Problems and solutions

Answer:

6, 16, 28

Most of the numbers on the cans are multiples of $3$. When $50$ is divided by $3$, the remainder is $2$, so the sum of the chosen cans must also leave a remainder of $2$ when divided by $3$. Therefore, at least one number that is not divisible by $3$ must be included. The only such numbers are $16$ and $28$.

Using only one such number is not possible: The number $16$ leaves a remainder of $1$ when divided by $3$, and $28$ also leaves a remainder of $1$. Adding any number of multiples of $3$ to either $16$ or $28$ does not change this remainder, so the total cannot equal $50$. Hence, both $16$ and $28$ must be used. Their sum is $16+28=44$, which leaves a remainder of $2$ when divided by $3$. To reach $50$, we have to add $50-44=6$, which is one of the cans and cannot be formed as a sum of any other cans. Therefore, the cans to knock down are $6$, $16$, and $28$.

Answer:

$21$

There are five possibilities to get a total of $8$ with three dice:

In the case $6+1+1$, the number $6$ may be shown on the red die or on the blue die or on the yellow die. Therefore we have three possibilities in the case $6+1+1$. By analogy, we also have three possibilities in the cases $4+2+2$ and $3+3+2$. In the case $5+2+1$ we have $3\cdot 2\cdot 1=6$ possibilities for the distribution of the numbers on the three colored dice. The same is true in the case $4+3+1$. Therefore there are $3\cdot 3+2\cdot 6=21$ possibilities in total.

Answer:

$12$

Let $a$ be the age of Adam and $c_1\le c_2\le c_3\le c_4$ the ages of his children. The problem statement implies the equations $a=c_2+c_3+c_4$ and $a+6=(c_1+6)+(c_2+6)+(c_3+6)$. Substituting the first equation into the second yields $c_2+c_3+c_4+6=c_1+c_2+c_3+18$, which can be simplified to $c_4=c_1+12$. Therefore the oldest child is 12 years older than the youngest child.

Answer:

$36$

The hour must be a two-digit number not exceeding $23$ with distinct digits picked from $1,\dots ,5$; the only possibilities are $12$, $13$, $14$, $15$, $21$, and $23$. For each such hour, two digits have been used, leaving three remaining digits from which the minute must be formed using two distinct digits; this can be done in $3\cdot 2=6$ ways, all of which are valid minutes. Hence each of the six allowable hours produces six valid times, giving $6\cdot 6=36$ in total.

Answer:

$67.5^{\circ }$

Let $X$ be the intersection point of the two diagonals. Instead of computing $\mathit{\angle CXE}$, as suggested by the diagram, we compute the measure of $\mathit{\angle FXG}$, which is the same. To this end, first observe that $\mathit{\angle GFX}=\mathit{\angle GFC}=90^{\circ }$, since $\mathit{BCFG}$ is a rectangle. Moreover, as the triangle $\mathit{EFG}$ is isosceles and $\mathit{\angle GFE}=135^{\circ }$, we have $\mathit{\angle XGF}=\mathit{\angle EGF}=22.5^{\circ }$. Therefore, from triangle $\mathit{FGX}$, we conclude that $\mathit{\angle FXG}=180^{\circ }-90^{\circ }-22.5^{\circ }=67.5^{\circ }$.

Answer:

$36$

Since

it follows that $a+b+c+d+e=80$. To maximize $d$, minimize the other terms subject to the ordering: The smallest possible values are $a=1$, $b=2$, $c=3$, and $e\ge d+1$. Hence

so $2d\le 73$ and therefore $d\le 36$ ($d$ is an integer). This bound is attained by $(a,b,c,d,e)=(1,2,3,36,38)$, which sums to $80$ and satisfies the ordering condition. Thus the maximum possible value of $d$ is $36$.

Answer:

$567$

By divisibility by $9$, the digit sum is $9$ or $18$. Since the number is odd, $6$ cannot be the units digit, so it is in the hundreds or tens place. It cannot be the hundreds digit, since then the smallest possible digit sum is $6+7+8=21>18$; hence $6$ is the tens digit. Therefore the units digit is at least $7$, so the digit sum must be $18$, and the hundreds and units digits must sum to $12$. The increasing possibilities are $(3,9)$, $(4,8)$, and $(5,7)$: the first is excluded because deleting the $9$ yields $36$, still divisible by $9$; the second is excluded because it gives an even number; thus $(5,7)$ is the only valid choice. Hence the sphinx’s number is $567$.

Answer:

$28$

The overlap creates three smaller and three larger congruent equilateral triangles outside the shaded hexagon. Set $\mathit{AB}=x$ and $\mathit{BC}=y$. Then the side length of the larger triangle is $2x+y=17$ and the side length of the smaller one is $x+2y=11$. Adding the two equations yields $3x+3y=28$, which is exactly the perimeter of the hexagon.

Alternative solution. This solution provides a proof without words by mapping the sides of the hexagon onto one side of each of the two given equilateral triangles. Therefore the perimeter of the hexagon is exactly the sum of the side lengths of the given equilateral triangles, namely $17+11=28$.

Answer:

$242$

We count all admissible outfits by cases. Anna must wear exactly one shirt, one pair of shorts, one pair of socks, and one pair of sneakers, and may optionally wear a ribbon.

First consider white socks. By the given condition, white socks may be worn only if the entire outfit is white. Thus the T-shirt, shorts, and sneakers must all be white, and the ribbon is either not worn or white. This gives $2$ outfits.

Now consider nonwhite socks (red or orange), giving $2$ choices. In this case there are no further restrictions. There are $5$ choices for the T-shirt, $3$ for the shorts, $2$ for the sneakers, and $4$ choices for the ribbon (none, white, black, or gray). Hence this case yields

outfits.

Adding both cases, the total number of possible outfits is $240+2=242$.

Answer:

$1239$

Let $\mathit{ABCD}$ denote the number, where $A$, $B$, $C$, and $D$ are its digits; the conditions give $B=2A$ and $D=3C$. Then

which simplifies to $A^2+2C^2=19$. Since the right-hand side is odd, $A$ must be odd; moreover, $A$ is a digit and $A^2\le 19$, leaving only $A\in \{1,3\}$. Testing these values yields only $A=1$, giving $2C^2=18$ and thus $C=3$. Hence $B=2$, $D=9$, and the number is $1239$.

Answer:

$16$

We first note that the sum $\pm 1\pm 2\pm 3\pm 4\pm 5$ can never be $0$; this is true since $1+2+3+4+5=15$ is odd, and changing any sign changes the total by an even number, so the result is always odd. Now, if we take any choice of signs and switch every plus to a minus and every minus to a plus, the value changes from positive to negative or vice versa, giving a perfect pairing between positive and negative results. For each of the five boxes we may independently choose either a plus or a minus sign, giving $2$ choices for each position and therefore $2^5=32$ total sign choices. Since none produce zero, exactly half must be positive, so Alice can obtain a positive result in $16$ ways.

Answer:

$619$

Since the first digit of all three numbers is the same, $A$ must satisfy $3A\le 20$ and $3A\ge 18$, as there cannot be a carry over of more than $2$ when adding three digits. Therefore, $A=6$. It follows that in the units column the digits $B$ and $C$ add up to $10$. Thus, there is a carry over of $1$ from the units column to the tens column. Therefore, we need $C=9$ and it follows that $B=1$. Hence $\mathit{ABC}=619$.

Answer:

$\frac{53}{3}=17\frac{2}{3}$

Since the square has area $4$, each rectangle must have area $1$. This immediately gives that the width of the right rectangle is $\frac{1}{2}$, so the width of the lower left rectangle is $2-\frac{1}{2}=\frac{3}{2}$, and hence its height is $1∕\frac{3}{2}=\frac{2}{3}$. Consequently, the common height of the two remaining rectangles is $2-\frac{2}{3}=\frac{4}{3}$. To find the total of the four perimeters, take the perimeter of the outer square, $4\cdot 2=8$, and add twice the total length of the interior segments, since each is counted in two rectangles. The interior lengths are $2$, $\frac{3}{2}$, and $\frac{4}{3}$, so their doubled sum is $2\cdot \left(2+\frac{3}{2}+\frac{4}{3}\right)=\frac{29}{3}$. Therefore, the desired sum of the perimeters is $8+\frac{29}{3}=\frac{53}{3}$.

Answer:

$12158$

The digit sum after writing $2026$ once is $10$. After the first one, every even numbered block will add the digits $202$, for an increase in the digit sum of $4$, and every odd numbered block will add $026$, increasing the sum by $8$.

This way, every time two blocks are added after the first one, which occurs $1012$ times ($1+2\cdot 1012=2025$), the sum is increased by $12$. At the end, we are left with the 2026th block, which is an even number, so it increases the sum by $4$. In total, we get $10+1012\cdot 12+4=12158$.

Answer:

$14$ = $2$ PM

Since a day has $24$ hours, the time of day repeats every $24$ hours. Thus it is enough to find the remainder when $N=260320261998$ is divided by $24$. This could be done by long division, but it is simpler to compute the remainders upon division by $3$ and by $8$ separately and combine the results.

Because the sum of the digits of $N$ is divisible by $3$, the number $N$ itself is divisible by $3$. For division by $8$, the remainder is determined by the final three-digit block; here it is $998$, and $998$ leaves remainder $6$ upon division by $8$, so $N$ also leaves remainder $6$ upon division by $8$. Therefore the remainder upon division by $24$ must be a number less than $24$ that is divisible by $3$ and leaves remainder $6$ when divided by $8$; the only possibility is $6$.

Hence after $N$ hours, the clock advances by the same amount as after $6$ hours. Starting from $8:00$ AM, adding $6$ hours gives $14:00$ (or $2$ PM).

Answer:

$16$

As it stands, the only grape that can be eaten is grape $F$, leaving grapes $A,B,C,D,$ and $E$.

At that point, we can either eat grape $D$ or $E$. The situation is obviously symmetrical, so let us start with $E$, leaving grapes $A,B,C,$ and $D$. Then, we can eat grapes $C$ or $D$:

• If we choose $C$, then we have to eat $D$, followed by either $A$ or $B$ ($2$ possibilities)
• If we choose $D$, we can eat $A,B,$ or $C$ next, followed by the remaining two in any order ($3\cdot 2$ possibilities)

Thus, we always start with $F$, then choose between $D$ or $E$, and then have $2+6=8$ distinct possibilities. This totals $1\cdot 2\cdot 8=16$ possible ways to eat the grapes.

Answer:

$12$

The number $x$ has to be of the form $2^a\cdot 3^b\cdot 5^c\cdot 7^d$ (no other primes can appear, or they would show up in the LCMs). The first condition is equivalent to $a=6$, $b\le 3$, $c\le 4$, and $d\le 2$, while the second condition translates to $a\le 8$, $b\le 4$, $c\le 3$, and $d=2$. This implies that in the prime factorization of $\gcd (x,2^2\cdot 3^4\cdot 5^2\cdot 7^3)$, the exponent of $2$ is equal to $2$, the exponent of $3$ can be any number between $0$ and $3$, the exponent of $5$ can be any number between $0$ and $2$, and the exponent of $7$ is equal to $2$. We conclude that there are $4\cdot 3=12$ such numbers.

Answer:

$71^{\circ }$

Computing complements to $90^{\circ }$ at the two vertices on the right, we conclude that the given interior point lies on the common axis of the two vertical sides of the rectangle. Therefore the whole configuration is symmetric with respect to this axis. We conclude that the desired angle equals $90^{\circ }-19^{\circ }=71^{\circ }$.

Answer:

$1078$

Let $c$ be the number of cheap beads and $p$ their (integer) price in florins. Then $\mathit{cp}=459$ and since there are $100$ beads in total with more cheap ones than expensive ones, we have $50<c<100$. Since the price is an integer, $c$ must be a divisor of $459=3^3\cdot 17$ lying strictly between $50$ and $100$, and the only such divisor is $c=51$. Thus $p=459∕51=9$, and each expensive bead costs $p+13=22$ florins. The number of expensive beads is $100-51=49$, so Alina paid $49\cdot 22=1078$ florins for the expensive beads.

Answer:

$1$

There are $10$ distinct letters ($M$, $A$, $T$, $H$, $N$, $B$, $O$, $J$, $G$, $E$), so the digits $0$–$9$ are all used exactly once, meaning one letter equals $0$. It follows that both sides of the equality are necessarily equal to $0$. In particular, $0$ appears on both sides while not occurring in the denominator of the fraction. The only such letter is $M$, hence $M=0$. Therefore, $M\cdot A\cdot N\cdot G\cdot O=0$ regardless of the values assigned to the remaining letters, so the product can take exactly one value.

Answer:

$24$

Let $S$ denote the position of the ship, let $A$ and $B$ be the two end lighthouses, and let $M$ be the middle lighthouse. Since $\mathit{MS}=\mathit{MA}=\mathit{MB}=13$, the points $A$, $B$, and $S$ all lie on a circle with center $M$. Moreover, the segment $\mathit{AB}$ is a diameter of this circle. By Thales’ theorem, the triangle $\mathit{ABS}$ is therefore right-angled at $S$. Since $\mathit{AB}=\mathit{MA}+\mathit{MB}=26$ and $\mathit{AS}=10$, the Pythagorean theorem yields

Hence, the distance from the ship to the other end lighthouse is $24\mathrm{km}$.

Answer:

$81$

Under the given conditions, the coloring of the pyramid is completely determined by the coloring of the bottom row, since every coloring of the bottom row can be uniquely completed to a coloring of the whole pyramid. Because three colors can be chosen for each building block, there are a total of $3^4=81$ possible colorings.

Answer:

$93$

Let $n$ be a positive integer for which such an arrangement exists. Then the number $1$ must be paired with $n+1$ (in the sense of being assigned to two opposite vertices; we will refer to such pairs simply as pairs in the sequel). Next, $2$ must be paired with $n+2$, and continuing in this way, we see that $k$ must be paired with $n+k$ for all $k=1,2,\dots ,n$. In particular, $n$ is paired with $2n$.

Thus every number from the block $\{1,2,\dots ,2n\}$ is paired with another number from the same block, and this block can be separated from the remaining numbers $\{2n+1,\dots ,100\}$. If $2n=100$, the process terminates. Otherwise, we repeat the same argument starting from $2n+1$, which must be paired with $3n+1$, and continue analogously.

Eventually, the set $\{1,2,\dots ,100\}$ splits into disjoint blocks of length $2n$. This is possible if and only if $100$ is divisible by $2n$, that is, if and only if $n$ is a positive divisor of $50$.

It is easy to see that this construction yields a valid pairing for every such divisor (by assembling $\frac{100}{2n}$ blocks as above). Therefore, the answer is the sum of all positive divisors of $50$, which is

Answer:

$\frac{6}{5}=1.2$

The grey triangle is similar to the large right triangle with the length of the shortest side equal to $13-12=1$, since both are right triangles and share another angle (the top one in our diagram).

Therefore, the similarity ratio between the two right triangles is $1:5$ and the second leg in the grey triangle has a length of $\frac{12}{5}$. As a consequence, the area sought is

Answer:

$7$, $14$

Let the dwarves’ numbers be $a_1,a_2,\dots ,a_7$. We first observe that the liar must be either the 6th or the 7th dwarf.

Indeed, if the 7th dwarf told the truth, then $a_7=a_1+\cdots +a_6$, so the total sum satisfies $a_1+\cdots +a_7=2a_7=46$, hence $a_7=\frac{46}{2}=23$. If the 6th dwarf were also telling the truth, then similarly $a_6=a_1+\cdots +a_5$, so $a_7=a_1+\cdots +a_6=2a_6$, which would force $a_6=\frac{23}{2}$, a contradiction since $a_6$ must be an integer. Thus, in this situation the 6th dwarf must be the liar.

Therefore the first five statements are true, giving

so $a_1+\cdots +a_5=16a_1$. Moreover, at least one of $a_6$ or $a_7$ (namely the one corresponding to a truthful dwarf) is greater than or equal to $a_1+\cdots +a_5=16a_1$, and hence

Since the total sum is $46$, this implies $32a_1\le 46$, so $a_1=1$ ($a_1$ is a positive integer), and consequently $a_2=1$, $a_3=2$, $a_4=4$, and $a_5=8$. In particular, we obtain $a_1+\cdots +a_5=16$.

Now consider the last two dwarves. If the 6th dwarf told the truth, then $a_6=16$, so to obtain the total sum $46$ we must have $a_7=46-(16+16)=14$. If instead the 6th dwarf lied, then the 7th dwarf is truthful and $a_7=a_1+\cdots +a_6=16+a_6$. By the earlier computation, $a_7=23$ in this case, which leads to $a_6=7$.

Both configurations satisfy the requirement that exactly one dwarf lies, so we conclude that the lying dwarf could have chosen either $7$ or $14$.

Answer:

$1708$

Two chosen channels differ by an odd number exactly when they have opposite parity. Rather than counting directly, we count the complement: we subtract the selections that have no odd differences from all possible selections.

A set of six channels has no odd differences precisely when every pair has even difference, which happens only if all six channels have the same parity. In $\{1,2,\dots ,13\}$ there are seven odd channels and six even channels, so the number of all-odd or all-even selections is $\left(\genfrac{}{}{0.0pt}{}{7}{6}\right)+\left(\genfrac{}{}{0.0pt}{}{6}{6}\right)=8$.

Since there are $\left(\genfrac{}{}{0.0pt}{}{13}{6}\right)=1716$ total choices of six channels, the number of sets with at least one odd difference is $1716-8=1708$.

Answer:

$36$

Since $N$ has seven distinct non-zero digits, at least one is even, so $N$ is even. If $5$ were a digit, then divisibility by $5$ would force $N$ to end in $0$, contradicting that all digits are non-zero; hence $5$ is excluded. Now $N$ uses seven digits from $\{1,2,3,4,6,7,8,9\}$, so it omits exactly one digit from this set. It cannot omit $9$: if $9$ were omitted, the digit set would be $\{1,2,3,4,6,7,8\}$ with sum $31$, so $N$ would fail the divisibility-by-$3$ test despite containing the digit $3$, a contradiction. Hence $9$ must appear among the digits of $N$, so $N$ (and thus its digit sum) is divisible by $9$, and since $1+2+3+4+6+7+8+9=40$, we must omit $4$ to obtain the digit sum $36$, the only possible multiple of $9$. One example of such a number is $N=9867312$.

Answer:

$\frac{8}{9}$

A regular octahedron with edge-length $a$ has volume $k{a}^3$ for some constant $k$ (in fact, it is not difficult to show that $k=\sqrt{2}∕3$, but this is not needed in the solution). To make the truncated octahedron we remove half an octahedron (a square-based pyramid) of edge-length $a∕3$ from each of the $6$ vertices of the octahedron. The volume removed is equivalent to that of three octahedra each with edge length $\frac{a}{3}$. The total volume removed is therefore $3\cdot k{\left(\frac{a}{3}\right)}^3=k{a}^3∕9$. So the volume of the truncated octahedron as a fraction of the volume of the original octahedron is

Answer:

$\pm 1$, $\pm 2$

Let $a=2^x+2^{-x}$. Then $4^x+4^{-x}=a^2-2$. Substituting into the equation gives

which simplifies to

Solving this quadratic equation leads to

with the results

In the first case, we have

which leads to $2^x=4$ or $2^x=\frac{1}{4}$, giving $x=2$ or $x=-2$. In the second case, we have

which leads to $2^x=2$ or $2^x=\frac{1}{2}$, giving $x=1$ or $x=-1$. Therefore the solutions to the original equation are $x=\pm 1,\pm 2$.

Answer:

$28$

Let us color the eight corner cities in two colors like a 3D checkerboard so that every railroad segment (whether it is an edge railroad or a tunnel) always goes from a city of one color to a city of the other color. Since the visitor starts at a city of one color and must end at a neighboring city of the other color in exactly five segments, the route must alternate colors and therefore visit exactly two intermediate cities of each color, with all visited cities distinct. Moreover, in the present configuration with tunnels, every pair of cities of different colors is directly connected by a railroad segment.

First we count all length-$5$ routes, disregarding the tunnel condition. Each route has the form

where $o_1$, $o_2$ are distinct cities of $B$’s color, and $e_2$, $e_3$ are distinct cities of $A$’s color. There are $3\cdot 2=6$ ways to choose the ordered pair $(o_1,o_2)$ from the three cities of $B$’s color other than $B$, and likewise there are $3\cdot 2=6$ ways to choose the ordered pair $(e_2,e_3)$ from the three cities of $A$’s color other than $A$. Hence there are $6\cdot 6=36$ length-$5$ routes in total.

Let us now subtract the routes that use no tunnel at all, i.e. routes using only edge railroads. The first move cannot go directly to $B$, so only two first moves are possible. From each of these two first moves there turn out to be exactly $4$ ways to complete a length-$5$ route to $B$ using only edges, giving $2\cdot 4=8$ edge-only routes.

Therefore, the number of length-$5$ routes that use at least one tunnel is $36-8=28$.

Answer:

$63$

Denote the individual numbers of votes by $a$, $b$, $c$, and $d$. These are nonnegative integers satisfying

If we add $c$ to both sides of the third relation, we obtain

Therefore $c=790$, from which it further follows that

Substituting for $b$, we get

Since $d<800$ and at the same time $d>c=790$, we have $791\le d\le 799$. Hence

The only number divisible by $7$ in the given interval is $441=7\cdot 63$. Therefore the answer is $a=63$. One can then compute that $b=378$, $c=790$, $d=795$, and all the conditions from the problem are indeed satisfied.

Answer:

$12∕7$

Let $a$ and $b$ be the side lengths of the squares $\mathit{EFGH}$ and $\mathit{KGIJ}$. From the problem statement, we get the equations $7=\mathit{AD}=\mathit{JK}+\mathit{HE}=a+b$ and $1=\mathit{HK}=\mathit{HG}-\mathit{KG}=a-b$. The solutions of this system of linear equations are $a=4$ and $b=3$.

The straight line passing through the centers of all the three squares divides the area of each square into two halves. Therefore, the area of the rectangle $\mathit{FBCI}$ can be computed by

The area of the rectangle $\mathit{FBCI}$ is also given by $\mathit{FB}\cdot (a+b)$. From these two equations, we obtain $\mathit{FB}=\frac{\mathit{ab}}{a+b}=\frac{12}{7}$.

Answer:

$10010$

Let the sum of the initial $n$ numbers be $S$. Then

These equations can be simplified to

Subtracting the second identity from the first gives

Since $n\ne 0$, it follows that $n=5$. Therefore, $S=2022n-4n^2=10010$.

Answer:

$80$

After $n$ jumps, the flea will have covered

degrees. We are looking for the smallest $n$ such that this number is a multiple of $360$; equivalently, we want $n(n+1)$ to be a multiple of $720=16\cdot 9\cdot 5$. Each of the three factors has to divide either $n$ or $n+1$, since any two consecutive numbers are coprime. Let us start looking at multiples of the largest factor $16$ – this gives candidates for $n$: $15$, $16$, $31$, $32$, $47$, $48$, $63$, $64$, $79$, and finally $80$, which is itself divisible by $5$ and $n+1=81$ is divisible by $9$. We conclude that $80$ is the sought number of jumps.

Answer:

$768$

Since four apprentices are selected and each guild must contribute at least one apprentice, the apprentice distribution among the guilds must be $(2,1,1)$ in some order. Similarly, four masters are selected and each guild must contribute at least one master, so the master distribution must also be $(2,1,1)$.

There are two cases, according to how the guild that contributes two apprentices relates to the guild that contributes two masters.

In the first case, a single guild supplies both two apprentices and two masters; let us denote this guild by $X$. There are $3$ possible choices for $X$. From the remaining two guilds, the apprentices can be chosen in $2\cdot 2=4$ ways, and likewise the masters can be chosen in $2\cdot 2=4$ ways.

For each such selection, the two apprentices from $X$ must be paired with the two masters coming from the other two guilds, which can be done in $2$ ways. Once this is fixed, the remaining two apprentices must be paired with the two masters from $X$, which can again be done in $2$ ways. Therefore, there are $2\cdot 2=4$ valid ways to pair masters with apprentices in this case. Hence this case contributes $3\cdot 4\cdot 4\cdot 4=192$ pairings.

In the second case, one guild ($X$) supplies two apprentices, while a different guild ($Y$) supplies two masters. There are $3$ choices for $X$ and, once it is fixed, $2$ choices for $Y$. The remaining apprentices can be chosen from the two guilds different from $X$ in $2\cdot 2=4$ ways, and the remaining masters can be chosen in the same number of ways. For any such selection, there are $3\cdot 2=6$ admissible ways to form the master–apprentice pairs: the two apprentices from $X$ may be assigned to any two of the three available masters, and once this choice is made the remaining apprentices and masters are forced to pair in a unique way. Consequently, this case contributes $3\cdot 2\cdot 4\cdot 4\cdot 6=576$ pairings.

Adding both cases yields $192+576=768$ possible battle pair formations.

Answer:

$3\sqrt[3]{2}$

Notice that

Further,

Hence,

Answer:

$38^{\circ }$

The angles $\mathit{\angle ADB}$ and $\mathit{\angle ACB}$ are equal since they subtend the same arc. Hence

Therefore $\mathit{CX}$ is the bisector of $\mathit{\angle ACB}$ and $X$ is in turn the common intersection of all three angle bisectors of $△\mathit{ABC}$ (known also as the incenter of triangle $\mathit{ABC}$).

Now we compute the desired angle:

where we used that $\mathit{\angle BAC}=\mathit{\angle BDC}$, which follows (analogously to the very first angle equality) from the fact that they both subtend the same circular arc.

Answer:

$29$

Let $n_1$ and $g_1$ denote the number of neurons and glia in one of the hemispheres and $n_2$ and $g_2$ in the other one. We know that $n_1{n}_2=168$, $g_1{g}_2=48$, and $n_1{g}_2+n_2{g}_1=191$. Therefore

so, since there must be at least one cell of each type on each hemisphere, without loss of generality, we can state that $n_1+g_1=11$. Likewise

which, together with the fact that $n_1$ is a divisor of $168$ and $g_1$ is a divisor of $48$, easily implies that $n_1=8$, $g_1=3$, hence $n_2=168∕8=21$, and the total number of neurons is $n_1+n_2=29$.

Alternative solution. Let us use the notation as in the previous solution. The number $n_1{g}_2+n_2{g}_1$ is odd, so one of the two summands is odd; without loss of generality, let it be $n_1{g}_2$. This implies that both $n_1$ and $g_2$ are odd. On the other hand, since $168=2^3\cdot 21$ and $48=2^4\cdot 3$, $n_2$ must be divisible by $2^3$ and $g_1$ by $2^4$. Therefore $n_2{g}_1$ is a multiple of $2^3\cdot 2^4=2^7=128$, and as it is smaller than $191$, it must equal $128$, with $n_2=2^3=8$ (and $g_1=2^4$). We conclude that $n_1=168∕8=21$, and the result follows.

Answer:

$75$

Because Agnes has only five triminoes, the path cannot ever move south or west: any such backward move would waste grid distance that cannot be recovered with only five pieces. Thus the endpoint of the growing path always progresses northward and/or eastward. We count the paths by filling a $7\times 9$ table: In each grid square we write the number of trimino paths whose current endpoint (the free end square of the last trimino) lies in that square.

After placing the first trimino from the bottom-left corner, the endpoint must be exactly two squares away: either two to the right (horizontal first trimino) or two up (vertical first trimino). Therefore we place a $1$ in the squares $(0,2)$ and $(2,0)$ (coordinates counted from the bottom-left).

Now fill the rest of the table moving from bottom-left toward top-right. To determine the value at a square $(x,y)$, consider the ways a trimino can end there. A trimino ending at $(x,y)$ may be vertical or horizontal, and in either case it must attach to the previous trimino at an end square. This gives four possible predecessor endpoints: $(x,y-3)$ and $(x-1,y-2)$ in the vertical case, and $(x-3,y)$ and $(x-2,y-1)$ in the horizontal case. Each of these corresponds to a valid way to extend an existing path by one trimino. Therefore, the value written in $(x,y)$ is the sum of the values already written in the squares

with any square outside the grid contributing $0$.

After filling all squares this way, the value 75 in the top-right corner square is exactly the number of valid trimino paths from the bottom-left corner to the top-right corner.

Alternative solution. Place the five I-triminoes end-to-end starting at the bottom-left corner $(0,0)$ in the following “skeletal” way: consecutive triminoes meet only at a corner (end-square corner to end-square corner). Each trimino advances the endpoint by either $(+1,+3)$ (an “up” trimino) or $(+3,+1)$ (a “right” trimino). After 5 triminoes the endpoint is $(x,y)$ with $x+y=20$. If $U$ is the number of up triminoes and $R$ the number of right triminoes, then $U+R=5$ and $(x,y)=(U+3R,3U+R)$.

To turn the corner-touching chain into a valid path where consecutive triminoes share an edge, we “fix” each of the 4 joints between consecutive triminoes. Let the $i$-th joint be the connection between triminoes $i$ and $i+1$; at each joint we choose one of two operations: either shift the entire suffix of triminoes $i+1,i+2,\dots ,5$ one unit to the left, or shift that suffix one unit downward. This ensures that triminoes $i$ and $i+1$ share a full grid edge instead of just a corner.

Let $\ell$ be the total number of joints at which we choose a left shift, and let $d$ be the total number of joints at which we choose a downward shift. Since there are exactly 4 joints, we have $\ell +d=4$. After performing all 4 shifts, the endpoint $(x,y)$ of the skeletal chain is translated to $(x-\ell ,y-d)$. To obtain a valid path from $(0,0)$ to the top-right corner $(9,7)$, this final endpoint must satisfy $(x-\ell ,y-d)=(9,7)$.

Let us now check the six $(U,R)$ cases:

• $(5,0)$: $(x,y)=(5,15)$ cannot reach $(9,7)$ by 4 left/down shifts.
• $(4,1)$: $(x,y)=(7,13)$ is impossible to fix, too.
• $(3,2)$: $(x,y)=(9,11)$ requires $\ell =0$, $d=4$ (all shifts down). There are $\left(\genfrac{}{}{0.0pt}{}{5}{3}\right)=10$ skeletal chains and only 1 fix, giving $10$ valid paths.
• $(2,3)$: $(x,y)=(11,9)$ requires $\ell =2$, $d=2$. There are $\left(\genfrac{}{}{0.0pt}{}{5}{2}\right)=10$ skeletal chains and $\left(\genfrac{}{}{0.0pt}{}{4}{2}\right)=6$ ways to choose which joints shift left (the rest shift down), giving $10\cdot 6=60$ valid paths.
• $(1,4)$: $(x,y)=(13,7)$ requires $\ell =4$, $d=0$ (all shifts left). There are $\left(\genfrac{}{}{0.0pt}{}{5}{1}\right)=5$ skeletal chains and 1 fix, giving $5$ valid paths.
• $(0,5)$: $(x,y)=(15,5)$ is impossible to fix.

Therefore, the total number of valid paths is $10+60+5=75$.

Answer:

$507$

Let $X=\frac{a^2}{a+2b}+\frac{b^2}{b+2c}+\frac{c^2}{c+2d}+\frac{d^2}{d+2a}=2026$ and $Y=\frac{b^2}{a+2b}+\frac{c^2}{b+2c}+\frac{d^2}{c+2d}+\frac{a^2}{d+2a}$.

Since we have $a+2b$ in the denominator, we can try to simplify it if we have a $(a+2b)(a-2b)=a^2-4b^2$ as the numerator, and so on for the remaining fractions. Then:

From this we obtain $X-4Y=-2$, hence $Y=\frac{1}{4}(X+2)=\frac{2026+2}{4}=507$.

Answer:

$2+\sqrt{2}$

Denote by $C_1$ and $C_2$ the centers of the two balls, and let $d$ be the horizontal distance between the two centers, i.e. the length of the projection of the segment $C_1{C}_2$ onto the top (or bottom) face of the box. Viewing the configuration from above, we obtain a square of side length $3$, where $C_1$ is at distance $1$ from two adjacent sides and $C_2$ is at distance $1$ from the other two adjacent sides.

Hence, in this top view, the projections of $C_1$ and $C_2$ lie at opposite corners of a square of side length $1$. Therefore, $d=\sqrt{2}$.

Next, consider the vertical plane perpendicular to the base that passes through $C_1$ and $C_2$. The intersection of this plane with the box is a rectangle of the sought height $a$, and the segment $C_1{C}_2$ lies in this plane. Let $x$ denote the vertical distance between $C_1$ and $C_2$.

Since the two balls touch each other and each has radius $1$, the distance between their centers is $C_1{C}_2=2$. In the right triangle with horizontal leg $d$ and vertical leg $x$, we have $C_1{C}_2^2=d^2+x^2$, so $x=\sqrt{2^2-d^2}=\sqrt{2}$.

Finally, the first ball touches the bottom face and the second ball touches the top face, so the height of the box is the sum of the radius of the first ball, the vertical distance between the centers, and the radius of the second ball: $a=1+x+1=2+\sqrt{2}$.

Answer:

$259200$

Since $0+1+\cdots +9=45$ is odd, omitting an odd digit yields an even sum over the whole table and hence all the row and column sums must be even. That can be arranged using $4$ odd and $5$ even digits only if one row and one column (5 cells) contain exactly the even numbers. Hence we can count all such tables by choosing the omitted odd digit ($5$ ways), then choosing the row and column with only even digits ($3\cdot 3=9$ ways), then placing the odd digits in any order ($4!=24$ ways) and finally placing the even digits in any order ($5!=120$ ways).

Similarly, if we omit an even digit, the sum of all three rows will be odd, and hence all rows and columns must yield odd sums. The only way to place five odd digits in a $3\times 3$ table so that each row or column contains one or three of them is to put them in one row and one column. The number of tables in this case is computed analogously as in the previous case, only with interchanged roles of even and odd digits. Therefore the total number equals $2\cdot 5\cdot 9\cdot 24\cdot 120=259200$.