Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly.  During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175.  In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars.  What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution: The difference in the money that Tom paid and Dorothy paid is $20$ .  In order for them both to have paid the same amount, Tom must pay $20$ more than Dorothy.  The answer is $\boxed{20}$

Answer: B 20
Problem: Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly.  During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175.  In order to share costs equally, Tom gave Sammy $t$ dollars, and Dorothy gave Sammy $d$ dollars.  What is $t-d$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$

Solution: The meaning of sharing costs equally is meaning that, after the vacation, they are equally dividing the money in a way such that, each person would have the same amount left. As each person spends an amount of money, greater than 100, let it be that they all had $200$ dollars to spend. This means that after the vacation we want the amount of money, they currently have. After the trip, Tom would've $95$ dollars, Dorothy would've $75$ dollars, and Sammy had $25$ dollars. This gives us a total of $95+75+25=195$ dollars.
We want to equally split this money, as that is what happens after splitting the cost equally. This means that we want Dorothy, Tom, and Sammy to each have $65$ dollars. We know that Tom gave Sammy $t$ dollars meaning that we want to split this money first. As Tom gives money to no one else, we want him to reach $65$ dollars in this trade, meaning that as Tom has $95$ dollars and Sammy has $25$ dollars, we can do a trade of $30$ so $t=30$ . After this trade, we get that Tom has $65$ dollars, Sammy has $55$ dollars, and Dorothy has $75$ dollars.
Next trade is where Dorothy gives $d$ dollars to Sammy. Dorothy has $75$ dollars and Sammy has $55$ dollars. As both of these don't have $65$ dollars and this is the last trade, we need to make sure both have $65$ dollars at the end. This is possible if $d=10$
We want to find $t-d=30-10=\boxed{20}$

Answer: B 20
Problem: Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ .  One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home.  How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7  \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution: Because the $5$ -year-old stayed home, we know that the $11$ -year-old did not go to the movies, as the $5$ -year-old did not and $11+5=16$ .  Also, the $11$ -year-old could not have gone to play baseball, as he is older than $10$ .  Thus, the $11$ -year-old must have stayed home, so Joey is $\boxed{11}$

Answer: D 11
Problem: Joey and his five brothers are ages $3$ $5$ $7$ $9$ $11$ , and $13$ .  One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$ -year-old stayed home.  How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7  \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$

Solution: There are only $4$ kids who are under $10$ but since the $5$ -year old stayed home, the only possible ages who went to play baseball are the brothers who are $3,7,9$ , either $13+3$ or $7+9$ is $16$ but since we need $2$ kids to go to baseball who are under $10$ $13,3$ must have been the pair to go to the movies and $9,7$ must have went to baseball, so only the $11$ -year old is left, which is answer choice $\boxed{11}$

Answer: D 11
Problem: A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$

Solution: Let us split this up into two cases.
Case $1$ : The student chooses both algebra and geometry.
This means that $3$ courses have already been chosen.  We have $3$ more options for the last course, so there are $3$ possibilities here.
Case $2$ : The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by $2$ , and then add to the previous.
Assume the mathematics course is algebra.  This means that we can choose $2$ of History, Art, and Latin, which is simply $\dbinom{3}{2} = 3$ .  If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen.
Thus, overall, we can choose a program in $6 + 3 = \boxed{9}$ ways

Answer: C 9
Problem: A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$

Solution: We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving $3$ remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing $3$ classes out of the $5$ is $\binom{5}3$ . The total number of ways of choosing only non-mathematical classes is $\binom{3}3$ . Therefore the amount of ways you can pick classes with at least one math class is $\binom{5}3-\binom{3}3=10-1=\boxed{9}$ ways.

Answer: C 9
Problem: A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$

Solution: Similar to Solution 1, note that for Case 1 of solution the answer is simply $\binom{4}2$ and for the second case it is $\binom{3}2$ hence $\boxed{9}$ ways

Answer: C 9
Problem: In a recent basketball game, Shenille attempted only three-point shots and two-point shots.  She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots.  Shenille attempted $30$ shots.  How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18  \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$

Solution: Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$ .  We know that $x+y=30$ , and we need to evaluate $3(0.2x) + 2(0.3y)$ , as we know that the three-point shots are worth $3$ points and that she made $20$ % of them and that the two-point shots are worth $2$ and that she made $30$ % of them.
Simplifying, we see that this is equal to $0.6x + 0.6y = 0.6(x+y)$ .  Plugging in $x+y=30$ , we get $0.6(30) = \boxed{18}$

Answer: B 18
Problem: In a recent basketball game, Shenille attempted only three-point shots and two-point shots.  She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots.  Shenille attempted $30$ shots.  How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18  \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$

Solution: The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes $0.2 \cdot 30 = 6$ shots, which are worth $6 \cdot 3 = \boxed{18}$ points. If we assume Shenille only attempts two-pointers, then she makes $0.3 \cdot 30 = 9$ shots, which are worth $9 \cdot 2 = 18$ points.

Answer: B 18
Problem: A flower bouquet contains pink roses, red roses, pink carnations, and red carnations.  One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink.  What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30  \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70$

Solution: Let the total amount of flowers be $x$ .  Thus, the number of pink flowers is $0.6x$ , and the number of red flowers is $0.4x$ .  The number of pink carnations is $\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\frac{3}{4}(0.4x) = 0.3x$ .  Summing these, the total number of carnations is $0.4x+0.3x=0.7x$ .  Dividing, we see that $\frac{0.7x}{x} = 0.7 = \boxed{70}$

Answer: E 70
Problem: A flower bouquet contains pink roses, red roses, pink carnations, and red carnations.  One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink.  What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30  \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70$

Solution: We have $\dfrac15$ for pink roses, $1-\dfrac6{10}=\dfrac4{10}=\dfrac25$ red flowers, $\dfrac6{10}-\dfrac15=\dfrac35-\dfrac15=\dfrac25$ pink carnations, $\dfrac25\cdot \dfrac34=\dfrac6{20}=\dfrac3{10}$ red carnations we add them up to get $\dfrac25+\dfrac3{10}=\dfrac7{10}=70\%$ so our final answer is 70% or $\boxed{70}$

Answer: E 70
Problem: A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$

Solution: Let the number of students on the council be $x$ .  To select a two-person committee, we can select a "first person" and a "second person." There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ ways to choose a two-person committee. However, this accounts for the order of committees. To understand this, suppose that Alice and Bob are two students in the council. If we choose Alice and then Bob, that is the same as choosing Bob and then Alice and so latter and former arrangements would be considered the same. Therefore, we have to divide by $2$ to account for overcounting. Thus, there are $\dfrac{x(x-1)} 2=10$ ways to choose the two-person committee. Solving this equation, we find that $5$ and $-4$ are integer solutions. $-4$ is a ridiculous situation, so there are $5$ people on the student council. The solution is $\dbinom 5 3=10\implies \boxed{10}$

Answer: A 10
Problem: A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$

Solution: To choose $2$ people from $n$ total people and get $10$ as a result, we can establish the equation $\binom{n}{2}=10$ which we can easily see $n=5$ , so there are $5$ people. The question asks how many ways to choose $3$ people from the $5$ , so there are $\binom{5}{3}=\boxed{10}$ ways.

Answer: A 10
Problem: How many three-digit numbers are not divisible by $5$ , have digits that sum to less than $20$ , and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60  \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution: We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$ .( $\overline{abc}$ denotes the number $100a+10b+c$ ).  We see that $x\neq 0$ and $x\neq 5$ , as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.
The second condition is that the sum $2x+y<20$ .  When $x$ is $1$ $2$ $3$ , or $4$ $y$ can be any digit from $0$ to $9$ , as $2x<10$ .  This yields $10(4) = 40$ numbers.
When $x=6$ , we see that $12+y<20$ so $y<8$ .  This yields $8$ more numbers.
When $x=7$ $14+y<20$ so $y<6$ .  This yields $6$ more numbers.
When $x=8$ $16+y<20$ so $y<4$ .  This yields $4$ more numbers.
When $x=9$ $18+y<20$ so $y<2$ .  This yields $2$ more numbers.
Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{60}$

Answer: B 60
Problem: A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ .  How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$

Solution: We can use Euler's polyhedron formula that says that $F+V=E+2$ .  We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$ .  In addition, each cube creates $7$ new vertices while taking away the original $8$ , yielding $8(7) = 56$ vertices.  Thus $E+2=56+30$ , so $E=\boxed{84}$

Answer: D 84
Problem: A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ .  How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$

Solution: The removal of each cube adds nine additional edges to the solid. Since a cube initially has $12$ edges and there are eight vertices, the number of edges will be $12 + 9 \times 8 = \boxed{84}$

Answer: D 84
Problem: Two sides of a triangle have lengths $10$ and $15$ .  The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides.  How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution: The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
therefore the length of the side perpendicular to that altitude will be between $10$ and $15$ . The only answer choice that meets this requirement is $\boxed{12}$

Answer: D 12
Problem: Two sides of a triangle have lengths $10$ and $15$ .  The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides.  How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution: Let the height to the side of length $15$ be $h_{1}$ , the height to the side of length 10 be $h_{2}$ , the area be $A$ , and the height to the unknown side be $h_{3}$
Because the area of a triangle is $\frac{bh}{2}$ , we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$ , so, setting them equal, $h_{2} = \frac{3h_{1}}{2}$ .  From the problem, we know that $2h_{3} = h_{1} + h_{2}$ .  Substituting, we get that \[h_{3} = 1.25h_{1}.\] Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{12}$

Answer: D 12
Problem: Two sides of a triangle have lengths $10$ and $15$ .  The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides.  How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution: Let $x$ be the height of triangle when the base is $10$ and $y$ is the height of the triangles when the base is $15$ . This means the height for when the triangles has the $3$ rd side length, the height would be $\dfrac{(x+y)}{2}$ giving us the following equation:
$\dfrac{10x}{2}=\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ For $n$ , we can plug in the answer choices and check when the ratio of $\dfrac{x}{y}$ for $\dfrac{10x}{2}=\dfrac{n(x+y)}{2}$ and $\dfrac{15y}{2}=\dfrac{n(x+y)}{2}$ is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that $\boxed{12}$ meaning that $12$ is the $3$ rd base.
~ Batmanstark

Answer:  12
Problem: Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire.  Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day.  All three friends visited Daphne yesterday.  How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72$

Solution: The $365$ -day time period can be split up into $6$ $60$ -day time periods, because after $60$ days, all three of them visit again (Least common multiple of $3$ $4$ , and $5$ ).
You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60.
Remember to subtract $1$ , because you do not wish to count the $60$ th day, when all three visit.
A and B visit $\frac{60}{3 \cdot 4}-1 = 4$ times.
A and C visit $\frac{60}{3 \cdot 5}-1 = 3$ times.
B and C visit $\frac{60}{4 \cdot 5}-1 = 2$ times.
This is a total of $9$ visits per $60$ day period.
Therefore, the total number of $2$ -person visits is $9 \cdot 6 = \boxed{54}$

Answer: B 54
Problem: Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire.  Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day.  All three friends visited Daphne yesterday.  How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72$

Solution: From the information above, we get that $A=3x$ $B=4x$ $C=5x$
Now, we want the days in which exactly two of these people meet up
The three pairs are $(A,B)$ $(B,C)$ $(A,C)$
Notice that we are trying to find the LCM of each pair.
Hence, $LCM(A,B)=12x$ $LCM(B,C)=20x$ $LCM(A,C)=15x$
Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
Hence, $LCM(A,B,C)=60x$
Now, we add all of the days up(including overcount).
We get $30+18+24=72$ .
Now, because $60(6)=360$ , we have to subtract $6$ days from every pair.
Hence, our answer is $72-18=\boxed{54}$

Answer: B 54
Problem: Let points $A = (0, 0)$ $B = (1, 2)$ $C=(3, 3)$ , and $D = (4, 0)$ .  Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ .  This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms.  What is $p+q+r+s$
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

Solution: First, we shall find the area of quadrilateral $ABCD$ . This can be done in any of three ways:
Pick's Theorem $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$
Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$
Shoelace Theorem : The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$ , or $\dfrac{15}{2}$
$[ABCD] = \frac{15}{2}$ .  Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$
Call the point where the line through $A$ intersects $\overline{CD}$ $E$ .  We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$ .  Furthermore, we know that $b = 4$ , as $AD = 4$ .  Thus, solving for $h$ , we find that $2h = \frac{15}{4}$ , so $h = \frac{15}{8}$ .  This gives that the y coordinate of E is $\frac{15}{8}$
Line CD can be expressed as $y = -3x+12$ , so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$ .  Solving for $x$ , we find that $x = \frac{27}{8}$
From this, we know that $E = \left(\frac{27}{8}, \frac{15}{8}\right)$ $27 + 15 + 8 + 8 = \boxed{58}$

Answer: B 58
Problem: Let points $A = (0, 0)$ $B = (1, 2)$ $C=(3, 3)$ , and $D = (4, 0)$ .  Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ .  This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms.  What is $p+q+r+s$
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

Solution: Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$ .
Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$ , and from there split the base into two parts, $x$ , and $4-x$ , such that $x$ is the segment from the origin to the point $F$ , and $4-x$ is the segment from point $F$ to point $D$ . Then, by the Pythagorean Theorem $x=\frac{27}{8}$ , and the answer is $\boxed{58}$

Answer: B 58
Problem: In base $10$ , the number $2013$ ends in the digit $3$ .  In base $9$ , on the other hand, the same number is written as $(2676)_{9}$ and ends in the digit $6$ .  For how many positive integers $b$ does the base- $b$ -representation of $2013$ end in the digit $3$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18$

Solution: We want the integers $b$ such that $2013\equiv 3\pmod{b} \Rightarrow b$ is a factor of $2010$ . Since $2010=2 \cdot 3 \cdot 5 \cdot 67$ , it has $(1+1)(1+1)(1+1)(1+1)=16$ factors. Since $b$ cannot equal $1, 2,$ or $3$ , as these cannot have the digit $3$ in their base representations, our answer is $16-3=\boxed{13}$

Answer: C 13
Problem: A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows.  The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest.  The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins.  How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$

Solution: Let $x$ be the number of coins.  After the $k^{\text{th}}$ pirate takes his share, $\frac{12-k}{12}$ of the original amount is left.  Thus, we know that
$x \cdot \frac{11}{12} \cdot \frac{10}{12} \cdot \frac{9}{12} \cdot \frac{8}{12} \cdot \frac{7}{12} \cdot \frac{6}{12} \cdot \frac{5}{12} \cdot \frac{4}{12} \cdot \frac{3}{12} \cdot \frac{2}{12} \cdot \frac{1}{12}$ must be an integer.  Simplifying, we get
$x \cdot \frac{11}{12} \cdot \frac{5}{6} \cdot \frac{1}{2}  \cdot \frac{7}{12} \cdot \frac{1}{2} \cdot \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{12}$ .  Now, the minimal $x$ is the denominator of this fraction multiplied out, obviously.  We mentioned before that this product must be an integer.  Specifically, it is an integer and it is the amount that the $12^{\text{th}}$ pirate receives, as he receives $\frac{12}{12} = 1 =$ all of what is remaining.
Thus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, $11 \cdot 5 \cdot 7 \cdot 5 = \boxed{1925}$

Answer: D 1925