, the probability to be chosen is clearly minimal if it is in a box with as many balls as possible.

Answer:

$80$

Let $t$ be the common time interval between the moments the vehicles pass the observers. The motorcycle reaches the first observer $t$ hours after the truck and reaches the second observer $t$ hours before the truck. Therefore, the motorcycle covers the distance between the two observers $2t$ hours faster than the truck does. The motorcycle is travelling twice as fast as the truck, so it covers the distance between the two observers in half the time. Consequently, the truck must take $4t$ hours to cover the distance between the observers while the motorcycle takes $2t$ hours.

Now the bus passes the first observer $t$ hours ahead of the truck and it passes the second observer $2t$ hours ahead of the truck. The truck takes $4t$ hours to go between the observers, so the bus takes $3t$ hours to go the same distance. Consequently, the bus is going $\frac{4}{3}$ the speed of the truck and this is $80\text{km/h}$.

Answer:

$50$, $50$, $60$

Clearly, there cannot be more than ten digits in the statement. Therefore, using the digits we already know, we observe that the first two gaps should be filled with at least $20$ and the third one with at least $40$. Thus there are altogether ten digits and all numbers in the gaps should end with $0$, so we can denote them by $\overline{a0}$, $\overline{b0}$, and $\overline{c0}$, respectively, with digits $a$, $b$, and $c$ such that $a+b=10$.

Because there are at least two digits greater than $4$ and at least five digits less or equal $4$, we can conclude that $5\ge a\ge 2$. Moreover, at least one of the digits $a$ and $b$ is greater than $4,$ so we get $5\ge a\ge 3$.

Having already four digits equal to either $4$ or to $5$ means $c\ge 4$. Since the definition of $\overline{c0}\%$ excludes the case $c=4$, we must have $c\ge 5$ which yields $5\ge a\ge 4$. In case $a=4$, we get $b=6$, but then for every $c\ge 5$, it is not possible to obtain a true statement as requested. Now let $a=5$. This leads to $b=5$ and the statement is true for $c=6$.

Answer:

$30\sqrt{\frac{2}{7}}$

Since triangle $\mathit{PBC}$ has greater area than $\mathit{ABP}$, the endpoint of the new fence on side $\mathit{AC}$ (denoted by $X$) lies between $C$ and $P$, and the other endpoint (denoted by $Y$) lies on side $\mathit{BC}$. Triangles $\mathit{PBC}$ and $\mathit{XY}C$ are similar, so $\mathit{XY}:\mathit{XC}=\mathit{PB}:\mathit{PC}$.

Further, as the area of triangle $\mathit{XY}C$ is one half of the area of triangle $\mathit{ABC}$,

or

therefore

Answer:

$-144$

Denote the initial numbers by $a\le b\le c\le d\le e$. Among the ten calculated sums, the smallest is $a+b=1$, the second smallest is $a+c=2$, the largest is $d+e=10$, and the second largest is $c+e=9$. Since all ten sums add up to

we get $a+b+c+d+e=14$ and $c=14-1-10=3$. It follows that $a=2-c=-1$, $b=1-a=2$, $e=9-c=6$, and $d=10-e=4$. For these values we can easily check that the remaining six sums exactly fit into the list of numbers on the blackboard. Therefore, the desired product is $-1\cdot 2\cdot 3\cdot 4\cdot 6=-144$.

Answer:

$3^2+5^2+7^2+9^2+13^2$

Since $17^2=289<333<361=19^2$, only the numbers $1^2,3^2,\dots ,17^2$ (altogether nine distinct numbers) may appear as summands. Further, whenever an odd integer is squared, the result gives remainder $1$ when divided by $8$—since $333$ has remainder $5$ after dividing by $8$, the number of summands in the desired sum has to be five.

Since $1^2+3^2+5^2+7^2+17^2>333$, $17^2$ cannot appear in the sum. Let us take a look at the leftover summands modulo $5$: Two of them are divisible by $5$ ($5^2$, $15^2$), three of them give remainder $1$ ($1^2$, $9^2$, $11^2$) and three give remainder $-1$ ($3^2$, $7^2$, $13^2$). Since $333$ gives remainder $3$ (or $-2$), there are two cases to consider. Firstly, we may sum up all the numbers with remainder $0$ or $1$; however, this turns out to exceed $333$. Secondly, to achieve $-2$, we have to sum all the numbers with remainder $-1$, one with remainder $0$ and one with $1$. It is easy to see that the results containing $11^2$ or $15^2$ are too large, and out of the two remaining possibilities, only $3^2+5^2+7^2+9^2+13^2$ is equal to $333$.

Answer:

$4$

From $0=0\odot u=\mathit{bu}$ we get $b=0$ ($u$ is non-zero). The equations in the statement now may be rewritten as

with solution $a=5$, $c=-1$. Finally $1=1\odot u=5-u$ yields $u=4$.

Answer:

$\frac{1+\sqrt{5}}{2}$

Denote certain points of contact by $A$, $B$, and $T$ and centers of circles by $X$, $Y$, and $Z$ as shown in the following picture:

Applying the Pythagorean theorem, we get

As $\mathit{BY}\parallel \mathit{TZ}$ and $\mathit{BY}=\mathit{TZ}=r$, the quadrilateral $\mathit{BY}\mathit{ZT}$ is a parallelogram and $\mathit{BT}=YZ=2r$. Since the Pythagorean theorem in triangle $\mathit{ABT}$ yields $A{B}^2+A{T}^2=B{T}^2$, we now obtain $4r+2^2={(2r)}^2$, which reduces to

The only positive solution of this quadratic equation is $r=\frac{1+\sqrt{5}}{2}$.

Answer:

$72$

Let $a_1,\dots ,a_5$ be the numbers on the wall. We are seeking the smallest possible value of

where $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$. We can rewrite this expression as

where $\{x\}$ stands for the fractional part of $x$ (i.e. $\{x\}=x-\lfloor x\rfloor$). Therefore our goal is to maximize the sum

This can be further split into two sums

(letting $a_6=a_1$ and $a_7=a_2$). Now the equalities

imply that both $S_1$, $S_2$ are integers. Further, they are both sums of five numbers less than $1$, and so both are at most $4$; consequently, $S\le 8$. For the original sum we thus obtain $W\ge 72$ and the equality is achievable by letting all the fractional parts of $a_1,\dots ,a_5$ be $0.4$, for example $a_1=2.4$, $a_2,\dots ,a_5=4.4$.

Answer:

$864$

The smallest divisor of $n$ is 1. Let the second and third smallest divisors of $n$ be $p$, $q$ respectively (so $p$ is a prime and $q$ is either a prime or $q=p^2$). Then $\xi (n)=1+p+q$ and $𝜗(n)=n+n∕p$. So after multiplying by $p$, the condition from the statement can be rewritten as

The right-hand side is divisible by $p+1$ and since $p$ and $p+1$ are coprime, $p+1\mid {(1+p+q)}^4$. If both $p$, $q$ were odd, $p+1$ would be even and ${(1+p+q)}^4$ odd, which is not possible, therefore $p=2$ ($2$ has to be a divisor of $n$ and $p$ is the smallest prime divisor). Now expanding ${(1+p+q)}^4={(3+q)}^4$ via the binomial formula and omitting the terms divisible by $3$ shows that $3\mid q^4$, therefore $3\mid q$. The option $q=p^2$ being clearly impossible, we conclude that $q$ is a prime and so $q=3$. Finally, $n\cdot 3=2\cdot 6^4$, so $n=2^5\cdot 3^3=864$.

Answer:

$51$

Firstly, let us consider the case when we fill the centre square. In such a case the mouse can reach the cheese only by moving along one of the “border corridors”, so we may place further obstacles to only one of these corridors. Each corridor consists of three squares, thus there are $2^3=8$ ways to fill one of them and $8+8-1=15$ ways to leave at least one corridor free (the $-1$ is due to the situation with both corridors free being counted twice).

Let us now focus on the case with the centre square free. The path to cheese exists if and only if at least one of the squares adjacent to the mouse and at least one adjacent to the cheese is free. This gives three options for the two squares adjacent to the mouse and three options for the two squares adjacent to the cheese (i.e. at most one filled square) and two options for each of the remaining two corners, so the number of placements plausible for the mouse in this case is $3\cdot 3\cdot 2\cdot 2=36$.

We conclude that there are $15+36=51$ ways in total.

Answer:

$-3$

Firstly, we observe that if we add $10$ to both sides, the binomial $x^2+1$ can be factored out from the left-hand side, so

The expression on the left-hand side can be viewed as a product of two quadratic functions $f(x)=x^2+1$ and $g(y)=y^2+6y+10$. Now since the graphs of quadratic functions are symmetric and in this case, both functions are first decreasing and then increasing, the minimal value $x_0$ gives the maximal value of $f$, which in turn forces $g(y_0)$ to be minimal possible (the product $f(x)g(y)$ is a positive constant). So it suffices to find where $g(y)={(y+3)}^2+1$ attains its minimum and we conclude that the answer is $y_0=-3$.

Answer:

$\frac{4\sqrt{3}}{11}$

Let $P$ be a point on $\mathit{BC}$ such that $\mathit{DP}\parallel \mathit{AC}$. As the triangles $\mathit{ABC}$ and $\mathit{DBP}$ are similar, $\mathit{AC}:\mathit{BC}=\mathit{DP}:\mathit{BP}$.

Since $\mathit{CE}$ is the angle bisector in triangle $\mathit{BCD}$, the Angle bisector theorem implies $\mathit{CD}:\mathit{CB}=\mathit{ED}:\mathit{EB}=8:15$. Further, $\mathit{DP}=\mathit{CD}\cdot \sin 60^{\circ }$ and $\mathit{BP}=\mathit{CB}-\mathit{CP}=\mathit{CB}-\mathit{CD}\cdot \cos 60^{\circ }$, therefore

Answer:

$232$

Denote by $N_k$ the maximal $N$ such that with $k$ Euros Mike can always find the integer from the range $1,\dots ,N$ (or, equivalently, any set of integers of length $N$); our goal is to compute $N_{10}$. Clearly $N_0=1$ and $N_1=2$. Let us show that the numbers satisfy the recurrence relation

Indeed, if Mike has $k+2$ euros and picks the number $N_{k+1}+1$, then either it is the correct one, or it is too large (in which situation he proceeds with $k+1$ money and a set of length $N_{k+1}$) or too small (leading to $k$ money and $N_k$ choices). This shows $N_{k+2}\ge N_{k+1}+N_k+1$. On the other hand, if there are more than $N_{k+1}+N_k+1$ numbers to choose from, then selecting a number greater than $N_{k+1}+1$ may lead to more than $N_{k+1}$ choices but only $k+1$ money left (too large choice) and similarly, selecting a number less than $N_{k+1}+2$ may result in more than $N_k$ choices but only $k$ money left (too small choice). Thus the recurrence relation is proved.

The number $N_{10}=232$ now may be computed via the recurrence in a straightforward way.

Answer:

$134$

Observe that if we multiply the left-hand side by $2$ and add $1$, we get

The same operation applied on the right-hand side yields $2\cdot 2017+1=4035$. Factorising into primes gives $4035=3\cdot 5\cdot 269$, and since $a\ge b\ge c\ge 1$, the equalities $2a+1=269$, $2b+1=5$, and $2c+1=3$ must hold. Therefore the result is $a=134$.

Answer:

$14$

Denote lower ends of the legs by $A$, $B$, $C$ and their upper ends by $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, respectively. Let $O$ be the center of triangle $\mathit{ABC}$ (i.e. the point of tangency the ball to the ground), $S$ be the center of the spacecraft and $D=\mathit{AO}\cap \mathit{BC}$. We have $\mathit{AO}=\mathit{AB}\cdot \frac{\sqrt{3}}{3}=3\sqrt{3}\text{af}$, $A{A}^{\prime }=1\text{af}$ and $\mathit{SO}=S{A}^{\prime }=R$. Pythagorean theorem yields

from which we get $R=14\text{af}$.

Answer:

$1021$

Let us add three additional, “fake” nuts at the beginning of the night. Then the number of all nuts is a multiple of four. It is enough to observe that after Allan’s supper the number of the remaining nuts in the bag is also a multiple of four and there are still the three fake nuts among them. By repeating this reasoning we get that the number of nuts with the fake ones included is equal to $4^{5}k$ for some positive integer $k$, so the actual number of nuts is $4^{5}k-3$. Plugging $k=1$ gives the desired minimum $4^5-3=1021$.

Answer:

$19$

We begin with a general observation: If $x$ is a real number with $x>1∕2$, then there is exactly one (integer) power of $2$ in the half-open interval $[x,2x)$.

Further, call a positive integer strongly convenient if all its prime divisors are among $2$ and $3$. The number of strongly convenient numbers in the interval $[x,2x)$ can be computed as follows: Firstly, there is exactly one power of two in the interval $[x,2x)$. Further, in the interval $[x,2x)$ there is at most one strongly convenient number (say $c_1$) divisible by $3$, but not by $3^2$, for $c_1∕3$ is the only power of two in the interval $[x∕3,2x∕3)$, provided that $x>1∕6$. This way we proceed and find a strongly convenient number $c_2$ divisible by $3^2$ but not by $3^3$ etc., until $[x∕3^k,2x∕3^k)\cap \mathbb{N}=\varnothing$, i.e., $x<3^{-k}∕2$. We infer that the number of strongly convenient numbers in the interval $[x,2x)$ is $1+$the greatest $k$ satisfying $3^k<2x$; denote this number by ${\ell }_3(x)$.

To obtain the number of convenient numbers in the interval $[x,2x)$, we employ a similar technique: The sought number is the sum of numbers of strongly convenient numbers in the intervals $[x,2x)$, $[x∕7,2x∕7)$, $[x∕7^2,2x∕7^2)$ etc. Thus the quantity can be computed as ${\ell }_3(x)+{\ell }_3(x∕7)+{\ell }_3(x∕7^2)+\dots$, the sum having ${\ell }_7(x)$ terms, with the symbol ${\ell }_7$ defined in a similar fashion as ${\ell }_3$.

Finally, as $2000$ is not a convenient number, our task is to compute

which is the desired quantity.

Answer:

$321$

Observe that all $k$-digit numbers (in Decimus notation) are represented exactly by the strings

Therefore, to count the total number of $D$’s occurring in the $k$-digit numbers, we can group together the numbers having $D$ on the first position, the second etc. up to the $k$-th position. There are exactly $10^{k-1}$ such numbers for each position—we have to fill the remaining $k-1$ positions with symbols $1,\dots ,9,D$. Since for each number we count all occurrences of $D$, we may distribute the ones with multiple occurrences into all corresponding groups and the result stays correct, so there are $k\cdot 10^{k-1}$ digits $D$ among the $k$-digit numbers.

It is easy to see that only numbers consisting of one, two, or three digits appear in the list $1,\dots ,\mathrm{DDD}$. We conclude that among them the digit $D$ appears $1\cdot 10^0+2\cdot 10^1+3\cdot 10^2=321$ times.

Answer:

$21$

By the (limit case of the) Inscribed angle theorem, $\mathit{\angle EXF}=\mathit{\angle EY}X$. Consequently, triangles $\mathit{FEX}$ and $\mathit{FY}X$ are similar (they also share the angle at $F$), therefore $\mathit{EF}:\mathit{XF}=\mathit{XF}:YF$ or $X{F}^2=\mathit{EF}\cdot YF=5\cdot 12=60$ (a fact also known as the Power of a point theorem).

Let $t=\frac{1}{2}\mathit{AB}$ be one half of the side length of the square. Then the Pythagorean theorem gives

hence the sought area of triangle can be computed as

Answer:

$60724$

Since in the given cryptogram ten different letters show up, all ten digits $0,1,\dots ,9$ are involved. The sum of all digits is $45$. By adding up the three given digital sums, we get $11+23+19=53$, which is equal to $45+E$. This yields $E=8$, and by squaring it, $J=4$. Using $S(WE)=11$, we also get $W=3$.

Next, observe that

Therefore, we have $L=1$ or $L=2$. In case $L=2$, the digital sum of $\mathit{LIKE}$ leads to $I+K=13$, which means that $(I,K)$ or $(K,I)$, respectively, can be one of the pairs $(4,9)$, $(5,8)$, or $(6,7)$. But none of them is possible any more, since the digits $4$ and $8$ are already taken and in case of $(6,7)$, the factor $\mathit{LIKE}$ would exceed the upper bound above. From that we conclude $L=1$. Then the digital sum of $\mathit{LIKE}$ leads to $I+K=14$, which restricts the options for $(I,K)$ or $(K,I)$, respectively, to the pair $(5,9)$. An easy calculation shows that only $(I,K)=(5,9)$ solves the cryptogram. From $38\cdot 1598=60724$ we obtain the desired number $N\mathit{ABOJ}=60724$.