Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Tamara has three rows of two $6$ -feet by $2$ -feet flower beds in her garden. The beds are separated and also surrounded by $1$ -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?
[asy] draw((0,0)--(0,10)--(15,10)--(15,0)--cycle); fill((0,0)--(0,10)--(15,10)--(15,0)--cycle, lightgray); draw((1,1)--(1,3)--(7,3)--(7,1)--cycle); fill((1,1)--(1,3)--(7,3)--(7,1)--cycle, white); draw((1,4)--(1,6)--(7,6)--(7,4)--cycle); fill((1,4)--(1,6)--(7,6)--(7,4)--cycle, white); draw((1,7)--(1,9)--(7,9)--(7,7)--cycle); fill((1,7)--(1,9)--(7,9)--(7,7)--cycle, white);  draw((8,1)--(8,3)--(14,3)--(14,1)--cycle); fill((8,1)--(8,3)--(14,3)--(14,1)--cycle, white); draw((8,4)--(8,6)--(14,6)--(14,4)--cycle); fill((8,4)--(8,6)--(14,6)--(14,4)--cycle, white); draw((8,7)--(8,9)--(14,9)--(14,7)--cycle); fill((8,7)--(8,9)--(14,9)--(14,7)--cycle, white);  defaultpen(fontsize(8, lineskip=1)); label("2", (1.2, 2)); label("6", (4, 1.2)); defaultpen(linewidth(.2)); draw((0,8)--(1,8), arrow=Arrows); draw((7,8)--(8,8), arrow=Arrows); draw((14,8)--(15,8), arrow=Arrows); draw((11,0)--(11,1), arrow=Arrows); draw((11,3)--(11,4), arrow=Arrows); draw((11,6)--(11,7), arrow=Arrows); label("1", (.5,7.8)); label("1", (7.5,7.8)); label("1", (14.5,7.8)); label("1", (10.8,.5)); label("1", (10.8,3.5)); label("1", (10.8,6.5)); [/asy]
$\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150$

Solution: Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.
Since the width of Tamara's garden contains three margins, the total width is $2\cdot 6+3\cdot 1 = 15$ feet.
Similarly, the height of Tamara's garden is $3\cdot 2+4\cdot 1 = 10$ feet.
Therefore, the total area of the garden is $15\cdot 10 =150$ square feet.
Finally, since the six flower beds each have an area of $2\cdot 6 = 12$ square feet, the area we seek is $150 - 6\cdot 12$ , and our answer is $\boxed{78}$

Answer: B 78
Problem: Tamara has three rows of two $6$ -feet by $2$ -feet flower beds in her garden. The beds are separated and also surrounded by $1$ -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?
[asy] draw((0,0)--(0,10)--(15,10)--(15,0)--cycle); fill((0,0)--(0,10)--(15,10)--(15,0)--cycle, lightgray); draw((1,1)--(1,3)--(7,3)--(7,1)--cycle); fill((1,1)--(1,3)--(7,3)--(7,1)--cycle, white); draw((1,4)--(1,6)--(7,6)--(7,4)--cycle); fill((1,4)--(1,6)--(7,6)--(7,4)--cycle, white); draw((1,7)--(1,9)--(7,9)--(7,7)--cycle); fill((1,7)--(1,9)--(7,9)--(7,7)--cycle, white);  draw((8,1)--(8,3)--(14,3)--(14,1)--cycle); fill((8,1)--(8,3)--(14,3)--(14,1)--cycle, white); draw((8,4)--(8,6)--(14,6)--(14,4)--cycle); fill((8,4)--(8,6)--(14,6)--(14,4)--cycle, white); draw((8,7)--(8,9)--(14,9)--(14,7)--cycle); fill((8,7)--(8,9)--(14,9)--(14,7)--cycle, white);  defaultpen(fontsize(8, lineskip=1)); label("2", (1.2, 2)); label("6", (4, 1.2)); defaultpen(linewidth(.2)); draw((0,8)--(1,8), arrow=Arrows); draw((7,8)--(8,8), arrow=Arrows); draw((14,8)--(15,8), arrow=Arrows); draw((11,0)--(11,1), arrow=Arrows); draw((11,3)--(11,4), arrow=Arrows); draw((11,6)--(11,7), arrow=Arrows); label("1", (.5,7.8)); label("1", (7.5,7.8)); label("1", (14.5,7.8)); label("1", (10.8,.5)); label("1", (10.8,3.5)); label("1", (10.8,6.5)); [/asy]
$\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150$

Solution: The long horizontal walkways have an area of $15 ft^{2}$ each, and there are $4$ of them $\longrightarrow 60 ft^{2}$
The short vertical walkways have an area of $2 ft^{2}$ each, and there are $9$ of them $\longrightarrow 18 ft^{2}$
Adding the areas together, we have $60 ft^{2} + 18 ft^{2} = 78 ft^{2} \Longrightarrow \boxed{78}$ .
~JH. L

Answer: B 78
Problem: Mia is "helping" her mom pick up $30$ toys that are strewn on the floor. Mia’s mom manages to put $3$ toys into the toy box every $30$ seconds, but each time immediately after those $30$ seconds have elapsed, Mia takes $2$ toys out of the box. How much time, in minutes, will it take Mia and her mom to put all $30$ toys into the box for the first time?
$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5$

Solution: Every $30$ seconds, $3$ toys are put in the box and $2$ toys are taken out, so the number of toys in the box increases by $3-2=1$ every $30$ seconds. Then after $27 \times 30 = 810$ seconds (or $13 \frac{1}{2}$ minutes), there are $27$ toys in the box. Mia's mom will then put the remaining $3$ toys into the box after $30$ more seconds, so the total time taken is $27\times 30+30=840$ seconds, or $\boxed{14}$ minutes.

Answer: B 14
Problem: The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution: Let the two real numbers be $x,y$ . We are given that $x+y=4xy,$ and dividing both sides by $xy$ $\frac{x}{xy}+\frac{y}{xy}=4.$
\[\frac{1}{y}+\frac{1}{x}=\boxed{4}.\]

Answer: C 4
Problem: The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution: Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.
See for yourself. And by looking into fractions, we immediately see that $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{4}.$

Answer: C 4
Problem: The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution: Notice that from the information given above, $x+y=4xy$
Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\frac{x+y}{xy}$
We can solve this by substituting $x+y\implies 4xy$
Our answer is simply $\frac{4xy}{xy}\implies4$
Therefore, the answer is $\boxed{4}$

Answer: C 4
Problem: Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%$

Solution: Let $j$ represent how far Jerry walked, and $s$ represent how far Silvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a $45$ $45$ $90$ triangle with leg length $1$ . Thus, $s = \sqrt{2} = 1.414...$ We can then take $\frac{j-s}{j} \approx \frac{2 - 1.4}{2}=0.3 \implies \boxed{30}$

Answer: A 30
Problem: At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution: Each one of the ten people has to shake hands with all the $20$ other people they don’t know. So $10\cdot20 = 200$ . From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands from $10$ , or $\binom{10}{2} = 45$ . Thus the answer is $200 + 45 = \boxed{245}$

Answer: B 245
Problem: At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution: We can also use complementary counting. First of all, $\dbinom{30}{2}=435$ handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from $435$ to find the handshakes. Hugs only happen between the $20$ people who know each other, so there are $\dbinom{20}{2}=190$ hugs. $435-190= \boxed{245}$

Answer: B 245
Problem: At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution: We can focus on how many handshakes the $10$ people who don't know anybody get.
The first person gets $29$ handshakes with other people not him/herself, the second person gets $28$ handshakes with other people not him/herself and not the first person, ..., and the tenth receives $20$ handshakes with other people not him/herself and not the first, second, ..., ninth person. We can write this as the sum of an arithmetic sequence:
$\frac{10(20+29)}{2}\implies 5(49)\implies 245.$ Therefore, the answer is $\boxed{245}$

Answer: B 245
Problem: At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?
$\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490$

Solution: First, we can find out the number of handshakes that the $10$ people who don't know anybody share with the $20$ other people. This is simply $10 \cdot 20 = 200$ . Next, we need to find out the number of handshakes that are shared within the $10$ people who don't know anybody. Here, we can use the formula $\frac{n(n-1)}{2}$ , where $n$ is the number of people being counted. The reason we divide by $2$ is because $n(n-1)$ counts the case where the $1^{st}$ person shakes hands with the $2^{nd}$ person $and$ the case where the $2^{nd}$ shakes hands with the $1^{st}$ (and these 2 cases are the same). Thus, plugging $n=10$ gives us $\frac{10 \cdot 9}{2} \implies 45$ . Adding up the 2 cases gives us $200+45=\boxed{245}$

Answer: B 245
Problem: Minnie rides on a flat road at $20$ kilometers per hour (kph), downhill at $30$ kph, and uphill at $5$ kph. Penny rides on a flat road at $30$ kph, downhill at $40$ kph, and uphill at $10$ kph. Minnie goes from town $A$ to town $B$ , a distance of $10$ km all uphill, then from town $B$ to town $C$ , a distance of $15$ km all downhill, and then back to town $A$ , a distance of $20$ km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the $45$ -km ride than it takes Penny?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 65\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 95$

Solution: The distance from town $A$ to town $B$ is $10$ km uphill, and since Minnie rides uphill at a speed of $5$ kph, it will take her $2$ hours. Next, she will ride from town $B$ to town $C$ , a distance of $15$ km all downhill. Since Minnie rides downhill at a speed of $30$ kph, it will take her half an hour. Finally, she rides from town $C$ back to town $A$ , a flat distance of $20$ km. Minnie rides on a flat road at $20$ kph, so this will take her $1$ hour. Her entire trip takes her $3.5$ hours.
Secondly, Penny will go from town $A$ to town $C$ , a flat distance of $20$ km. Since Penny rides on a flat road at $30$ kph, it will take her $\frac{2}{3}$ of an hour. Next Penny will go from town $C$ to town $B$ , which is uphill for Penny. Since Penny rides at a speed of $10$ kph uphill, and town $C$ and $B$ are $15$ km apart, it will take her $1.5$ hours. Finally, Penny goes from Town $B$ back to town $A$ , a distance of $10$ km downhill. Since Penny rides downhill at $40$ kph, it will only take her $\frac{1}{4}$ of an hour. In total, it takes her $29/12$ hours, which simplifies to $2$ hours and $25$ minutes.
Finally, Penny's $2$ hr $25$ min trip was $\boxed{65}$ minutes less than Minnie's $3$ hr $30$ min trip.

Answer: C 65
Problem: Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$

Solution: The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $15<x+3+7$ yields $5<x<25$ . Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $19-2=\boxed{17}$

Answer: B 17
Problem: The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution: In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ .  However, we need to find the region containing all points within $3$ units of a segment.  It can be seen that our region is a cylinder with two hemispheres/endcaps on either end.  We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \pi$ ):
$\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi$ , where $x$ is equal to the length of our line segment.
Solving, we find that $x = \boxed{20}$

Answer: D 20
Problem: The region consisting of all points in three-dimensional space within $3$ units of line segment $\overline{AB}$ has volume $216\pi$ . What is the length $\textit{AB}$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24$

Solution: Because this is just a cylinder and $2$ hemispheres ("half spheres"), and the radius is $3$ , the volume of the $2$ hemispheres is $\frac{4(3^3)\pi}{3} = 36 \pi$ . Since we also know that the volume of this whole thing is $216 \pi$ , we do $216-36$ to get $180 \pi$ as the volume of the cylinder. Thus the height is $180 \pi$ divided by the area of the base, or $\frac{180 \pi}{9\pi}=20$ , so our answer is $\boxed{20}.$

Answer: D 20
Problem: Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the sequence, which is $\boxed{9}$

Answer: D 9
Problem: Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) }  23\%\qquad \textbf{(E) } 25\%$

Solution: Let $m$ = cost of movie ticket Let $s$ = cost of soda
We can create two equations:
\[m = \frac{1}{5}(A - s)\] \[s  = \frac{1}{20}(A - m)\]
Substituting we get:
\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\] which yields: \[m = \frac{19}{99}A\]
Now we can find s and we get:
\[s = \frac{4}{99}A\]
Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:
\[\frac{19}{99}A + \frac{4}{99}A \implies \boxed{23}\]

Answer: D 23
Problem: Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) }  23\%\qquad \textbf{(E) } 25\%$

Solution: We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$ .
Solving, we get $S=\frac{400}{99}$ and $M=\frac{1900}{99}$ .
Adding, we get $\frac{2300}{99}$ , which is closest to $23$ which is $\boxed{23}$

Answer: D 23
Problem: Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?
$\textbf{(A) } 9\%\qquad \textbf{(B) } 19\%\qquad \textbf{(C) } 22\%\qquad \textbf{(D) }  23\%\qquad \textbf{(E) } 25\%$

Solution: Let $m$ be the price of a movie ticket and $s$ be the price of a soda.
Then,
\[m=\frac{A-s}{5}\] and \[s=\frac{A-m}{20}\] Then, we can turn this into \[5m=A-s\] \[20s=A-m\]
Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$ . Assume WLOG that $s=4$ $m=19$ , thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$ . Hence, $\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{23}$

Answer: D 23
Problem: There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$

Solution: If we have horses, $a_1, a_2, \ldots, a_n$ , then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12$ . Finally, $1+2 = \boxed{3}$

Answer: B 3
Problem: There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$

Solution: We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$ $2$ $3$ , or $4$ . We quickly consider $12$ since it is the smallest number that is the LCM of $1$ $2$ $3$ and $4$ . Since $12$ has $5$ single-digit divisors, namely $1$ $2$ $3$ $4$ , and $6$ , our answer is $1+2 = \boxed{3}$

Answer: B 3
Problem: There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$

Solution: First, for 5 horses to simultaneously pass the starting line after $T$ seconds, $T$ must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of $T$ , and therefore meaning $T$ must have at least $5$ $1$ -digit divisors. Since we want to minimize $T$ , we will start by guessing the lowest natural number, $1$ $1$ has only $1$ factor, so it does not work, we now repeat the process for the numbers between $2$ and $12$ (This should not take more than a minute) to get that $12$ is the first number to have $5$ or more single-digit divisors ( $1, 2, 3, 4, 6$ ). The sum of the digits of $12$ is $1+2 = \boxed{3}$ , which is our answer.

Answer: B 3
Problem: There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$

Solution: By inspection, $(1, 2, 3, 4, 6)$ yields the lowest answer of $12$ and the sum of the digits is $1+2 \Longrightarrow \boxed{3}$

Answer: B 3