Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\square\square, \square ST\square, \square\square ST$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$ ). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{12}$

Answer: C 12
Problem: Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \[4!-2\cdot3!=\boxed{12}\]

Answer: C 12
Problem: Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are $2!$ ways to arrange Steelie and Tiger within this "super marble." Then there are $3!$ ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are $4!$ ways to arrange the marbles without any restrictions, the answer is given by $4!-2!\cdot 3!=\boxed{12}$

Answer: C 12
Problem: Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
$\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: We will use the following
$\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects.
$\textit{Proof. (Created by AoPS user RedFireTruck)}$
Let our group of $a$ objects be represented like so $1$ $2$ $3$ , ..., $a-1$ $a$ . Let the last $b$ objects be the ones we can't have together.
Then we can organize our objects like so $\square1\square2\square3\square...\square a-b-1\square a-b\square$
We have $(a-b)!$ ways to arrange the objects in that list.
Now we have $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together.
By fundamental counting principle our answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$
Proof by RedFireTruck talk ) 12:09, 1 February 2021 (EST)
Back to the problem. By the Georgeooga-Harryooga Theorem , our answer is $\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{12}$

Answer: C 12
Problem: After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7});  for (int i = 0; i < 6; ++i) {     for (int j = 0; j < 6; ++j) {         draw((i,0)--(i,6), grey);         draw((0,j)--(6,j), grey);     } }  for (int i = 1; i <= 6; ++i) {     draw((-0.1,i)--(0.1,i),linewidth(1.25));     draw((i,-0.1)--(i,0.1),linewidth(1.25));     label(string(5*i), (i,0), 2*S);     label(string(i), (0, i), 2*W);  }  draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));  label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S);  dot("Naomi", (2,6), 3*dir(305)); dot((6,6));  label("Maya", (4.45,3.5));  draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]
$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

Solution: Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$ , her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{24}$

Answer: E 24
Problem: After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?
[asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7});  for (int i = 0; i < 6; ++i) {     for (int j = 0; j < 6; ++j) {         draw((i,0)--(i,6), grey);         draw((0,j)--(6,j), grey);     } }  for (int i = 1; i <= 6; ++i) {     draw((-0.1,i)--(0.1,i),linewidth(1.25));     draw((i,-0.1)--(i,0.1),linewidth(1.25));     label(string(5*i), (i,0), 2*S);     label(string(i), (0, i), 2*W);  }  draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));  label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S);  dot("Naomi", (2,6), 3*dir(305)); dot((6,6));  label("Maya", (4.45,3.5));  draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy]
$\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

Solution: Naomi's speed of $6$ miles in $10$ minutes is equivalent to $6 \cdot 6 = 36$ miles per hour, while Maya's speed of $6$ miles in $30$ minutes (i.e. half an hour) is equivalent to $6 \cdot 2 = 12$ miles per hour. The difference is consequently $36-12=\boxed{24}$

Answer: E 24
Problem: For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Solution: We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{10}$

Answer: A 10
Problem: For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Solution: Since $5! = 120$ , we obtain $120\cdot 9!=12\cdot N!$ , which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$ . We therefore deduce $N=\boxed{10}$

Answer: A 10
Problem: For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\]
$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Solution: We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$
We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$
Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{10}.$

Answer: A 10
Problem: Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$ , so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$ , the answer is $\boxed{9}$

Answer: B 9
Problem: Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: As in Solution 1, we have the equation $\frac{18+x}{36+x}=\frac{3}{5}$ . Cross-multiplying yields $90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9$ . Thus, Jamal added $\boxed{9}$ purple socks.

Answer: B 9
Problem: Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Solution: $6$ green socks and $12$ orange socks together should be $100\%-60\% = 40\%$ of the new total number of socks, so that new total must be $\frac{6+12}{0.4}= 45$ . Therefore, $45-6-18-12=\boxed{9}$ purple socks were added.

Answer: B 9
Problem: There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities?
[asy] // made by SirCalcsALot  size(300);  pen shortdashed=linetype(new real[] {6,6});  for (int i = 2000; i < 9000; i = i + 2000) {     draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);     label(string(i), (0,i), W); }   for (int i = 500; i < 9300; i=i+500) {     draw((0,i)--(150,i),linewidth(1.25));     if (i % 2000 == 0) {         draw((0,i)--(250,i),linewidth(1.25));     } }  int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20;  int r = 550; for (int i = 0; i < data_length; ++i) {     fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);     draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); }  draw((0,4750)--(11450,4750),shortdashed);  label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W);  // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); [/asy]
$\textbf{(A) }65000 \qquad \textbf{(B) }75000 \qquad \textbf{(C) }85000 \qquad \textbf{(D) }95000 \qquad \textbf{(E) }105000$

Solution: We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \cdot 20 = \boxed{95000}$

Answer: D 95000
Problem: Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution: Since $20\% = \frac{1}{5}$ , multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{75}$ percent of $x$

Answer: C 75
Problem: Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution: Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$ . Clearly, it follows that $y$ is $75\%$ of $x$ , so the answer is $\boxed{75}$

Answer: C 75
Problem: Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution: We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$ , so $\frac{3}{20}x=\frac{1}{5}y$ . Solving for $y$ , we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$

Answer: C 75
Problem: Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution: We are given $0.15x = 0.20y$ , so we may assume without loss of generality that $x=20$ and $y=15$ . This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$ , and thus the answer is $\boxed{75}$

Answer: C 75
Problem: Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution: $15\%$ of $x$ is $0.15x$ , and $20\%$ of $y$ is $0.20y$ . We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$ . Solving for $y$ , dividing $0.2$ to both sides, we get $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$

Answer: C 75
Problem: Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Solution: $15\%$ of $x$ can be written as $\frac{15}{100}x$ , or $\frac{15x}{100}$ $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$ , or $\frac{20y}{100}$ . So now, $\frac{15x}{100} = \frac{20y}{100}$ . Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$ . Dividing both sides by $500$ , we get: $3x = 4y$ $\frac{3}{4}$ is the same thing as $75\%$ , so the answer is $\boxed{75}$

Answer: C 75
Problem: Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$
[asy] size(200); dotfactor = 10;  pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1));  pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1));  pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1));  pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1));  pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1));  pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W);  pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW);  pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW);  pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE);  pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE);  pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy]
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$

Solution: We can form the following expressions for the sum along each line: \[\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}\] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$ , i.e. $2(A+B+C+D+E+F)+B=47$ . Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$ , we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$ . This means $B = \boxed{5}$ .
~RJ5303707

Answer: E 5
Problem: Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$
[asy] size(200); dotfactor = 10;  pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1));  pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1));  pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1));  pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1));  pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1));  pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W);  pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW);  pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW);  pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE);  pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE);  pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy]
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$

Solution: Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$ , we get $43=47$ which is not true. Testing $B=3$ , we get $45=47$ , which is also not true. Therefore, we have $B = \boxed{5}$

Answer: E 5
Problem: How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution: Since $2020 = 2^2 \cdot 5 \cdot 101$ , we can simply list its factors: \[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\] There are $12$ factors; only $1, 2, 4, 5, 101$ don't have over $3$ factors, so the remaining $12-5 = \boxed{7}$ factors have more than $3$ factors.

Answer: B 7
Problem: How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution: As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$ $5$ , and $101$ . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$ . Thus, there are $5$ factors of $2020$ which themselves have $1$ $2$ , or $3$ factors (namely $1$ $2$ $4$ $5$ , and $101$ ), so the number of factors of $2020$ that have more than $3$ factors is $12-5=\boxed{7}$

Answer: B 7
Problem: How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$
$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution: Let $d(n)$ be the number of factors of n. We know by prime factorization that $d(2020) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$ . Since $d(2020) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ with more than $3$ factors. However, this argument has exceptions where $a$ and $b$ share a nontrivial common factor, which in this case can only be two. There are two cases - One in which $5$ and $101$ divide the same factor, WLOG assumed to be $a$ , so that $d(a) = 2^3 > 3$ and $d(b) = 2$ , as otherwise. In the other case, $a = 5\cdot2$ and $b = 101\cdot2$ , so that $d(a) = d(b) = 4$ . This adds one factor with more than $3$ factors, so the answer is $\boxed{7}$

Answer: B 7
Problem: Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$
[asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N);  [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Solution: [asy]  draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));[/asy]
Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$ , so $OC = 17$ . By symmetry, $O$ is the midpoint of $DA$ , so $OD=OA=\frac{16}{2}= 8$ . By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$ ), we have that $CD$ (or $AB$ ) is $\sqrt{17^2-8^2}=15$ . Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{240}$

Answer: A 240