Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41$

Solution: Let the number of hours Danica drove be $k$ . Then we know that $100a + 10b + c + 55k$ $100c + 10b + a$ . Simplifying, we have $99c - 99a = 55k$ , or $9c - 9a = 5k$ . Thus, k is divisible by $9$ . Because $55 * 18 = 990$ $k$ must be $9$ , and therefore $c - a = 5$ . Because $a + b + c \leq{7}$ and $a \geq{1}$ $a = 1$ $c = 6$ and $b = 0$ , and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$ , or $\boxed{37}$

Answer: D 37
Problem: A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$

Solution: We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.
Since the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$ . In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$ . If we let the list be $1,2,3,8,8,9,10,11,12,13,j$ , then $j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$
Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$ . Following the same process as above, we get $j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$ . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{35}$

Answer: E 35
Problem: A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution: Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \ge b \ge c$ $b+c > a$ , and $a, b, c < 5$ . Now we enumerate the elements of $T$
$(4, 4, 4)$
$(4, 4, 3)$
$(4, 4, 2)$
$(4, 4, 1)$
$(4, 3, 3)$
$(4, 3, 2)$
$(3, 3, 3)$
$(3, 3, 2)$
$(3, 3, 1)$
$(3, 2, 2)$
$(2, 2, 2)$
$(2, 2, 1)$
$(1, 1, 1)$
It should be clear that $|S|$ is simply $|T|$ minus the larger "duplicates" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$ ). Since $|T|$ is $13$ and the number of higher duplicates is $4$ , the answer is $13 - 4$ or $\boxed{9}$

Answer: B 9
Problem: Let $P$ be a cubic polynomial with $P(0) = k$ $P(1) = 2k$ , and $P(-1) = 3k$ .  What is $P(2) + P(-2)$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ k\qquad\textbf{(C)}\ 6k\qquad\textbf{(D)}\ 7k\qquad\textbf{(E)}\ 14k$

Solution: Let $P(x) = Ax^3+Bx^2 + Cx+D$ .   Plugging in $0$ for $x$ , we find $D=k$ , and plugging in $1$ and $-1$ for $x$ , we obtain the following equations: \[A+B+C+k=2k\] \[-A+B-C+k=3k\] Adding these two equations together, we get \[2B=3k\] If we plug in $2$ and $-2$ in for $x$ , we find that \[P(2)+P(-2) = 8A+4B+2C+k+(-8A+4B-2C+k)=8B+2k\] Multiplying the third equation by $4$ and adding $2k$ gives us our desired result, so \[P(2)+P(-2)=12k+2k=\boxed{14}\]

Answer: E 14
Problem: Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$

Solution: Let $y = m(x - 20) + 14$ . Equating them:
$x^2 = mx - 20m + 14$
$x^2 - mx + 20m - 14 = 0$
For there to be no solutions, the discriminant must be less than zero:
$m^2 - 4(20m - 14) < 0$
$m^2 - 80m + 56 < 0$
So $m < 0$ for $r < m < s$ where $r$ and $s$ are the roots of $m^2 - 80m + 56 = 0$ and their sum by Vieta's formulas is $\boxed{80}$

Answer: E 80
Problem: Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$

Solution: The line will begin to intercept the parabola when its slope equals that of the parabola at the point of tangency.  Taking the derivative of the equation of the parabola, we get that the slope equals $2x$ .  Using the slope formula, we find that the slope of the tangent line to the parabola also equals $\frac{14-x^2}{20-x}$ .  Setting these two equal to each other, we get \[2x = \frac{14-x^2}{20-x} \implies x^2-40x+14 = 0\] Solving for $x$ , we get \[x= 20\pm \sqrt{386}\] The sum of the two possible values for $x$ where the line is tangent to the parabola is $40$ , and the sum of the slopes of these two tangent lines is equal to $2x$ , or $\boxed{80}$

Answer: E 80
Problem: Let $P$ be the parabola with equation $y=x^2$ and let $Q = (20, 14)$ . There are real numbers $r$ and $s$ such that the line through $Q$ with slope $m$ does not intersect $P$ if and only if $r$ $m$ $s$ . What is $r + s$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 52\qquad\textbf{(E)}\ 80$

Solution: The smaller solution is basically negligible in comparison with the solution with the larger slope. Try some values like of $x^2 = y$ where $x = 40$ => $y = 1600$ , and slope ~80. Trying a few values leads us to conclude the least possible value is around $80$ , so the answer is $\boxed{80}$

Answer: E 80
Problem: For how many positive integers $x$ is $\log_{10}(x-40) + \log_{10}(60-x) < 2$
$\textbf{(A) }10\qquad \textbf{(B) }18\qquad \textbf{(C) }19\qquad \textbf{(D) }20\qquad \textbf{(E) }\text{infinitely many}\qquad$

Solution: The domain of the LHS implies that \[40<x<60\] Begin from the left hand side \[\log_{10}[(x-40)(60-x)]<2\] \[(x-40)(60-x)<100\] \[-x^2+100x-2500<0\] \[(x-50)^2>0\] \[x \not = 50\] Hence, we have integers from 41 to 49 and 51 to 59. There are $\boxed{18}$ integers.

Answer: B 18
Problem: The number $2017$ is prime.  Let $S = \sum \limits_{k=0}^{62} \dbinom{2014}{k}$ .  What is the remainder when $S$ is divided by $2017?$
$\textbf{(A) }32\qquad \textbf{(B) }684\qquad \textbf{(C) }1024\qquad \textbf{(D) }1576\qquad \textbf{(E) }2016\qquad$

Solution: Note that $2014\equiv -3 \mod2017$ .  We have for $k\ge1$ \[\dbinom{2014}{k}\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mod 2017\] \[\equiv (-1)^k\dbinom{k+2}{k} \mod 2017\] Therefore \[\sum \limits_{k=0}^{62} \dbinom{2014}{k}\equiv \sum \limits_{k=0}^{62}(-1)^k\dbinom{k+2}{2} \mod 2017\] This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-21....$ After finding the first few sums of the series, it becomes apparent that \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv -\left(\frac{n+1}{2} \right) \left(\frac{n+1}{2}+1 \right) \mod 2017 \textnormal{  if n is odd}\] and \[\sum \limits_{k=1}^{n}(-1)^k\dbinom{k+2}{2}\equiv \left(\frac{n}{2}+1 \right)^2 \mod 2017 \textnormal{  if n is even}\] Obviously, $62$ falls in the second category, so our desired value is \[\left(\frac{62}{2}+1 \right)^2 = 32^2 = \boxed{1024}\]

Answer: C 1024
Problem: Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$

Solution: Rewrite $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ .  Now let $a = \cos{2x}$ , and let $b = \cos{\left( \frac{2014\pi^2}{x} \right) }$ .  We have: \[2a(a - b) = 2a^2 - 2\]
Therefore, \[ab = 1\]
Notice that either $a = 1$ and $b = 1$ or $a = -1$ and $b = -1$ .  For the first case, $a = 1$ only when $x = k\pi$ and $k$ is an integer. $b = 1$ when $\frac{2014\pi^2}{k\pi}$ is an even multiple of $\pi$ , and since $2014 = 2*19*53$ $b =1$ only when $k$ is an odd divisor of $2014$ .  This gives us these possible values for $x$ \[x= \pi, 19\pi, 53\pi, 1007\pi\] For the case where $a = -1$ $\cos{2x} = -1$ , so $x = \frac{m\pi}{2}$ , where m is odd. $\frac{2014\pi^2}{\frac{m\pi}{2}}$ must also be an odd multiple of $\pi$ in order for $b$ to equal $-1$ , so $\frac{4028}{m}$ must be odd.  We can quickly see that dividing an even number by an odd number will never yield an odd number, so there are no possible values for $m$ , and therefore no cases where $a = -1$ and $b = -1$ . Therefore, the sum of all our possible values for $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{1080}\]

Answer: D 1080
Problem: Find the sum of all the positive solutions of
$2\cos2x \left(\cos2x - \cos{\left( \frac{2014\pi^2}{x} \right) } \right) = \cos4x - 1$
$\textbf{(A)}\ \pi \qquad\textbf{(B)}\ 810\pi \qquad\textbf{(C)}\ 1008\pi \qquad\textbf{(D)}\ 1080 \pi \qquad\textbf{(E)}\ 1800\pi$

Solution: Rewriting $\cos{4x} - 1$ as $2\cos^2{2x} - 2$ and transposing $2\cos{2x}$ from the LHS to the RHS, we get,
\[\cos{2x} - \cos{\left(\frac{2014\pi^2}{x}\right)} = 1 - \frac{1}{\cos{2x}}\] \[\implies \underbrace{\cos{2x} + \frac{1}{\cos{2x}}}_{\text{LHS}} = \underbrace{1 + \cos{\left(\frac{2014\pi^2}{x}\right)}}_{\text{RHS}}\]
By the AM-GM Inequality
\[\cos{2x} + \frac{1}{\cos{2x}} \in (-\infty, -2] \cup [2, \infty)\]
Also, because of the range of $\cos$
\[1 + \cos{\left(\frac{2014\pi^2}{x}\right)} \in [0, 2]\]
Hence, $\text{LHS} = \text{RHS} = 2$ , and we get ( $m, n \in \mathbb{Z}$ ),
From $(1)$ and $(2)$ \[m = \frac{1007}{n}\] \[\implies m \in \{1, 19, 53, 1007\}\] \[\implies x \in \{1\pi, 19\pi, 53\pi, 1007\pi\}\]
Therefore, sum of values of $x$ is \[\pi + 19\pi + 53\pi + 1007\pi = \boxed{1080}\]

Answer: D 1080
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 the 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: Simply write down two algebraic equations. We know that Tom gave $t$ dollars and Dorothy gave $d$ dollars. In addition, Tom originally paid $105$ dollars and Dorothy paid $125$ dollars originally. Since they all pay the same amount, we have: \[105 + t = 125 + d.\] Rearranging, we have \[t-d = \boxed{20}.\]

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 the 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: Add up the amounts that Tom, Dorothy, and Sammy paid to get $ $405$ , and divide by 3 to get $ $135$ , the amount that each should have paid.
Tom, having paid $ $105$ , owes Sammy $ $30$ , and Dorothy, having paid $ $125$ , owes Sammy $ $10$
Thus, $t - d = 30 - 10 = 20$ , which is $\boxed{20}$

Answer: B 20
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: (similar to Solution 1, however a slightly more obvious way)
Say that
x = # of 2-pt shots
y = # of 3-pt shots
Because the total number of shots is $30$ $x + y = 30$
However, Shenille was only successful on $20\%$ of the 3-pt shots, and $30\%$ of the 2-pt shots, so
$0.2x + 0.3y$ = #number of successful shots
For each successful shot, there is an associated number of points with it.
Therefore, $0.2(x)(3) + (0.3)(y)(2)$ = her score
this evaluates to $0.6 (x + y)$ = her score
$x + y$ is already determined to be 30, so her score is $0.6 (30) = \boxed{18}$

Answer: B 18
Problem: The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two.  That is, \[S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.\] Suppose that $S_9 = 110$ and $S_7 = 42$ .  What is $S_4$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad$

Solution: $S_9 = 110$ $S_7 = 42$
$S_8 = S_9 - S_ 7 = 110 - 42 = 68$
$S_6 = S_8 - S_7 = 68 - 42 = 26$
$S_5 = S_7 - S_6 = 42 - 26 = 16$
$S_4 = S_6 - S_5 = 26 - 16 = 10$
Therefore, the answer is $\boxed{10}$

Answer: C 10
Problem: Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$

Solution: $x+\tfrac{2}{x}= y+\tfrac{2}{y}$
Since $x\not=y$ , we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$
Cross multiply in either equation, giving us $xy=2$
$\boxed{2}$

Answer: D 2
Problem: Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$

Solution: Let $A = x + \frac{2}{x} = y + \frac{2}{y}.$ Consider the equation \[u + \frac{2}{u} = A.\] Reorganizing, we see that $u$ satisfies \[u^2 - Au + 2 = 0.\] Notice that there can be at most two distinct values of $u$ which satisfy this equation, and $x$ and $y$ are two distinct possible values for $u.$ Therefore, $x$ and $y$ are roots of this quadratic, and by Vieta’s formulas we see that $xy$ thereby must equal $\boxed{2}.$

Answer:  2
Problem: Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$ , what is $xy$
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$

Solution: \[x + \frac{2}{x} = y + \frac{2}{y}.\]
Multiply both sides by xy to get
\[x^2y + 2y = y^2x +2x\]
Rearrange to get
\[x^2y - y^2x = 2x - 2y\]
Factor out $xy$ on the left side and $2$ on the right side to get
\[xy(x-y) = 2(x-y)\]
Divide by $(x-y)$ {You can do this since x and y are distinct} to get
$\boxed{2}$ ~ e__ (the goat)

Answer: D 2
Problem: In $\triangle ABC$ $AB=AC=28$ and $BC=20$ .  Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively.  What is the perimeter of parallelogram $ADEF$
[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]
$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$

Solution: Note that because $\overline{DE}$ and $\overline{EF}$ are parallel to the sides of $\triangle ABC$ , the internal triangles $\triangle BDE$ and $\triangle EFC$ are similar to $\triangle ABC$ , and are therefore also isosceles triangles.
It follows that $BD = DE$ . Thus, $AD + DE = AD + DB = AB = 28$
Since opposite sides of parallelograms are equal, the perimeter is $2 * (AD + DE) =   \boxed{56}$

Answer: C 56
Problem: In $\triangle ABC$ $AB=AC=28$ and $BC=20$ .  Points $D,E,$ and $F$ are on sides $\overline{AB}$ $\overline{BC}$ , and $\overline{AC}$ , respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB}$ , respectively.  What is the perimeter of parallelogram $ADEF$
[asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real r=5/7; pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r); pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y)); pair E=extension(D,bottom,B,C); pair top=(E.x+D.x,E.y+D.y); pair F=extension(E,top,A,C); draw(A--B--C--cycle^^D--E--F); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,S); label("$F$",F,dir(0)); [/asy]
$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }60\qquad \textbf{(E) }72\qquad$

Solution: We can set point $F$ to be on point $C$ , and point $D$ to be on point $A$
This makes a degenerate parallelogram with sides of length $28$ and $0$ , so it has a perimeter of $28 + 28 =  \boxed{56}$

Answer: C 56
Problem: Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits.  What is the sum of the elements of $S$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$

Solution: Let us begin by working with the condition $0.\overline{ab} = 0.ababab\cdots,$ . Let $x = 0.ababab\cdots$ . So, $100x-x = ab \Rightarrow x = \frac{ab}{99}$ . In order for this fraction $x$ to be in the form $\frac{1}{n}$ $99$ must be a multiple of $ab$ . Hence the possibilities of $ab$ are $1,3,9,11,33,99$ . Checking each of these, $\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},$ and $\frac{99}{99} = 1$ . So the only values of $n$ that have distinct $a$ and $b$ are $11,33,$ and $99$ . So, $11+33+99= \boxed{143}$

Answer: D 143
Problem: Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits.  What is the sum of the elements of $S$
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$

Solution: Notice that we have $\frac{100}{n}= ab.\overline{ab}$
We can subtract $\frac{1}{n}=00.\overline{ab}$ to get \[\frac{99}{n}=ab\]
From this we determine $n$ must be a positive factor of $99$
The factors of $99$ are $1,3,9,11,33,$ and $99$
For $n=1,3,$ and $9$ however, they yield $ab=99,33$ and $11$ which doesn't satisfy $a$ and $b$ being distinct.
For $n=11,33$ and $99$ we have $ab=09,03$ and $01$ .  (Notice that $a$ or $b$ can be zero)
The sum of these $n$ are $11+33+99=143$
$\boxed{143}$

Answer: D 143
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: If you have graph paper, use Pick's Theorem to quickly and efficiently find the area of the quadrilateral. If not, just find the area by other methods.
Pick's Theorem states that
$A$ $I$ $+$ $\frac{B}{2}$ $1$ , where $I$ is the number of lattice points in the interior of the polygon, and $B$ is the number of lattice points on the boundary of the polygon.
In this case,
$A$ $5$ $+$ $\frac{7}{2}$ $1$ $7.5$
so
$\frac{A}{2}$ $3.75$
The bottom half of the quadrilateral makes a triangle with base $4$ and half the total area, so we can deduce that the height of the triangle must be $\frac{15}{8}$ in order for its area to be $3.75$ . This height is the y coordinate of our desired intersection point.
Note that segment CD lies on the line $y = -3x + 12$ . Substituting in $\frac{15}{8}$ for y, we can find that the x coordinate of our intersection point is $\frac{27}{8}$
Therefore the point of intersection is ( $\frac{27}{8}$ $\frac{15}{8}$ ), and our desired result is $27+8+15+8= \boxed{58}$

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 of intersection be $E$ , with coordinates $(x, y)$ . Then, $ABCD$ is cut into $ABCE$ and $AED$
Since the areas are equal, we can use Shoelace Theorem to find the area. This gives $3 + 3x - 3y = 4y$
The line going through $CD$ is $y = -3x + 12$ . Since $E$ is on $CD$ , we can substitute this in, giving $3 + 3x = -21x + 84$ . Solving for $x$ gives $\frac{27}{8}$ . Plugging this back into the line equation gives $y = \frac{15}{8}$ , for a final answer of $\boxed{58}$

Answer:  58
Problem: Rabbits Peter and Pauline have three offspring—Flopsie, Mopsie, and Cotton-tail. These five rabbits are to be distributed to four different pet stores so that no store gets both a parent and a child. It is not required that every store gets a rabbit. In how many different ways can this be done?
$\textbf{(A)} \ 96 \qquad  \textbf{(B)} \ 108 \qquad  \textbf{(C)} \ 156 \qquad  \textbf{(D)} \ 204 \qquad  \textbf{(E)} \ 372$

Solution: We tackle the problem by sorting it by how many stores are involved in the transaction.
1) 2 stores are involved. There are $\binom{4}{2} = 6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6 \cdot 2 = 12$ total arrangements.
2) 3 stores are involved. There are $\binom{4}{3} = 4$ ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately.
Separately: All children must be in one store. There are $3!$ ways to arrange this. $6$ ways in total.
Together: Both parents are in one store and the 3 children are split between the other two. There are $\binom{3}{2}$ ways to split the children and $3!$ ways to choose to which store each group will be sold. $3! \cdot \binom{3}{2} = 18$
$(6 + 18) \cdot 4 = 96$ total arrangements.
3) All 4 stores are involved. We break down the problem as previously shown.
Separately: All children must be split between two stores. There are $\binom{3}{2} = 3$ ways to arrange this. We can then arrange which group is sold to which store in $4!$ ways. $4! \cdot 3 = 72$
Together: Both parents are in one store and the 3 children are each in another store. There are $4! = 24$ ways to arrange this.
$24 + 72 = 96$ total arrangements.
Final Answer: $12 + 96 + 96 = \boxed{204}$

Answer: D 204