Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$

Solution: Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\sqrt{3}$ , the altitude of the top vertex of the newly placed cube is $3\sqrt{3}$ . The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\frac{3\sqrt{3}}{2}$ , so that the height of the plane is $\frac{3\sqrt{3}}{2}$
By symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.
We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$ . The base of the pyramid is an equilateral triangle with side length of $\sqrt{3^2+3^2}=3\sqrt{2}$ . The height of the triangle is $\frac{ \sqrt{3} }{2} \cdot 3\sqrt{2}$ . The distance of the center of the triangle to the vertex is $\frac23 \cdot \frac{ \sqrt{3} }{2} \cdot 3\sqrt{2} = \sqrt {6}$ . Therefore, $h = \sqrt{3^2-(\sqrt {6})^2} = \sqrt{9-6}=\sqrt{3}$ . The altitude of the pyramid at the bottom of the cube is also $h$ . The altitude in the middle is $3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}$
The altitude of the vertex at the top is $3\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $2\sqrt{3}$ . The altitude of the third highest $3$ vertices are all $\sqrt{3}$ . The altitude of the bottom-most vertex is $0$
By scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $\frac{2\sqrt{3}}{3}$ . The altitude of the third highest $3$ vertices are all $\frac{\sqrt{3}}{3}$ . The altitude of the bottom-most vertex is $0$
The length of the space diagonal of a unit cube is $\sqrt{3}$ . The highest vertex of the bottom-most unit cube has an altitude of $\sqrt{3}$ . As $\sqrt{3} < \frac{3\sqrt{3}}{2}$ , therefore, the plane will not pass through the unit cube at the bottom.
For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{\sqrt{3}}{3} + \sqrt{3} = \frac{4\sqrt{3}}{3}$ . As $\frac{4\sqrt{3}}{3} < \frac{3\sqrt{3}}{2}$ , therefore, the plane will not pass through the next $3$ unit cubes.
For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}$ . As $\frac{5\sqrt{3}}{3} > \frac{3\sqrt{3}}{2}$ , therefore, the plane will pass through the next $3$ unit cubes.
So at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.
Thus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\boxed{19}$ unit cubes.

Answer: D 19
Problem: A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle? [asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,2.5)); label("14", (6.5,6)); label("35", (6.5,2.5)); [/asy]
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 25$

Solution: [asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path BG=shift(0,-0.5)*(B--G); path BF=shift(0.5,0)*(B--F); path FC=shift(0.5,0)*(F--C); path DH=shift(0,0.5)*(D--H); draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(BG,L=Label("$7$",position=MidPoint,align=(0,-1)),arrow=Arrows(),bar=Bars,red); draw(BF,L=Label("$5$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(FC,L=Label("$2$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(DH,L=Label("$3$",position=MidPoint,align=(0,1)),arrow=Arrows(),bar=Bars,red); label("$6$", (1.5,6)); label("$15$", (1.5,2.5),blue); label("$14$", (6.5,6)); label("$35$", (6.5,2.5)); [/asy]
We can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\times 5=\boxed{15}$

Answer: B 15
Problem: How many of the following are equal to $x^x+x^x$ for all $x>0$
$\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textbf{IV:}\ (2x)^{2x}$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: We look at each statement individually.
$\textbf{I:}\ 2x^x$ . We note that $x^x+x^x=x^x(1+1)=2x^x$ . So statement $\textbf{I}$ is true.
$\textbf{II:}\ x^{2x}$ . We find a counter example which is $x=1$ $2\neq 1$ . So statement $\textbf{II}$ is false.
$\textbf{III:}\ (2x)^x$ . We see that this statement is equal to $2^xx^x$ $x=2$ is a counter example. $8\neq 16$ . So statement $\textbf{III}$ is false.
$\textbf{IV:}\ (2x)^{2x}$ . We see that $x=1$ is again a counter example. $2\neq 4$ . So statement $\textbf{IV}$ is false.
Therefore, our answer is $\boxed{1}$

Answer: B 1
Problem: In the $xy$ -plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$

Solution: We see that the center of this circle is at $\left(\frac{-5+25}{2},0\right)=(10,0)$ . The radius is $\frac{30}{2}=15$ . So the equation of this circle is \[(x-10)^2+y^2=225.\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\boxed{10}$

Answer: A 10
Problem: Pat intended to multiply a number by $6$ but instead divided by $6$ . Pat then meant to add $14$ but instead subtracted $14$ . After these mistakes, the result was $16$ . If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$

Solution: We reverse the operations that he did and then use the correct operations. His end result is $16$ . Before that, he subtracted $14$ which means that his number after the first operation was $30$ . He divided by $6$ so his number was $180$
Now, we multiply $180$ by $6$ to get $1080$ . Finally, $1080+14=1094$ . Since $1094>1000$ , our answer is $\boxed{1000}$

Answer: E 1000
Problem: In the sequence \[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\] each term is the sum of the two terms to its left. Find $a$
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$

Solution: We work backwards to find $a$
\[d+0=1\implies d=1\] \[c+1=0\implies c=-1\] \[b+(-1)=1\implies b=2\] \[a+2=-1\implies a=\boxed{3.}\]

Answer: A 3.
Problem: Squares $ABCD$ and $EFGH$ are congruent, $AB=10$ , and $G$ is the center of square $ABCD$ . The area of the region in the plane covered by these squares is [asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle); label("A", (0,0), W); label("B", (10,0), E); label("C", (10,10), NE); label("D", (0,10), NW); label("G", (5,5), N); label("F", (12,-2), E); label("E", (5,-9), S); label("H", (-2,-2), W); dot((-2,-2)); dot((5,-9)); dot((12,-2)); dot((0,0)); dot((10,0)); dot((10,10)); dot((0,10)); dot((5,5)); [/asy] $\textbf{(A)}\ 75 \qquad\textbf{(B)}\ 100 \qquad\textbf{(C)}\ 125 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 175$

Solution: The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\triangle ABG$ and subtract this from $200$ , the total area of the two squares.
Since $G$ is the center of $ABCD$ $BG$ is half of the diagonal of the square. The diagonal of $ABCD$ is $10\sqrt{2}$ so $BG=5\sqrt{2}$ . Since $EFGH$ is a square, $\angle G=90^\circ$ . So $\triangle ABG$ is an isosceles right triangle. Its area is $\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25$ . Therefore, the area of the region is $200-25=\boxed{175.}$

Answer: E 175.
Problem: In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is $56$ . The area of the region bounded by the polygon is [asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); [/asy] $\textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196$

Solution: Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$ . We break the figure into squares as shown below.
[asy] unitsize(0.8cm); draw(shift(3,4)*((0,-1)--(1,-1))); draw(shift(3,4)*((-1,-2)--(2,-2))); draw(shift(3,4)*((-2,-3)--(3,-3))); draw(shift(3,4)*((-2,-4)--(3,-4))); draw(shift(3,4)*((-1,-5)--(2,-5))); draw(shift(3,4)*((0,-6)--(1,-6))); draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,-1)--(2,2)); draw((3,-2)--(3,3)); draw((4,-2)--(4,3)); draw((5,-1)--(5,2)); draw((6,0)--(6,1)); [/asy]
We see that there are a total of $2(1+3+5)+7=25$ squares with side length $2$ . Therefore, the total area is $4\cdot 25=\boxed{100.}$

Answer: C 100.
Problem: If $\angle A$ is four times $\angle B$ , and the complement of $\angle B$ is four times the complement of $\angle A$ , then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$

Solution: Let $\angle A=x$ and $\angle B=y$ . From the first condition, we have $x=4y$ . From the second condition, we have \[90-y=4(90-x).\] Substituting $x=4y$ into the previous equation and solving yields \begin{align*}90-y=4(90-4y)&\implies 90-y=360-16y\\&\implies 15y=270\\&\implies y=\boxed{18}

Answer: D 18
Problem: If $i^2=-1$ , then $(i-i^{-1})^{-1}=$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}$

Solution: We simplify step by step as follows: \begin{align*}(i-i^{-1})^{-1}&=\frac{1}{i-i^{-1}}\\&=\frac{1}{i-\frac{1}{i}}\\&=\frac{1}{\left(\frac{i^2-1}{i}\right)}\\&=\frac{i}{i^2-1}\\&=\boxed{2}

Answer: D 2
Problem: Find the sum of the arithmetic series \[20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40\] $\textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000$

Solution: Our first term is $20$ and our last term is $40$ . To find the number of terms, $n$ , we note that the common difference between each term is $\frac{1}{5}$ . So we have \[20+\frac{1}{5}(n-1)=40\implies n-1=100\implies n=101.\] Using our formula, our sum is \[101\left(\frac{20+40}{2}\right)=101\times 30=\boxed{3030.}\]

Answer: B 3030.
Problem: For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$

Solution: Let $n=10a+b$ . So $n^2=(10a+b)^2=100a^2+20ab+b^2$ . The term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd.  The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ which is even.  So we see this term also does not affect whether the tens digit is odd.  This means only $b^2$ can affect whether the tens digit is odd.  We can quickly check $1^2=1, \dots, 9^2=81$ and discover only for $b=4$ or $b=6$ does $b^2$ have an odd tens digit. The total number of positive integers less than or equal to $100$ that have $4$ or $6$ as the units digit is \[10\times 2=\boxed{20.}\]

Answer: B 20.
Problem: Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71$

Solution: Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So \[\frac{r-1}{r+b-1}=\frac{1}{7}.\] When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total marbles left in the bag. So \[\frac{r}{r+b-2}=\frac{1}{5}.\] Cross multiplying for each yields \begin{align*}7r-7=r+b-1&\implies 7r-6=r+b\\ 5r=r+b-2&\implies 5r+2=r+b.\end{align*} We can equate each of these expressions to yields \[7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.\] Therefore, the total number of marbles is \[r+b=4+18=\boxed{22.}\]

Answer: B 22.
Problem: Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$ . If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$ [asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$

Solution: We solve for $\angle A$ as follows: \[4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.\] That means that minor arc $\widehat{BC}$ has measure $40^\circ$ . We can fit a maximum of $\frac{360}{40}=\boxed{9}$ of these arcs in the circle.

Answer: C 9
Problem: Label one disk " $1$ ", two disks " $2$ ", three disks " $3$ $, ...,$ fifty disks " $50$ ". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501$

Solution: We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle. 
We can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks.
From disks numbered from 10 to 50, we can pick up at most 9 disks to prevent picking up 10. There are 50-10+1 = 41 different numbers from 10 to 50. We pick up 9 from each number, therefore, we multiply $41 \cdot 9 = 369$ . In total, the maximum number we can pick up without picking up 10 of the same kind is $369+45=414$ . We need one more disk to guarantee a complete set of 10. Therefore, the answer is $\boxed{415}$

Answer:  415
Problem: A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$ . The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$

Solution: The minimum range occurs in the set $\{7,7,12,12,12\}$ , so the answer is $\boxed{5}$

Answer: C 5
Problem: If $x$ and $y$ are non-zero real numbers such that \[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\] then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$

Solution: We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$
Solving for $y$ in the top equation gives us $3-x$ . Plugging this in gives us: $x^3-x^2+3x=0$ .  Since we're told $x$ is not zero, we can divide by $x$ , giving us: $x^2-x+3=0$
The discriminant of this is $(-1)^2-4(1)(3)=-11$ , which means the equation has no real solutions.
We conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$ .  Negating the top equation gives us $x-y=-3$ . We seek $x-y$ , so the answer is $\boxed{3}$

Answer: A 3
Problem: A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$ .) If $m=10$ , what is the value of $n$ [asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 26$

Solution: To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\frac{(n-2)*180}{n}$ , the measure of the decagon's interior angle is $\frac{8*180}{10} = 144$ degrees.
The regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\boxed{5}$

Answer: A 5
Problem: In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: The line with $x$ -intercept $a$ and $y$ -intercept $b$ is given by the equation $\frac{x}{a} + \frac{y}{b} = 1$ .  We are told $(4,3)$ is on the line so
\[\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12\]
Since $a$ and $b$ are integers, this equation holds only if $(a-4)$ is a factor of $12$ .  The factors are $1, 2, 3, 4, 6, 12$ which means $a$ must be one of $5, 6, 7, 8, 10, 16$ .  The only members of this list which are prime are $a=5$ and $a=7$ , so the number of solutions is $\boxed{2}$

Answer: C 2
Problem: In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: Let $C = (4,3)$ $DF=a$ , and $AD=b$ . As stated in the problem, the $x$ -intercept $DF=a$ is a positive prime number, and the $y$ -intercept $AD=b$ is a positive integer.
Through similar triangles, $\frac{AB}{BC}=\frac{CE}{EF}$ $\frac{b-3}{4}=\frac{3}{a-4}$ $(a-4)(b-3)=12$
The only cases where $a$ is prime are: \[\begin{cases}  a-4=1 & a=5 \\ b-3=12 & b=15 \end{cases}\]
\[and\]
\[\begin{cases}  a-4=3 & a=7 \\ b-3=4 & b=5 \end{cases}\]
So the number of solutions are $\boxed{2}$

Answer: C 2
Problem: Given $0\le x_0<1$ , let \[x_n=\left\{ \begin{array}{ll} 2x_{n-1} &\text{ if }2x_{n-1}<1 \\ 2x_{n-1}-1 &\text{ if }2x_{n-1}\ge 1 \end{array}\right.\] for all integers $n>0$ . For how many $x_0$ is it true that $x_0=x_5$
$\text{(A) 0} \quad \text{(B) 1} \quad \text{(C) 5} \quad \text{(D) 31} \quad \text{(E) }\infty$

Solution: We are going to look at this problem in binary.
$x_0 = (0.a_1 a_2 \cdots )_2$
$2x_0 = (a_1.a_2 a_3 \cdots)_2$
If $2x_0 < 1$ , then $x_0 < \frac{1}{2}$ which means that $a_1 = 0$ and so $x_1 = (.a_2 a_3 a_4 \cdots)_2$
If $2x_0 \geq 1$ then $x \geq \frac{1}{2}$ which means that $x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2$
Using the same logic, we notice that this sequence cycles and that since $x_0 = x_5$ we notice that $a_n = a_{n+5}$
We have $2$ possibilities for each of $a_1$ to $a_5$ but we can't have $a_1 = a_2 = a_3 = a_4 = a_5 = 1$ so we have $2^5 - 1 = \boxed{31}$

Answer: D 31
Problem: Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$ -axis. The value of $m+b$ is
$\text{(A) -6} \quad \text{(B) } -5\quad \text{(C) } -4\quad \text{(D) } -3\quad \text{(E) } -2$

Solution: $\fbox{C}$ First we want to put this is slope-intercept form, so we get $y=\dfrac{1}{3}x+\dfrac{11}{3}$ . When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since $m+b$ is the sum of the slope and the y-intercept, we get $-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{4}$

Answer:  4
Problem: If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$ , all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$

Solution: Since \[\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},\] we can say that \[\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = \frac {y+(x+y)+x}{(x-z)+z+y} = \boxed{2}$

Answer: E 2
Problem: For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$ . If the Cesáro sum of
the 99-term sequence $(a_1,...,a_{99})$ is 1000,  what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$
$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$

Solution: Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \[S_1 + S_2 + S_3 + ... + S_n\] \[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)\] \[= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.\] If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is $1000 * 99 = 99,000,$ than the Cesáro total of the second sequence is $n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.$ Thus the Cesáro sum of the second sequence is $\frac{99,100}{100} = \boxed{991,}\, .$

Answer: A 991,
Problem: Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$ . Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$ , respectively, with $AE=BF=AG=2$ . Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$ . The area of quadrilateral $EFHG$ is
$\text{(A) } 4\quad \text{(B) } 4.5\quad \text{(C) } 5\quad \text{(D) } 5.5\quad \text{(E) } 6$

Solution: We note that $ABFG$ is a parallelogram because $AG = BF = 2$ and $AG \parallel BF$ . Using the same reasoning, $GFCD$ is also a parallelogram.
Assume that the height of parallelogram $ABFG$ with respect to base $AB$ is $x$ . Then, the area of parallelogram $ABFG$ is $AB * x$ . The area of triangle $EFG$ is $\frac{AB * x}{2}$ , which is half of the area of parallelogram $ABFG$
Likewise, the area of triangle $FGH$ is half the area of parallelogram $GFCD$
Thus, $[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}$

Answer:  5