Answer: 401.0

The numbers in the sequence $101$ $104$ $109$ $116$ $\ldots$ are of the form $a_n=100+n^2$ , where $n=1,2,3,\ldots$ For each $n$ , let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$ . Find the maximum value of $d_n$ as $n$ ranges through the positive integers

Solution:
If $(x,y)$ denotes the greatest common divisor of $x$ and $y$ , then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$ . Now assuming that $d_n$ divides $100+n^2$ , it must divide $2n+1$ if it is going to divide the entire expression $100+n^2+2n+1$
Thus the equation turns into $d_n=(100+n^2,2n+1)$ . Now note that since $2n+1$ is odd for integral $n$ , we can multiply the left integer, $100+n^2$ , by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there.
So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$ . It simplified the way we wanted it to!
Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$ .  Thus $d_n$ must divide $\boxed{401}$ for every single $n$ .  This means the largest possible value for $d_n$ is $401$ , and we see that it can be achieved when $n = 200$

Solution:
We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$ . Since we want to find the GCD of $a_n$ and $a_{n+1}$ , we can use the Euclidean algorithm
$a_{n+1}-a_n = 2n+1$
Now, the question is to find the GCD of $2n+1$ and $100+n^2$ .
We subtract $2n+1$ $100$ times from $100+n^2$ \[(100+n^2)-100(2n+1)\] \[=n^2+100-200n-100\] This leaves us with \[n^2-200n.\] Factoring, we get \[n(n-200)\] Because $n$ and $2n+1$ will be coprime, the only thing stopping the GCD from being $1$ is $n-200.$ We want this to equal 0, because that will maximize our GCD.
Solving for $n$ gives us $n=200$ . The last remainder is 0, thus $200*2+1 = \boxed{401}$ is our GCD.


Answer: 220.0

The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.

Solution: In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\] We are given that \begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*} The second equation implies that $a=5d.$ Substituting this into the first equation, we get \begin{align*} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{align*} It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\boxed{220}.$

Solution:
Let the terms in the sequence be defined as \[a_1, a_2, ..., a_6.\]
Since this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, \[\sum_{i=1}^6 a_i=3(a_1+a_6)=990.\] Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}.$


Answer: 81.0

The numbers on the faces of this cube are consecutive whole numbers.  The sum of the two numbers on each of the three pairs of opposite faces are equal.  The sum of the six numbers on this cube is
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); label("$15$",(1.5,1.2),N); label("$11$",(4,2.3),N); label("$14$",(2.5,3.7),N); [/asy]
$\text{(A)}\ 75 \qquad \text{(B)}\ 76 \qquad \text{(C)}\ 78 \qquad \text{(D)}\ 80 \qquad \text{(E)}\ 81$

Solution:
The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$
In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$ , so $11$ must be paired with $14$ , which
it isn't.
Thus, the only possibility is the first case and the sum of the six numbers is $81\rightarrow \boxed{81}$


Answer: 28.0

The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$

Solution:
We can execute Dijkstra's algorithm by hand to find the shortest path from $A$ to every other town, including $Z$ . This effectively proves that, assuming we execute the algorithm correctly, that we will have found the shortest distance. The distance estimates for each step of the algorithm (from $A$ to each node) are shown below:
\[\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Current node} & A & M & C & X & Y & Z \\ \hline A & 0 & 8 & \infty & 5 & \infty & \infty \\ X & 0 & 7 & \infty & 5 & 15 & \infty \\ M & 0 & 7 & 21 & 5 & 13 & 32 \\ Y & 0 & 7 & 18 & 5 & 13 & 30 \\ C & 0 & 7 & 18 & 5 & 13 & 28 \\ Z & 0 & 7 & 18 & 5 & 13 & \textbf{28} \\ \hline \end{array}\] The steps are as follows: starting with the initial node $A$ , set $d(A)=0$ and $d(v)=\infty$ for all $v \in \{M,C,X,Y,Z\}$ where $d(v)$ indicates the distance from $A$ to $v$ . Consider the outgoing edges $(A,X)$ and $(A,M)$ and update the distance estimates $d(X)=5$ and $d(M)=8$ , completing the first row of the table.
The node $X$ is the unvisited node with the lowest distance estimate, so we will consider $X$ and its outgoing edges $(X,Y)$ and $(X,M)$ . The distance estimate $d(Y)$ equals $d(X)+10=15$ , and the distance estimate $d(M)$ updates to $d(X)+2=7$ , because $7 < 8$ . This completes the second row of the table. Repeating this process for each unvisited node (in order of its distance estimate) yields the correct distance $d(Z) = \boxed{28}$ once the algorithm is complete.

Solution: From $A$ , we want to find the shortest route to $Z$ , so we want to try to find the shortest path through each node (not necessarily all of them). We should follow the arrows, since all of them lead to $Z$ . From $A$ , there are $2$ paths we can take, to $M$ $(8)$ , or to $X$ $(5)$ . We travel to $X$ , since $5 < 8$ . From $X$ , we go to $M$ $(2 < 10)$ , we go to $Y$ $6 < 14 < 25$ , we go to $C$ $(5 < 17)$ and finally go to $Z$ . Adding up gives $5 + 2 + 6 + 5 + 10 = 28 = \boxed{28}$

Solution:
We can cross off a few routes:
Finally, we are left with a single path AXMYCZ from A to Z which adding it up gives $28 = \boxed{28}.$


Answer: 28.0

Solution: We can simply see that path $A \rightarrow X \rightarrow M \rightarrow Y \rightarrow C \rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10 = \boxed{28}$ . This is nice as it’s also the smallest value, solidifying our answer.


Answer: 33.0

The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$ . When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$ . What was the number of the page that was added twice?

Solution: Denote the page number as $x$ , with $x < n$ . The sum formula shows that $\frac{n(n + 1)}{2} + x = 1986$ . Since $x$ cannot be very large, disregard it for now and solve $\frac{n(n+1)}{2} = 1986$ . The positive root for $n \approx \sqrt{3972} \approx 63$ . Quickly testing, we find that $63$ is too large, but if we plug in $62$ we find that our answer is $\frac{62(63)}{2} + x = 1986 \Longrightarrow x = \boxed{033}$

Solution:
Use the same method as above where you represent the sum of integers from $1$ to $n$ expressed as $\frac{n(n + 1)}{2}$ , plus the additional page number $k$ . We can establish an upper and lower bound for the number of pages contained in the book, where the upper bound maximizes the sum of the pages and the lower bound maximizes the extra page (and thus minimizes the total sum of the pages.)
$\frac{n(n + 1)}{2} = 1986$ is the quadratic you must solve to obtain the upper bound of $n$ and $\frac{n(n + 1)}{2} + n = 1986$ is the quadratic you must solve to obtain the lower bound of $n$
Solving the two equations gives values that are respectively around $62.5$ and $61.5$ with the quadratic formula, and the only integer between the two is $62$
This implies that we can plug in $62$ and come to the same conclusion as the above solution where $x = \boxed{033}$


Answer: 40.0

The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$ .  For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$
$\textbf{(A) }38\qquad \textbf{(B) }40\qquad \textbf{(C) }42\qquad \textbf{(D) }44\qquad \textbf{(E) }46\qquad$

Solution: The axis of $P$ is inclined at an angle $\theta$ relative to the coordinate axis, where $\tan\theta = \tfrac 34$ . We rotate the coordinate axis by angle $\theta$ anti-clockwise, so that the parabola now has a vertical symmetry axis relative to the rotated coordinates. Let $(\widetilde{x}, \widetilde{y})$ be the coordinates in the rotated system. Then $(x,y)$ and $(\widetilde{x}, \widetilde{y})$ are related by \begin{align}   \nonumber x = \widetilde{x}\cos\theta -\widetilde{y}\sin\theta &= \tfrac 45 \widetilde{x} - \tfrac 35 \widetilde{y}, \\    y = \widetilde{x}\sin\theta +\widetilde{y}\cos\theta &= \tfrac 35 \widetilde{x} + \tfrac 45 \widetilde{y} \end{align} In the rotated coordinate system, the parabola has focus at $(0,0)$ and the two points on it are at $(5,0)$ and $(-5,0)$ . Therefore, the directrix is $\widetilde{y}=\pm 5$ ; we can, WLOG, choose $\widetilde{y}=-5$ . For a point on the parabola, it is equidistant from the focus and directrix, so the equation of the parabola is \begin{align}\tag{2}   \widetilde{x}^2+\widetilde{y}^2 = (\widetilde{y}+5)^2 \qquad  &\Longrightarrow\qquad \widetilde{y} = \tfrac{1}{10}(\widetilde{x}^2-25) \end{align} From $(1)$ we have $|4x+3y|=5\widetilde{x}$ , so we need $|\widetilde{x}|<200$ . Substituting $(2)$ in $(1)$ , we get \begin{align*}   50x &= 40 \widetilde{x} - 3 \widetilde{x}^2 + 75, \\   50y &= 30 \widetilde{x} + 4 \widetilde{x}^2 - 100 \end{align*} For $x$ to be an integer $\widetilde{x}$ must be a multiple of 5; setting $\widetilde{x}=5a$ we get \[2x = 8a - 3 a^2 + 3\] Now we need $a$ to be odd, i.e. $\widetilde{x}=5a$ is an odd multiple of $5$ , in which case we get $y = 3 a + 2 a^2 - 2$ , which is also an integer. The values that satisfy the given conditions correspond to $\widetilde{x}= \{\pm 5\cdot (2k-1)\mid k = 0, 1, \ldots , 19 \}={-195, -185, -175, ..., 195}$ , and there are $\boxed{40}$ such numbers.


Answer: 4.0

The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$ -intercept $(0,-p)$ , where $p\ne 0$ . What is $b$
$\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p$

Solution: A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $p\ne 0$ , it follows that $a=-2/p$ . Thus \[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\boxed{4}$


Answer: 1.5

The parabolas $y=ax^2 - 2$ and $y=4 - bx^2$ intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area $12$ . What is $a+b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3$

Solution: Clearly, the parabolas must intersect the x-axis at the same two points. Their distance multiplied by $4 - (-2)$ (the distance between the y-intercepts), all divided by 2 is equal to 12, the area of the kite (half the product of the diagonals). That distance is thus 4, and so the x-intercepts are $(2, 0), (-2, 0).$ Then $0 = 4a - 2 \rightarrow a = 0.5$ , and $0 = 4 - 4b \rightarrow b = 1.$ Then $a + b = \boxed{1.5}$

Solution: The parabolas must intersect the x-axis at the same two points for the kite to form. We find the x values at which they intersect by equating them and solving for x as shown below. $y = ax^2-2$ and $y = 4 - bx^2\rightarrow ax^2-2 = 4-bx^2\rightarrow (a+b)x^2 = 6 \rightarrow x = +\sqrt{\dfrac{6}{a+b}}$ or $-\sqrt{\dfrac{6}{a+b}}$ . The x-values of the y-intercepts is 0, so we plug in zero in each of them and get $-2$ and $4$ . The area of a kite is $\dfrac{d_1*d_2}{2}$ . The $d_1$ is $2+4 = 6$ . The $d_2$ is $2\sqrt{\dfrac{6}{a+b}}$ . Solving for the area $\rightarrow \dfrac{1}{2}*(6)*(2*\sqrt{\dfrac{6}{a+b}}) = 12 \rightarrow (2*\sqrt{\dfrac{6}{a+b}}) = 4 \rightarrow (\sqrt{\dfrac{6}{a+b}}) = 2 \rightarrow \dfrac{6}{a+b} = 4 \rightarrow  \dfrac{6}{4} = (a+b)$ , therefore $a + b = \boxed{1.5}$


Answer: 16.0

The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution:
Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$ . Because $72= 4\cdot 18$ , we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$ , which gives $c=3d$ (consider a homothety , or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$ , it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$ ). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$ , so substituting $c = 3d$
Thus $3|d$ , and we verify that $d = 3$ $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$ . Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{16}$

Solution: The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$ . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-c}{b-a}$ $\frac{(d-c)^2}{b-a}=18$ , and the area contained by the latter is $\frac{(c+d)^2}{b-a}=72$ . Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$ $c^2=18(b-a)$ so $c$ is a multiple of $3$ . Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$ , so the sum is $\boxed{16}$

Solution:
Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$ ) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$ . Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is $\frac{bh}{2}$ , but since a parallelogram is two such triangles, the area becomes $bh$
Let $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$ , or $(c-d)$ . Also notice that $h_1$ is the distance from the $x$ -axis to the intersection of lines $ax+c$ and $bx+d$ . This is equivalent to the value of the $x$ -coordinate of intersection, so we solve for $x$
$ax+c=bx+d$
$\Rightarrow bx-ax = c-d$
$\Rightarrow (b-a)x = c-d$
$\Rightarrow x = \frac{c-d}{b-a}$
The area of the smaller parallelogram is $b_1*h_1$ , or
$(c-d) * \frac{c-d}{b-a}$
$\Rightarrow \frac{(c-d)^2}{b-a}$
$b_2$ is the distance from $(0, -d)$ to $(0, c)$ , or $(c+d)$ $h_2$ is the $x$ -coordinate of the intersection of the lines $ax+c$ and $bx-d$ . Again, we solve for x:
$ax+c=bx-d$
$\Rightarrow bx-ax = c+d$
$\Rightarrow (b-a)x = c+d$
$\Rightarrow x = \frac{c+d}{b-a}$
The area of the larger parallelogram is $b-1*h_1$ , or
$(c+d) * \frac{c+d}{b-a}$
$\Rightarrow \frac{(c+d)^2}{b-a}$
The areas of the parallelograms are given to us: $18$ and $72$ . Therefore we can set up a ratio:
$\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}$
$\Rightarrow 18(c+d)^2 = 72(c-d)^2$
$\Rightarrow (c+d)^2 = 4(c-d)^2$
$\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2$
$\Rightarrow 3c^2 - 10cd +3d^2 = 0$
$\Rightarrow (3c-d)(c-3d)=0$
$\Rightarrow c=3d, c=\frac{d}{3}$
We established earlier that $c>d$ , so $c=3d$ . Plugging this into the intial equations yields
$\frac{16d^2}{b-a} = 72$
and
$\frac{4d^2}{b-a} = 18$
Solving for $d$ , we get
$d = 3\sqrt{\frac{b-a}{2}}$
We want the sum of $a$ $b$ $c$ , and $d$ . We can now rewrite this
$a + b + 12\sqrt{\frac{b-a}{2}}$
We are told that $a$ $b$ $c$ , and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this
$\frac{b-a}{2} = k^2$
where $k$ is some positive integer.
Rearranging, we get
$b = a + 2k^2$
Now we can rewrite the sum as
$a + a + 2k^2 +12k$
Since both $a$ and $k$ must be at least $1$ , the minimum value is
$1 + 1 + 2(1)^2 + 12(1) = 16 \Rightarrow \boxed{16}$


Answer: 100.0

The percent that $M$ is greater than $N$ is:
$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$

Solution: $M-N$ is the amount by which $M$ is greater than $N$ . We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\boxed{100}$


Answer: 9.0

The perimeter of one square is $3$ times the perimeter of another square.  The area of the larger square is how many times the area of the smaller square?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 9$

Solution:
Let $a$ be the sidelength of one square, and $b$ be the sidelength of the other, where $a>b$ . If the perimeter of one is $3$ times the other's, then $a=3b$ . The area of the larger square over the area of the smaller square is
\[\frac{a^2}{b^2} = \frac{(3b)^2}{b^2} = \frac{9b^2}{b^2} = \boxed{9}\]


Answer: 28.0

The perimeter of the polygon shown is
$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$
$\text{(E)}\ \text{cannot be determined from the information given}$

Solution:
You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines!

So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\boxed{28}$


Answer: 98.0

The perimeter of triangle $APM$ is $152$ , and the angle $PAM$ is a right angle . A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$ . Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$

Solution: Let the circle intersect $\overline{PM}$ at $B$ . Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point . Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have \[\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}\] Solving, $AM = 38$ . So the ratio of the side lengths of the triangles is 2. Therefore, \[\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2\] so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$ , we see that $4OP-76 = OP+19$ , so $OP = \frac{95}3$ and the answer is $\boxed{098}$

Solution:
Reflect triangle $PAM$ across line $AP$ , creating an isoceles triangle. Let $x$ be the distance from the top of the circle to point $P$ , with $x + 38$ as $AP$ . Given the perimeter is 152, subtracting the altitude yields the semiperimeter $s$ of the isoceles triangle, as $114 - x$ . The area of the isoceles triangle is:
$[PAM] = r \cdot s$
$[PAM] = 19 \cdot (114 - x)$
Now use similarity, draw perpendicular from $O$ to $PM$ , name the new point $D$ . Triangle $PDO$ is similar to triangle $PAM$ , by AA Similarity. Equating the legs, we get:
$\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}$
Solving for $AM$ ,  it yields $19 \cdot \sqrt{\frac{x + 38}{x}}$
$19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}$
The $x^3$ cancels, yielding a quadratic. Solving yields $x = \frac{38}{3}$ .
Add $19$ to find $OP$ , yielding $\frac{95}{3}$ or $\boxed{098}$

Solution:
Let the foot of the perpendicular from $O$ to $PM$ be $D;$ now $OD=19.$ Also let $AM=x$ and $PM=y.$ This means that $OP=\frac{y}{x}\cdot 19$ , since $O$ is on the angle bisector of $\angle M.$
We have that $\tan(\angle AMO)=\frac{19}{x},$ so \[\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.\]
However $\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}$ , so \[\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}\] \[\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1\] \[\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.\] \[x\cdot \frac{x^{2}+361}{x^{2}-361}=y\]
We now use the fact that the perimeter of $\triangle PAM$ is $152$ \[PO+OA+AM+MP=152\] \[\frac{y}{x}\cdot 19+19+x+y=152\] \[19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152\] \[(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152\] \[\frac{2x^{2}}{x-19}=152\] \[x^{2}-76x+19\cdot 76=0.\] This quadratic factors as $(x-38)^{2}=0,$ so $x=38$ , and \[\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}\] \[OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}\]


Answer: 32.0

Solution:
In the first school, $2000 \cdot 22\% = 2000 \cdot 0.22 = 440$ students prefer tennis.
In the second school, $2500 \cdot 40\% = 2500 \cdot 0.40 = 1000$ students prefer tennis.
In total, $440 + 1000 = 1440$ students prefer tennis out of a total of $2000 + 2500 = 4500$ students
This means $\frac{1440}{4500}\cdot 100\% = \frac{32}{100} \cdot 100\% = 32\%$ of the students in both schools prefer tennis, giving answer $\boxed{32}$


Answer: 13.0

The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $61$ points. How many free throws did they make?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

Solution:
For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \[2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1\]
Set that equal to $61$ , we get $x = 4$ , and therefore the number of free throws they made $3 \times 4 + 1 = 13 \Rightarrow \boxed{13}$

Solution: Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \Rightarrow x=13$ , giving us $\boxed{13}$ for an answer.

Solution:
We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see
\begin{align*} a&=x-1 \\ 2a &= 2x-2 \\ 3a &= 2x-2. \\ \end{align*}
Plugging this into $2a+3b+1x=61$ , we see
\begin{align*} 2x-2+2x-2+x &= 61 \\ 5x-4 &= 61 \\ 5x &= 65 \\ x &= \boxed{13}


Answer: 13.0

Solution:
For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \[2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1\]
Set that equal to $61$ , we get $x = 4$ , and therefore the number of free throws they made $3 \times 4 + 1 = 13 \Rightarrow \boxed{13}$

Solution: Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \Rightarrow x=13$ , giving us $\boxed{13}$ for an answer.

Solution:
We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see
\begin{align*} a&=x-1 \\ 2a &= 2x-2 \\ 3a &= 2x-2. \\ \end{align*}
Plugging this into $2a+3b+1x=61$ , we see
\begin{align*} 2x-2+2x-2+x &= 61 \\ 5x-4 &= 61 \\ 5x &= 65 \\ x &= \boxed{13}


Answer: 5.0

The point $(-1, -2)$ is rotated $270^{\circ}$ counterclockwise about the point $(3, 1)$ . What are the coordinates of its new position?
$\textbf{(A)}\ (-3, -4) \qquad \textbf{(B)}\ (0,5) \qquad \textbf{(C)}\ (2,-1) \qquad \textbf{(D)}\ (4,3) \qquad \textbf{(E)}\ (6,-3)$

Solution: $(-1,-2)$ is $4$ units west and $3$ units south of $(3,1)$ . Performing a counterclockwise rotation of $270^{\circ}$ , which is equivalent to a clockwise rotation of $90^{\circ}$ , the answer is $3$ units west and $4$ units north of $(3,1)$ , or $\boxed{0,5}$

Solution:
We write $(-1, -2)$ as $-1-2i.$ We'd like to rotate about $(3, 1),$ which is $3+i$ in the complex plane, by an angle of $270^{\circ} = \frac{3\pi}{2} \text{ rad}$ counterclockwise.
The formula for rotating the complex number $z$ about the complex number $w$ by an angle of $\theta$ counterclockwise is given as $(z-w)e^{\theta i} + w.$ Plugging in our values $z = -1-2i, w = 3+i, \theta = \frac{3\pi}{2}$ , we evaluate the expression as $((-1-2i) - (3+i))e^{\frac{3\pi i}{2}} + (3+i) = (-4-3i)(-i) + (3+i) = 4i-3i^2+3+i = 4i-3+3+i = 5i,$ which corresponds to $\boxed{0,5}$ on the Cartesian plane.


Answer: 32.0

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: 50.0

The point $O$ is the center of the circle circumscribed about $\triangle ABC,$ with $\angle BOC=120^\circ$ and $\angle AOB=140^\circ,$ as shown. What is the degree measure of $\angle ABC?$
$\textbf{(A) } 35 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 50 \qquad\textbf{(E) } 60$

Solution:
Because all the central angles of a circle add up to $360^\circ,$
\begin{align*} \angle BOC + \angle AOB + \angle AOC &= 360\\ 120 + 140 + \angle AOC &= 360\\ \angle AOC &= 100. \end{align*}
Therefore, the measure of $\text{arc}AC$ is also $100^\circ.$ Since the measure of an inscribed angle is equal to half the measure of the arc it intercepts, $\angle ABC = \boxed{50}$


Answer: 7.0

The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution:
The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$
By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$
Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \boxed{7}$

Solution:
Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$ .
A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$ , and then translating the origin back to the point $(1, 5)$ \[a+b\cdot{i}  \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = 5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$ .
Hence, we have \[6-b+(a+4)i = -3+6i \implies b=9, a=2,\] from which $b-a = 9-2 = \boxed{7}$

Solution:
Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ .
If we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ .
We know that two vectors are perpendicular if their dot product is equal to $0$ , and that both points are the same distance ( $\sqrt {17}$ ) from $(1,5)$
Therefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)
Solving, we simplify the second equation to $y=4x+1$ , and plug it into the first equation.
We obtain $(x-1)^2 + (4x-4)^2 = 17$ . We can simplify this to the quadratic $17x^2-34x=0$ .
When we factor out $17x$ , we find that $x = 2$ or $x = 0$ . However, $x$ cannot equal $0$ .
Therefore, $x = 2$ , and plugging this into the second equation gives us that $y = 9$ .
Since the point is $(9, 2)$ , we compute $9-2 = \boxed{7}$

Solution:
Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$
Let the original point be $\langle x,y \rangle.$ From point $(1,5),$ we form the vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \[\langle -4,1 \rangle \cdot \langle x-1, y-5 \rangle = 0 \implies -4x+y-1= 0.\] Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \pmod 3$ so that brings our potential answers down to $\text{\textbf{(A)}}$ and $\text{\textbf{(C)}}.$ If $A = 1$ from $\text{\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\boxed{7}$ is the answer.


Answer: 7.0

Solution:
The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$
By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$
Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \boxed{7}$

Solution:
Let us reconstruct that coordinate plane as the complex plane. Then, the point $P(a, b)$ becomes $a+b\cdot{i}$ .
A $90^\circ$ rotation around the point $(1, 5)$ can be done by translating the point $(1, 5)$ to the origin, rotating around the origin by $90^\circ$ , and then translating the origin back to the point $(1, 5)$ \[a+b\cdot{i}  \implies (a-1)+(b-5)\cdot{i} \implies ((a-1)+(b-5)\cdot{i})\cdot{i} = 5-b+(a-1)i \implies 5+1-b+(a-1+5)i = 6-b+(a+4)i.\] By basis reflection rules, the reflection of $(-6, 3)$ about the line $y = -x$ is $(-3, 6)$ .
Hence, we have \[6-b+(a+4)i = -3+6i \implies b=9, a=2,\] from which $b-a = 9-2 = \boxed{7}$

Solution:
Using the same method as in Solution 1, we can obtain that the point before the reflection is $(-3,6)$ .
If we let the original point be $(x, y)$ , then we can use that the starting point is $(1,5)$ to obtain two vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ .
We know that two vectors are perpendicular if their dot product is equal to $0$ , and that both points are the same distance ( $\sqrt {17}$ ) from $(1,5)$
Therefore, we can write two equations using these vectors: $(x-1)^2 + (y-5)^2 = 17$ (from distance and pythagorean theorem) and $-4x+y-1 = 0$ (from dot product)
Solving, we simplify the second equation to $y=4x+1$ , and plug it into the first equation.
We obtain $(x-1)^2 + (4x-4)^2 = 17$ . We can simplify this to the quadratic $17x^2-34x=0$ .
When we factor out $17x$ , we find that $x = 2$ or $x = 0$ . However, $x$ cannot equal $0$ .
Therefore, $x = 2$ , and plugging this into the second equation gives us that $y = 9$ .
Since the point is $(9, 2)$ , we compute $9-2 = \boxed{7}$

Solution:
Using the same method as in Solution 1 reflecting $(-6,3)$ about the line $y = -x$ gives us $(-3,6).$
Let the original point be $\langle x,y \rangle.$ From point $(1,5),$ we form the vectors $\langle -4,1 \rangle$ and $\langle x-1, y-5 \rangle$ that extend out from the initial point. If they are perpendicular, we know that their dot product has to equal zero. Therefore, \[\langle -4,1 \rangle \cdot \langle x-1, y-5 \rangle = 0 \implies -4x+y-1= 0.\] Now, we have to do some guess and check from the multiple choices. Let $y - x = A$ where $A$ is one of the answer choices. Then, $A -3x = 1.$ By intuition and logical reasoning we deduce that $A$ must be $1 \pmod 3$ so that brings our potential answers down to $\text{\textbf{(A)}}$ and $\text{\textbf{(C)}}.$ If $A = 1$ from $\text{\textbf{(A)}},$ then $x = 0,$ which we can quickly rule out since we know thar $P$ rotated counterclockwise not clockwise. Hence, $\boxed{7}$ is the answer.


Answer: 10001988.0

The point in the $xy$ -plane with coordinates $(1000, 2012)$ is reflected across the line $y=2000$ .  What are the coordinates of the reflected point?
$\textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012)$

Solution: The line $y = 2000$ is a horizontal line located $12$ units beneath the point $(1000, 2012)$ . When a point is reflected about a horizontal line, only the $y$ - coordinate will change. The $x$ - coordinate remains the same. Since the $y$ -coordinate of the point is $12$ units above the line of reflection, the new $y$ - coordinate  will be $2000 - 12 = 1988$ . Thus, the coordinates of the reflected point are $(1000, 1988)$ $\boxed{1000,1988}$


Answer: 315.0

The points $(0,0)\,$ $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle.  Find the value of $ab\,$

Solution:
Consider the points on the complex plane . The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:
\[(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]
Equating the real and imaginary parts, we have:
\begin{align*}b&=\frac{a}{2}-\frac{11\sqrt{3}}{2}\\37&=\frac{11}{2}+\frac{a\sqrt{3}}{2} \end{align*}
Solving this system, we find that $a=21\sqrt{3}, b=5\sqrt{3}$ . Thus, the answer is $\boxed{315}$

Solution:
Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\sqrt{3}$ and perpendiculars inspires this solution:
First, drop a perpendicular from $O$ to $AB$ . Call this midpoint of $AB M$ . Thus, $M=\left(\frac{a+b}{2}, 24\right)$ . The vector from $O$ to $M$ is $\left[\frac{a+b}{2}, 24\right]$ . Meanwhile from point $M$ we can use a vector with $\frac{\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]$ . (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?)
We see this displacement from $M$ to $A$ is $\left[\frac{a-b}{2}, 13\right]$ as well. Equating the two vectors, we get $a+b=26\sqrt{3}$ and $a-b=16\sqrt{3}$ . Therefore, $a=21\sqrt{3}$ and $b=5\sqrt{3}$ . And the answer is $\boxed{315}$

Solution:
Plot this equilateral triangle on the complex plane.
Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\left(\frac{a+b}{3}, 16i\right)$ . The new coordinates of the equilateral triangle are $\left(-\frac{a+b}{3}-16i\right), \left(a-\frac{a+b}{3}-5i\right), \left(b-\frac{a+b}{3}+21i\right)$ . These three vertices are solutions of a cubic polynomial of form $x^3 + C$ . By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots $r_1, r_2,$ and $r_3,$ $\, r_1r_2 + r_2r_3 + r_3r_1 = 0$ .) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$ . Use your previously obtained equation to plug in for $a$ and solve for $b$ , which should yield $5\sqrt{3}$ $a$ is then $\frac{21}{5}\sqrt{3}$ . Multiplying $a$ and $b$ yields $\boxed{315}$

Solution:
Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 = 37^2$
$a^2 + 121 = b^2 + 1369 ==> a^2 = b^2 + 1248$ . Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$ $b^2 = 2ab - 555 ==> a^2 = 2ab + 693$
We have $b^2 + 1248 = 2b\sqrt{b^2+1248} + 693$
Moving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$
$3b^4 + 3882b^2 - 308025 = 0$
This factors to $(3b^2-225)(b^2+1369)$ , so $3b^2 = 225, b = 5\sqrt 3$
Plugging it back in, we find that a = $\sqrt{1323}$ which is $21\sqrt3$ , so the product $ab$ is $\boxed{315}$


Answer: 33.0

The points $(6,12)$ and $(0,-6)$ are connected by a straight line. Another point on this line is:
$\textbf{(A) \ }(3,3)  \qquad \textbf{(B) \ }(2,1) \qquad \textbf{(C) \ }(7,16) \qquad \textbf{(D) \ }(-1,-4) \qquad \textbf{(E) \ }(-3,-8)$

Solution: The slope of this line is $\frac{y_2-y_1}{x_2-x_1}=\frac{12+6}{6-0}=3$ . Hence, its equation is $y=3x-6$ . The only given point which satisfies these conditions is $\boxed{3,3}$


Answer: 118.0

The points $A$ $B$ and $C$ lie on the surface of a sphere with center $O$ and radius $20$ . It is given that $AB=13$ $BC=14$ $CA=15$ , and that the distance from $O$ to $\triangle ABC$ is $\frac{m\sqrt{n}}k$ , where $m$ $n$ , and $k$ are positive integers, $m$ and $k$ are relatively prime, and $n$ is not divisible by the square of any prime. Find $m+n+k$

Solution:
Let $D$ be the foot of the perpendicular from $O$ to the plane of $ABC$ . By the Pythagorean Theorem on triangles $\triangle OAD$ $\triangle OBD$ and $\triangle OCD$ we get:
\[DA^2=DB^2=DC^2=20^2-OD^2\]
It follows that $DA=DB=DC$ , so $D$ is the circumcenter of $\triangle ABC$
By Heron's Formula the area of $\triangle ABC$ is (alternatively, a $13-14-15$ triangle may be split into $9-12-15$ and $5-12-13$ right triangles ):
\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-15)(21-14)(21-13)} = 84\]
From $R = \frac{abc}{4K}$ , we know that the circumradius of $\triangle ABC$ is:
\[R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}\]
Thus by the Pythagorean Theorem again,
\[OD = \sqrt{20^2-R^2} = \sqrt{20^2-\frac{65^2}{8^2}} = \frac{15\sqrt{95}}{8}.\]
So the final answer is $15+95+8=\boxed{118}$

Solution: We know the radii to $A$ $B$ , and $C$ form a triangular pyramid $OABC$ . We know the lengths of the edges $OA = OB = OC = 20$ . First we can break up $ABC$ into its two component right triangles $5-12-13$ and $9-12-15$ . Let the $y$ axis be perpendicular to the base and $x$ axis run along $BC$ , and $z$ occupy the other dimension, with the origin as $C$ . We look at vectors $OA$ and $OC$ . Since $OAC$ is isoceles we know the vertex is equidistant from $A$ and $C$ , hence it is $7$ units along the $x$ axis. Hence for vector $OC$ , in form $<x,y,z>$ it is $<7, h, l>$ where $h$ is the height (answer) and $l$ is the component of the vertex along the $z$ axis. Now on vector $OA$ , since $A$ is $9$ along $x$ , and it is $12$ along $z$ axis, it is $<-2, h, 12- l>$ . We know both vector magnitudes are $20$ . Solving for $h$ yields $\frac{15\sqrt{95} }{8}$ , so Answer = $\boxed{118}$


Answer: 6.0

The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined edge -to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?
2003amc10a10.gif
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution:
2003amc10a10solution.gif
Let the squares be labeled $A$ $B$ $C$ , and $D$
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$ , and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.
Therefore, squares $1$ $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.
Squares $4$ $5$ $6$ $7$ $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.
Thus the answer is $\boxed{6}$

Solution: Another way to think of it is that a cube missing one face has $5$ of its $6$ faces.  Since the shape has $4$ faces already, we need another face.  The only way to add another face is if the added square does not overlap any of the others. $1$ $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\boxed{6}$

Solution: If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\boxed{6}$ . ~Sophia866


Answer: 6.0

Solution:
2003amc10a10solution.gif
Let the squares be labeled $A$ $B$ $C$ , and $D$
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$ , and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing.
Therefore, squares $1$ $2$ , and $3$ will prevent the polygon from becoming a cube with one face missing.
Squares $4$ $5$ $6$ $7$ $8$ , and $9$ will allow the polygon to become a cube with one face missing when folded.
Thus the answer is $\boxed{6}$

Solution: Another way to think of it is that a cube missing one face has $5$ of its $6$ faces.  Since the shape has $4$ faces already, we need another face.  The only way to add another face is if the added square does not overlap any of the others. $1$ $2$ , and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\boxed{6}$

Solution: If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\boxed{6}$ . ~Sophia866


Answer: 816.0

The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$ , where $y=x+1$ and the $a_i$ 's are constants . Find the value of $a_2$

Solution: Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$ . Since $x = y - 1$ , this becomes $\frac {1-(y - 1)^{18}}{y}$ . We want $a_2$ , which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the numerator by $1$ ). By the Binomial Theorem , this is $(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}$

Solution:
Again, notice $x = y - 1$ . So
\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.
We want the coefficient of the $y^2$ term of each power of each binomial, which by the binomial theorem is ${2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}$ . The Hockey Stick Identity tells us that this quantity is equal to ${18\choose 3} = \boxed{816}$

Solution: Again, notice $x=y-1$ . Substituting $y-1$ for $x$ in $f(x)$ gives: \begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}. From binomial theorem, the coefficient of the $y^2$ term is ${2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}$ . This is actually the sum of the first 16 triangular numbers, which evaluates to $\frac{(16)(17)(18)}{6} = \boxed{816}$

Solution:
Let $f(x)=1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ and $g(y)=a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$
Then, since $f(x)=g(y)$ \[\frac{d^2f}{dx^2}=\frac{d^2g}{dy^2}\] $\frac{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}$ by the power rule.
Similarly, $\frac{d^2g}{dy^2} = a_2(2\cdot 1) + a_3(3\cdot 2y)+\cdots+a_{17}(17\cdot 16y^{15})$
Now, notice that if $x = -1$ , then $y = 0$ , so $f^{''}(-1) = g^{''}(0)$
$g^{''}(0)= 2a_2$ , and $f^{''}(-1) = 2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17$
Now, we can use the hockey stick theorem to see that $2\cdot 1 + 3\cdot 2 +\cdots + 16\cdot 17 = 2\binom{18}{3}$
Thus, $2a_2 = 2\binom{18}{3}\rightarrow a_2 = \binom{18}{3}=\boxed{816}$

Solution:
Let $V$ be the vector space of polynomials of degree $\leq 17,$ and let $B = \{1, x, x^2, ..., x^{17} \}$ and $C = \{1, (x+1), (x+1)^2, ..., (x+1)^{17} \}$ be two bases for $V$ .
Let $\vec{v} \in V$ be the polynomial given in the problem, and it is easy to see that $[ \vec{v} ]_B = \langle 1, -1, 1, -1, ... , 1, -1 \rangle.$
Note that the transformation matrix from $C$ to $B$ can be easily found to be $P_{C \to B} = [ [\vec{c_1}]_B [\vec{c_2}]_B ... [\vec{c_3}]_B ] = \begin{bmatrix} \tbinom{0}{0} & \tbinom{1}{0} & \tbinom{2}{0} & \cdots & \tbinom{17}{0} \\ 0 & \tbinom{1}{1} & \tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & \tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$
I claim that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} ,$ where the term $\dbinom{n}{k}$ is negated if $n+k$ is odd.
One can prove that the $i$ th row of $P_{C \to B}$ dotted with the $j$ th column of $P_{B \to C}$ is $\delta_{i, j}$ by using combinatorial identities, which is left as an exercise for the reader. Thus, since the two matrices multiply to form $\mathbb{I}_{18},$ we have proved that $P_{B \to C} = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} .$
To find the coordinates of $\vec{v}$ under basis $C,$ we compute the product $[ \vec{v} ]_C = P_{B \to C} [\vec{v} ]_B = \begin{bmatrix} \tbinom{0}{0} & -\tbinom{1}{0} & \tbinom{2}{0} & \cdots & -\tbinom{17}{0} \\ 0 & \tbinom{1}{1} & -\tbinom{2}{1} & \cdots & \tbinom{17}{1} \\ 0 & 0 & \tbinom{2}{2} & \cdots & -\tbinom{17}{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \tbinom{17}{17} \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 1 \\ \vdots \\ -1 \end{bmatrix} = \begin{bmatrix} \sum_{n=0}^{17} \tbinom{n}{0} \\ -\sum_{n=1}^{17} \tbinom{n}{1} \\  \sum_{n=2}^{17} \tbinom{n}{2} \\ \vdots \\  -\sum_{n=17}^{17} \tbinom{n}{17} \end{bmatrix} = \begin{bmatrix} \tbinom{18}{1} \\ - \tbinom{18}{2} \\ \tbinom{18}{3} \\ \vdots \\ -\tbinom{18}{18} \end{bmatrix},$ where the last equality was obtained via Hockey Stick Identity.
Thus, our answer is $a_2 =  [ [ \vec{v} ]_C ]_3 = \dbinom{18}{3} = \boxed{816}.$


Answer: 30.0

The polynomial $P(x)$ is cubic .  What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$

Solution:
We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic.
Let this root be $a$
We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \frac{-k}{5}$
We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$ , so $k=30$ is the highest.
We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $\boxed{030}$

Solution:
Again, let the common root be $a$ ; let the other two roots be $m$ and $n$ . We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$
Therefore, we can write four equations (and we have four variables ), $a + m = 29 - k$ $a + n = \frac{43}{2} - k$ $am = -k$ , and $an = \frac{k}{2}$
The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$ . The last two equations show that $\frac{m}{n} = -2$ . Solving these show that $m = 5$ and that $n = -\frac{5}{2}$ . Substituting back into the equations, we eventually find that $k = \boxed{030}$

Solution:
Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$ , which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$ , where $R(x) = ax + b$ and $S(x) = cx + d$ . Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$ . Equating coefficients, we get the following system of equations:
\begin{align}  a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck  \end{align}
Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$ , we see that the $k$ 's drop out and we're left with $d = -5c$ . Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed{030}$

Solution:
Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$
Thus, we see that we have \[r^2 + (k - 29)r - k = 0,\] \[2r^2 + (2k - 43)r + k = 0.\] Adding the two equations gives us \[3r^2 + (3k - 72)r = 0 \implies r = 0, 24 - k.\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.\] This can easily be checked to see that it does indeed work, and we're done!


Answer: 482.0

The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$ with $0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1$ and $r_k>0.$ Given that $a_1 + a_2 + a_3 + a_4 + a_5 = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution:
We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series
\begin{align*} P(x) &= \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\ &= \frac{x^{36} - x^{19} - x^{17} + 1}{(x - 1)^2} = \frac{(x^{19} - 1)(x^{17} - 1)}{(x - 1)^2} \end{align*}
This expression has roots at every $17$ th root and $19$ th roots of unity , other than $1$ .  Since $17$ and $19$ are relatively prime , this means there are no duplicate roots.  Thus, $a_1, a_2, a_3, a_4$ and $a_5$ are the five smallest fractions of the form $\frac m{19}$ or $\frac n {17}$ for $m, n > 0$
$\frac 3 {17}$ and $\frac 4{19}$ can both be seen to be larger than any of $\frac1{19}, \frac2{19}, \frac3{19}, \frac 1{17}, \frac2{17}$ , so these latter five are the numbers we want to add.
$\frac1{19}+ \frac2{19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}$ and so the answer is $159 + 323 = \boxed{482}$


Answer: 53.0

The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$ , and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$ . Find the remainder when $f(1)$ is divided by $1000$

Solution:
We have $\frac{1+\sqrt{3}i}{2} = \omega$ where $\omega = e^{\frac{i\pi}{3}}$ is a primitive 6th root of unity. Then we have
\begin{align*} f(\omega) &= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\ &= a\omega^2 + b\omega + c \end{align*}
We wish to find $f(1) = a+b+c$ . We first look at the real parts. As $\text{Re}(\omega^2) = -\frac{1}{2}$ and $\text{Re}(\omega) = \frac{1}{2}$ , we have $-\frac{1}{2}a + \frac{1}{2}b + c = 2015 \implies -a+b + 2c = 4030$ . Looking at imaginary parts, we have $\text{Im}(\omega^2) = \text{Im}(\omega) = \frac{\sqrt{3}}{2}$ , so $\frac{\sqrt{3}}{2}(a+b) = 2019\sqrt{3} \implies a+b = 4038$ . As $a$ and $b$ do not exceed 2019, we must have $a = 2019$ and $b = 2019$ . Then $c = \frac{4030}{2} = 2015$ , so $f(1) = 4038 + 2015 = 6053 \implies f(1) \pmod{1000} = \boxed{053}$

Solution:
Denote $\frac{1+\sqrt{3}i}{2}$ with $\omega$
By using the quadratic formula ( $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ) in reverse, we can find that $\omega$ is the solution to a quadratic equation of the form $ax^2+bx+c=0$ such that $2a=2$ $-b=1$ , and $b^2-4ac=-3$ . This clearly solves to $a=1$ $b=-1$ , and $c=1$ , so $\omega$ solves $x^2-x+1=0$
Multiplying $x^2-x+1=0$ by $x+1$ on both sides yields $x^3+1=0$ . Muliplying this by $x^3-1$ on both sides yields $x^6-1=0$ , or $x^6=1$ . This means that $\omega^6=1$
We can use this to simplify the equation $a\omega^{2018}+b\omega^{2017}+c\omega^{2016}=f(\omega)=2015+2019\sqrt{3}i$ to $a\omega^2+b\omega+c=2015+2019\sqrt{3}i.$
As in Solution 1, we use the values $\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\omega^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ to find that $-\frac{1}{2}a+\frac{1}{2}b+c=2015$ and $\frac{\sqrt{3}}{2}a+\frac{\sqrt{3}}{2}b=2019\sqrt{3} \implies a+b=4038.$ Since neither $a$ nor $b$ can exceed $2019$ , they must both be equal to $2019$ . Since $a$ and $b$ are equal, they cancel out in the first equation, resulting in $c=2015$
Therefore, $f(1)=a+b+c=2019+2019+2015=6053$ , and $6053\bmod1000=\boxed{053}$ . ~ emerald_block

Solution:
Calculate the first few powers of $\frac{1+\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^1=\frac{1+\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^2=\frac{-1+\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^3=-1$
$(\frac{1+\sqrt{3}i}{2})^4=\frac{-1-\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^5=\frac{1-\sqrt{3}i}{2}$
$(\frac{1+\sqrt{3}i}{2})^6=1$
We figure that the power of $\frac{1+\sqrt{3}i}{2}$ repeats in a cycle 6.
$f(\frac{1+\sqrt{3}i}{2})=(a(\frac{1+\sqrt{3}i}{2})^2+b(\frac{1+\sqrt{3}i}{2})+c)(\frac{1+\sqrt{3}i}{2})^{2016}$
Since 2016 is a multiple of 6, $(\frac{1+\sqrt{3}i}{2})^{2016}=1$
$f(\frac{1+\sqrt{3}i}{2})=a(\frac{-1+\sqrt{3}i}{2})+b(\frac{1+\sqrt{3}i}{2})+c$
$f(\frac{1+\sqrt{3}i}{2})=(-\frac{1}{2}a+\frac{1}{2}b+c)+(\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b)=2015+2019\sqrt{3}i$
Therefore, $-\frac{1}{2}a+\frac{1}{2}b+c=2015$ and $\frac{\sqrt{3}i}{2}a+\frac{\sqrt{3}i}{2}b=2019\sqrt{3}i$
Using the first equation, we can get that $-a+b+2c=4030$ , and using the second equation, we can get that $a+b=4038$
Since all coefficients are less than or equal to $2019$ $a=b=2019$
Therefore, $2c=4030$ and $c=2015$
$f(1)=a+b+c=2019+2019+2015=6053$ , and the remainder when it divides $1000$ is $\boxed{053}.$


Answer: 250500.0

The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two.  How many values of $n$ are possible?
$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$

Solution:
Let the roots be $r,s,r + s$ , and let $t = rs$ . Then
and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$ . Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we need the number of possible products $t = rs = r(1002 - r)$ . Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer.
Since $r > 0$ and $t > 0$ , it follows that $0 < t = r(1002-r) < 501^2 = 251001$ , with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \boxed{250,500}$ possible values of $n = -1002t$

Solution: Letting the roots be $r$ $s$ , and $t$ , where $t = r+s$ , we see that by Vieta's Formula's, $2004 = r+s+t = t + t = 2t$ , and so $t = 1002$ . Therefore, $x-1002$ is a factor of $x^3 - 2004x^2 + mx + n$ . Letting $x = 0$ gives that $1002 \mid n$ because $x - 1002 \mid x^3 - 2004x^2 + mx + n$ . Letting $n = -1002a$ and noting that $x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)$ for some $b$ , we see that $b$ is the sum of the roots of $x^2 - bx + a$ $r$ and $s$ , and so $b = 1002$ . Now, we have that $x^2 - 1002x + a$ has roots $r$ and $s$ , and we wish to find the number of possible values of $a$ . By the quadratic formula, we see that \[\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}\] are the two values of noninteger positive real numbers $r$ and $s$ , neither of which is equal to $1002$ . This information gives us that $0 < 501^2 - a < 501^2$ , and so since $501^2 - a$ is evidently not a square, we have $501^2 - 1 - 500 = 251,001 - 500 - 1 = \boxed{250,500}$ possible values of $n = 1002a$


Answer: 78.0

The polynomial $x^3-ax^2+bx-2010$ has three positive integer roots. What is the smallest possible value of $a$
$\textbf{(A)}\ 78 \qquad \textbf{(B)}\ 88 \qquad \textbf{(C)}\ 98 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 118$

Solution:
By Vieta's Formulas , we know that $a$ is the sum of the three roots of the polynomial $x^3-ax^2+bx-2010$ . Again Vieta's Formulas tell us that $2010$ is the product of the three integer roots. Also, $2010$ factors into $2\cdot3\cdot5\cdot67$ . But, since the polynomial has only three roots, two of the four prime factors must be multiplied so that we are left with three roots. To minimize $a$ $2$ and $3$ should be multiplied, which means $a$ will be $6+5+67=78$ and the answer is $\boxed{78}$ .
~JimPickens

Solution:
We can expand $(x+a)(x+b)(x+c)$ as $(x^2+ax+bx+ab)(x+c)$
$(x^2+ax+bx+ab)(x+c)=x^3+abx+acx+bcx+abx^2+acx^2+bcx^2+abc=x^3+x^2(a+b+c)+x(ab+ac+bc)+abc$
We do not care about $+bx$ in this case, because we are only looking for $a$ .  We know that the constant term is $-2010=-(2\cdot 3\cdot 5\cdot 67)$ We are trying to minimize a, such that we have $-ax^2$ Since we have three positive solutions, we have $(x-a)(x-b)(x-c)$ as our factors.  We have to combine two of the factors of $2\cdot 3\cdot 5\cdot 67$ , and then sum up the $3$ resulting factors.  Since we are minimizing, we choose $2$ and $3$ to combine together.  We get $(x-6)(x-5)(x-67)$ which gives us a coefficient of $x^2$ of $-6-5-67=-78$ Therefore $-a=-78$ or $a=\boxed{78}$

Solution:
We want the polynomial $x^3-ax^2+bx-2010$ to have POSITIVE integer roots. That means we want to factor it in to the form $(x-a)(x-b)(x-c).$ We therefore want the prime factorization for $2010$ . The prime factorization of $2010$ is $2 \cdot 3 \cdot 5 \cdot 67$ . We want the smallest difference of the $3$ roots since by Vieta's formulas $a$ is the sum of the $3$ roots.
We proceed to factorize it in to $(x-5)(x-6)(x-67)$ . Therefore, our answer is $5+6+67$ $\boxed{78}$


Answer: 7.0

The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$ , the population was one more than a perfect square. Now, with an additional increase of $100$ , the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$

Solution:
Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count
We first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$ .
We then factor the right side getting $99$ $=$ $(b-a)(b+a)$ .
Since we can only have an nonnegative integral population, clearly $b+a$ $>$ $b-a$ and both factor $99$ .
We factor $99$ into $3^2 \cdot 11$ $=$ $(b-a)(b+a)$ There are a few cases to look at: $1)$ $b+a$ $=$ $11$ and $b-a$ $=$ $9$ .
Adding the two equations we get $2b$ $=$ $20$ or $b$ $=$ $10$ , which means $a$ $=$ $1$ .
But looking at the restriction that the second population + $100$ $=$ third population... $10^2$ $+$ $1$ $+$ $100$ $=$ $201$ $\neq$ a perfect square.
$2)$ $b+a$ $=$ $33$ and $b-a$ $=$ $3$ .
Adding the two equations we get $2b$ $=$ $36$ or $b$ $=$ $18$ , which means $a$ $=$ $15$ .
Looking at the same restriction, we get $18^2$ $1$ $100$ $=$ $324$ $101$ $=$ $425$ , which is NOT a perfect square.
Finally, $b+a$ $=$ $99$ and $b-a$ $=$ $1$ $2b$ $=$ $100$ or $b$ $=$ $50$ , which means $a$ $=$ $49$ .
Looking at the same restriction, we get $50^2$ $1$ $100$ $=$ $2500$ $101$ $=$ $2601$ $=$ $51^2$ . Thus we find that the original population is $a^2$ $=$ $49^2$ $=$ $7^4$ . Or $a^2$ is a multiple of $\boxed{7}$

Solution:
Let $P$ $=$ original population. Translating the word problem into a system of equations, we got: \begin{align} P &= x^2 \\ P + 100 &= y^2 + 1 \\ P + 200 &= z^2 \end{align} for some positive integers $x$ $y$ and $z$ .
Now, by subtracting $(2)$ from $(3)$ (i.e. $(3) - (2)$ ), we got: \begin{align*} 100 &= z^2 - y^2 - 1 \\ 101 &= z^2 - y^2 \\ 101 &= (z - y)(z + y) \end{align*} Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is $y = 50$ and $z = 51$ .
Plugging that back to $(2)$ and simplify, we got $P = 2401 = (49)^2 = x^2$ , a multiple of $7$ .
Therefore, the answer is $\boxed{7}$ . -nullptr07


Answer: 160.0

The population of a small town is $480$ .  The graph indicates the number of females and males in the town, but the vertical scale-values are omitted.  How many males live in the town?
[asy] draw((0,13)--(0,0)--(20,0));  draw((3,0)--(3,10)--(8,10)--(8,0)); draw((3,5)--(8,5)); draw((11,0)--(11,5)--(16,5)--(16,0));  label("$\textbf{POPULATION}$",(10,11),N); label("$\textbf{F}$",(5.5,0),S);  label("$\textbf{M}$",(13.5,0),S); [/asy]
$\text{(A)}\ 120 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 240 \qquad \text{(E)}\ 360$

Solution: The graph show that the ratio of men to total population is $\frac{1}{3}$ , so the total number of men is $\frac{1}{3} \times 480= \boxed{160}$


Answer: 160.0

Solution: The graph shows $3$ equal squares, each with value $x$ . So $3x = 480$ , so $x = \boxed{160}$


Answer: 500000.0

The population of the United States in $1980$ was $226,504,825$ . The area of the country is $3,615,122$ square miles. There are $(5280)^{2}$ square feet in one square mile. Which number below best approximates the average number of square feet per person?
$\textbf{(A)}\ 5,000\qquad \textbf{(B)}\ 10,000\qquad \textbf{(C)}\ 50,000\qquad \textbf{(D)}\ 100,000\qquad \textbf{(E)}\ 500,000$

Solution: With about $230$ million people and under $4$ million square miles, there are about $60$ people per square mile. Since a square mile is about $(5000 \ \text{ft})^{2} = 25$ million square feet, that gives approximately $\frac{25}{60}$ of a million square feet per person. $\frac{25}{60}$ is approximately half, so the answer is approximately half a million, which is $\boxed{500,000}$


Answer: 16.0

The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$

Solution:
Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$
Now let's introduce another variable. Instead of using $a$ and $b$ , we can express the primes as $x+2$ and $x$ where $a$ is $x+2$ and b is $x$ . Plugging $x$ and $x+2$ in, we would have $(x+2)^{2}+x(x+2)+x^{2}$ . When we expand the parenthesis, it would become $x^{2}+4x+4+x^{2}+2x+x^{2}$ . Then we combine like terms to get $3x^{2}+6x+4$ which equals $15553$ . Then we subtract 4 from both sides to get $3x^{2}+6x=15549$ . Since all three numbers are divisible by 3, we can divide by 3 to get $x^{2}+2x=5183$
Notice how if we add 1 to both sides, the left side would become a perfect square trinomial: $x^{2}+2x+1=5184$ which is $(x+1)^{2}=5184$ . Since $2$ is too small to be a valid number, the two primes must be odd, therefore $x+1$ is the number in the middle of them. Conveniently enough, $5184=72^{2}$ so the two numbers are $71$ and $73$ . The next prime number is $79$ , and $7+9=16$ so the answer is $\boxed{16}$

Solution: Let the two primes be $a$ and $b$ , with $a$ being the larger prime. We have $a - b = 2$ , and $a^3 - b^3 = 31106$ . Using difference of cubes, we obtain $a^2 + ab + b^2 = 15553$ . Now, we use the equation $a - b = 2$ to obtain $a^2 - 2ab + b^2 = 4$ . Hence, \[a^2 + ab + b^2 - (a^2 - 2ab + b^2) = 3ab = 15553 - 4 = 15549\] \[ab = 5183.\] Because we have $b = a+2$ $ab = (a+1)^2 - (1)^2$ . Thus, $(a+1)^2 = 5183 + 1 = 5184$ , so $a+1 = 72$ . This implies $a = 71$ $b = 73$ , and thus the next biggest prime is $79$ , so our answer is $7 + 9 = \boxed{16}$

Solution:
Let the two primes be $p$ and $q$ such that $p-q=2$ and $p^{3}-q^{3}=31106$
By the difference of cubes formula, $p^{3}-q^{3}=(p-q)(p^{2}+pq+q^{2})$
Plugging in $p-q=2$ and $p^{3}-q^{3}=31106$
$31106=2(p^{2}+pq+q^{2})$
Through the givens, we can see that $p \approx q$
Thus, $31106=2(p^{2}+pq+q^{2})\approx 6p^{2}\\p^2\approx \tfrac{31106}{6}\approx 5200$
Recall that $70^2=4900$ and $80^2=6400$ . It follows that our primes must be only marginally larger than $70$ , where we conveniently find $p=73, q=71$
The least prime greater than these two primes is $79 \implies 7 + 9 \implies \boxed{16}$


Answer: 16.0

Solution: Let the two primes be $x + 1$ and $x - 1$ . Then, plugging it into the second condition, we get $(x + 1)^3 - (x - 1)^3 = 31106.$ Expanding the left side, \[6x^2 + 2 = 31106 \implies x^2 = 5184.\] Taking the square root of both sides, we get that $x = 72$ and the larger prime is $73$ . The smallest prime larger than $73$ is $79$ , which has a digit sum of $7 + 9 = \boxed{16}.$


Answer: 937.0

The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$

Solution:
We have that $N^2 - N = N(N - 1)\equiv 0\mod{10000}$
Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or $5$ . Thus, one of them is divisible by $2^4 = 16$ , and the other is divisible by $5^4 = 625$ . Noting that $625 \equiv 1\mod{16}$ , we see that $625$ would work for $N$ , except the thousands digit is $0$ . The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$ . In order for this to happen, \[N-1 \equiv -1 \pmod {16}.\] Since $625 \equiv 1 \pmod{16}$ , we know that $15 \cdot 625 = 9375 \equiv 15 \equiv -1 \mod{16}$ . Thus, $N-1 = 9375$ , so $N = 9376$ , and our answer is $\boxed{937}$

Solution:
let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$ .
When we square $N$ we get that \begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \end{align*}
However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \[2000ad+2000bc+100c^2+200bd+20cd+d^2.\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10.
we first consider the ones digits:
$d^2\equiv d \pmod{10}.$
This can happen for only 3 values : 1, 5 and 6.
We can try to solve each case:
Considering the tenths place,
we have that:
$20cd=20c\equiv 10c \pmod {100}$ so $c= 0$
Considering the hundreds place we have that
$200bd+100c^2= 200b \equiv 100b \pmod{1000}$ so again $b=0$
now considering the thousands place we have that
$2000ad+2000bc = 2000a \equiv 1000a \pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$
considering the tenths place
we have that:
$20cd+20=100c+20\equiv 20 \equiv 10c \mod  {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )
so $c=2$
considering the hundreds place we have that
$200bd+100c^2+100c= 1000b+600 \equiv600\equiv 100b \pmod{1000}$ ( the extra $100c$ is carried from the tenths place)
so $b=6$
now considering the thousands place we have that
$2000ad+2000bc +1000b= 10000a+24000+ 6000\equiv0\equiv 1000a \pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place)
so a is equal 0 again
considering the tenths place
we have that:
$20cd+30=120c+30\equiv 30+20c \equiv 10c \pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )
if $c=7$ then we have
$30+20 \cdot 7 \equiv 70\equiv7 \cdot 10 \pmod{100}$
so $c=7$
considering the hundreds place we have that
$200bd+100c^2+100c+100= 1200b+4900+800 \equiv200b+700\equiv 100b \pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place)
if $b=3$ then we have
$700+200 \cdot 3 \equiv 300\equiv3 \cdot 100 \pmod {1000}$
so $b=3$
now considering the thousands place we have that
$2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\equiv0\equiv 2000a+1000\equiv 1000a \pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place)
if $a=9$ then we have
$2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}$
so $a=9$
so we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $\boxed{937}$

Solution:
By the Chinese Remainder Theorem, the equation $N(N-1)\equiv 0\pmod{10000}$ is equivalent to the two equations: \begin{align*} N(N-1)&\equiv 0\pmod{16},\\ N(N-1)&\equiv 0\pmod{625}. \end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\mod{16},N\mod{625})\in\{(0,0),(0,1),(1,0),(1,1)\}$
Let \[\varphi:\mathbb Z/10000\mathbb Z\to\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] \[x\mapsto (x\mod{16},x\mod{625}).\] The statement of the Chinese Remainder theorem is that $\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\varphi^{-1}(0,0)$ $\varphi^{-1}(0,1)$ $\varphi^{-1}(1,0)$ , and $\varphi^{-1}(1,1)$ . Now we easily see that \[\varphi^{-1}(0,0)=0\] and \[\varphi^{-1}(1,1)=1.\] Noting that $625\equiv 1\pmod{16}$ , it follows that \[\varphi^{-1}(1,0)=625.\] To compute $\varphi^{-1}(0,1)$ , note that \[(0,1)=15(1,0)+(1,1)\] in \[\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] so since $\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \[\varphi^{-1}(0,1)=15\varphi^{-1}(1,0)+\varphi^{-1}(1,1)=15\times 625+1=9376.\]
The four candidate digit strings $abcd$ are then $0000,0001,0625,9376$ . Of those, only $9376$ has nonzero first digit, and therefore the answer is $\boxed{937}$

Solution:
WLOG, we can assume that $N$ is a 4-digit integer $\overline{abcd}$ . Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\equiv d\pmod{10}$ , as the units digit of $\overline{abcd}^2$ is affected only by $d$ , regardless of $a$ $b$ , or $c$
By checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$
Now, we seek to find $c$ . Note that the only $\overline{cd}$ that will satisfy $N$ will also satisfy $\overline{cd}^2 \equiv \overline{cd}\pmod{100}$ , by the same reasoning as above - the last two digits of $\overline{abcd}^2$ are only affected by $c$ and $d$ . As we already have narrowed choices for $d$ , we start reasoning out.
First, we note that if $d=0$ , then $c=0$ , as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$ , then recursively, $a$ and $b$ must be $0$ , as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$
Now, let's assume that $d=1$ $\overline{cd}$ is equal to $10c + d = 10c + 1$ . Squaring this gives $100c^2 + 20c + 1$ , and when modulo 100 is taken, it must equal $10c + 1$ . As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \equiv 20c + 1\pmod{100}$ , which must be equivalent to $10c + 1$ . Note that this is really $\overline{(2c)1}$ and $\overline{c1}$ , and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$ .However, if we extend this process to find $b$ and $a$ , we'd find that they are also 0. The only solution in this case is $1$ , and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case.
Let's assume that $d=5$ . Note that $(10c + 5)^2 = 100c^2 + 100c + 25$ , and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$ , which works for now.
Now, let $d=6$ . Note that $(10c + 6)^2 = 100c^2 + 120c + 36$ . Taking modulo 100, this reduces to $20c+36$ , which must be equivalent to $10c+6$ . Again, this is similar to $\overline{(2c+3)6}$ and $\overline{c6}$ , so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$ . Checking quickly $76^2 = 5776$ , which works for now.
Now, we find $b$ , given two possibilities for $\overline{cd}$
Start with $\overline{cd} = 25$ $\overline{bcd} = 100b + \overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$ , which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$ . Therefore, the only $\overline{bcd}$ that works here is $625$ $625^2 = 390625$
Now, assume that $\overline{cd}=76$ . We have $100b + 76$ , and when squared, becomes $10000b^2 + 15200b + 5776$ , which, modulo 1000, should be equivalent to $100b+76$ . Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$ . Using the same technique as before, we must equate the hundreds digit of $\overline{(2b+7)76}$ to $\overline{b76}$ , or equate the units digit of $2b+7$ and $b$ . Since $2b+7$ is necessarily odd, any possible $b$ 's must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$ $376^2 = 141376$
Finally, we solve for $a$ . Start with $\overline{bcd}=625$ . We have $1000a + 625$ , which, squared, gives \[1000000a^2 + 1250000a + 390625,\] and reducing modulo 10000 gives simply 625. So $\overline{abcd}=625$ . However, that makes $a=0$ . Therefore, no solutions exist in this case.
We turn to our last case, $\overline{bcd}=376$ . We have \[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\] and reducing modulo $10000$ gives $2000a + 1376$ , which must be equivalent to $1000a + 376$ . So we must have $\overline{(2a+1)376}$ being equivalent to $\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$ . Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$ . Therefore, $\overline{abcd}=9376$ , so our answer is $\boxed{937}$


Answer: 503.0

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$

Solution:
Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).
Break up the problem into two cases: an even number of sides $2n$ , or an odd number of sides $2n-1$ . For polygons with $2n$ sides, the circumdiameter has endpoints on $2$ vertices. There are $n-1$ points on one side of a diameter, plus $1$ of the endpoints of the diameter for a total of $n$ points. For polygons with $2n - 1$ points, the circumdiameter has $1$ endpoint on a vertex and $1$ endpoint on the midpoint of the opposite side. There are also $n - 1$ points on one side of the diameter, plus the vertex for a total of $n$ points on one side of the diameter.
Case 1: $2n$ -sided polygon. There are clearly $\binom{2n}{3}$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously $2n$ choices for this point. From there, the other two points must be within the $n-1$ points remaining on the same side of the diameter. So our desired probability is $\frac{2n\binom{n-1}{2}}{\binom{2n}{3}}$ $=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}$ $=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}$ $=\frac{3(n-2)}{2(2n-1)}$
so $\frac{93}{125}=\frac{3(n-2)}{2(2n-1)}$
$186(2n-1)=375(n-2)$
$372n-186=375n-750$
$3n=564$
$n=188$ and so the polygon has $376$ sides.
Case 2: $2n-1$ -sided polygon. Similarly, $\binom{2n-1}{3}$ total triangles. Again choose the leftmost point, with $2n-1$ choices. For the other two points, there are again $\binom{n-1}{2}$ possibilities.
The probability is $\frac{(2n-1)\binom{n-1}{2}}{\binom{2n-1}{3}}$
$=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}$
$=\frac{3(n-2)}{2(2n-3)}$
so $\frac{93}{125}=\frac{3(n-2)}{2(2n-3)}$
$186(2n-3)=375(n-2)$
$375n-750=372n-558$
$3n=192$
$n=64$ and our polygon has $127$ sides.
Adding, $127+376=\boxed{503}$

Solution:
We use casework on the locations of the vertices, if we choose the locations of vertices $v_a, v_b, v_c$ on the n-gon (where the vertices of the n-gon are $v_0, v_1, v_2, ... v_{n-1},$ in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that $a<b<c$
By symmetry, we can assume W/O LOG that the location of vertex A is vertex $v_0$
Now, vertex B can be any of $v_1, v_2, ... v_{n-2}$ .  We start in on casework.
Case 1: vertex B is at one of the locations $v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}$ . (The ceiling function is necessary for the cases in which n is odd.)
Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.
There are $\lceil n/2 \rceil - 2$ choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n).  If vertex B is placed at $v_m$ , there are $n - m - 1$ possible places for vertex C.
Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil) - 1}{2}$
Case 2: vertex B is at one of the locations not covered in the first case.
Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2.  This is because fixing vertex B in $v_0$ , then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at $v_0$ , then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at $v_0$ , and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case.
Therefore, there are $\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}$ total obtuse triangles obtainable.
The total number of triangles obtainable is $1+2+3+...+(n-2) = \frac{(n-2)(n-1)}{2}$
The ratio of obtuse triangles obtainable to all triangles obtainable is therefore
$\frac{\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}}{\frac{(n-2)(n-1)}{2}} = \frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)}  = \frac{93}{125}$
So, $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)}  = \frac{31}{125}$
Now, we have that $(n-2)(n-1)$ is divisible by $125 = 5^3$ .  It is now much easier to perform trial-and-error on possible values of n, because we see that $n \equiv 1,2 \pmod{125}$
We find that $n = 127$ and $n = 376$ both work, so the final answer is $127 + 376 = \boxed{503}$


Answer: 74.0

The product $(1.8)(40.3+.07)$ is closest to
$\text{(A)}\ 7 \qquad \text{(B)}\ 42 \qquad \text{(C)}\ 74 \qquad \text{(D)}\ 84 \qquad \text{(E)}\ 737$

Solution: Approximating $40.37$ instead of $1.8$ is more effective because larger numbers are less affected by absolute changes (e.g $1001$ is much closer relatively to $1000$ than $2$ is to $1$ ). $74$ is the closest to $72$ , so the answer is $\boxed{74}$


Answer: 12.0

The product $8\times .25\times 2\times .125 =$
$\text{(A)}\ \frac18 \qquad \text{(B)}\ \frac14 \qquad \text{(C)}\ \frac12 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution: Converting the decimals to fractions , this is \begin{align*} 8\times \frac{1}{4} \times 2\times \frac{1}{8} &= \frac{8\times 2}{4\times 8} \\ &= \frac{16}{32} \\ &= \frac{1}{2} \rightarrow \boxed{12}


Answer: 336.0

The product $N$ of three positive integers is $6$ times their sum , and one of the integers is the sum of the other two. Find the sum of all possible values of $N$

Solution: Let the three integers be $a, b, c$ $N = abc = 6(a + b + c)$ and $c = a + b$ .  Then $N = ab(a + b) = 6(a + b + a + b) = 12(a + b)$ .  Since $a$ and $b$ are positive, $ab = 12$ so $\{a, b\}$ is one of $\{1, 12\}, \{2, 6\}, \{3, 4\}$ so $a + b$ is one of $13, 8, 7$ so the sum of all possible values of $N$ is $12 \cdot (13 + 8 + 7) = 12(28) = \boxed{336}$


Answer: 150.0

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
$\textbf{(A)} \: 105 \qquad\textbf{(B)} \: 120 \qquad\textbf{(C)} \: 135 \qquad\textbf{(D)} \: 150 \qquad\textbf{(E)} \: 165$

Solution:
Let the lengths of the two congruent sides of the triangle be $x$ , then the product desired is $x^2$
Notice that the product of the base and twice the height is $4$ times the area of the triangle.
Set the vertex angle to be $a$ , we derive the equation:
$x^2=4\left(\frac{1}{2}x^2\sin(a)\right)$
$\sin(a)=\frac{1}{2}$
As the triangle is obtuse, $a=150^\circ$ only. We get $\boxed{150}.$

Solution:
Denote by $a$ the length of each congruent side. Denote by $\theta$ the degree measure of each acute angle.
Denote by $\phi$ the degree measure of the obtuse angle.
Hence, this problem tells us the following relationship: \[ a^2 = 2 a \cos \theta \cdot 2 a \sin \theta . \]
Hence, \begin{align*} 1 & = 2 \cdot 2 \sin \theta \cos \theta \\ & = 2 \sin 2 \theta \\ & = 2 \sin \left( 180^\circ - 2 \theta \right) \\ & = 2 \sin \phi . \end{align*}
Hence, $\phi = 150^\circ$
Therefore, the answer is $\boxed{150}$


Answer: 8.0

The product of the two $99$ -digit numbers
$303,030,303,...,030,303$ and $505,050,505,...,050,505$
has thousands digit $A$ and units digit $B$ . What is the sum of $A$ and $B$
$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 5 \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 8 \qquad \mathrm{(E)}\ 10$

Solution:
We can first make a small example to find out $A$ and $B$ . So,
$303\times505=153015$
The ones digit plus thousands digit is $5+3=8$
Note that the ones and thousands digits are, added together, $8$ . (and so on...) So the answer is $\boxed{8}$ This is a direct multiplication way.


Answer: 77.0

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares
$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$

Solution: Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$ $24=4\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{77}$


Answer: 77.0

Solution: Let the three consecutive positive integers be $a-1$ $a$ , and $a+1$ . Since the mean is $a$ , the sum of the integers is $3a$ . So $8$ times the sum is just $24a$ . With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$ $24=4\times6$ , so $a=5$ . Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{77}$


Answer: 1.0

The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$

Solution: \[a^x=b\] \[b^y=a\] \[{a^x}^y=a\] \[xy=1\] \[\log_a b\log_b a=1\] As a result, the answer should be $\boxed{1}$

Solution:
Apply the change of base formula to $\log_a b$ and $\log_b a$ . For simplicity, let us convert to base-10 log.
By change of base, the expression becomes $\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{1}$


Answer: 8.0

The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$

Solution:
Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$ . Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{8}$

Solution: If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \[2a + 2b = -m\] \[a + b = -p\] \[2a(2b) = n\] \[a(b) = m\] Therefore, substituting the second equation into the first equation gives \[m = 2(p)\] and substituting the fourth equation into the third equation gives \[n = 4(m)\] Therefore, $n = 8p$ , so $\frac{n}{p}= \boxed{8}$


Answer: 8.0

Solution:
Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$ . Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{8}$

Solution: If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \[2a + 2b = -m\] \[a + b = -p\] \[2a(2b) = n\] \[a(b) = m\] Therefore, substituting the second equation into the first equation gives \[m = 2(p)\] and substituting the fourth equation into the third equation gives \[n = 4(m)\] Therefore, $n = 8p$ , so $\frac{n}{p}= \boxed{8}$


Answer: 20.0

The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click.
The speed of the train in miles per hour is approximately the number of clicks heard in:
$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad \textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}$

Solution: We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$ , we can convert it to clicks per minute: \[\frac{x\text{ mile}}{\text{hr}}\cdot\left(\frac{1\text{ hr}}{60\text{ min}}\right)\cdot\left(\frac{5280\text{ ft}}{1\text{ mile}}\right)\cdot\left(\frac{1\text{ click}}{30\text{ ft}}\right)=\frac{5280x}{1800}\cdot\frac{\text{click}}{\text{min}}.\] Therefore every minute, on average, $5280x/1800$ clicks are heard. The number of clicks heard in $y$ minutes is $y\cdot 5280x/1800$ , so the number of clicks heard in $1800/5280$ minutes is equal to $x$ . In other words, the speed of the train in miles per hour is equal to the number of clicks heard in $1800/5280$ minutes, which is approximately one-third of a minute, or $\boxed{20}$


Answer: 5.0

The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$

Solution: We factor $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\frac{101}{20}$ . As $\frac{101}{20}=5.05$ , our answer is $\boxed{5}$


Answer: 18.0

The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\  50$

Solution: Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{18}.$

Solution: We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct.


Answer: 18.0

Solution: Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{18}.$

Solution: We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct.


Answer: 6.0

The ratio of boys to girls in Mr. Brown's math class is $2:3$ .  If there are $30$ students in the class, how many more girls than boys are in the class?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$

Solution:
Let the number of boys be $2x$ .  It follows that the number of girls is $3x$ . These two values add up to $30$ students, so \[2x+3x=5x=30\Rightarrow x=6\]
The difference between the number of girls and the number of boys is $3x-2x=x$ , which is $6$ , so the answer is $\boxed{6}$


Answer: 3.0

The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$ . How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$

Solution: The formula for the measure of the interior angle of a regular polygon with $n$ -sides is $180 - \frac{360}{n}$ . Letting our two polygons have side length $r$ and $k$ , we have that the ratio of the interior angles is $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$ .  Cross multiplying both sides, we have $2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0$ . Using Simon's Favorite Factoring Trick, we have $-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24$ . Because $k$ and $r$ are both more than $2$ , we know that $6-k < r+4$ . Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$ $2$ $3$ , or $4$ and $r+4$ to $24$ $12$ $8$ , or $6$ respectively to get the following pairs for $(k, r)$ $(5, 20)$ $(4, 8)$ $(3, 4)$ , and $(2, 2)$ . However, we have to take out the solution with $k = 2$ , because $k$ and $r$ are both more than $2$ , leaving us with $\boxed{3}$ as the correct answer.


Answer: 135.0

The ratio of the measures of two acute angles is $5:4$ , and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

Solution:
We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$ . Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$ . Solving for $y$ in the first equation and substituting into the second equation yields \[\begin{split} 90 - x & = 2 (90 - 1.25x) \\ 1.5x & = 90 \\ x & = 60 \end{split}\]
Substituting this $x$ value back into the first equation yields $y$ $=$ $75$ , leaving $x$ $+$ $y$ equal to $\boxed{135}$

Solution:
We let the measures be $5x$ and $4x$ giving us the ratio of $5:4$ . We know $90-4x>90-5x$ since this inequality gives $x>0$ , which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so
$90-4x=2(90-5x)$
$90-4x=180-10x$
$6x=90$
$x=15$
Therefore, the angles are $75$ and $60$ , which sum to $\boxed{135}$


Answer: 135.0

Solution:
By: dragonfly
We set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that
$4x=5y$
and
$2\times(90-x)=90-y$
since the larger the original angle, the smaller the complement.
We then find that $x=75$ and $y=60$ , and their sum is $\boxed{135}$

Solution: We can visualize the problem like so:  \[5x+1y = 90^\circ = 4x+2y\] Moving like terms to the same side gets $x = y$ , and substituting this back gets $6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ$ , so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \boxed{135}$ . ~ emerald_block


Answer: 27.0

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11/4$ . To the nearest whole percent, what percent of its games did the team lose?
$\text{(A)}\ 24\% \qquad \text{(B)}\ 27\% \qquad \text{(C)}\ 36\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ 73\%$

Solution: The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\dfrac{4x}{15x}\times100=\dfrac{4}{15}\times 100=26.\overline{6}\%\approx\boxed{27}$


Answer: 27.0

Solution: The Won/Lost ratio is 11/4 so, for some number $N$ , the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\dfrac{4N}{15N}=\dfrac{4}{15}\approx 0.27=\boxed{27}.$ -Clara Garza


Answer: 23.0

The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$

Solution: If the radius of the circle is $r$ , then the perimeter of the first triangle is $3\left(\frac{2r}{\sqrt3}\right)=2r\sqrt3$ , and the perimeter of the second is $3r\sqrt3$ . So the ratio is $\boxed{23}$