Squares $ABCD$ and $EFGH$ are congruent, $AB=10$, and $G$ is the center of square $ABCD$. The area of the region in the plane covered by these squares is

1. **Identify the given information**: We are given two congruent squares $ABCD$ and $EFGH$ with side length $AB = 10$. The point $G$ is the center of square $ABCD$.

2. **Calculate the area of each square**:
   - The area of square $ABCD$ is $AB^2 = 10^2 = 100$.
   - Since $EFGH$ is congruent to $ABCD$, its area is also $100$.

3. **Calculate the total area if there were no overlap**:
   - If there were no overlap, the total area would be $100 + 100 = 200$.

4. **Determine the overlap**:
   - Since $G$ is the center of $ABCD$, it lies at the midpoint of the diagonal of $ABCD$.
   - The length of the diagonal of $ABCD$ is $AB \sqrt{2} = 10\sqrt{2}$.
   - Therefore, $BG$, being half the diagonal, is $5\sqrt{2}$.

5. **Analyze triangle $ABG$**:
   - $\triangle ABG$ is formed by the line segment $AB$, $BG$, and $AG$.
   - Since $G$ is the midpoint of the diagonal, and the diagonals of a square bisect each other at right angles, $\triangle ABG$ is a right isosceles triangle with $\angle BGA = 90^\circ$.

6. **Calculate the area of $\triangle ABG$**:
   - The legs of $\triangle ABG$ are each $5\sqrt{2}$.
   - The area of $\triangle ABG$ is $\frac{1}{2} \times \text{leg} \times \text{leg} = \frac{1}{2} \times (5\sqrt{2})^2 = \frac{1}{2} \times 50 = 25$.

7. **Calculate the area of the region covered by the squares**:
   - Subtract the area of $\triangle ABG$ from the total area of the two squares to account for the overlap.
   - The area of the region is $200 - 25 = 175$.

Thus, the area of the region in the plane covered by these squares is $\boxed{\textbf{(E)}\ 175}$.

Answer: 175


Given that the last initial of Mr. and Mrs. Alpha's baby's monogram is 'A', determine the number of possible monograms in alphabetical order with no letter repeated.

Answer: 300


Given $\sin\alpha= \frac {2 \sqrt {2}}{3}$, $\cos(\alpha+\beta)=- \frac {1}{3}$, and $\alpha, \beta\in(0, \frac {\pi}{2})$, determine the value of $\sin(\alpha-\beta)$.

Answer: \frac {10 \sqrt {2}}{27}


In acute triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given that $b=2$, $B= \frac{π}{3}$, and $c\sin A= \sqrt{3}a\cos C$, find the area of $\triangle ABC$.

Answer: \sqrt{3}


If \( A \) is a positive integer such that \( \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \cdots + \frac{1}{(A+1)(A+3)} = \frac{12}{25} \), find the value of \( A \).

Answer: 22


The sequence \(\{a_n\}\) has consecutive terms \(a_n\) and \(a_{n+1}\) as the roots of the equation \(x^2 - c_n x + \left(\frac{1}{3}\right)^n = 0\), with initial term \(a_1 = 2\). Find the sum of the infinite series \(c_1, c_2, \cdots, c_n, \cdots\).

Answer: \frac{9}{2}


Given two circles $O\_1$: $(x-2)^2 + y^2 = 16$ and $O\_2$: $x^2 + y^2 = r^2 (0 < r < 2)$, a moving circle $M$ is tangent to both circles $O\_1$ and $O\_2$. The locus of the center of the moving circle $M$ forms two ellipses, with eccentricities $e\_1$ and $e\_2 (e\_1 > e\_2)$. Find the minimum value of $e\_1 + 2e\_2$.

Answer: \frac{3+2\sqrt{2}}{4}


Two cars, Car A and Car B, start from point A and point B, respectively, and move towards each other simultaneously. They meet after 3 hours, after which Car A turns back towards point A and Car B continues moving forward. Once Car A reaches point A, it turns towards point B again and meets Car B half an hour later. How many minutes does Car B take to travel from point A to point B?

Answer: 432


Suppose $\cos S = 0.5$ in the diagram below. What is $ST$?

[asy]
pair P,S,T;
P = (0,0);
S = (6,0);
T = (0,6*tan(acos(0.5)));
draw(P--S--T--P);
draw(rightanglemark(S,P,T,18));
label("$P$",P,SW);
label("$S$",S,SE);
label("$T$",T,N);
label("$10$",S/2,S);
[/asy]

Answer: 20


Let $f(x)$ have a domain of $R$, $f(x+1)$ be an odd function, and $f(x+2)$ be an even function. When $x\in [1,2]$, $f(x)=ax^{2}+b$. If $f(0)+f(3)=6$, then calculate the value of $f\left(\frac{9}{2}\right)$.

Answer: \frac{5}{2}


An ant starts at the origin of a coordinate plane. Each minute, it either walks one unit to the right or one unit up, but it will never move in the same direction more than twice in the row. In how many different ways can it get to the point $(5,5)$ ?

We can change the ant's sequence of moves to a sequence $a_{1}, a_{2}, \ldots, a_{10}$, with $a_{i}=0$ if the $i$-th step is up, and $a_{i}=1$ if the $i$-th step is right. We define a subsequence of moves $a_{i}, a_{i+1}, \ldots, a_{j}$, ( $i \leq j$ ) as an up run if all terms of the subsequence are equal to 0 , and $a_{i-1}$ and $a_{j+1}$ either do not exist or are not equal to 0 , and define a right run similarly. In a sequence of moves, up runs and right runs alternate, so the number of up rights can differ from the number of right runs by at most one. Now let $f(n)$ denote the number of sequences $a_{1}, a_{2}, \ldots, a_{n}$ where $a_{i} \in\{1,2\}$ for $1 \leq i \leq n$, and $a_{1}+a_{2}+\cdots+a_{n}=5$. (In essence, we are splitting the possible 5 up moves into up runs, and we are doing the same with the right moves). We can easily compute that $f(3)=3, f(4)=4, f(5)=1$, and $f(n)=0$ otherwise. For each possible pair of numbers of up runs and right runs, we have two choices of which type of run is first. Our answer is then $2\left(f(3)^{2}+f(3) f(4)+f(4)^{2}+f(4) f(5)+f(5)^{2}\right)=2(9+12+16+4+1)=84$.

Answer: 84


Find the shortest distance from the line $3 x+4 y=25$ to the circle $x^{2}+y^{2}=6 x-8 y$.

The circle is $(x-3)^{2}+(y+4)^{2}=5^{2}$. The center $(3,-4)$ is a distance of $$ \frac{|3 \cdot 3+4 \cdot-4-25|}{\sqrt{3^{2}+4^{2}}}=\frac{32}{5} $$ from the line, so we subtract 5 for the radius of the circle and get $7 / 5$.

Answer: 7 / 5


Maximum difference in weights of two bags is achieved by taking the largest and smallest possible values of the two different brands. Given the weights of three brands of flour are $\left(25\pm 0.1\right)kg$, $\left(25\pm 0.2\right)kg$, and $\left(25\pm 0.3\right)kg$, calculate the maximum possible difference in weights.

Answer: 0.6


Let $a, b, c$ be positive real numbers such that $a \leq b \leq c \leq 2 a$. Find the maximum possible value of $$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$

Fix the values of $b, c$. By inspecting the graph of $$f(x)=\frac{b}{x}+\frac{x}{c}$$ we see that on any interval the graph attains its maximum at an endpoint. This argument applies when we fix any two variables, so it suffices to check boundary cases in which $b=a$ or $b=c$, and $c=b$ or $c=2 a$. All pairs of these conditions determine the ratio between $a, b, c$, except $b=c$ and $c=b$, in which case the boundary condition on $a$ tells us that $a=b$ or $2 a=b=c$. In summary, these cases are $$(a, b, c) \in\{(a, a, a),(a, a, 2 a),(a, 2 a, 2 a)\}$$ The largest value achieved from any of these three is $\frac{7}{2}$.

Answer: \frac{7}{2}


Determine the set of all real numbers $p$ for which the polynomial $Q(x)=x^{3}+p x^{2}-p x-1$ has three distinct real roots.

First, we note that $x^{3}+p x^{2}-p x-1=(x-1)(x^{2}+(p+1)x+1)$. Hence, $x^{2}+(p+1)x+1$ has two distinct roots. Consequently, the discriminant of this equation must be positive, so $(p+1)^{2}-4>0$, so either $p>1$ or $p<-3$. However, the problem specifies that the quadratic must have distinct roots (since the original cubic has distinct roots), so to finish, we need to check that 1 is not a double root-we will do this by checking that 1 is not a root of $x^{2}+(p+1)x+1$ for any value $p$ in our range. But this is clear, since $1+(p+1)+1=0 \Rightarrow p=-3$, which is not in the aforementioned range. Thus, our answer is all $p$ satisfying $p>1$ or $p<-3$.

Answer: p>1 \text{ and } p<-3


Let $n$ be the integer such that $0 \le n < 31$ and $3n \equiv 1 \pmod{31}$. What is $\left(2^n\right)^3 - 2 \pmod{31}$?

Express your answer as an integer from $0$ to $30$, inclusive.

Answer: 6


What is the number of units in the distance between $(2,5)$ and $(-6,-1)$?

Answer: 10


Let $\mathrm {Q}$ be the product of the roots of $z^8+z^6+z^4+z^3+z+1=0$ that have a positive imaginary part, and suppose that $\mathrm {Q}=s(\cos{\phi^{\circ}}+i\sin{\phi^{\circ}})$, where $0<s$ and $0\leq \phi <360$. Find $\phi$.

Answer: 180


What is the smallest integral value of $k$ such that
$2x(kx-4)-x^2+6=0$ has no real roots?

1. **Expand and simplify the given quadratic equation**:
   \[
   2x(kx-4) - x^2 + 6 = 0
   \]
   Expanding the terms:
   \[
   2kx^2 - 8x - x^2 + 6 = 0
   \]
   Combine like terms:
   \[
   (2k-1)x^2 - 8x + 6 = 0
   \]

2. **Condition for no real roots**:
   A quadratic equation $ax^2 + bx + c = 0$ has no real roots if its discriminant $b^2 - 4ac$ is negative. For the equation $(2k-1)x^2 - 8x + 6 = 0$, we identify $a = 2k-1$, $b = -8$, and $c = 6$.

3. **Calculate the discriminant and set it to be less than zero**:
   \[
   \text{Discriminant} = (-8)^2 - 4(2k-1)(6)
   \]
   Simplify the expression:
   \[
   64 - 4(2k-1)(6) = 64 - (48k - 24) = 88 - 48k
   \]
   Set the discriminant to be less than zero:
   \[
   88 - 48k < 0
   \]

4. **Solve the inequality for $k$**:
   \[
   88 - 48k < 0 \implies 88 < 48k \implies \frac{88}{48} < k \implies \frac{11}{6} < k
   \]
   Since $k$ must be an integer, the smallest integer greater than $\frac{11}{6}$ is $2$.

5. **Conclusion**:
   The smallest integral value of $k$ such that the quadratic equation has no real roots is $2$.

   \[
   \boxed{\text{B}}
   \]

Answer: 2


Let $p$ be the largest prime with 2010 digits. What is the smallest positive integer $k$ such that $p^2 - k$ is divisible by 12?

Answer: k = 1


Starting at 1:00 p.m., Jorge watched three movies. The first movie was 2 hours and 20 minutes long. He took a 20 minute break and then watched the second movie, which was 1 hour and 45 minutes long. He again took a 20 minute break and then watched the last movie, which was 2 hours and 10 minutes long. At what time did the final movie end?

Starting at 1:00 p.m., Jorge watches a movie that is 2 hours and 20 minutes long. This first movie ends at 3:20 p.m. Then, Jorge takes a 20 minute break. This break ends at 3:40 p.m. Then, Jorge watches a movie that is 1 hour and 45 minutes long. After 20 minutes of this movie, it is 4:00 p.m. and there is still 1 hour and 25 minutes left in the movie. This second movie thus ends at 5:25 p.m. Then, Jorge takes a 20 minute break which ends at 5:45 p.m. Finally, Jorge watches a movie that is 2 hours and 10 minutes long. This final movie ends at 7:55 p.m.

Answer: 7:55 \text{ p.m.}


Given 365 cards, in which distinct numbers are written. We may ask for any three cards, the order of numbers written in them. Is it always possible to find out the order of all 365 cards by 2000 such questions?

Answer: 1691


In a certain country, there are 100 senators, each of whom has 4 aides. These senators and aides serve on various committees. A committee may consist either of 5 senators, of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5 committees, and every aide serves on 3 committees. How many committees are there altogether?

If each senator gets a point for every committee on which she serves, and every aide gets $1 / 4$ point for every committee on which he serves, then the 100 senators get 500 points altogether, and the 400 aides get 300 points altogether, for a total of 800 points. On the other hand, each committee contributes 5 points, so there must be $800 / 5=160$ committees.

Answer: 160


\(\triangle ABC\) is isosceles with base \(AC\). Points \(P\) and \(Q\) are respectively in \(CB\) and \(AB\) and such that \(AC=AP=PQ=QB\).
The number of degrees in \(\angle B\) is:

1. **Identify the properties of the triangles**: Given that $\triangle ABC$ is isosceles with base $AC$, and $AC = AP = PQ = QB$, we can infer that $\triangle APQ$ and $\triangle BQP$ are also isosceles. Additionally, $\triangle APC$ and $\triangle BQC$ are isosceles because $AP = AC$ and $BQ = BC$.

2. **Analyze $\triangle BQP$**: Since $\triangle BQP$ is isosceles with $PQ = QB$, the base angles are equal. Let $\angle BPQ = \angle BQP = b$. By the Triangle Angle Sum Property, the sum of angles in a triangle is $180^\circ$. Therefore,
   \[
   \angle BPQ + \angle BQP + \angle PQB = 180^\circ \implies b + b + \angle PQB = 180^\circ \implies \angle PQB = 180^\circ - 2b.
   \]

3. **Analyze $\triangle APQ$**: Since $\triangle APQ$ is isosceles with $AP = PQ$, the base angles are equal. Let $\angle PAQ = \angle PQA = 2b$ (since $\angle PQA$ is an exterior angle to $\triangle BQP$ and equal to $2b$). Again, using the Triangle Angle Sum Property,
   \[
   \angle PAQ + \angle PQA + \angle QPA = 180^\circ \implies 2b + 2b + \angle QPA = 180^\circ \implies \angle QPA = 180^\circ - 4b.
   \]

4. **Analyze $\triangle APC$**: Since $\triangle APC$ is isosceles with $AP = AC$, the base angles are equal. Let $\angle PAC = \angle PCA = b$. Since $\angle APC$ is an exterior angle to $\triangle APQ$,
   \[
   \angle APC = 180^\circ - \angle QPA = 180^\circ - (180^\circ - 4b) = 4b.
   \]

5. **Analyze $\triangle ABC$**: Since $\triangle ABC$ is isosceles with $AB = BC$, the base angles are equal. Let $\angle BAC = \angle BCA = 3b$. Using the Triangle Angle Sum Property,
   \[
   \angle BAC + \angle BCA + \angle B = 180^\circ \implies 3b + 3b + b = 180^\circ \implies 7b = 180^\circ.
   \]

6. **Solve for $b$**: Solving for $b$,
   \[
   b = \frac{180^\circ}{7} = 25\frac{5}{7}^\circ.
   \]

7. **Conclusion**: Therefore, the measure of $\angle B$ in $\triangle ABC$ is $\boxed{25\frac{5}{7}^\circ}$, corresponding to choice $\textbf{(A)}\ 25\frac{5}{7}$.

Answer: 25\frac{5}{7}


Given that the sum of two numbers is 15 and their product is 36, find the sum of their reciprocals and the sum of their squares.

Answer: 153


Given $f(x)= \sqrt {3}\sin \dfrac {x}{4}\cos \dfrac {x}{4}+ \cos ^{2} \dfrac {x}{4}+ \dfrac {1}{2}$.
(1) Find the period of $f(x)$;
(2) In $\triangle ABC$, sides $a$, $b$, and $c$ correspond to angles $A$, $B$, and $C$ respectively, and satisfy $(2a-c)\cos B=b\cos C$, find the value of $f(B)$.

Answer: \dfrac{\sqrt{3}}{2} + 1


Find the area of the  $MNRK$  trapezoid with the lateral side   $RK = 3$  if the distances from the vertices  $M$  and  $N$  to the line  $RK$  are  $5$  and  $7$ , respectively.

Answer: 18


An assembly line produces, on average, 85% first grade products. How many products need to be sampled so that, with a probability of 0.997, the deviation of the proportion of first grade products from 0.85 in absolute value does not exceed 0.01?

Answer: 11475


Find a six-digit number where the first digit is 6 times less than the sum of all the digits to its right, and the second digit is 6 times less than the sum of all the digits to its right.

Answer: 769999


There are 12 sprinters in an international track event final, including 5 Americans. Gold, silver, and bronze medals are awarded to the first, second, and third place finishers respectively. Determine the number of ways the medals can be awarded if no more than two Americans receive medals.

Answer: 1260


In an opaque bag, there are four small balls labeled with the Chinese characters "阳", "过", "阳", and "康" respectively. Apart from the characters, the balls are indistinguishable. Before each draw, the balls are thoroughly mixed.
$(1)$ If one ball is randomly drawn from the bag, the probability that the character on the ball is exactly "康" is ________;
$(2)$ If two balls are drawn from the bag by person A, using a list or a tree diagram, find the probability $P$ that one ball has the character "阳" and the other ball has the character "康".

Answer: \frac{1}{3}


If the numbers $x$ and $y$ are inversely proportional and when the sum of $x$ and $y$ is 54, $x$ is three times $y$, find the value of $y$ when $x = 5$.

Answer: 109.35


Given that the roots of the polynomial $81x^3 - 162x^2 + 81x - 8 = 0$ are in arithmetic progression, find the difference between the largest and smallest roots.

Answer: \frac{4\sqrt{6}}{9}


If $k$ and $\ell$ are positive 4-digit integers such that $\gcd(k,\ell)=3$, what is the smallest possible value for $\mathop{\text{lcm}}[k,\ell]$?

Answer: 335{,}670


If $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = a \cdot d - b \cdot c$, what is the value of $\begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix}$?

Given the determinant formula for a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is calculated as:
\[ \text{det} = ad - bc \]

For the specific matrix $\begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$, we identify the elements as:
- $a = 3$
- $b = 4$
- $c = 1$
- $d = 2$

Using the determinant formula:
1. Calculate $ad$:
   \[ ad = 3 \times 2 = 6 \]
2. Calculate $bc$:
   \[ bc = 4 \times 1 = 4 \]
3. Subtract $bc$ from $ad$ to find the determinant:
   \[ \text{det} = ad - bc = 6 - 4 = 2 \]

Thus, the value of $\begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$ is $2$.

$\boxed{\text{E}}$

Answer: $2$


Given in $\triangle ABC$, $\tan A$ and $\tan B$ are the two real roots of the equation $x^2 + ax + 4 = 0$:
(1) If $a = -8$, find the value of $\tan C$;
(2) Find the minimum value of $\tan C$, and specify the corresponding values of $\tan A$ and $\tan B$.

Answer: \frac{4}{3}


Given that the sum of the first three terms of a geometric sequence $\{a_n\}$ is $3$ and the sum of the first nine terms is $39$, calculate the value of the sum of the first six terms.

Answer: 12


Suppose you have $6$ red shirts, $7$ green shirts, $9$ pairs of pants, $10$ blue hats, and $10$ red hats, all distinct. How many outfits can you make consisting of one shirt, one pair of pants, and one hat, if neither the hat nor the pants can match the shirt in color?

Answer: 1170


50 students from fifth to ninth grade collectively posted 60 photos on Instagram, with each student posting at least one photo. All students in the same grade (parallel) posted an equal number of photos, while students from different grades posted different numbers of photos. How many students posted exactly one photo?

Answer: 46


What is equivalent to $\sqrt{\frac{x}{1-\frac{x-1}{x}}}$ when $x < 0$?

To solve the problem, we first simplify the expression inside the square root:

\[
\sqrt{\frac{x}{1-\frac{x-1}{x}}}
\]

1. Simplify the denominator:
   \[
   1 - \frac{x-1}{x} = 1 - \left(\frac{x}{x} - \frac{1}{x}\right) = 1 - 1 + \frac{1}{x} = \frac{1}{x}
   \]

2. Substitute back into the original expression:
   \[
   \sqrt{\frac{x}{\frac{1}{x}}} = \sqrt{x^2} = |x|
   \]

   Since $x < 0$, $|x| = -x$.

3. Now, we compare this result with the given options:
   - $\mathrm{(A) \ } -x$
   - $\mathrm{(B) \ } x$
   - $\mathrm{(C) \ } 1$
   - $\mathrm{(D) \ } \sqrt{\frac{x}{2}}$
   - $\mathrm{(E) \ } x\sqrt{-1}$

   Given that $x < 0$, $|x| = -x$ matches with option $\mathrm{(A)}$.

Thus, the equivalent expression for $\sqrt{\frac{x}{1-\frac{x-1}{x}}}$ when $x < 0$ is $\boxed{\textbf{(A)} -x}$.

Answer: -x


Compute the number of ways there are to assemble 2 red unit cubes and 25 white unit cubes into a $3 \times 3 \times 3$ cube such that red is visible on exactly 4 faces of the larger cube. (Rotations and reflections are considered distinct.)

We do casework on the two red unit cubes; they can either be in a corner, an edge, or the center of the face. - If they are both in a corner, they must be adjacent - for each configuration, this corresponds to an edge, of which there are 12. - If one is in the corner and the other is at an edge, we have 8 choices to place the corner. For the edge, the red edge square has to go on the boundary of the faces touching the red corner square, and there are six places here. Thus, we get $8 \cdot 6=48$ configurations. - If one is a corner and the other is in the center of a face, we again have 8 choices for the corner and 3 choices for the center face (the faces not touching the red corner). This gives $8 \cdot 3=24$ options. - We have now completed the cases with a red corner square! Now suppose we have two edges: If we chose in order, we have 12 choices for the first cube. For the second cube, we must place the edge so it covers two new faces, and thus we have five choices. Since we could have picked these edges in either order, we divide by two to avoid overcounting, and we have $12 \cdot 5 / 2=30$ in this case. Now, since edges and faces only cover at most 2 and 1 face respectively, no other configuration works. Thus we have all the cases, and we add: $12+48+24+30=114$.

Answer: 114


The equations $x^3 + Ax + 10 = 0$ and $x^3 + Bx^2 + 50 = 0$ have two roots in common.  Then the product of these common roots can be expressed in the form $a \sqrt[b]{c},$ where $a,$ $b,$ and $c$ are positive integers, when simplified.  Find $a + b + c.$

Answer: 12


Let $\pi$ be a uniformly random permutation of the set $\{1,2, \ldots, 100\}$. The probability that $\pi^{20}(20)=$ 20 and $\pi^{21}(21)=21$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$. (Here, $\pi^{k}$ means $\pi$ iterated $k$ times.)

We look at the cycles formed by $\pi$ Let $\operatorname{ord}_{\pi}(n)$ denote the smallest $m$ such that $\pi^{m}(n)=n$. In particular, the condition implies that $\operatorname{ord}_{\pi}(20) \mid 20$ and $\operatorname{ord}_{\pi}(21) \mid 21$. Claim 1. 20 and 21 cannot be in the same cycle. Proof. If 20 and 21 were in the same cycle, then $x=\operatorname{ord}_{\pi}(20)=\operatorname{ord}_{\pi}(21)$ for some $x$. Then $x>1$ since the cycle contains both 20 and 21, but $x|20, x| 21$ implies $x=1$, a contradiction. Claim 2. The probability that $a=\operatorname{ord}_{\pi}(20), b=\operatorname{ord}_{\pi}(21)$ for some fixed $a, b$ such that $a+b \leq 100$ is $\frac{1}{99 \cdot 100}$. Proof. We can just count these permutations. We first choose $a-1$ elements of $[100] \backslash\{20,21\}$ to be in the cycle of 20, then we similarly choose $b-1$ to be in the cycle of 21. We then have $(a-1)$! ways to reorder within the cycle of $20,(b-1)$! ways to reorder within the cycle of 21, and $(100-a-b)$! ways to permute the remaining elements. The total number of ways is just $$\frac{98!}{(a-1)!(b-1)!(100-a-b)!} \cdot(a-1)!(b-1)!(100-a-b)!=98!$$ so the probability this happens is just $\frac{98!}{100!}=\frac{1}{9900}$. Now, since $\operatorname{ord}_{\pi}(20) \mid 20$ and $\operatorname{ord}_{\pi}(21) \mid 21$, we have 6 possible values for $\operatorname{ord}_{\pi}(20)$ and 4 for $\operatorname{ord}_{\pi}(21)$, so in total we have a $\frac{6 \cdot 4}{9900}=\frac{2}{825}$ probability that the condition is satisfied.

Answer: 1025


Harry Potter is creating an enhanced magical potion called "Elixir of Life" (this is a very potent sleeping potion composed of powdered daffodil root and wormwood infusion. The concentration of the "Elixir of Life" is the percentage of daffodil root powder in the entire potion). He first adds a certain amount of wormwood infusion to the regular "Elixir of Life," making its concentration $9 \%$. If he then adds the same amount of daffodil root powder, the concentration of the "Elixir of Life" becomes $23 \%$. What is the concentration of the regular "Elixir of Life"? $\qquad$ $\%$.

Answer: 11


On an infinite tape, numbers are written in a row. The first number is one, and each subsequent number is obtained from the previous one by adding the smallest nonzero digit of its decimal representation. How many digits are in the decimal representation of the number that is in the $9 \cdot 1000^{1000}$-th position in this row?

Answer: 3001


Find the ordered pair $(x,y)$ if

\begin{align*}
x+y&=(5-x)+(5-y),\\
x-y&=(x-1)+(y-1).
\end{align*}

Answer: (4,1)


In triangle \( ABC \) with sides \( AB = 8 \), \( AC = 4 \), \( BC = 6 \), the angle bisector \( AK \) is drawn, and on the side \( AC \) a point \( M \) is marked such that \( AM : CM = 3 : 1 \). Point \( N \) is the intersection point of \( AK \) and \( BM \). Find \( AN \).

Answer: \frac{18\sqrt{6}}{11}


Given that the function $f(x)$ defined on $\mathbb{R}$ satisfies $f(4)=2-\sqrt{3}$, and for any $x$, $f(x+2)=\frac{1}{-f(x)}$, find $f(2018)$.

Answer: -2-\sqrt{3}


Given vectors $\overrightarrow{m}=(\sin x,-1)$ and $\overrightarrow{n}=(\sqrt{3}\cos x,-\frac{1}{2})$, and the function $f(x)=(\overrightarrow{m}+\overrightarrow{n})\cdot\overrightarrow{m}$.
- (I) Find the interval where $f(x)$ is monotonically decreasing;
- (II) Given $a$, $b$, and $c$ are respectively the sides opposite to angles $A$, $B$, and $C$ in $\triangle ABC$, with $A$ being an acute angle, $a=2\sqrt{3}$, $c=4$, and $f(A)$ is exactly the maximum value of $f(x)$ on the interval $\left[0, \frac{\pi}{2}\right]$, find $A$, $b$, and the area $S$ of $\triangle ABC$.

Answer: 2\sqrt{3}


Find the largest positive integer  $N$  so that the number of integers in the set  $\{1,2,\dots,N\}$  which are divisible by 3 is equal to the number of integers which are divisible by 5 or 7 (or both).

Answer: 65


If \( p \) and \( q \) are positive integers, \(\max (p, q)\) is the maximum of \( p \) and \( q \) and \(\min (p, q)\) is the minimum of \( p \) and \( q \). For example, \(\max (30,40)=40\) and \(\min (30,40)=30\). Also, \(\max (30,30)=30\) and \(\min (30,30)=30\).

Determine the number of ordered pairs \((x, y)\) that satisfy the equation

$$
\max (60, \min (x, y))=\min (\max (60, x), y)
$$

where \(x\) and \(y\) are positive integers with \(x \leq 100\) and \(y \leq 100\).

Answer: 4100


How many 5-digit positive numbers contain only odd numbers and have at least one pair of consecutive digits whose sum is 10?

Answer: 1845


How many three-digit numbers are there in which each digit is greater than the digit to its right?

Answer: 84


In the geometric sequence $\{a_n\}$ with a common ratio greater than $1$, $a_3a_7=72$, $a_2+a_8=27$, calculate $a_{12}$.

Answer: 96


There are two hourglasses - one for 7 minutes and one for 11 minutes. An egg needs to be boiled for 15 minutes. How can you measure this amount of time using the hourglasses?

Answer: 15


Point $(x,y)$ is chosen randomly from the rectangular region with vertices at $(0,0)$, $(3036,0)$, $(3036,3037)$, and $(0,3037)$. What is the probability that $x > 3y$? Express your answer as a common fraction.

Answer: \frac{506}{3037}


The average age of 8 people in Room C is 35. The average age of 6 people in Room D is 30. Calculate the average age of all people when the two groups are combined.

Answer: \frac{460}{14}


How many $4$-digit positive integers (that is, integers between $1000$ and $9999$, inclusive) having only even digits are divisible by $5?$

1. **Identify the range and conditions**: We are looking for 4-digit integers between 1000 and 9999, inclusive, that have only even digits and are divisible by 5.

2. **Digits must be even**: The possible even digits are 0, 2, 4, 6, and 8.

3. **Divisibility by 5**: For a number to be divisible by 5, its units digit must be either 0 or 5. Since we are restricted to even digits, the units digit must be 0.

4. **Thousands digit**: The thousands digit must be one of 2, 4, 6, or 8, as it cannot be 0 (otherwise, it would not be a 4-digit number). This gives us 4 choices for the thousands digit.

5. **Middle two digits**: Each of the middle two digits (hundreds and tens) can independently be any of the 5 even digits (0, 2, 4, 6, 8). This gives us 5 choices for each of these digits.

6. **Calculate the total number of such numbers**:
   - 1 choice for the units digit (0).
   - 5 choices for the hundreds digit.
   - 5 choices for the tens digit.
   - 4 choices for the thousands digit.

   Multiplying these choices together gives the total number of valid numbers:
   \[
   1 \times 5 \times 5 \times 4 = 100
   \]

7. **Conclusion**: There are 100 such numbers that meet all the conditions.

Thus, the number of 4-digit positive integers having only even digits and divisible by 5 is $\boxed{\textbf{(B) } 100}$.

Answer: 100