Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005 , how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$

Solution: This is very analogous to base $9$ . But, in base $9$ , we don't have a $9$ . So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$ $1463 - 1 = 1462$
Therefore, our answer is $\boxed{1462}$

Answer:  1462
Problem: A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$

Solution: Box P has dimensions $l$ $w$ , and $h$ . 
Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]
The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\] so the diameter is \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{10}.\]

Answer: B 10
Problem: A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$

Solution: As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$ , and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$
Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$ . (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$ , and we are solving for $\sqrt{l^2 + w^2}$ . Because \[(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,\] this means that \[l^2 + w^2 = 28^2 - 384 = 400,\] so the space diagonal is $\sqrt{400} = 20$ . Since the diameter of the sphere is $20$ , the radius is $\boxed{10}$ . ~ emerald_block

Answer: B 10
Problem: Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$
$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$

Solution: We can write the problem as
Since $\deg P(x) = 3$ and $\deg R(x) = 3$ $\deg P(x)\cdot R(x) = 6$ . Thus, $\deg P(Q(x)) = 6$ , so $\deg Q(x) = 2$
Hence, we conclude $Q(1)$ $Q(2)$ , and $Q(3)$ must each be $1$ $2$ , or $3$ . Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$ $Q(2)$ , and $Q(3)$ are chosen.
However, we have included $Q(x)$ which are not quadratics: lines. Namely,
Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is y-value of the midpoint of $(1, Q(1))$ and $(3, Q(3))$ . So we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = \boxed{22}$

Answer: B 22
Problem: Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$
$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$

Solution: We see that \[P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x).\] Therefore, $P(x) | P(Q(x))$ .  Since $\deg Q = 2,$ we must have $x-1, x-2, x-3$ divide $P(Q(x))$ . So, we pair them off with one of $Q(x)-1, Q(x)-2,$ and $Q(x)-3$ to see that there are $3!+3 \cdot 2 \cdot \binom{3}{2} = 24$ without restrictions. (Note that this count was made by pairing off linear factors of $P(x)$ with $Q(x)-1, Q(x)-2,$ and $Q(x)-3$ , and also note that the degree of $Q$ is 2.) However, we have two functions which are constant, which are $Q(x) = x$ and $Q(x) = 4-x.$ So, we subtract $2$ to get a final answer of $\boxed{22}$

Answer: B 22
Problem: Let $S$ be the set of all points with coordinates $(x,y,z)$ , where $x$ $y$ , and $z$ are each chosen from the set $\{0,1,2\}$ . How many equilateral triangles all have their vertices in $S$
$(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88$

Solution: For this solution, we will just find as many solutions as possible, until it becomes intuitive that there are no more size of triangles left.
First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any vertex , and connecting the three adjacent vertices into a triangle. This triangle will have a side length of $\sqrt{2}$ ; a quick further examination of this cube will show us that this is the only possible side length (red triangle in diagram). Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube (green triangle), giving us 9 cubes and $9 \cdot 8 = 72$ equilateral triangles.
NOTE: Connecting the centers of the faces will actually give an octahedron , not a cube, because it only has $6$ vertices.
Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle).  Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\boxed{80}$

Answer: C 80
Problem: What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6      \qquad \mathrm{(B)}\ 12      \qquad \mathrm{(C)}\ 16      \qquad \mathrm{(D)}\ 24      \qquad \mathrm{(E)}\ 25$

Solution: If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$ , then $a$ is either $b$ or $-b$ ):
\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}
We can then put these equations in slope-intercept form in order to graph them.
\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}
Now you can graph the lines to find the shape of the graph:
[asy] Label f;  f.p=fontsize(6);  xaxis(-8,8,Ticks(f, 4.0));  yaxis(-6,6,Ticks(f, 3.0));  fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey); draw((-4,-6)--(8,3), Arrows(4)); draw((4,-6)--(-8,3), Arrows(4)); draw((-4,6)--(8,-3), Arrows(4)); draw((4,6)--(-8,-3), Arrows(4));[/asy]
We can easily see that it is a rhombus with diagonals of $6$ and $8$ . The area is $\dfrac{1}{2}\times 6\times8$ , or $\boxed{24}$

Answer: D 24
Problem: What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6      \qquad \mathrm{(B)}\ 12      \qquad \mathrm{(C)}\ 16      \qquad \mathrm{(D)}\ 24      \qquad \mathrm{(E)}\ 25$

Solution: The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is
\[\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{24}\]

Answer: D 24
Problem: For how many values of $a$ is it true that the line $y = x + a$ passes through the
vertex of the parabola $y = x^2 + a^2$
$\mathrm{(A)}\ 0      \qquad \mathrm{(B)}\ 1      \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 10      \qquad \mathrm{(E)}\ \text{infinitely many}$

Solution: We see that the vertex of the quadratic function $y = x^2 + a^2$ is $(0,\,a^2)$ . The y-intercept of the line $y = x + a$ is $(0,\,a)$ . We want to find the values (if any) such that $a=a^2$ . Solving for $a$ , the only values that satisfy this are $0$ and $1$ , so the answer is $\boxed{2}$

Answer: C 2
Problem: The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1      \qquad \mathrm{(B)}\ 2      \qquad \mathrm{(C)}\ 3      \qquad \mathrm{(D)}\ 4      \qquad \mathrm{(E)}\ 5$

Solution: $221$ can be written as the sum of four two-digit numbers, let's say $\overline{ae}$ $\overline{bf}$ $\overline{cg}$ , and $\overline{dh}$ . Then $221= 10(a+b+c+d)+(e+f+g+h)$ . The last digit of $221$ is $1$ , and $10(a+b+c+d)$ won't affect the units digits, so $(e+f+g+h)$ must end with $1$ . The smallest value $(e+f+g+h)$ can have is $(1+2+3+4)=10$ , and the greatest value is $(6+7+8+9)=30$ . Therefore, $(e+f+g+h)$ must equal $11$ or $21$
Case 1: $(e+f+g+h)=11$
The only distinct positive integers that can add up to $11$ is $(1+2+3+5)$ . So, $a$ $b$ $c$ , and $d$ must include four of the five numbers $(4,6,7,8,9)$ . We have $10(a+b+c+d)=221-11=210$ , or $a+b+c+d=21$ . We can add all of $4+6+7+8+9=34$ , and try subtracting one number to get to $21$ , but none of them work. Therefore, $(e+f+g+h)$ cannot add up to $11$
Case 2: $(e+f+g+h)=21$
Checking all the values for $e$ $f$ $g$ ,and $h$ each individually may be time-consuming, instead of only having $1$ solution like Case 1. We can try a different approach by looking at $(a+b+c+d)$ first. If $(e+f+g+h)=21$ $10(a+b+c+d)=221-21=200$ , or $(a+b+c+d)=20$ . That means $(a+b+c+d)+(e+f+g+h)=21+20=41$ . We know $(1+2+3+4+5+6+7+8+9)=45$ , so the missing digit is $45-41=\boxed{4}$

Answer: D 4
Problem: The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1      \qquad \mathrm{(B)}\ 2      \qquad \mathrm{(C)}\ 3      \qquad \mathrm{(D)}\ 4      \qquad \mathrm{(E)}\ 5$

Solution: Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d$ , where $d$ is some digit. Clearly, $\boxed{4}$

Answer:  4
Problem: How many distinct four-tuples $(a,b,c,d)$ of rational numbers are there with
\[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]
$\mathrm{(A)}\ 0      \qquad \mathrm{(B)}\ 1      \qquad \mathrm{(C)}\ 17     \qquad \mathrm{(D)}\ 2004   \qquad \mathrm{(E)}\ \text{infinitely many}$

Solution: Using the laws of logarithms , the given equation becomes
\[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\] \[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\] \[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]
As $a,b,c,d$ must all be rational, and there are no powers of $3$ or $7$ in $10^{2005}$ $b=d=0$ . Then $2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005$
Only the four-tuple $(2005,0,2005,0)$ satisfies the equation, so the answer is $\boxed{1}$

Answer: B 1
Problem: Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$
$\mathrm{(A)}\ 25     \qquad \mathrm{(B)}\ 39     \qquad \mathrm{(C)}\ 51     \qquad \mathrm{(D)}\ 60      \qquad \mathrm{(E)}\ 80$

Solution: [asy] Label f;  f.p=fontsize(6); xaxis(-1,15,Ticks(f, 2.0));  yaxis(-1,15,Ticks(f, 2.0));  pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE); D(A--B); filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray); filldraw(CP(0.5(A+B),A),white); D(A); D(B); [/asy]
For angle $A$ and $B$ to be acute, $C$ must be between the two lines that are perpendicular to $\overline{AB}$ and contain points $A$ and $B$ . For angle $C$ to be acute, first draw a $45-45-90$ triangle with $\overline{AB}$ as the hypotenuse. Note $C$ cannot be inside this triangle's circumscribed circle or else $\angle C > 90^\circ$ . Hence, the area of $R$ is the area of the large triangle minus the area of the small triangle minus the area of the circle, which is $\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}$ , which is approximately $51$ . The answer is $\boxed{51}$

Answer: C 51
Problem: Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30     \qquad \mathrm{(B)}\ 32     \qquad \mathrm{(C)}\ 34     \qquad \mathrm{(D)}\ 40     \qquad \mathrm{(E)}\ 50$

Solution: The sum of the set is $-7-5-3-2+2+4+6+13=8$ , so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$ . Considering the set of four terms containing $13$ , this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $13-7-5-3=-2$ , and with two odd terms then its minimum value is $13-7+2-2=6$ , so we cannot achieve two sums of $4$ . The closest we could have to $4$ and $4$ is $3$ and $5$ , which can be achieved through $13-7-5+2$ and $6-3-2+4$ . So the minimum possible value is $3^2+5^2=34\Rightarrow\boxed{34}$

Answer: C 34
Problem: Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30     \qquad \mathrm{(B)}\ 32     \qquad \mathrm{(C)}\ 34     \qquad \mathrm{(D)}\ 40     \qquad \mathrm{(E)}\ 50$

Solution: Trying out the values for a bit leads to us getting $34$ with $(-3, -2, 2, 6)$ in 1 set for a total of 9 and the other numbers giving a total of 25.
We start off with trying to get a $2(4^2) = 32$ solution, so we only need to find one set with a sum of 4 (the other will automatically have a sum of 4).
We can add 7 to each number. We are now trying to get a set with 32.
\[(0, 2, 4, 5, 9, 11, 13, 20)\]
Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers $a,b,c,d$ , the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26).
If $a,b,c,d$ has 2 odd numbers and 2 even numbers we can divide them into two cases: whether $a,b,c,d$ has 20 or doesn't have 20.
Case 1: $a,b,c,d$ doesn't have 20.
The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need.
Case 2: $a,b,c,d$ has 20.
The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need.
We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest value is $\boxed{34}$

Answer: C 34
Problem: A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors.  What is the greatest integer $k$ such that $7^k$ divides $n$
$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$

Solution: We may let $n = 7^k \cdot m$ , where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$ . Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$ . These numbers are in the ratio 3:4, so $\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{2}$

Answer: C 2
Problem: All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$

Solution: Let the three points be at $A = (x_1, x_1^2)$ $B = (x_2, x_2^2)$ , and $C = (x_3, x_3^2)$ , such that the slope between the first two is $2$ , and $A$ is the point with the least $y$ -coordinate.
Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$ . Similarly, the slope of $BC$ is $x_2 + x_3$ , and the slope of $AB$ is $x_1 + x_2 = 2$ . The desired sum is $x_1 + x_2 + x_3$ , which is equal to the sum of the slopes divided by $2$
To find the slope of $AC$ , we note that it comes at a $60^{\circ}$ angle with $AB$ . Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \sqrt{3}i$ . What this does is generate the complex number that is at a $60^{\circ}$ angle with the complex number $1 + 2i$ . Then, we can find the slope of the line between this new complex number and the origin: \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\] The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ}$ angle with $AB$ \[(1+2i)(-1+\sqrt{3}i)\] \[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\] \[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\]
At this point, we start to notice a pattern: This expression is equal to $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$ , except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.
Our sum is simply $2 - 2\cdot\frac{8}{11} = \frac{6}{11}$ , and thus we can divide by $2$ to obtain $\frac{3}{11}$ , which gives the answer $\boxed{14}$

Answer: A 14
Problem: All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$

Solution: Using the slope formula and differences of squares, we find:
$a+b$ = the slope of $AB$
$b+c$ = the slope of $BC$
$a+c$ = the slope of $AC$
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by $2$ . Without loss of generality, let $AB$ be the side that has the smallest angle with the positive $x$ -axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$ . Translate the triangle so $A$ is at the origin. Then $\tan(BOJ) = 2$ . Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is $\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}$
Using $\tan(BOJ) = 2$ , and the tangent addition formula, this simplifies to $\dfrac{3}{11}$ , so the answer is $3 + 11 = \boxed{14}$

Answer: A 14
Problem: For how many ordered pairs of positive integers $(x,y)$ is $x + 2y = 100$
$\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100$

Solution: If $x$ and $2y$ must each be positive integers, then we can say that $x$ is at least 1 and $2y$ is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because $y$ must be an integer, $2y$ must be even; thus only $\left\lfloor \frac{99}{2} \right\rfloor = \boxed{49}$ ways exist to distribute these ones.

Answer: B 49
Problem: For each integer $n\geq 4$ , let $a_n$ denote the base- $n$ number $0.\overline{133}_n$ . The product $a_4a_5\cdots a_{99}$ can be expressed as $\frac {m}{n!}$ , where $m$ and $n$ are positive integers and $n$ is as small as possible. What is $m$
$\text {(A)} 98 \qquad \text {(B)} 101 \qquad \text {(C)} 132\qquad \text {(D)} 798\qquad \text {(E)}962$

Solution: Note that \[0.\overline{133}_n = \frac{n^2+3n+3}{n^3-1},\] by geometric series.
Thus, we're aiming to find the value of \[\prod_{k=4}^{99} \frac{k^2+3k+3}{k^3 - 1}.\] Expanding the product out, this is equivalent to \[\frac{4^2+3(4)+3}{4^3 - 1} \cdot \frac{5^2+3(5)+3}{5^3 - 1} \cdot \frac{6^2+3(6)+3}{6^3 - 1} \cdot ... \cdot \frac{99^2+3(99)+3}{99^3 - 1}.\] Note that the numerator of the $a$ th fraction and the denominator of the $a+1$ th fraction for $1 \leq a \leq 95$ cancel out to be $\frac{1}{a+3},$ since \[\frac{k^2 + 3k + 3}{(k+1)^3 - 1} = \frac{k^2 + 3k + 3}{k^3 + 3k^2 + 3k} = \frac{1}{k},\] by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to \[\frac{99^2 + 3(99) + 3}{4^3 - 1} \cdot \prod_{k=4}^{98} \frac{1}{k},\] which, after simplification gives $\frac{6}{98!} \cdot \frac{10101}{63} = \frac{962}{98!},$ giving an answer of $\boxed{962}.$

Answer:  962
Problem: At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice?
$(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15$

Solution: Each day Jenny makes half as many free throws as she does at the next practice. Hence on the fourth day she made $\frac{1}{2} \cdot 48 = 24$ free throws, on the third $12$ , on the second $6$ , and on the first $3 \Rightarrow \mathrm{(A)}$
Because there are five days, or four transformations between days (day 1 $\rightarrow$ day 2 $\rightarrow$ day 3 $\rightarrow$ day 4 $\rightarrow$ day 5), she makes $48 \cdot \frac{1}{2^4} = \boxed{3}$

Answer: A 3
Problem: If $x$ and $y$ are positive integers for which $2^x3^y=1296$ , what is the value of $x+y$
$(\mathrm {A})\ 8 \qquad (\mathrm {B})\ 9 \qquad (\mathrm {C})\ 10 \qquad (\mathrm {D})\ 11 \qquad (\mathrm {E})\ 12$

Solution: $1296 = 2^4 3^4$ and $4+4=\boxed{8}$

Answer: A 8
Problem: The point $(-3,2)$ is rotated $90^\circ$ clockwise around the origin to point $B$ . Point $B$ is then reflected over the line $x=y$ to point $C$ . What are the coordinates of $C$
$\mathrm{(A)}\ (-3,-2) \qquad \mathrm{(B)}\ (-2,-3) \qquad \mathrm{(C)}\ (2,-3) \qquad \mathrm{(D)}\ (2,3) \qquad \mathrm{(E)}\ (3,2)$

Solution: The entire situation is in the picture below. The correct answer is $\boxed{3,2}$

Answer: E 3,2
Problem: All the students in an algebra class took a $100$ -point test. Five students scored $100$ , each student scored at least $60$ , and the mean score was $76$ . What is the smallest possible number of students in the class?
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$

Solution: Let the number of students be $n\geq 5$ . Then the sum of their scores is at least $5\cdot 100 + (n-5)\cdot 60$ . At the same time, we need to achieve the mean $76$ , which is equivalent to achieving the sum $76n$
Hence we get a necessary condition on $n$ : we must have $5\cdot 100 + (n-5)\cdot 60 \leq 76n$ .
This can be simplified to $200 \leq 16n$ . The smallest integer $n$ for which this is true is $n=13$
To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\cdot 76 = 988$ points to divide among them. The five $100$ s make $500$ , hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them.
Hence the smallest possible number of students is $\boxed{13}$

Answer: D 13