Problems and solutions

Answer:

$58$

Clearly, the ages of the children are $21$, $21+4$, $\text{…}$, $21+4\cdot 4$. Grandma is $21$ years older than her oldest offspring, namely she is $21+4\cdot 4+21=58$ years old.

Answer:

$56$

The ten angles adjacent to point $S$ form five pairs of opposite angles of equal size and the five of them which are the interior angles at the vertex $S$ in the five isosceles triangles sum up to $180^{\circ }$. Hence the sum of the five marked angles equals $\frac{5\cdot 180^{\circ }-180^{\circ }}{2}=360^{\circ }$ and it follows that the size of the missing angle is $360^{\circ }-67^{\circ }-80^{\circ }-85^{\circ }-72^{\circ }=56^{\circ }$.

Answer:

$32$

Add the number of participants of the single competitions and subtract the number of students who have chosen two competitions from the result. Hence, the very motivated students who have taken part in all three competitions are subtracted once too often, their number has to be added. Therefore, the final result is $22+13+15-8-7-6+3=32$.

Answer:

$90$

There are five even one-digit numbers. To guarantee that the result is super-even, any two digits added during the traditional addition algorithm must lead to a sum less than $10$. Bearing in mind that a number cannot start with zero, we have three possibilities ($2$, $4$ and $6$) for the first digit, another three possibilities for the second digit, two possibilities for the third digit, one possibility at the fourth digit and five possibilities for the last one. Since these choices are independent, we get $3\cdot 3\cdot 2\cdot 1\cdot 5=90$ such numbers in total.

Answer:

$\frac{25}{4}$

We know that the sum of the perimeters of the four circular walls will be the perimeter of new circular wall. Denote the radius of the new circular wall by $r$. Then

implies that $r$ is the sum of the given radii, i.e. $r=500$. Hence, the desired ratio is $\frac{\pi \cdot 500^2}{\pi \cdot 200^2}=\frac{25}{4}$.

Answer:

$9316$

Since the sought number is divisible by the primes $137$ and $17$, it must be a multiple of $137\cdot 17=2329$. Observe that this number does not comply with the conditions and that this number only can be multiplied by $2$, $3$, or $4$ to stay a four-digit number. We compute $2329\cdot 2=4658$, $2329\cdot 3=6987$ and $2329\cdot 4=9316$, which have sums of digits $23$, $30$, and $19$, respectively. Because $19$ is the smallest occurring prime, the number $9316$ opens the lock.

Answer:

$27$

Let us assume the congruent polygons have $n$ sides each. Then the resulting shape has $3(n-3)$ sides, since three sides of all polygons including the triangle are shared. We only need to determine $n$. Since the outer angle of the equilateral triangle is $300^{\circ }$, the inner angle of the congruent polygons must be $150^{\circ }$. Since the sum of inner angles of any $n-$gon equals $(n-2)\cdot 180^{\circ }$, we have to solve the equation $150n=180(n-2)$, which holds for $n=12$. Plugging into the above formula, we obtain that the resulting polygon has $3\cdot 9=27$ sides.

Answer:

$8081$

Let us refer to equilateral triangles just as to triangles. There is the original triangle, then $3\cdot 2020$ triangles sharing exactly one vertex with the original triangle ($2020$ for each vertex) and there are $2020$ rotated triangles (as on the picture from the statement) sharing no vertex with the original one. It is easy to see that these triangles are all distinct and there is no other such triangle. Altogether, we have $1+3\cdot 2020+2020=8081$ desired triangles.

Answer:

$\sqrt{300}\approx 17.32051$

The interior angles of regular octagon are $135^{\circ }$. Therefore Veronica cuts off four right-angled isosceles triangles which may be assembled in a square with the side length of the regular octagon. As a consequence, the side length of the regular octagon equals $\sqrt{300}$.

Answer:

$853$

From the second condition we obtain that at least one digit of $n$ has to be $5$ and at least one digit has to be even. However, none of the digits may be $0$. Taking this into consideration, we derive the possibilities $5,2,9$ or $5,4,7$ or $5,6,5$ or $5,8,3$ from the first condition. From these values only $5,8,3$ fulfill the last condition and hence the largest three-digit integer satisfying the conditions is $853$.

Answer:

$75928$

The largest ten-thousand-digit that can be achieved by removing at most five digits from the left is $7$, so we need to remove the first three digits. By analogous arguments we conclude that the optimal choice is then finished by removing digits $0$ and $1$. Therefore, $75928$ is the sought answer.

Answer:

$12$

The number $2021$ appears as a term in the sequence $S_n$ if and only if $2021=1+\mathit{an}$ for some positive integer $a$. In other words, $2020=\mathit{an}$, so $n$ has to be a divisor of $2020$. The prime factorization of $2020$ is $2020=2^2\cdot 5\cdot 101$. Any divisor of $2020$ is obtained by multiplying some of these primes. We can take the prime $2$ zero, one or two times, that gives $3$ possibilities. The prime $5$ can be taken or not – giving $2$ possibilities and the same holds for $101$. In total, we have $3\cdot 2\cdot 2=12$ possibilities how to choose a divisor of $2020$. Hence $2021$ is a term in $12$ sequences $S_n$.

Answer:

$510$

For a given robot $A$ we can determine the window and direction of a robot which would meet $A$ in the longest possible time – it is the window right behind $A$ heading in the opposite direction (if placed by one of the end windows, we assume that a robot is heading out of the corridor in this argument). The time to the first meeting is then easily computed to be $8.5$ minutes which is equal to $510$ seconds.

Answer:

$42$

From the formula for the area of a triangle “side length times the corresponding altitude divided by two” we deduce that triangles $\mathit{APD}$ and $\mathit{BCP}$ cover half of the area of the parallelogram. Therefore the area of triangle $\mathit{ABP}$ equals

Answer:

$107$

The distances between the three numbers are $1486-1058=428$ and $2021-1486=535$. Since the numbers given leave the same remainder when divided by $d$, the distances have to be multiples of $d$. The greatest common divisor of $428$ and $535$ is the prime $107$, which is the sought integer $d$.

Answer:

$3024$

Imagine the ten substitute players standing in a row. There are nine gaps between the ten players and each gap can be occupied by at most one member of the trainer team. Therefore they have $9\cdot 8\cdot 7\cdot 6=3024$ possibilities to sit in the desired way.

Answer:

$\frac{\sqrt{3}}{6}\approx 0.288675$

The base has sides of size $1$. As the whole pyramid has surface area $3$, each of its four triangular faces has the area $\frac{1}{2}$ and height equal to $1$. Thus a cut through its vertex and heights of two opposite faces creates an equilateral triangle of side $1$ whose height $\frac{1}{2}\sqrt{3}$ is the same as the height of the pyramid. The volume of the pyramid equals one third of the base surface area times the height, namely $\frac{1}{3}\cdot 1\cdot \frac{1}{2}\sqrt{3}=\frac{1}{6}\sqrt{3}$.

Answer:

$2250$

Denote by $x$ the amount of wine, and by $z$ the amount of water in litres before the Cana machine was used. We know that $0.06z$ litres of water gets converted into wine and $0.1x$ litres of wine to water and that $x$ is the same before and after running the machine. Hence, we must have $0.06z=0.1x$. Given that $x+z=6000$, we can write $0.06(6000-x)=0.1x$. Solving for $x$ leads to $6000\cdot \frac{3}{8}=2250$ litres of wine.

Answer:

$35$

It is easy to compute that the radius of the incircle of an equilateral triangle is half the radius of its circumcircle. Thus the area of the incircle equals $\frac{140}{4}=35$. Furthermore, the shaded area is exactly one third of the annulus given by the two circles, i.e. $35$ as well.

Answer:

$4044$

Let us denote the positive integers as $c_1\ge c_2\ge \cdots \ge c_{2020}\ge c_{2021}\ge 1$. We wish to determine the largest possible value of $c_1$ assuming that the equation

holds. Dividing by the left-hand side and estimating the denominators from below by setting some of the numbers $c_i$ to $1$ yields

Multiplying by $c_1{c}_2$ and rearranging then gives

If $c_2\ge 3$ we get $c_1\le 4044$ and the choice $c_1=4044$, $c_2=3$ and $c_3=\cdots =c_{2021}=1$ satisfying the equation (1) shows that this value can be attained. On the other hand, if $c_2\le 2$ then the numbers $c_2\ge c_3\ge \cdots \ge c_{2021}$ consist of $k\ge 0$ twos and $2020-k$ ones and the equation (1) reads as

It can be further simplified to $c_1(2^{k-1}-1)=2020+k$ and we observe that for $k\le 1$ there is no $c_1$ satisfying this equation and for $k\ge 2$ we have

so the result is $4044$.

Answer:

$59$

First of all, observe that there are three cells with one neighbour, three ones with two and three ones with three neighbours. This means, if a prime $p$ is one of the nine numbers, at least one multiple of $p$ has also to be among the nine numbers. Secondly, note that $1$ can not be one of the nine numbers. Denote the sum of a valid filling by $S$.

If there is a prime $p\ge 11$ in a valid filling, then at least one multiple $k\cdot p$ with $k\ge 2$ must also be present. Fill the remaining seven cells by the smallest available numbers, regardless whether they comply with the rules. Then

Now we assume no prime $p\ge 11$ is present in a valid filling and consider four subcases:

• Both the numbers $5$ and $7$ are present:

  $$S\ge 5+k_5\cdot 5+7+k_7\cdot 7+2+3+4+6+8=(k_5+1)\cdot 5+(k_7+1)\cdot 7+23\ge 15+21+23=59.$$

• Number $5$ is present, but there is no $7$:

  $$S\ge 5+k\cdot 5+2+3+4+6+8+9+\{\begin{array}{c}10\ge 20+32+10=62\text{for}k\ge 3,\hfill \\ 12=15+32+12=59\text{for}k=2.\hfill \end{array}$$

• There is neither $5$ nor $7$:

  $$S\ge 2+3+4+6+8+9+10+12+14=68.$$

• Number $7$ is present, but there is no $5$:

  $$S\ge 2+3+4+6+7+8+9+10+k\cdot 7\ge 49+14=63.$$

Altogether, we can conclude that $59$ is a candidate for the minimal sum. In fact, both sets of numbers from above having sum $59$ can be filled in according to the rules:

Therefore, the answer is $59$.

Answer:

$93$

Define the area of one house to be 3. In the first house, there are 8 triangles of area $\frac{1}{4}$, 8 triangles of area $\frac{1}{2}$, and 3 triangles of area 1. These add up to 19. In the second house, we can find the same number of triangles as in a single house plus 2 triangles of area 1 which reach from one house into the other. Therefore, the second house provides 21 triangles.

Starting at house number 3, each additional house contributes the number of triangles of the second house plus one triangle of area 4 which ranges over three houses. So, for each additional house, there are 22 triangles. Since $2021-19-21=1981$ and $1981=90\cdot 22+1$, Lotta has to draw $2+90+1=93$ houses.

Answer:

$\frac{15}{4}$

The area ratio between the triangles $\mathit{BEG}$ and $\mathit{BEF}$ is $4:3$. Since these triangles have the same base line $\mathit{BE}$, the respective heights must have the ratio $4:3$, too. Since the triangles $\mathit{ABG}$ and $\mathit{CDF}$ have the same area, we conclude that $\mathit{AB}=\frac{3}{4}\mathit{CD}=\frac{15}{4}$.

Answer:

$6375$

Anna may cut off strips of odd width as long as

Since

the strip of width $89$ is the last one possible having odd width. Furthermore, she may cut off strips of even width as long as

Due to

the strip of width $90$ is the last one possible having even width. Therefore, as she can cut off all 90 strips, the remaining rectangle has the area

Answer:

$\frac{10}{3}$

Let us denote the desired width by $w$ and look at the figure depicting the vertical projection of the two rings onto the table.

The altitude of the lowest point(s) of the vertical ring above the table is then $1=w-x$. Hence

and thus $w=\frac{10}{3}$.

Answer:

$29030400$

The polynomial can be written as $c\cdot (x-a_1)\cdot (x-a_2)\cdots (x-a_{14})$ for some pairwise distinct integers $a_1,a_2,\dots ,a_{14}$ and a number $c$. The leading coefficient is then equal to $c$, so $c$ is positive. The value at $0$ equals the constant coefficient, that is the product $c\cdot a_1\cdot a_2\cdots a_{14}$. As we want to minimize it, we set $c$ as small as possible, that is $c=1$. To minimize the remaining product of the roots, we have to take them as close to zero as possible, that is $1,-1,2,-2,\dots$ However, we have to choose an even number of negative integers, which leads to the result $6!\cdot 8!=29030400$.

Answer:

$8783$

When she writes the integers from $1$ to $2021$, she writes $9$ one-digit numbers, $90$ two-digit numbers, $900$ three-digit numbers and $2021-(9+90+900)=1022$ four-digit numbers. In total, she writes

digits. For each digit, she does one stroke, while she does one additional stroke for each digit which equals $4$, $5$ or $7$. So it is sufficient to count how many of these digits are written. Since the number $2021$ contains none of the digits $4$, $5$ or $7$, we consider only integers up to $2020$.

Let us count the digits $4$ she writes. There is $\frac{1}{10}$ of all numbers up to $2020$ which has the digit $4$ in the ones place, i.e. $202$ numbers. The digit $4$ occurs in the tens place in $\frac{1}{10}$ of the numbers up to $2000$ and in no number between $2001$ and $2020$. Altogether that is $200$ times. Similarly, the digit $4$ occurs in hundreds place $200$ times. In total, the digit $4$ is written $202+200+200=602$ times. The same goes for the digits $5$ and $7$.

To conclude, she writes $6977$ digits while $3\cdot 602=1806$ of them are $4$, $5$ or $7$. Therefore, she does

strokes.

Answer:

$3845$

For the ease of notation we treat the table given as an array $f$ with sixteen entries. From the three entries given, $f(6)=6$, $f(7)=7$, $f(9)=2$, we uniquely get $f(13)=6$, $f(15)=7$, and $f(12)=2$.

In principle there are two different strategies: either to look, where a specific pair of numbers may be placed, or to consider which numbers are possible in a specific cell like in Sudoku.

For example we can look for the possibilities to fill in the pair of $3$. There are three possibilities: either $f(1)=f(5)=3$ or $f(4)=f(8)=3$ or $f(10)=f(14)=3$.

It is easy to see that $f(10)=f(14)=3$ leaves the only possibility $f(16)=4$ to fill $f(16)$ and, as a consequence, we get $f(11)=4$. But now we cannot place the pair of $8$ anymore. The case $f(4)=f(8)=3$ results in two possibilities for the pair of $5$, namely $f(5)=f(11)=5$ or $f(10)=f(16)=5$. Both alternatives produce a contradiction immediately, as we can’t place the pair of 4 in neither case.

Now $f(1)=f(5)=3$ leaves the unique possibility $f(2)=f(11)=8$ and the only way to fill the remaining cells according to the rules is $f(3)=f(8)=4$, $f(14)=f(16)=1$ and finally $f(4)=f(10)=5$.

Therefore the unique solution asked for is $3845$.

Answer:

$5.7=\frac{57}{10}$

Since $50^{\circ }+65^{\circ }+65^{\circ }=180^{\circ }$ and due to the two pairs of parallel segments, this is exactly the configuration obtained by folding the parallelogram $A{E}^{\prime }{D}^{\prime }F$ along a suitable segment $\mathit{BC}$ (it is indeed a parallelogram as $\mathit{\angle BAF}=\mathit{\angle CDE}$; see the sketch). Equivalently, one gets the same result by mapping the vertices $D$, $E$ along the line $\mathit{BC}$. Observing moreover that the triangle $\mathit{BCG}$ is isosceles, the length conditions yield $\mathit{CD}=\mathit{AB}+B{E}^{\prime }-\mathit{CF}=\mathit{AB}+\mathit{EG}-\mathit{FG}=5.7$.

Answer:

$36$

Consider a $4\times 2$ block. As can be seen in the first two pictures, choosing exactly one square does not guarantee a hit, since there is still an option to place the battle cruiser without being hit. Furthermore, it is clear that the choice of any square in one row and another square in the other row prevents hiding the battle cruiser in this block. The third picture shows an example for that. Therefore, two shots into a $4\times 2$ block are necessary, yielding at least $18\cdot 2=36$ shots in total.

On the other hand, the following diagonal pattern of the $12\times 12$ grid demonstrates that $36$ shots are also sufficient to ensure at least one hit on the battle cruiser.

Answer:

$13$

As $\mathit{ABC}$ is acute-angled, we obtain $h_b<a$ and $h_c<a$. On the other hand, it is easy to see that in order to maximize the area $S=\frac{1}{2}a{h}_a$, both altitudes $h_b$ and $h_c$ should be as long as possible (indeed, prolongation of, say, the altitude $h_b$ keeping the condition $h_b<a$ with $h_c$ fixed makes $h_a$ longer as well). Therefore $h_b=h_c=a-1$ and $\mathit{ABC}$ is isosceles. Let us denote the midpoint of $\mathit{BC}$ by $M$ and the foot of the altitude $h_c$ by $C_0$. From the Pythagorean theorem for similar right triangles $\mathit{ABM}\sim \mathit{CB}{C}_0$ we compute

Since $101.4$ is a rational number, the integer $2a-1$ must be an odd square, say ${(2k+1)}^2$ for some integer $k\ge 0$, and hence $a=\frac{{(2k+1)}^2+1}{2}\in \{1,5,13,25,41,\dots \}$. Checking the first few values and realizing that the area is strictly increasing in $a$, we quickly find the correct answer $a=13$.

Answer:

$3099$

Let $101=n_1,n_2,\dots ,n_{1000}$ denote the positive integers. Assuming that the difference between two consecutive numbers would always be $2$, we get

Therefore, the difference between Ludwig’s sum and the minimal possible sum is $100500=20100\cdot 5$. If $n_{i+1}-n_i=7$ instead of $2$, then the sum is increased by $(1000-i)\cdot 5$ and $n_{1000}$ is increased by $5$. As a consequence, in order to maximize $n_{1000}$ considering a constant sum of $1200500$, the distances of $7$ should be put at the highest indices possible. Luckily, the value $20100$ is the triangular number $\frac{1}{2}\cdot 200\cdot 201$. Therefore, if Ludwig changes the distances $n_{i+1}-n_i$ for $i=800,\dots ,999$ to $7$ instead of $2$ compared to the minimal sum above, he gets the largest possible

Answer:

$29292929292$

Our solution requires a little modular arithmetic. Observe that $100\mathit{mod}11=1$. This means that multiplying by 100 does not change a number’s remainder modulo 11. Now suppose a number’s decimal expansion contains the same digit $a$ twice in a row. This number has the form $x\cdot 10^{n+2}+11a\cdot 10^n+y$. Crossing out the consecutive $a$’s we get the number $x\cdot 10^n+y$ which has the same remainder modulo 11. All of this means that our answer cannot contain a pair of consecutive 2’s or 9’s, for otherwise we could cross them out and get a smaller positive integer, still with an odd number of digits, and still divisible by 11. Now we simply try longer and longer strings of the form $2929\dots 292$ or $9292\dots 929$ until we find the smallest such number that is divisible by 11.

The solution is even easier to find if one knows the divisibility criterion for 11. A number is divisible by 11 if and only if the sum of the digits in even positions minus the sum of the digits in odd positions is divisible by 11. This immediately implies that our answer cannot contain a pair of consecutive 2’s or 9’s, for otherwise we could cross them out and get a smaller solution, as before. Since 2’s and 9’s must alternate, we look for the smallest $n$ such that $2n-9(n+1)$ is a multiple of $11$, or $9n-2(n+1)$ is. In each case the solution is $n=5$, which yields the candidates $29292929292$ and $92929292929$. We take the smaller of these two numbers.

Answer:

$\frac{3+\sqrt{5}}{2}\approx 2.61803$

Observe that the two pentagons, $p$ and $q$, are homothetic with center $F$, therefore the line segment between $X$ and $Z$, one of the two possible pairs of points of $p$ and $q$ maximizing the distance, passes through $F$.

Straightforward angle chasing reveals the following facts:

• $\mathit{\angle AFE}=108^{\circ }$, hence $\mathit{AE}=\frac{1}{2}(1+\sqrt{5})$ by the Law of Cosines,
• triangles $\mathit{AFZ}$, $\mathit{XFD}$ and $\mathit{DFE}$ are isosceles.

Therefore

and $\mathit{ZF}=\mathit{AF}=1$. The sought distance is $\mathit{XF}+\mathit{ZF}=\frac{1}{2}(3+\sqrt{5})$.

Answer:

$281$

We can transform the equation to $a(\mathit{bc}-11b-175)=\mathit{bc}$. From this we see that either $a=b$ or $a=c$ since all the variables should be primes. In the first case, we get

giving only the solution $(2,2,197)$ in primes. The second case yields

and the other two solutions in primes are $(47,5,47)$ and $(37,7,37)$. Therefore the value sought is $197+47+37=281$.

Answer:

$\frac{21}{40}$

For any ranking of Mary, Tom needs to have exactly one house of Mary’s top 3 in his top 3 and the remaining two of Mary’s top 3 among his rankings 4 to 10. So for any ranking of Mary, Tom has a chance of

Answer:

$-2$

Since the polynomial $p(x)$ can be factorized into $(\mathit{ax}+b)(x^{2020}+x^{2018}+\cdots +x^2+1)$ where the second factor is positive, the only real root of $p(x)$ is $x=\frac{-b}{a}$. Given that the quadratic polynomial $q(x)$ has only one real root, it must be a double root. Hence we get $b^2-4a^2=0$ for the discriminant, and since $a,b$ are positive we can say $b=2a$. Therefore, the only real root equals $x=\frac{-b}{a}=\frac{-2a}{a}=-2$.

Answer:

$312$

We can assume that $n<p$ and that the decimal expansion of $\frac{n}{p}$ is $5$-periodic starting right after the decimal point. Indeed, if $\frac{n}{p}$ is only eventually periodic, one can shift the decimal point by multiplying $n$ by a suitable power of $10$ and then, if $n\ge p$, we can replace it by $n^{\prime }<p$ such that $n=\mathit{kp}+n^{\prime }$: this only “erases” the part in front of the decimal point.

If $0.\overline{\mathit{ABCDE}}$ is the periodic decimal expansion of $\frac{n}{p}$ then $99999\cdot \frac{n}{p}=10^5\cdot \frac{n}{p}-\frac{n}{p}=\mathit{ABCDE}$ is an integer. Since $n<p$ and $p$ is a prime, it follows that $p\mid 99999$, that is $p\mid 3^2\cdot 41\cdot 271$. Both $\frac{1}{3}$ and $\frac{2}{3}$ have the shortest period $1$, but $\frac{1}{41}=0.\overline{02439}$ and $\frac{1}{271}=0.\overline{00369}$ are of the desired type. Therefore, the result is $41+271=312$.

Answer:

$5680$

We will use the inclusion–exclusion principle to compute the result. Let us number the languages in some order by $1,\dots ,5$. Denoting by $A_i$ the set of assignments of the languages to the people such that all of them speak the $i$-th language, we are asked for $\left|{\bigcup }_{i=1}^5{A}_i\right|$, the size of the union of the sets. Firstly, $\left|A_i\right|={{\displaystyle \left(\genfrac{}{}{0.0pt}{}{4}{2}\right)}}^4=6^4$ since we can choose the two of the four remaining languages for all the four people independently. There are $5$ possible choices of $i$. For any fixed $i\ne j$, we similarly obtain that $\left|A_i\cap A_j\right|$, the number of language assignments for which everybody speaks both the $i$-th and the $j$-th language equals $3^4$. Clearly, there are ${\displaystyle \left(\genfrac{}{}{0.0pt}{}{5}{2}\right)}=10$ such pairs of indices $i\ne j$. Finally, there is exactly one assignment where all the people speak the same fixed three languages and there are ${\displaystyle \left(\genfrac{}{}{0.0pt}{}{5}{3}\right)}$ such choices of three different languages (in our notation, we just observed that $\left|A_i\cap A_j\cap A_k\right|=1$ for any of the $10$ choices of pairwise different indices $i,j,k$ ). The aforementioned inclusion–exclusion principle then yields

Answer:

$\frac{65}{98}$

We should try fractions with denominators as large as possible, because if $\frac{a}{b}$ is less than $\frac{2}{3}$, then $\frac{a+2}{b+3}$ is greater than $\frac{a}{b}$ and still less than $\frac{2}{3}$. In particular, this means that we should try the denominators 98, 99, and 100. Thus, the possible solutions are $\frac{65}{98}$, $\frac{65}{99}$, and $\frac{66}{100}=\frac{33}{50}$. Comparing these fractions, we find that

Answer:

$130$

The perimeter equals the sum of the perimeters of the black unit cubes minus twice the number of faces shared by a pair of black cubes. Such a pair can be oriented in three directions, let us call them “up-down”, “front-back” and “left-right”. If the “left-right” case occurs, it must appear also as a pair of two black squares sharing a vertical side separating the same two columns in both of the projections. Checking column by column, we easily check that this never happens. The “front-back” cases must affect the left square – it happens twice in the first column. As the first column of the front overview contains a single black cell, the positions of the black cubes in the leftmost layer of the big cube are uniquely determined and there are indeed two vertical faces shared by two black cubes. The last case “up-down” can be treated similarly revealing that it contributes two shared faces (due to the fifth columns of the projections). We conclude that the desired surface area is $6\cdot 23-2\cdot 4=130$.

Answer:

$204$

We have $720=2^4\cdot 3^2\cdot 5$. The exponents of any prime $p$ in $d_1,d_2,\dots ,d_k$ form a non-decreasing sequence, which sums up to the exponent of $p$ in $720$. For the prime $2$, this sequence might be $(1,1,1,1)$, $(1,1,2)$, $(2,2)$, $(1,3)$, $(4)$ or any of these preceded by some number of zeros. For $3$, the sequences end $(1,1)$ or $(2)$. For $5$, there is just $(1)$. Any combination of these sequences gives us a dividing decomposition. Therefore, there are $10$ dividing decompositions of $720$ and the arithmetic mean of their leaders is $\frac{2+4+4+8+16}{5}\cdot \frac{3+9}{2}\cdot 5=204$.

Answer:

$9450$

Denote by $n$, $a$, $b$, $o$, $j$ the numbers of tiles with letters $N$, $A$, $B$, $O$, $J$, respectively. We are looking for the number of 5-tuples $(n,a,b,o,j)$ with

Exponents of each prime can be independently distributed among the numbers $n$, $a$, $b$, $o$, $j$ and different distributions yield different 5-tuples. For example for the prime $2$, we need to divide $5$ objects into $5$ boxes. This can be done in $\left(\genfrac{}{}{0.0pt}{}{9}{4}\right)$ ways, indeed we can choose which of $9$ things are objects and which ones are dividers between the boxes. Similarly, for $3$ there are $\left(\genfrac{}{}{0.0pt}{}{6}{4}\right)$ ways and for $5$ there are $\left(\genfrac{}{}{0.0pt}{}{5}{4}\right)$ ways. Therefore there are

different Scrabboj sets.

Answer:

$4978$

Assume that the largest such integer $n$ is at least $2021$. Then, after dividing $4^{2021}+4^n+4^{3500}$ by $4^{2021}={\left(2^{2021}\right)}^2$, we get another square

where $m=n-2021$. Moreover, it is a square of a number larger than $2^m$: let us write

for some positive integer $x$. Then $x\cdot 2^{m+1}+x^2=2^{2958}+1$. The left-hand side is increasing in both $m$ and $x$ while the right-hand side is a constant, so the solution with the largest $m$ will have the smallest possible $x$. If we try to take $x=1$, then $m=2957$ solves this equation and by reverting the previous argument we see that $n=m+2021=4978$ works. This justifies our initial assumption that the largest admissible $n$ is at least $2021$ and we conclude that $4978$ is the desired maximal value.

Answer:

$1021616$

Consider the polynomial

and notice that all its nonzero coefficients are positive. We claim that these are exactly the ones corresponding to the powers $x^k$ for $k$ between the minimum possible one, i.e. $0$, and the maximum possible one, i.e.

except for exactly the two numbers $1$ and $S-1$. When we imagine the product of the $2020$ factors defining $Q$ expanded, it is clear that there will be no linear terms and it is also easy to see that the same holds for $x^{S-1}$: if we choose the power of $x$ from every bracket, we obtain $x^S$, otherwise the exponent is at most $S-2$.

Now we prove that every other exponent from the range above is present with a positive coefficient or, equivalently, that the smallest number $m$ larger than $1$ that cannot be written as a sum of a subset of $\left\{2,3,\dots ,2021\right\}$ equals $S-1$. We claim that

for some $k\in \{2,3,\dots ,2021\}$. Indeed, clearly $m\ge 3$ and hence it must be possible to write $m-1$ as a sum of some subset of $\left\{2,3,\dots ,2021\right\}$. Moreover, for every subset different from $\{k,k+1,\dots ,2021\}$ we can just increase one of the numbers in the sum by $1$ and obtain a valid representation of $m$. Moreover, $k\le 3$ as otherwise

is a way to express $m$. Hence $m=1+3+4+\cdots +2021=S-1$.

Therefore $Q(x)$ has $\frac{S}{2}+1$ positive coefficients at even powers of $x$ and $\frac{S}{2}-2$ positive coefficients at odd powers of $x$. The original polynomial $P(x)$ has the signs at the odd coefficients flipped, and hence it has exactly $\frac{S}{2}+1=1021616$ positive coefficients.

Answer:

$26550$

Firstly, we will calculate how many shortest paths there are without condition. We have to go from $(0,0,0)$ to $(4,4,4)$. We have to take four roads in $x$ direction, four in $y$ direction and four in $z$ direction, in any order. If we go back, then the path will not be the shortest. That is $\frac{12!}{4!\cdot 4!\cdot 4!}$ possible paths.

Now we have to subtract those paths that go through crossroad $(2,2,2)$. Due to symmetry we know that the number of paths from $(0,0,0)$ to $(2,2,2)$ is equal to the number of paths from $(2,2,2)$ to $(4,4,4)$ and that is equal to $\frac{6!}{2!\cdot 2!\cdot 2!}$. For any path from $(0,0,0)$ to $(2,2,2)$, there are $\frac{6!}{2!\cdot 2!\cdot 2!}$ possible continuations from $(2,2,2)$ to $(4,4,4)$. Hence the number of all roads through crossroad $(2,2,2)$ is $\frac{6!}{2!\cdot 2!\cdot 2!}\cdot \frac{6!}{2!\cdot 2!\cdot 2!}=\frac{6!\cdot 6!}{2^6}$. The total number of paths is thus $\frac{12!}{4{!}^3}-\frac{6{!}^2}{2^6}=26550$.

Answer:

$98$

The relevant points are labeled as in the picture.

Since triangle $\mathit{ABC}$ is equilateral, we get $\mathit{\angle BDE}+\mathit{\angle DEB}=120^{\circ }=\mathit{\angle BDE}+\mathit{\angle FDA}$, i.e. $\mathit{\angle DEB}=\mathit{\angle FDA}$ and therefore $\mathit{ADF}\sim \mathit{BED}$. Since the areas of these triangles have the ratio $100:64$, the respective sides have the ratio $5:4$. If we set $r=\mathit{DB}$, $s=\mathit{BE}$, $t=\mathit{ED}$, we get the lengths as labeled in the following picture.

We can derive the following equations using $a$ for the side length of the equilateral triangle:

The linear equations (1)–(3) yield $r=\frac{a}{3}$ and $s=\frac{8a}{15}$. Inserting these values in the equation (4) then gives

where $S$ is the area of the equilateral triangle. It follows that the desired area $A$ satisfies $100+64+2A=360$ and hence $A=98$.

Answer:

$\frac{5}{8}$

Let us denote by $E$ the event that sequence “heads-tails-heads”, or simply $\mathit{HTH}$, occurs before $\mathit{THTH}$ and by $P(E)$ its probability. For a fixed finite sequence $s$ of heads and tails denote by $P(E\mid s)$ the probability that $E$ occurs in a sequence starting by $s$ and continuing randomly. Set $x=P(E\mid H)$ and $y=P(E\mid T)$. Since our coin is fair, moving one or two steps ahead (we always write the new outcome to the right end of the current sequence) we compute

and analogously by moving up to three steps ahead we obtain