Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ .  Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$

Solution: By multiplying both sides by $7$ we get
\[5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
since $0 \le a_2 < 2$ , if $a_2 = 0$ the rest of the right hand side will not add up to be $5$ , so $a_2 = 1$
\[\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_3 = 2$ $\frac76 a_3 = \frac73 > \frac32$ , so $a_3 = 1$
\[\frac13 = \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_4 = 2$ $\frac{7}{24} a_4 = \frac{7}{12} > \frac13$ , so $a_4 = 1$
\[\frac{1}{24} = \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_5 = 1$ $\frac{7}{120} a_5 = \frac{7}{120} > \frac{1}{24}$ , so $a_5 = 0$
\[\frac{1}{24} =  \frac{7}{720} a_6 + \frac{a_7}{720}\]
$30 = 7 a_6 + a_7$ . Since $a_7 < 7$ $a_7 = 2$ and $a_6=4$
Therefore, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \boxed{9}$

Answer: B 9
Problem: Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$

Solution: We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\circ}$
Let the number of sides in our polygons be $3\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$
The integral angle of a regular $k$ -gon is $180 \frac{k-2}k$ . Therefore we are looking for integer solutions to:
\[360 = 180\left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Which can be simplified to:
\[2 = \left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Furthermore, we know that two of the polygons are congruent, thus WLOG $a=c$ . Our equation now becomes
\[2 = \left( 2\cdot\frac{a-2}a + \frac{b-2}b \right)\]
Multiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$
Using the standard technique for Diophantine equations , we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$
Remembering that $a,b\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$ $(2,4)$ $(4,2)$ , and $(8,1)$
These correspond to the following pairs $(a,b)$ $(5,10)$ $(6,6)$ $(8,4)$ , and $(12,3)$
The perimeters of the resulting polygon for these four cases are $14$ $12$ $14$ , and $\boxed{21}$

Answer:  21
Problem: Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$

Solution: We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\boxed{21}$ sides.

Answer:  21
Problem: Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$

Solution: Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in\{-1,1,2\}$
Also, order of the $x_i$ does not matter, so we are only interested in the counts of the variables of each type. Let $a$ of the $x_i$ be equal to $-1$ $b$ equal to $1$ , and $c$ equal to $2$
The conditions (ii) and (iii) simplify to: (ii) $2c + b - a = 19$ (iii) $4c + a + b = 99$ and we want to find the maximum and minimum of $2^3c + 1^3b + (-1)^3a = 8c + b - a$ over all non-negative solutions of the above two equations.
Subtracting twice (ii) from (iii) we get $b = 3a-61$ . By entering that into one of the two equations and simplifying we get $c=40-a$
Thus all the solutions of our system of equations have the form $(a,3a-61,40-a)$
As all three variables must be non-negative integers, we have $3a-61 \geq 0 \Rightarrow a \geq 21$ and $40-a\geq 0 \Rightarrow a \leq 40$
For $(a,b,c)$ of the form $(a,3a-61,40-a)$ the expression we are maximizing/minimizing simplifies to $320-8a+3a-61-a = 259 - 6a$ . Clearly, the maximum is achieved for $a=21$ and the minimum for $a=40$ . Their values are $M=133$ and $m=19$ , and therefore $\frac Mm = \boxed{7}$

Answer:  7
Problem: Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$

Solution: Let $a =$ number of $2$ s, $b =$ number of $1$ s, $c =$ number of $-1$
\[4a+b+c=99\] \[2a+b-c=19\]
Multiplying the second equation by $2$ gives $4a+2b-2c=38$ . By subtracting this equation from the first equation we get $3c-b=61$ $3c=61+b$ . As we need to minimize the value of $c$ $c = 21$ $b = 2$ $a = 19$
$x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133$
Therefore, $\frac Mm = \frac{133}{19}=\boxed{7}$

Answer: E 7
Problem: A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The probability that $P$ lies inside one of the five small spheres is closest to
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.1 \qquad \mathrm{(C) \ }0.2 \qquad \mathrm{(D) \ }0.3 \qquad \mathrm{(E) \ }0.4$

Solution: Let the side length of the tetrahedron be $1$
The altitude of the equilateral triangle base is $\frac{\sqrt{3}}{2}$ . Thus, the distance from the center of the equilateral triangle to its vertex is $\frac{\sqrt{3}}{2} \cdot \frac 23 = \frac{\sqrt{3}}{3}$ . Therefore, the altitude of the tetrahedron is $\sqrt{1^2-(\frac{\sqrt{3}}{3})^2} = \frac{\sqrt{6}}{3}$
Let the radius of the large sphere be $R$ \[(\frac{\sqrt{3}}{3})^2 + (\frac{\sqrt{6}}{3} - R)^2 = R^2\] \[R = \frac{\sqrt{6}}{4}\]
The distance from the center of the tetrahedron to the center of one of the bases is $\frac{\sqrt{6}}{3} - R = \frac{\sqrt{6}}{3} - \frac{\sqrt{6}}{4} = \frac{\sqrt{6}}{12}$
The distance from the center of the tetrahedron to the center of one of the bases is also the same as the radius of the small sphere in the center of the tetrahedron.
The radius of the smaller spheres tangent to the large sphere and the tetrahedron is $\frac{\frac{\sqrt{6}}{4}-\frac{\sqrt{6}}{12}}{2} = \frac{\sqrt{6}}{12}$
Therefore, the probability that $P$ lies inside one of the five small spheres is $\frac{5 \cdot (\frac{\sqrt{6}}{12})^3 \cdot \frac43 \cdot \pi }{(\frac{\sqrt{6}}{4})^3 \cdot \frac43 \cdot \pi} = \frac{5}{27} \approx \boxed{0.2}$

Answer: C 0.2
Problem: Letters $A,B,C,$ and $D$ represent four different digits selected from $0,1,2,\ldots ,9.$ If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$
$\mathrm{(A) \  }13 \qquad \mathrm{(B) \  }14 \qquad \mathrm{(C) \  } 15\qquad \mathrm{(D) \  }16 \qquad \mathrm{(E) \  } 17$

Solution: If we want $\frac{A+B}{C+D}$ to be as large as possible, we want to try to maximize the numerator $A+B$ and minimize the denominator $C+D$ .  Picking $A=9$ and $B=8$ will maximize the numerator, and picking $C=0$ and $D=1$ will minimize the denominator.
Checking to make sure the fraction is an integer, $\frac{A+B}{C+D} = \frac{17}{1} = 17$ , and so the values are correct, and $A+B = 17$ , giving the answer $\boxed{17}$

Answer: E 17
Problem: If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which
then $\texttt{a+b+c =}$
$\mathrm{(A) \  }14 \qquad \mathrm{(B) \  }15 \qquad \mathrm{(C) \  }16 \qquad \mathrm{(D) \  }17 \qquad \mathrm{(E) \  }18$

Solution: Working from right to left, we see that $2 - b = 3$ .  Clearly if $b$ is a single digit integer, this cannot be possible.  Therefore, there must be some borrowing from $a$ .  Borrow $1$ from the digit $a$ , and you get $12 - b = 3$ , giving $b = 9$
Since $1$ was borrowed from $a$ , we have from the tens column $(a-1) - 8 = 7$ .  Again for single digit integers this will not work.  Again, borrow $1$ from $7$ , giving $10 + (a-1) - 8 = 7$ .  Solving for $a$
$10 + a - 1 - 8 = 7$
$1 + a = 7$
$a = 6$
Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$ , giving $c = 2$
As a check, the problem is $762 - 489 = 273$ , which is a true sentence.
The desired quantity is $a + b + c = 6 + 9 + 2 = 17$ , and the answer is $\boxed{17}$

Answer: D 17
Problem: If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$
$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$

Solution: Divide both sides of the original equation by $2^{1995}$ , giving:
$2^3 - 2^2 - 2^1 + 1 = k$
$k = 3$ , and the answer is $\boxed{3}$

Answer: C 3
Problem: If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$

Solution: If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.
Since $45^2 = 2025$ , the factors should be as close to $44$ or $45$ as possible.
Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$
$37$ is relatively close to $44$ , and no numbers between $38$ and $44$ are factors of $1998$ .  Thus, the two factors are $37$ and $2\cdot 3^3 = 54$ , and the difference is $54 - 37 = 17$ , and the answer is $\boxed{17}$

Answer: C 17
Problem: A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac  78$

Solution: The area of the trapezoid is $\frac{1}{3}$ , and the shorter base and height are both $\frac{1}{2}$ .  Therefore, \[\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{56}\]

Answer: D 56
Problem: A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac  78$

Solution: Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}$ , we have \[b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})\] \[x+\frac{1}{2}=2((1-x)+\frac{1}{2})\] \[x=\frac{5}{6}\boxed{56}\]

Answer: D 56
Problem: A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by members of the audience?
$\mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36$

Solution: Assume that there are $100$ people in the audience.
$20$ people heard $60$ minutes of the talk, for a total of $20\cdot 60 = 1200$ minutes heard.
$10$ people heard $0$ minutes.
$\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\cdot 20 = 700$ minutes.
$35$ people heard $40$ minutes of the talk, for a total of $1400$ minutes.
Altogether, there were $1200 + 0 + 700 + 1400 = 3300$ minutes heard among $100$ people.
Thus, the average is $\frac{3300}{100} = 33$ minutes, and the answer is $\boxed{33}$

Answer: D 33
Problem: A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$

Solution: Let the length of the longer side be $x$ , and the length of the shorter side be $y$ .  We are given that $2x+2y=14\implies x+y=7$ .  However, note that $x+y$ is also the length of a side of the larger square.  Thus the area of the larger square is $(x+y)^2=7^2=\boxed{49}$

Answer: A 49
Problem: A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$

Solution: Expand the small square so it basically equals the area of the large square.  Two of the sides of the rectangles shrink to zero.  The other two sides expand to equal the length of the large outer square, and have a length of $\frac{14}{2} = 7$ .  Thus, the area of the larger square is $7^2=\boxed{49}$

Answer: A 49
Problem: Let $R$ be a rectangle. How many circles in the plane of $R$ have a diameter both of whose endpoints are vertices of $R$
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$

Solution: There are $6$ pairs of vertices of $R$ . However, both diagonals determine the same circle, therefore the answer is $\boxed{5}$

Answer:  5
Problem: A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?
$\mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6$

Solution: $A_{\triangle} = \frac{s_t^2\sqrt{3}}{4}$
$A_{hex} = \frac{6s_h^2\sqrt{3}}{4}$ since a regular hexagon is just six equilateral triangles.
Setting the areas equal, we get:
$s_t^2 = 6s_h^2$
$\left(\frac{s_t}{s_h}\right)^2 = 6$
$\frac{s_t}{s_h} = \sqrt{6}$ , and the answer is $\boxed{6}$

Answer: C 6
Problem: How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution: The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$ . The length of the base is $10$ , thus the height must be $2$ . The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant.

Answer:  4
Problem: How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution: Alternatively, we use shoelace to get: \[\frac {1}{2}|(-5*0+5*5\sin (\theta)+5 \cos (\theta) *0)-(0*5+0*5 \cos (\theta)-5*5\sin (\theta))|=10 \implies \frac {1}{2}|50\sin (\theta)|=10\] This means $\sin (\theta)=\pm \frac {2}{5}$ . We see that if it equals $\frac {2}{5}$ , then $\cos (\theta)=\pm \frac {\sqrt {21}}{5}$ . Likewise, we see that if $\sin (\theta)=-\frac {2}{5}$ , then $\cos (\theta)$ has $2$ solutions. Thus, there are $\boxed{4}$ unique points such that the triangle has an area of $10$ , or $C$

Answer:  4
Problem: Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers."  Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers."  Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers."  Assume that each person knows that the other two reason perfectly and hears their comments.  What number is on the middle card?
$\textrm{(A)}\ 2 \qquad \textrm{(B)}\ 3 \qquad \textrm{(C)}\ 4 \qquad \textrm{(D)}\ 5 \qquad \textrm{(E)}\ \text{There is not enough information to determine the number.}$

Solution: Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$ $(1,3,9)$ $(1,4,8)$ $(1,5,7)$ $(2,3,8)$ $(2,4,7)$ $(2,5,6)$ , and $(3,4,6)$
If Casey saw the number $3$ , she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$
At this moment, if the last card contained a $10$ $9$ , or a $6$ , Tracy would know the other two numbers. (Note that Tracy is aware of the fact that $(3,4,6)$ was eliminated. If she saw the number $6$ , she would know that the other two are $2$ and $5$ .) This eliminates three more possibilities.
Thus before Stacy took her look, we are left with four possible cases: $(2,4,7)$ $(1,4,8)$ $(1,5,7)$ , and $(2,3,8)$ . As Stacy could not find out the exact combination, the middle number must be $\boxed{4}$

Answer:  4
Problem: In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?
$\mathrm{(A) \ } \frac dh \qquad \mathrm{(B) \ } 0 \qquad \mathrm{(C) \ } \frac {d^2}h \qquad \mathrm{(D) \ } \frac {h^2}d \qquad \mathrm{(E) \ } \frac{d^2}{h-d}$

Solution: Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\frac sw = \frac h{h-d}$ . In the second race, Sunny's track length is $h+d$ . She will finish this track in $\frac{h+d}s$ . In this time, Windy will run the distance $w\cdot \frac{h+d}s = \frac{(h+d)(h-d)}h$ . This is less than $h$ , therefore Sunny is ahead. The exact distance between Windy and the finish is $h - \frac{(h+d)(h-d)}h = \frac{ h^2 - (h^2-d^2) }h = \boxed{2}$

Answer:  2
Problem: The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$ , and for no other values of $k$ . Find $b-a$
$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$

Solution: Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$
The first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$ , which is a circle centered at $(6,3)$ with radius $7$
The second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$ , which is a circle centered at $(2,6)$ with radius $r=\sqrt{40+k}$
The distance between the two centers is $5$ , and therefore the two circles intersect if $2\leq r \leq 12$
From $\sqrt{40+k} \geq 2$ we get that $k\geq -36$ . From $\sqrt{40+k}\leq 12$ we get $k\leq 104$
Therefore $b-a = 104 - (-36) = \boxed{140}$

Answer:  140
Problem: Call a $7$ -digit telephone number $d_1d_2d_3-d_4d_5d_6d_7$ memorable if the prefix sequence $d_1d_2d_3$ is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digits $0,1,2, \ldots, 9$ , the number of different memorable telephone numbers is
$\mathrm{(A)}\ 19,810 \qquad\mathrm{(B)}\ 19,910 \qquad\mathrm{(C)}\ 19,990 \qquad\mathrm{(D)}\ 20,000 \qquad\mathrm{(E)}\ 20,100$

Solution: In this problem, we only need to consider the digits $\overline{d_4d_5d_6d_7}$ . Each possibility of $\overline{d_4d_5d_6d_7}$ gives $2$ possibilities for $\overline{d_1d_2d_3}$ , which are $\overline{d_1d_2d_3}=\overline{d_4d_5d_6}$ and $\overline{d_1d_2d_3}=\overline{d_5d_6d_7}$ with the exception of the case of $d_4=d_5=d_6=d_7$ , which only gives one sequence. After accounting for overcounting, the answer is $(10 \times 10 \times 10 \times 10) \times 2 - 10=19990 \Rightarrow \boxed{19,990}$

Answer: C 19,990
Problem: A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$ , and $(7,3)$ is matched with $(m,n)$ . Find $m+n$
$\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$

Solution: The line of the fold is the perpendicular bisector of the segment that connects $(0,2)$ and $(4,0)$ . 
The point $(m,n)$ is the image of the point $(7,3)$ according to this axis.
The situation looks as follows.
Now, we will compute the coordinates of the point $D$ , using the following facts:
As the triangles $SCT$ and $SDT$ are congruent, their areas are equal. The area of the triangle $SCT$ is $1/2$ of the size of the vector product $\overrightarrow{SC}\times\overrightarrow{ST}$ , and the area of $SDT$ is $1/2$ of the size of $\overrightarrow{ST}\times\overrightarrow{SD}$
We get that $\overrightarrow{SC}\times\overrightarrow{ST} = \overrightarrow{ST}\times\overrightarrow{SD}$
The equality remains valid if we multiply the vector $\overrightarrow{ST}$ by any constant. In other words, instead of $\overrightarrow{ST}$ we can use any vector with the same direction.
The axis of symmetry is perpendicular to $\overrightarrow{AB}$ . Thus its direction is $(2,4)=(1,2)$
We get that $\overrightarrow{SC}\times (1,2) = (1,2) \times\overrightarrow{SD}$
Substituting the coordinates $\overrightarrow{SC}=(5,2)$ and $\overrightarrow{SD}=(m-2,n-1)$ we get $5\cdot 2 - 2\cdot 1 = 1\cdot (n-1) - 2\cdot(m-2)$ . This simplifies to $n=2m+5$
We just discovered that the coordinates of $D$ are $(m,2m+5)$ . We will now use the second two facts mentioned above to find $m$
We have $|SD| = |SC|$ and therefore $|SD|^2 = |SC|^2$ . We know that $|SC|^2 = 5^2 + 2^2 = 29$ , and $|SD|^2 = (m-2)^2 + (2m+4)^2$ . Simplifying, we get the equation $5m^2 + 12m - 9 = 0$ . This has exactly one positive root $m=0.6$
It follows that $D=(0.6,6.2)$ , and that $m+n = 6.2 + 0.6 = \boxed{6.8}$

Answer:  6.8