Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$

Solution: Notice that Samia walks $\frac {17}{5}$ times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is \[\frac{\frac{17}{5}}{\frac{17}{5}+\frac{5}{5}} = \frac{17}{22}\] Then, multiply this by the time \[\frac{17}{22} \cdot 44 \text{minutes} = 34 \text{minutes}\] 34 minutes is a little greater than $\frac {1}{2}$ of an hour so Samia traveled \[\sim \frac {1}{2} \cdot 5 = 2.5 \text{kilometers}\] The answer choice a little greater than 2.5 is $\boxed{2.8}$ by $5$ and gotten the exact answer as well)

Answer: C 2.8
Problem: Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$

Solution: Since the distance between biking and walking is equal, we can use the given rates' harmonic mean to find the average speed.
\[\frac{2ab}{a+b}\implies\frac{2 \cdot 17 \cdot 5}{17+5}=\frac{85}{11}\text{kph}\]
Then, multiplying by $\frac{11}{15}$ hours gives the overall distance $\frac{17}{3}$ kilometers. Samia only walks half of that, so $\frac{17}{6}\approx \boxed{2.8}$ kilometers.

Answer: C 2.8
Problem: Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?
$\textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.2\qquad\textbf{(C)}\ 2.8\qquad\textbf{(D)}\ 3.4\qquad\textbf{(E)}\ 4.4$

Solution: Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\frac{5}{17}$ of the time traveled by foot, or $f$
Accordingly, we have $\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ minutes on bicycle
Because $10$ minutes on bike yields $\frac{17}{6}$ kilometers, and distance on bike = distance on foot, we have the final answer of $\frac{17}{6} \approx \boxed{2.8}$

Answer:  2.8
Problem: Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

Solution: Since $AB = AC$ , then $\triangle ABC$ is isosceles, so $BD = CD$ . Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{4,9}$

Answer: C 4,9
Problem: Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

Solution: Calculating the equation of the line running between points $B$ and $D$ $y = -2x + 1$ . The only coordinate of $C$ that is also on this line is $\boxed{4,9}$

Answer: C 4,9
Problem: Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

Solution: Similar to the first solution, because the triangle is isosceles, then the line drawn in the middle separates the triangle into two smaller congruent triangles. To get from $B$ to $D$ , we go to the right $3$ and up $6$ . Then to get to point $C$ from point $D$ , we go to the right $3$ and up $6$ , getting us the coordinates $\boxed{4,9}$

Answer: C 4,9
Problem: Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

Solution: As stated in solution 1, the triangle is isosceles.
[asy] pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B)); label("$A$",A,E); label("$B$",B,S); label("$D$",D,W); label("$C$",C,N); [/asy]
This means that $D(-1, 3)$ is the midpoint of $B(2, -3)$ and $C(x,y)$ . So $\frac{x+2}{2}$ $= -1$ and so $x = -4$ . Similarly for $y$ , we have $\frac{y-3}{2}$ $=3$ and so $y=9$ . So our final answer is $\boxed{4,9}$

Answer: C 4,9
Problem: The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5)$ . What is $c$
$\textbf{(A)}\ -13\qquad\textbf{(B)}\ -8\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution: Writing each equation in slope-intercept form, we get $y=\frac{a}{2}x-\frac{1}{2}c$ and $y=-\frac{2}{b}x-\frac{c}{b}$ . We observe the slope of each equation is $\frac{a}{2}$ and $-\frac{2}{b}$ , respectively. Because the slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , we see that $\frac{a}{2}=-\frac{1}{-\frac{2}{b}}$ because it is given that the two lines are perpendicular. This equation simplifies to $a=b$
Because $(1, -5)$ is a solution of both equations, we deduce $a \times 1-2 \times -5=c$ and $2 \times 1+b \times -5=-c$ . Because we know that $a=b$ , the equations reduce to $a+10=c$ and $2-5a=-c$ . Solving this system of equations, we get $c=\boxed{13}$

Answer: E 13
Problem: At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$

Solution: $60\% \cdot 20\% = 12\%$ of the people that claim that they like dancing actually dislike it, and $40\% \cdot 90\% = 36\%$ of the people that claim that they dislike dancing actually dislike it. Therefore, the answer is $\frac{12\%}{12\%+36\%} = \boxed{25}$

Answer: D 25
Problem: At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$

Solution: Assume WLOG that there are 100 people. Then 60 of them like dancing, 40 dislike dancing. Of the ones that like dancing, 48 say they like dancing and 12 say they dislike it. Of the ones who dislike dancing, 36 say they dislike dancing and 4 say they like it. We want the ratio of students like it but say they dislike it to the total amount of students that say they dislike it. This is $\frac{12}{12+36}=\frac{12}{48}=\frac{1}{4}$ . We choose $\boxed{25}$

Answer: D 25
Problem: At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 12\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 25\%\qquad\textbf{(E)}\ 33\frac{1}{3}\%$

Solution: WLOG, assume that there are a total of $100$ students at Typico High School. We make a chart:
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & & & \\\hline \text{Says they don't like dancing} & & & \\\hline \text{Total} & & & 100 \\\hline \end{tabular}\]
We know that $60$ of the students like dancing (since $60\%$ of $100$ is $60$ ), so we fill that in:
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & & & \\\hline \text{Says they don't like dancing} & & & \\\hline \text{Total} & 60 & & 100 \\\hline \end{tabular}\]
$80\%$ of those $60$ kids say that they like dancing, so that's $48$ kids who like dancing and say that they like dancing. The other $12$ kids like dancing and say that they do not.
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & 48 & & \\\hline \text{Says they don't like dancing} & 12 & & \\\hline \text{Total} & 60 & & 100 \\\hline \end{tabular}\]
$40$ students do not like dancing. $90\%$ of those $40$ students say that they do not like it, which is $36$ of them.
\[\begin{tabular}[t]{|c|c|c|c|}\hline & \text{Likes dancing} & \text{Doesn't like dancing} & \text{Total} \\\hline \text{Says they like dancing} & 48 & & \\\hline \text{Says they don't like dancing} & 12 & 36 & \\\hline \text{Total} & 60 & 40 & 100 \\\hline \end{tabular}\]
At this point, one can see that there are $12+36=48$ total students who say that they do not like dancing. $12$ of those actually like it, so that is $\dfrac{12}{48}=\dfrac14=\boxed{25}.$

Answer: D 25
Problem: Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$

Solution: Suppose that his old car runs at $x$ km per liter. Then his new car runs at $\frac{3}{2}x$ km per liter, or $x$ km per $\frac{2}{3}$ of a liter. Let the cost of the old car's fuel be $c$ , so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $\frac{2}{3}\cdot\frac{6}{5}xc = \frac{4}{5}xc$ . He saves $\frac{\frac{1}{5}xc}{xc} = \boxed{20}$

Answer: A 20
Problem: Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$

Solution: Because they do not give you a given amount of distance, we'll just make that distance $3x$ miles. Then, we find that the new car will use $2*1.2=2.4x$ . The old car will use $3x$ . Thus the answer is $(3-2.4)/3=.6/3=20/100=  \boxed{20}$

Answer: A 20
Problem: Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$

Solution: You can find that the ratio of fuel used by the old car and the new car for the same amount of distance is $3 : 2$ , and the ratio between the fuel price of these two cars is $5 : 6$ . Therefore, by multiplying these two ratios, we get that the costs of using these two cars is \[15 : 12 = 5 : 4\] So the percentage of money saved is $1 - \frac{4}{5} = \boxed{20}$

Answer: A 20
Problem: Elmer's new car gives $50\%$ percent better fuel efficiency, measured in kilometers per liter, than his old car. However, his new car uses diesel fuel, which is $20\%$ more expensive per liter than the gasoline his old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?
$\textbf{(A) } 20\% \qquad \textbf{(B) } 26\tfrac23\% \qquad \textbf{(C) } 27\tfrac79\% \qquad \textbf{(D) } 33\tfrac13\% \qquad \textbf{(E) } 66\tfrac23\%$

Solution: Assume WLOG that Elmer's old car's range is $100$ miles. So, Elmer's new car's range is $100 \times 1.5 = 150$ miles. Also, assume that the gas Elmer's old car uses is $$10$ , which means that diesel will cost $$12$ . Now we can deduce that Elmer's old car uses $10 \div 100 = $0.10$ per mile, and Elmer's new car uses $12 \div 150 = $0.08$ per mile. Therefore, Elmer's new car saves $\boxed{20}$ more money than his old car.

Answer: A 20
Problem: There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: By PIE (Property of Inclusion/Exclusion), we have
$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.$ Number of people in at least two sets is $\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.$ So, $20 = (10 + 13 + 9) - (9 + 2x) + x,$ which gives $x = \boxed{3}.$

Answer: C 3
Problem: There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: The total number of classes taken among the 20 students is $10 + 13 + 9 = 32$ . Each student is taking at least one class so let's subtract the $20$ classes ( $1$ per each of the $20$ students) from $32$ classes to get $12$ $12$ classes is the total number of extra classes taken by the students who take $2$ or $3$ classes. Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \boxed{3}$ students that are taking all $3$ classes.

Answer: C 3
Problem: There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Total class count is 32. Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$ . Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$ . There are $9$ students taking two or three classes, so $b+c = 9$ . Solving this system of equations gives us $c=\boxed{3}$

Answer: C 3
Problem: There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: We are told that there are $20$ students in all, and $10$ take yoga, $13$ take bridge, and $9$ take painting.
Representing each student as a number from $1$ to $20$ , we can then make a list of which classes they are taking. Students 1-10 take yoga, and students $11$ to $20$ take bridge. However, this means that only $10$ students are taking bridge. To make up for it, we go back to the top of the list and start over from student $1$ . Students $1$ $2$ , and $3$ will also take bridge giving the desired count of $13$ total bridge students.
Now, all that is left are the students who take painting. There are $9$ students who take painting, so students $4$ through students $9$ take painting. Note that because $9$ people take at least two classes, students $10$ through $20$ are unable to take more than one class. This means that we must once more start over from the top. Already $6$ painting slots have been filled, so students $1$ $2$ , and $3$ will also take painting. This gives a total of $3$ students (students $1$ $2$ , and $3$ ) who take all three classes. Therefore, our answer is $\boxed{3}$

Answer: C 3
Problem: An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$
$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$

Solution: Notice that we can rewrite $N^{16}$ as $(N^{4})^4$ . By Fermat's Little Theorem , we know that $N^{(5-1)} \equiv 1 \pmod {5}$ if $N \not \equiv 0 \pmod {5}$ . Therefore for all $N \not \equiv 0 \pmod {5}$ we have $N^{16} \equiv (N^{4})^4 \equiv 1^4 \equiv 1 \pmod 5$ . Since $1\leq N \leq 2020$ , and $2020$ is divisible by $5$ $\frac{1}{5}$ of the possible $N$ are divisible by $5$ . Therefore, $N^{16} \equiv 1 \pmod {5}$ with probability $1-\frac{1}{5},$ or $\boxed{45}$

Answer: D 45
Problem: An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$
$\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1$

Solution: Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ .
The pattern for $0$ is $0$ , no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$ . Doing the same for the rest of the digits, we find that the units digits of $1^{16}$ $2^{16}$ $3^{16}$ $4^{16}$ $6^{16}$ $7^{16}$ $8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$ , so $\boxed{45}$

Answer: D 45
Problem: How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$
$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$

Solution: We can use complementary counting. There are $2017$ positive integers in total to consider, and there are $9$ one-digit integers, $9 \cdot 9 = 81$ two digit integers without a zero, $9 \cdot 9 \cdot 9 = 729$ three digit integers without a zero, and $9 \cdot 9 \cdot 9 = 729$ four-digit integers starting with a 1 without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = \boxed{469}$

Answer: A 469
Problem: How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$
$\textbf{(A)}\ 469\qquad\textbf{(B)}\ 471\qquad\textbf{(C)}\ 475\qquad\textbf{(D)}\ 478\qquad\textbf{(E)}\ 481$

Solution: First, we notice there are no one-digit numbers that contain a zero. There are $9$ two-digit integers and $9 \cdot 9 + 9 \cdot 9 + 9 = 171$ three-digit integers containing at least one zero. Next, we consider the four-digit integers beginning with one. There are $3 \cdot 9 \cdot 9 = 243$ of these four-digit integers with one zero, ${3 \choose 2} \cdot 9 = 27$ with two zeros, and $1$ with three zeros $(1000)$ . Finally, we consider the numbers $2000$ to $2017$ which all contain at least one zero. Adding all of these together we get $9 + 171 + 243 + 27 + 1 + 18 = \boxed{469}$

Answer: A 469
Problem: Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ $23578$ , and $987620$ are monotonous, but $88$ $7434$ , and $23557$ are not. How many monotonous positive integers are there?
$\textbf{(A)}\ 1024\qquad\textbf{(B)}\ 1524\qquad\textbf{(C)}\ 1533\qquad\textbf{(D)}\ 1536\qquad\textbf{(E)}\ 2048$

Solution: Case 1: monotonous numbers with digits in ascending order
There are $\sum_{n=1}^{9} \binom{9}{n}$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\emptyset$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{9} \binom{9}{n} -\binom{9}{0} = 2^9 - 1 = 511.$
Case 2: monotonous numbers with digits in descending order
There are $\sum_{n=1}^{10} \binom{10}{n}$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\emptyset$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to $\sum_{n=0}^{10} \binom{10}{n} -\binom{10}{0} = 2^{10} - 1 = 1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here.
Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{1524}$ monotonous numbers.

Answer: B 1524