Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of $6$ meters, and it takes her $36$ seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?
$\textbf{(A)}\ \frac{\pi}{3} \qquad\textbf{(B)}\ \frac{2\pi}{3} \qquad\textbf{(C)}\ \pi \qquad\textbf{(D)}\ \frac{4\pi}{3} \qquad\textbf{(E)}\ \frac{5\pi}{3}$

Solution: It is basically the same as Solution 1 except you can completely disregard the straight edges of the track since it will take Keiko the same time to walk that length for both inner and outer circles. Instead, focus on the circular part. If the diameter of the smaller circle were $r,$ then the length of the smaller circle would be $r \pi$ and the length of the larger circle would be $(r+12) \pi.$ Since it still takes $36$ seconds longer,
\begin{align*} \frac{r \pi}{s} + 36 &= \frac{(r+12)\pi}{s}\\ r \pi + 36s &= r \pi + 12 \pi\\ s&=\boxed{3}

Answer: A 3
Problem: Two real numbers are selected independently at random from the interval $[-20, 10]$ . What is the probability that the product of those numbers is greater than zero?
$\textbf{(A)}\ \frac{1}{9} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{4}{9} \qquad\textbf{(D)}\ \frac{5}{9} \qquad\textbf{(E)}\ \frac{2}{3}$

Solution: We will use complementary counting. The probability that the product is negative can be found by finding the probability that one number is positive and the other number is negative. The probability of a positive number being selected is $\frac13$ , and the probability of a negative number being selected is $\frac23$ . So, we see that the probability of the product being negative is \[2 \cdot \frac23 \cdot \frac13 = \frac49.\]
We multiply by $2$ because you can either pick the negative or positive number first.
Thus, the probability of the product being positive is \[1-\frac49 = \boxed{59}\]

Answer: D 59
Problem: A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution: Let the sides of the rectangular parking lot be $a$ and $b$ . Then $a^2 + b^2 = 625$ and $ab = 168$ . Add the two equations together, then factor. \begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \boxed{62}$

Answer: C 62
Problem: A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution: We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy $a^2+b^2=25^2$ . In other words, $(a,b,25)$ is a set of Pythagorean triples. Guessing and checking, we have $(7,24,25)$ as the triplet, as the area is $7 \cdot 24 = 168$ as requested. Therefore, the perimeter is $2(7+24)=\boxed{62}$

Answer: C 62
Problem: In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

Solution: We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$ . When an inscribed angle contains the diameter , the inscribed angle is a right angle . Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$
$\angle ABE = \angle BED$ because they are alternate interior angles and $\overline{AB} \parallel \overline{ED}$ . Opposite angles in a cyclic quadrilateral are supplementary , so $\angle BED + \angle BCD = 180$ . Use substitution to get $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{130}$

Answer: C 130
Problem: In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

Solution: Note $\angle ABE = \angle BED=50$ as before. The sum of the interior angles for quadrilateral $EBCD$ is $360$ . Denote the center of the circle as $P$ $\angle PDE = \angle PED = 50$ . Denote $\angle PDC = \angle PCD = x$ and $\angle PBC = \angle PCB = y$ . We wish to find $\angle BCD = x+y$ . Our equation is $(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360$ . Our final equation becomes $2(x+y)+100 = 360$ . After subtracting $100$ and dividing by $2$ , our answer becomes $x+y=\boxed{130}$

Answer: C 130
Problem: In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$ , and $\overline{AB}$ is parallel to $\overline{ED}$ . The angles $AEB$ and $ABE$ are in the ratio $4 : 5$ . What is the degree measure of angle $BCD$
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$

Solution: Note that $\overset{\Large\frown} {BE}$ intercepts $\angle BAE$ . Since, $\overset{\Large\frown} {BE}=180$ , thus $\angle BAE=90°$ (courtesy of the Inscribed Angles Theorem).
Since we know that $\angle BAE=90°$ , then $\angle AEB + \angle ABE = 90°$ , (courtesy of the Triangle Sum Theorem) and also $5\angle AEB = 4\angle ABE$ . By solving this variation, $\angle AEB = 40$ and $\angle ABE = 50$ . After that, due to the Alternate Interior Angles Theorem, $\angle ABE \cong \angle BED$ , which means $\angle BED = 50$
After doing some angle chasing, then these following facts should be true, $\overset{\Large\frown} {AB}=80$ $\overset{\Large\frown} {BD}=100$ $\overset{\Large\frown} {AE}=100$
Note that the arcs have to equal 360, so, $360=\overset{\Large\frown} {AB}+\overset{\Large\frown} {BD}+\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}$
$360=80+100+100+\overset{\Large\frown} {DE}$
$\overset{\Large\frown} {DE}=80$
Notice how $\overset{\Large\frown} {DB}$ intercepts $\angle BCD$ and that $\overset{\Large\frown} {DB}=\overset{\Large\frown} {DE}+\overset{\Large\frown} {AE}+\overset{\Large\frown} {AB}$
$\overset{\Large\frown} {DB}=80+100+80$
$\overset{\Large\frown} {DB}=260$
According to the Inscribed Angles Theorem, $2\angle BCD=\overset{\Large\frown} {DB}$ , therefore the answer is $\frac{260}{2}= \boxed{130}$

Answer: C 130
Problem: Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$

Solution: It is given that $\angle AMD \sim \angle CMD$ . Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB} \parallel \overline{DC}$ $\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM$ . Use the Base Angle Theorem to show $\overline{DC} \cong \overline{MC}$ . We know that $ABCD$ is a rectangle , so it follows that $\overline{MC} = 6$ . We notice that $\triangle BMC$ is a $30-60-90$ triangle, and $\angle BMC = 30^{\circ}$ . If we let $x$ be the measure of $\angle AMD,$ then \begin{align*} 2x + 30 &= 180\\ 2x &= 150\\ x &= \boxed{75}

Answer: E 75
Problem: Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$

Solution: After finding $MC = 6,$ we can continue using trigonometry as follows.
We know that $\angle{BMC} = 180-2x$ and so $\sin (180-2x) = \frac{3}{6} = \frac{1}{2}$
It is obvious that $\sin (30) = \frac{1}{2}$ and so $180-2x=30.$
Solving, we have $x = \boxed{75}$

Answer:  75
Problem: Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$

Solution: Let $\angle{DMC} = \angle{AMD} = \theta$ . If we let $AM = x$ , we have that $MD = \sqrt{x^2 + 9}$ , by the Pythagorean Theorem, and similarily, $MC = \sqrt{x^2 - 12x + 45}$ . Applying the law of cosine, we see that \[2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36\] and \[\tan (\theta) = \frac{3}{x}\] YAY!!! We have two equations for two variables... that are relatively difficult to deal with. Well, we'll try to solve it. First of all, note that $\sin (\theta) = \frac{3}{\sqrt{x^2+9}}$ , so solving for $\cos (\theta)$ in terms of $x$ , we get that $\cos (\theta) = \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9}$ . The equation now becomes
\[2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9} = 36\] Simplifying, we get
\[4x^4 - 48x^3 + 216x^2 - 432x + 324\]
Now, we apply the quartic formula to get
\[x = 6 \pm 3 \sqrt{3}\]
We can easily see that $x = 6 + 3 \sqrt{3}$ is an invalid solution. Thus, $x = 6 - 3 \sqrt{3}$
Finally, since $\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}$ $\theta = \frac{5 + 12n}{12} \pi$ , where $n$ is any integer. Converting to degrees, we have that $\theta = 75 + 180n$ . Since $0 < \theta < 90$ , we have that $\theta = \boxed{75}$ $\square$

Answer:  75
Problem: Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$

Solution: We have $DC=CM=6$ . By the Pythagorean Theorem, $BM=\sqrt{6^2-3^2}=3\sqrt{3}$ , and thus $AM=6-3\sqrt{3}$ , We have $\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}$ , or $\angle AMD=\boxed{75}$ ~awsomek

Answer:  75
Problem: Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$

Solution: Let $\angle AMD=\angle DMC=\theta$ . Thus, $\angle BMC=180-2\theta\implies\angle MCB=2\theta-90$ . Since all angles should be positive, $\theta>45^\circ$ , narrowing the options to D and E. Trying $60^\circ$ (option D), $\Delta AMD$ is a 30-60-90 triangle. $AD=3$ , so it follows that $AM=\sqrt3$ . 
Since $\angle BMC=180-2\theta$ $\angle BMC=60^\circ$ , too. However, that would imply that $\Delta MBC$ is also a $30-60-90$ triangle, which would, in turn, imply that $MB=3\sqrt3$ , since $BC=3$ . We know that $AM+MB=AB$ and $AB=6,$ but we know that $AM=\sqrt3$ and $MB=3\sqrt3$ $AM+MB$ is clearly not $6$ , so this is not possible. 
Thus, the answer must be $\boxed{75}$ .
~ Technodoggo

Answer: E 75
Problem: What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$

Solution: First, square both sides, and isolate the absolute value. \begin{align*} 5|x|+8&=x^2-16\\ 5|x|&=x^2-24\\ |x|&=\frac{x^2-24}{5}. \\ \end{align*} Solve for the absolute value and factor.
Case 1: $x=\frac{x^2-24}{5}$
Multiplying both sides by $5$ gives us \[5x=x^2-24.\] Rearranging and factoring, we have \begin{align*} x^2-5x-24 &=0, \\ (x-8)(x+3) &= 0.\\ \end{align*}
Case 2: $x=\frac{-x^2+24}{5}$
As above, we multiply both sides by $5$ to find \[5x=-x^2+24.\] Rearranging and factoring gives us \begin{align*} x^2+5x-24 &=0, \\ (x+8)(x-3) &= 0. \\ \end{align*}
Combining these cases, we have $x= -8, -3, 3, 8$ . Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for $x$ back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways.
Trying $|x|=|3|$ , we have \begin{align*} \sqrt{5|3|+8}&=\sqrt{3^2-16}, \\ \sqrt{15+8}&=\sqrt{9-16}, \\ \sqrt{23} &\not= \sqrt{-7}.\\ \end{align*} Therefore, $x = 3$ and $x= -3$ are extraneous.
Checking $|x|=|8|$ , we have \begin{align*} \sqrt{5|8|+8}&=\sqrt{8^2-16}, \\ \sqrt{40+8}&=\sqrt{64-16}, \\ \sqrt{48}&=\sqrt{48}.\\ \end{align*}
The roots of our original equation are $-8$ and $8$ and product is $-8 \times 8 = \boxed{64}$

Answer: A 64
Problem: What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$

Solution: Square both sides, to get $5|x| + 8 = x^2-16$ . Rearrange to get $x^2 - 5|x| - 24 = 0$ . Seeing that $x^2 = |x|^2$ , substitute to get $|x|^2 - 5|x| - 24 = 0$ . We see that this is a quadratic in $|x|$ . Factoring, we get $(|x|-8)(|x|+3) = 0$ , so $|x| = \{8,-3\}$ . Since the radicand of the equation can't be negative, the sole solution is $|x| = 8$ . Therefore, $x$ can be $8$ or $-8$ . The product is then $-8 \times 8 = \boxed{64}$

Answer: A 64
Problem: What is the product of all the roots of the equation \[\sqrt{5 | x | + 8} = \sqrt{x^2 - 16}.\]
$\textbf{(A)}\ -64 \qquad\textbf{(B)}\ -24 \qquad\textbf{(C)}\ -9 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 576$

Solution: First we note that $x \in (-\infty,-4] \cup [4,\infty]$ . This will help us later with finding extraneous solutions. 
Next, we have two cases: \[\text{IF } x\leq 4:\] \[\text{ } \sqrt{-5x+8} = \sqrt{x^2-16} \implies x^2+5x-24=0 \rightarrow x = -8,3\] .
We note that $3$ is not in the range of possible $x$ 's and thus is not a solution.
\[\text{IF } x\geq 4:\] \[\text{ } \sqrt{5x+8} = \sqrt{x^2-16} \implies x^2-5x-24=0 \rightarrow x = -3,8\] .
We again not that $-3$ is an extraneous solution.
Thus, we have the two solutions $-8$ and $8$ . Therefore product is $-8 \cdot 8 = \boxed{64}$

Answer: A 64
Problem: Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$

Solution: The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences can't be right next to each other because they would make a difference of $8,$ which isn't given as a possibility in the problem. This means $1$ must be the difference between $y$ and $x.$ We can express the possible configurations as the lines.
If we look at the first number line, you can express $x$ as $w-5,$ $y$ as $w-6,$ and $z$ as $w-9.$ Since the sum of all these integers equal $44$ \begin{align*} w+w-5+w-6+w-9&=44\\ 4w&=64\\ w&=16 \end{align*} You can do something similar to this with the second number line to find the other possible value of $w.$ \begin{align*} w+w-3+w-4+w-9&=44\\ 4w&=60\\ w&=15 \end{align*} The sum of the possible values of $w$ is $16+15 = \boxed{31}$

Answer: B 31
Problem: Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$

Solution: First, like Solution 1, we know that $w-z=9 \ \text{(1)}$ , because no two numbers could have a larger difference. Next, we find the sum of all the differences; since $w$ is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes $3w$ . Continuing in this way, we find that \[3w+x-y-3z=28 \ \text{(2)}\] . Now, we can subtract $3w-3z=27$ from (2) to get $x-y=1 \ \text{(3)}$ . Also, adding (2) with $w+x+y+z=44$ gives $4w+2x-2z=72$ , or $2w+x-z=36$ . Subtracting (1) from this gives $w+x=27$ . Since we know $w-z$ and $x-y$ , we find that \[(w-z)+(x-y)=(w-y)+(x-z)=9+1=10\] . This means that $w-y$ and $x-z$ must be 4 and 6, in some order. If $w-y=6$ , then subtracting this from (3) gives $(w-y)-(x-y)=6-1=5$ , so $w-x=5$ . This means that $(w-x)+(w+x)=2w=27+5=32$ , so $w=16$ . Similarly, $w$ can also equal $15$
Now if you are in a rush, you most likely would have answered $16+15=\boxed{31}$ . But we do have to check if these work. In fact, they do, giving solutions $(w,x,y,z)=(16, 11, 10, 7)$ and $(w,x,y,z)=(15, 12, 11, 6)$

Answer: B 31
Problem: Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$

Solution: Let $w - x = a$ $w - y = b$ $w - z = c$ . As above, we know that $c = 9$ . Thus, $a < b < c$ . 
So, we have $w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44$ . This means $a + b + 9$ is a multiple of $4$ . Testing values of $a$ and $b$ , we find $(a, b, c) = (1, 6, 9), (3, 4, 9),$ and $(5, 6, 9)$ all satisfy this relation. The corresponding $(w, x, y, z)$ sets are $(15, 14, 9, 6), (15, 12, 11, 6),$ and $(16, 11, 10, 7)$ . The first set does not satisfy the given conditions, but the other two do. Thus, $w = 15$ and $w = 16$ are both possible solutions so the answer is $16+15=\boxed{31}$

Answer: B 31
Problem: Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$

Solution: From the problem, we know that $w+x+y+z=44$ . Since it is said that the pairwise positive differences between numbers are 1, 3, 4, 5, 6, 9, and we can figure that the pairwise positive differences are $(w-x)$ $(w-y)$ $(w-z)$ $(x-y)$ $(x-z)$ $(y-z)$ , the sum of $(w-x), (w-y), (w-z), (x-y), (x-z), (y-z)$ is equal to the sum of 1, 3, 4, 5, 6, 9, so $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ . Simplifying, we get $3w-3z+x-y=28$ . Adding $w+x+y+z=44$ and $(w-x)+(w-y)+(w-z)+(x-y)+(x-z)+(y-z)=28$ , we get $4w+2x-2z=72$ , and simplifying we get $2w+x-z=36$ . Since $x-z$ is one of our positive differences, we can start guessing values for $w$ , and if the equation simplifies to one of our numerical positive differences, that value of $w$ should work. We can start at $w=18$ and keep going down, because our sum has to be positive. For $w=18$ $x-z=0$ , which is not one of our sums. For $w=17$ $x-z=2$ ,which is not one of our sums. For $w=16$ $x-z=4$ , which is one of our sums, so 16 works. For $w=15$ $x-z=6$ , which is one of our sums, so 15 works. For $w=14$ $x-z=8$ , which is not one of our sums. If we keep going, $x-z$ will soon exceed 10 and exceed all our sums, so any value below $w=14$ will not work. Therefore, our only solutions for $w$ are 15 and 16, which means our sum is $\boxed{31}$ . You can check that 15 and 16 work by forming a string of 4 numbers as shown above.

Answer: B 31
Problem: Brian writes down four integers $w > x > y > z$ whose sum is $44$ . The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$ . What is the sum of the possible values for $w$
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$

Solution: Because we know that $w>x>y>z$ and that the positive differences are $1, 3, 4, 5, 6, 9$ , we can immediately come to the conclusion that $w-z= 9$ (because w is the largest integer and z is the smallest integer, so their difference must be the greatest). With this we have $3$ equations, we have that $x+y+z+w=44$ (from the problem), $w-z=9$ and because we can add up all the possible possible differences (as shown in the previous solutions), we get that $3w+x-y-3z=28$ . With these equations, we eventually manipulate these equations by doing the first equation minus 3 times the second to get $x-y=1$ . We can also add the first and third equation and subtract the second equation to get $w+x=27$ thus we know that $w-x$ can be $3, 4, 5,$ and $6$ (note: 1 and 9 are not possible because we know that $x-y=1$ and $w-z=9$ and the remaining differences can only be taken by 1 pair, so $w-x$ cannot be equal to 1 or 9). However, in order to get an integer value for x and z, we find that $w-z$ can only be equal to 3 and 5. Thus, by solving these, we see that $w= 15$ and $w=16$ $15+16=\boxed{31}$

Answer: B 31
Problem: What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

Solution: Since $2011 \equiv 11  \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011}  \pmod{1000}.$
To compute this, we use a clever application of the binomial theorem
$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}$
In all of the other terms, the power of $10$ is greater than $2$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:
$\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}$
Therefore, the hundreds digit is $\boxed{6}.$

Answer: D 6
Problem: What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

Solution: We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem , it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$
In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have \[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.\]
In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have $\begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}$
After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{6}.$

Answer: D 6