Answer: 11.0

parabola with equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$ .  What is $c$
$\textbf{(A) } 2\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$

Solution:
Substitute the points $(2,3)$ and $(4,3)$ into the given equation for $(x,y)$
Then we get a system of two equations:
$3=4+2b+c$
$3=16+4b+c$
Subtracting the first equation from the second we have:
$0=12+2b$
$b=-6$
Then using $b=-6$ in the first equation:
$0=1+-12+c$
$c=\boxed{11}$

Solution: Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely $(3,2)$ . Thus, the form of the equation of the parabola is $y - 2 = (x - 3)^2$ . Expanding this out, we find that $c=\boxed{11}$

Solution:
The points given have the same $y$ -value, so the vertex lies on the line $x=\frac{2+4}{2}=3$
The $x$ -coordinate of the vertex is also equal to $\frac{-b}{2a}$ , so set this equal to $3$ and solve for $b$ , given that $a=1$
$x=\frac{-b}{2a}$
$3=\frac{-b}{2}$
$6=-b$
$b=-6$
Now the equation is of the form $y=x^2-6x+c$ .  Now plug in the point $(2,3)$ and solve for $c$
$y=x^2-6x+c$
$3=2^2-6(2)+c$
$3=4-12+c$
$3=-8+c$
$c=\boxed{11}$

Solution:
Substituting y into the two equations, we get:
$3=x^2+bx+c$
Which can be written as:
$x^2+bx+c-3=0$
$4$ and $2$ are the solutions to the quadratic. Thus:
$c-3=4\times2$
$c-3=8$
$c=\boxed{11}$


Answer: 745.0

pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$ , although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$

Solution:
Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple . We know that $31$ and either $25,\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$ , to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \boxed{745}$


Answer: 144.0

point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ $9$ , and $49$ , respectively. Find the area of $\triangle ABC$

Solution: By the transversals that go through $P$ , all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \dfrac{ab\sin C}{2}$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\ 3x,\ 7x$ . Thus, the corresponding side on the large triangle is $12x$ , and the area of the triangle is $12^2 = \boxed{144}$

Solution: Alternatively, since the triangles are similar by $AA$ , then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\dfrac{base}{height} = \dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, the base of $t_{2}$ is 6, and the base of $t_{3}$ is 14. Since the quadrilaterals underneath $t_{1}$ and $t_{2}$ are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is $4 + 14 + 6 = 24$ . Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is $\dfrac{1}{2} \cdot 24 \cdot 12 = \boxed{144}$

Solution: The base of $\triangle{ABC}$ is $BC$ . Let the base of $t_1$ be $x$ , the base of $t_2$ be $y$ , and the base of $t_3$ be $z$ . Since $\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$ . We can relate $t_1,t_2,$ and $t_3$ by the square root of the ratio of their areas. $\sqrt{\frac{4}{9}}=\frac{2}{3}$ and $\sqrt{\frac{4}{49}}=\frac{2}{7}$ so $y=\frac{3x}{2}$ and $z=\frac{7x}{2}$ $x+\frac{7x}{2}+\frac{3x}{2}=6x$ , so $\triangle{ABC}$ has a base that is $6$ times $t_1$ $[\triangle{ABC}]=36[t_1]=36 \cdot 4=\boxed{144}$

Solution: Since the three lines through $P$ are parallel to the sides, $t_1$ $t_2$ $t_3$ , and $\triangle{ABC}$ are similar by $AA$ similarity. Suppose the area of $\triangle{ABC}$ is $x^2$ , so the ratio of the base of $t_1$ to the base of $t_2$ to the base of $t_3$ to the base of $\triangle{ABC}$ is $2:3:7:x$ . Because the quadrilaterals below $t_1$ and $t_2$ are parallelograms, the base of $\triangle{ABC}$ is equal to the sum of the bases of $t_1, t_2,$ and $t_3$ . Therefore, $x$ equals $2+3+7=12$ so the area of $\triangle{ABC}$ equals $x^2=12^2=\boxed{144}.$


Answer: 502.0

positive integer is called ascending if, in its decimal representation , there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?

Solution:
Note that an ascending number is exactly determined by its digits : for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.
So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $2^9=512$ potential ascending numbers, one for each subset of $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
However, we've counted one-digit numbers and the empty set , so we must subtract them off to get our answer, $512-10=\boxed{502}.$


Answer: 744.0

problem_id
10b89ea3f13aa2c628bafff65bf48904    Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904    It has been noted that this answer won't actua...
Name: Text, dtype: object

Solution:
We start by drawing a diagram;

We know that $\sin \alpha = \frac{1}{5}$ . Since $\sin \alpha = \cos (90- \alpha)$ \[\cos (90- \alpha) = \frac{1}{5} \implies  \cos (\beta + \gamma) = \frac{1}{5}\]
Using our angle sum identities, we expand this to $\cos \beta \cdot \cos \gamma - \sin \beta \cdot \sin \gamma = \frac{1}{5}$ .
We can now use the right triangle definition of cosine and sine to rewrite this equation as;
\[\frac{l}{40} \cdot \frac{w}{31} - \frac{x}{40} \cdot \frac{y}{31} = \frac{1}{5} \implies lw- xy = 8 \cdot 31 \implies lw = xy + 31 \cdot 8\]
Hang on; $lw$ is the area we want to maximize! Therefore, to maximize this area we must maximize $xy = 40 \sin \beta \cdot 31 \sin \gamma = 31 \cdot 40 \cdot \frac{1}{2}( \cos (\beta - \gamma) - \cos ( \beta + \gamma)) = 31 \cdot 20 \cdot (\cos(\beta-\gamma)-\frac{1}{5})$ .
Since $\cos(\beta-\gamma)$ is the only variable component of this expression, to maximize the expression we must maximize $\cos(\beta-\gamma)$ . The cosine function has a maximum value of 1, so our equation evaluates to $xy = 31 \cdot 20 \cdot (1-\frac{1}{5}) = 31 \cdot 20 \cdot \frac{4}{5} = 31 \cdot 16$ (Note that at this max value, since $\beta$ and $\gamma$ are both acute, $\beta-\gamma=0 \implies \beta=\gamma$ ).
Finally, $lw = xy + 31 \cdot 8 = 31 \cdot 16 + 31\cdot 8 = 31 \cdot 24 = \boxed{744}$

Solution: As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\text{cis} A$ , and let $z$ be a complex number with $|z|=1$ $\text{Arg}(z)\ge 0^\circ$ and $\text{Arg}(zw)\le90^\circ$ . Then we represent $B$ by $40z$ and $C$ by $31zw$ . The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$ . Thus \[[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).\] We can expand this, using the fact that $z\overline{z}=|z|^2$ , finding \[[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).\] Now as $w=\text{cis}A$ , we know that $\Im(w)=\frac15$ . Also, $|z^2w|=1$ , so the maximum possible imaginary part of $z^2w$ is $1$ . This is clearly achievable under our conditions on $z$ . Therefore, the maximum possible area of $AQRS$ is $620(1+\tfrac15)=\boxed{744}$

Solution:
Let $\theta$ be the angle $\angle BAQ$ . The height of the rectangle then can be expressed as $h = 31 \sin (A+\theta)$ , and the length of the rectangle can be expressed as $l = 40\cos \theta$ . The area of the rectangle can then be written as a function of $\theta$ $[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta$ . For now, we will ignore the $1240$ and focus on the function $f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta$
Taking the derivative, $f'(\theta) = \sin A \cdot -2\cos \theta \sin \theta + \cos A \cos 2\theta = \cos A \cos 2\theta - \sin A \sin 2\theta = \cos(2\theta + A)$ . Setting this equal to $0$ , we get $\cos(2 \theta + A) = 0 \Rightarrow 2\theta +A = 90, 270 ^\circ$ . Since we know that $A+ \theta < 90$ , the $270^\circ$ solution is extraneous. Thus, we get that $\theta = \frac{90 - A}{2} = 45 - \frac{A}{2}$
Plugging this value into the original area equation, $a(45 -  \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2})$ . Using a product-to-sum formula, we get that: \[1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) =\] \[1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2})))=\] \[620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}\]

Solution:
Let $\alpha$ be the angle $\angle CAS$ and $\beta$ be the angle $\angle BAQ$ . Then \[\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A\] \[\cos(\alpha + \beta) = \cos(90^\circ - \angle A)\] \[\cos(\alpha + \beta) = \sin(\angle A) = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sqrt{(1-\cos^2\alpha)(1-\cos^2\beta)} = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} = \frac{1}{5}\] However, by AM-GM: \[\cos^2\alpha+\cos^2\beta \ge 2\cos\alpha\beta\] Therefore, \[1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta \le 1-2\cos\alpha\beta+\cos^2\alpha\cos^2\beta = (1-\cos\alpha\cos\beta)^2\] \[\sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} \le 1-\cos\alpha\cos\beta\] So, \[\frac{1}{5} \ge \cos\alpha\cos\beta - (1-\cos\alpha\cos\beta) = 2\cos\alpha\cos\beta-1\] \[\frac{3}{5} \ge \cos\alpha\cos\beta\] .
However, the area of the rectangle is just $AS \cdot AQ = 31\cos\alpha \cdot 40\cos\beta \le 31 \cdot 40 \cdot \frac{3}{5} = \boxed{744}$


Answer: 6.0

problem_id
18bf638a0cb797d222d008b988c16b6d    Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d    From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object

Solution:
We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$
If we define a "move" as each time Erin crawls along a single edge from one vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be one unit away from $A$ (since Erin cannot crawl back to $A$ ), this vertex must be $4$
Therefore, we now just need to count the number of paths from $A$ to $4$ . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to $4$ , and 2 choices for which vertex Erin was at 2 moves before $4$ . All of Erin's previous moves were forced, so the total number of legal paths from $A$ to $4$ is $3 \cdot 2 = \boxed{6}$

Solution: Lets say that this cube is an unit cube and the given corner is $(0,0,0)$ . Because Erin cannot return back to her starting point, she cannot be on $(0,0,1)$ $(0,1,0)$ , or $(1,0,0)$ . She cannot be on $(1,1,0)$ $(1,0,1)$ , or $(0,1,1)$ , because after $7$ moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at $(1,1,1)$ . Now, we draw a net and see that there are $3$ choices for the first move, $2$ for the second, and the rest are forced. Thus the answer is $3\cdot2 = \boxed{6}$

Solution: Let's suppose the given corner on the cube is $(0,0,0)$ . Erin has 3 identical ways to proceed. Suppose she goes to $(0,1,0)$ . She now has two more identical ways to go. Let's say she goes to $(0,1,1)$ . She has to go to $(0,0,1)$ , otherwise, she will end up on that point after 7 moves. This is because once if she chooses the other path to $(1,1,1)$ , the endpoint is certain to be $(0,0,1)$ , which is directly connected to $(0,0,0)$ . After she goes to $(0,0,1)$ , the only option she has is to go to $(1,0,1)$ , then $(1,0,0)$ , after that $(1,1,0)$ , and finally $(1,1,1)$ . She is forced to go this way because she cannot end up on $(1,0,0)$ . At the start, she had 3 ways to choose, and after that 2 ways to choose, so $3\cdot2=\boxed{6}$


Answer: 801.0

problem_id
21b33a597f802b0a4ad56201b9aeba1e    How many positive integers not exceeding $2001...
21b33a597f802b0a4ad56201b9aeba1e    How many positive integers not exceeding $2001...
Name: Text, dtype: object

Solution:
We can solve this problem by finding the cases where the number is divisible by $3$ or $4$ , then subtract from the cases where none of those cases divide $5$ . To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\left\lfloor \frac{2001}{3} \right\rfloor$ $\left\lfloor \frac{2001}{4} \right\rfloor$ , and $\left\lfloor \frac{2001}{12} \right\rfloor$ . The first two floor functions were for calculating the number of individual cases for $3$ and $4$ . The third case was to find any overlapping numbers. The numbers were $667$ $500$ , and $166$ , respectively. We add the first two terms and subtract the third to get $1001$ . The first case is finished.
The second case is more or less the same, except we are applying $3$ and $4$ to $5$ . We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\left\lfloor \frac{2001}{3\cdot5} \right\rfloor$ $\left\lfloor \frac{2001}{4\cdot5} \right\rfloor$ , and $\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\rfloor$ yields the numbers $133$ $100$ , and $33$ . The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$ . The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$ . Subtracting this number from the original $1001$ numbers procures $\boxed{801}$

Solution:
First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$
There are $\frac45\cdot2000=1600$ numbers that are not multiples of $5$ $\frac23\cdot\frac34\cdot1600=800$ are not multiples of $3$ or $4$ , so $800$ numbers are. $800+1=\boxed{801}$

Solution: Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$ , so there are $\frac{10}{25}\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$ . Since $2001$ is a multiple of $3$ , add $1$ to $800$ to get $800+1=\boxed{801}$

Solution: By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$ $80\%$ of them will not be divisible by $5$ , and by far the closest number to $80\%$ of $1001$ is $\boxed{801}$

Solution:
Similar to some of the above solutions.
We can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$ . Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$ , and $500$ numbers that are divisible by $4$ . However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$ , since they're also divisible by 5 which we don't want. There are $133$ $100$ $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$ , so we have to add $33$ back to the tally (there are $33$ multiples of $60$ that does not exceed $2001$ ). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$ . This is tantamount to subtracting the number of multiples of $12$ . And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ $500$ $133$ $100$ $33$ $166$ $\boxed{801}$

Solution:
Similar to @above:
Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$ . Then we have that the desired number is: \[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)\]
Evaluating each of these we get: \[667+500-166-133-100+33 = 1100-299 = 801.\]
Thus, the answer is $\boxed{801}.$


Answer: 32000.0

problem_id
227cbd9a094a48b5f95a026123843b8c    The state income tax where Kristin lives is le...
227cbd9a094a48b5f95a026123843b8c    The state income tax where Kristin lives is le...
Name: Text, dtype: object

Solution: Let $A$ $T$ be Kristin's annual income and the income tax total, respectively. Notice that \begin{align*} T &= p\%\cdot28000 + (p + 2)\%\cdot(A - 28000) \\ &= [p\%\cdot28000 + p\%\cdot(A - 28000)] + 2\%\cdot(A - 28000) \\ &= p\%\cdot A + 2\%\cdot(A - 28000) \end{align*} We are also given that \[T = (p + 0.25)\%\cdot A = p\%\cdot A + 0.25\%\cdot A\] Thus, \[p\%\cdot A + 2\%\cdot(A - 28000) = p\%\cdot A + 0.25\%\cdot A\] \[2\%\cdot(A - 28000) = 0.25\%\cdot A\] Solve for $A$ to obtain $A = \boxed{32000}$


Answer: 4.0

problem_id
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
319997b31b053c975a9c9d5ff9b8fbcc    Five positive consecutive integers starting wi...
Name: Text, dtype: object

Solution:
Let $a=1$ . Our list is $\{1,2,3,4,5\}$ with an average of $15\div 5=3$ . Our next set starting with $3$ is $\{3,4,5,6,7\}$ . Our average is $25\div 5=5$
Therefore, we notice that $5=1+4$ which means that the answer is $\boxed{4}$

Solution:
We are given that \[b=\frac{a+a+1+a+2+a+3+a+4}{5}\] \[\implies b =a+2\]
We are asked to find the average of the 5 consecutive integers starting from $b$ in terms of $a$ . By substitution, this is \[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]
Thus, the answer is $\boxed{4}$

Solution: The list of numbers is $\left\{a,\ a+1,\ b,\ a+3,\ a+4\right\}$ so $b=a+2$ . The new list is $\left\{a+2,\ a+3,\ a+4,\ a+5,\ a+6\right\}$ and the average is $a+4 \Longrightarrow \boxed{4}$


Answer: 4.0

Solution: We know from experience that the average of $5$ consecutive numbers is the $3^\text{rd}$ one or the $1^\text{st} + 2$ . With the logic, we find that $b=a+2$ $b+2=(a+2)+2=\boxed{4}$


Answer: 3.0

problem_id
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
396f2f976638a27e600105ba37c3a839    On an algebra quiz, $10\%$ of the students sco...
Name: Text, dtype: object

Solution:
Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean
is $87$ and the median is $90$
Thus, the solution is \[90-87=3\implies\boxed{3}\]

Solution:
The percentage who scored $100$ points is $100\%-(10\%+35\%+30\%)=100\%-75\%=25\%$ . Now, we need to find the median, which is the score that splits the upper and lower $50\%$ .The lower $10\%+35\%=45\%$ scored $70$ or $80$ points, so the median is $90$ (since the upper $25\%$ is $100$ points and the lower $45\%$ is $70$ or $80$ ).The mean is $10\%\cdot70+35\%\cdot80+30\%\cdot90+25\%\cdot100=7+28+27+25=87$ .
So, our solution is $90-87=3\Rightarrow\boxed{3}$ ~sosiaops

Solution:
The $\le 80$ -point scores make up $10\%+35\% = 45\% < 50\%$ of the scores, but the $\le 90$ -point scores make up $45\%+30\% = 75\% > 50\%$ of the scores, so the median is $90$
$10\%$ of scores were $70-90 = -20$ more than the median, $35\%$ were $-10$ more, $100\%-75\% = 25\%$ were $10$ more, and the rest were equal. This means that the mean score is $10\%\cdot(-20)+35\%\cdot(-10)+25\%\cdot10 = -2 + (-3.5) + 2.5 = -3$ more than the median, so their difference is $\left|-3\right| = \boxed{3}$ .
~ emerald_block


Answer: 35.0

problem_id
44dac98b900fb2d03612e3e20d26762f    A box contains exactly five chips, three red a...
44dac98b900fb2d03612e3e20d26762f    A box contains exactly five chips, three red a...
Name: Text, dtype: object

Solution: Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: $W, W, R, R, R$ such that both $W$ 's appear in the first $4$ . We find the number of ways to arrange the white chips in the first $4$ and divide that by the total ways to choose all the chips. The probability of this occurring is $\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{35}$

Solution: The amount of ways to end with a white chip is by having $RRWW, RWW,$ and $WW$ . The amount of arrangements for $RRWW$ with $W$ at the end is $3$ , the number of arrangements of $RWW$ with $W$ at the end is $2$ , and the number of arrangements with $WW$ is just $1$ . This gives us $6$ total ways to end with white. Next, the cases to end with a red are $RWRR$ , and $RRR$ $RWRR$ gives us $3$ ways and $RRR$ gives us $1$ way. So the number of ways to end with a red is $4$ . Thus, our answer is simply $\frac{6}{4+6}$ $\boxed{35}$


Answer: 15.0

problem_id
4ca2bf599f1693b695997dd52d9e4774    A man walked a certain distance at a constant ...
4ca2bf599f1693b695997dd52d9e4774    A man walked a certain distance at a constant ...
Name: Text, dtype: object

Solution:
We can make three equations out of the information, and since the distances are the same, we can equate these equations.
\[\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{5}{2})(x-\frac{1}{2})\] where $x$ is the man's rate and $t$ is the time it takes him.
Looking at the first two parts of the equations,
\[\frac{4t}{5}(x+\frac{1}{2})=xt\]
we note that we can solve for $x$ . Solving for $x$ , we get $x=2.$
Now we look at the last two parts of the equation:
\[xt=(t+\frac{5}{2})(x-\frac{1}{2})\]
we note that we can solve for $t$ and we get $t=\frac{15}{2}.$ We want the find the distance, which is $xt= \boxed{15}.$


Answer: 991.0

problem_id
6a2cc5541e1749d3d0bb739aacabadf4    The product $(8)(888\dots8)$ , where the secon...
6a2cc5541e1749d3d0bb739aacabadf4    The product $(8)(888\dots8)$ , where the secon...
Name: Text, dtype: object

Solution:
We can list the first few numbers in the form $8 \cdot (8....8)$
(Hard problem to do without the multiplication, but you can see the pattern early on)
$8 \cdot 8 = 64$
$8 \cdot 88 = 704$
$8 \cdot 888 = 7104$
$8 \cdot 8888 = 71104$
$8 \cdot 88888 = 711104$
By now it's clear that the numbers will be in the form $7$ $k-2$ $1$ 's, and $04$ . We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$ . Solving, we get $k = 991$ , meaning the answer is $\fbox{(D)}$
Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$ , so we just do $1000-9=\boxed{991}$


Answer: 8.0

problem_id
74b973e4f94621e9337c1a9c0077ccfc    A telephone number has the form $\text{ABC-DEF...
74b973e4f94621e9337c1a9c0077ccfc    A telephone number has the form $\text{ABC-DEF...
Name: Text, dtype: object

Solution:
We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ . Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
Case 1: $G$ $H$ $I$ , and $J$ are $7$ $5$ $3$ , and $1$ respectively.
A cursory glance allows us to deduce that the smallest possible sum of $A + B + C$ is $11$ when $D$ $E$ , and $F$ are $8$ $6$ , and $4$ respectively, so this is out of the question.
Case 2: $G$ $H$ $I$ , and $J$ are $3$ $5$ $7$ , and $9$ respectively.
A cursory glance allows us to deduce the answer. Clearly, when $D$ $E$ , and $F$ are $6$ $4$ , and $2$ respectively, $A + B + C$ is $9$ when $A$ $B$ , and $C$ are $8$ $1$ , and $0$ respectively, giving us a final answer of $\boxed{8}$


Answer: 2.0

problem_id
77e0fabb4f8f99b4274e980292ddff5d    The numbers $1, 2, 3, 4, 5$ are to be arranged...
77e0fabb4f8f99b4274e980292ddff5d    The numbers $1, 2, 3, 4, 5$ are to be arranged...
Name: Text, dtype: object

Solution:
We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$ , or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$ . By choosing the full circle, we can obtain $15$ . By choosing everything except for $1, 2, 3, 4,$ and $5$ , we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$
This means that we now only need to check for $6, 7, 8,$ and $9$ . However, once we have found a set summing to $6$ , we can choose all remaining numbers and obtain a set summing to $15-6=9$ , and similarly for $7$ and $8$ . Thus, we only need to check each case for whether or not we can obtain $6$ or $7$
We can make $6$ by having $4, 2$ , or $3, 2, 1$ , or $5, 1$ . We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.

Now, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$ , because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.

We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$ , so we must place $4$ next to the $1$ and $5$ next to the $2$

This is the only solution to make $6$ "bad."
Next we move on to $7$ , which can be made by $3, 4$ , or $5, 2$ , or $4, 2, 1$ . We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.

This is the only solution to make $7$ "bad."
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\boxed{2}$


Answer: 125.0

problem_id
891fbd11f453d2b468075929a7f4cfd8    For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8    Warning: This solution doesn't explain why $43...
Name: Text, dtype: object

Solution:
We first claim that $n$ must be divisible by $42$ . Since $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \neq 0,1 \pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.
Then $\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)$ . In order for this expression to be divisible by $2021=43\cdot 47$ , a necessary condition is $a^n - 1 \equiv 0 \pmod{43}$ . By Fermat's Little Theorem $a^{42} \equiv 1 \pmod{43}$ . Moreover, if $a$ is a primitive root modulo $43$ , then $\text{ord}_{43}(a) = 42$ , so $n$ must be divisible by $42$
By similar reasoning, $n$ must be divisible by $46$ , by considering $a \not\equiv 0,1 \pmod{47}$
We next claim that $n$ must be divisible by $43$ . By Dirichlet, let $a$ be a prime that is congruent to $1 \pmod{43}$ . Then $\sigma(a^n) \equiv n+1 \pmod{43}$ , so since $\sigma(a^n)-1$ is divisible by $43$ $n$ is a multiple of $43$
Alternatively, since $\left(\frac{a(a^n - 1^n)}{a-1}\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,$ which simplifies to $v_{43}(n) \geq 1,$ which implies the desired result.
Similarly, $n$ is a multiple of $47$
Lastly, we claim that if $n = \text{lcm}(42, 46, 43, 47)$ , then $\sigma(a^n) - 1$ is divisible by $2021$ for all positive integers $a$ . The claim is trivially true for $a=1$ so suppose $a>1$ . Let $a = p_1^{e_1}\ldots p_k^{e_k}$ be the prime factorization of $a$ . Since $\sigma(n)$ is multiplicative , we have \[\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.\] We can show that $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for all primes $p_i$ and integers $e_i \ge 1$ , so \[\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in}),\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for this choice of $n$ , so suppose $p_i \not\equiv 0 \pmod{43}$ and $p_i \not\equiv 0 \pmod{47}$ . Each expression in parentheses equals $\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$ . If $p_i \not\equiv 1 \pmod{43}$ , then FLT implies $p_i^n - 1 \equiv 0 \pmod{43}$ , and if $p_i \equiv 1 \pmod{43}$ , then $p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}$ (since $n$ is also a multiple of $43$ , by definition). Similarly, we can show $\sigma(p_i^{e_in}) \equiv 1 \pmod{47}$ , and a simple CRT argument shows $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ . Then $\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}$
Then the prime factors of $n$ are $2,3,7,23,43,$ and $47,$ and the answer is $2+3+7+23+43+47 = \boxed{125}$

Solution:
$n$ only needs to satisfy $\sigma(a^n)\equiv 1 \pmod{43}$ and $\sigma(a^n)\equiv 1 \pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . The following three statements are the same:
We can show this by casework on $p$
Similar arguments for modulo $47$ lead to $46|n$ and $47|n$ . Therefore, we get $n=\operatorname{lcm}[42,43,46,47]$ . Following the last paragraph of Solution 1 gives the answer $\boxed{125}$

Solution:
We perform casework on $a:$
Finally, the least such positive integer $n$ for all cases is \begin{align*} n&=\operatorname{lcm}(42,43,46,47) \\ &=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\ &=2\cdot3\cdot7\cdot23\cdot43\cdot47, \end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\boxed{125}.$

Solution: Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem , we get that $n \equiv 0 \pmod{42}$ and $n \equiv 0 \pmod{46}.$ Then, we can look at $a$ being a $1\pmod{43}$ prime and a $1\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$ . There don't seem to be any other odd "numbers" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $\boxed{125}$


Answer: 279.0

problem_id
8dd58d71d622cdca7859918e42e278d8    The number $5^{867}$ is between $2^{2013}$ and...
8dd58d71d622cdca7859918e42e278d8    The number $5^{867}$ is between $2^{2013}$ and...
Name: Text, dtype: object

Solution:
The problem is asking for between how many consecutive powers of $5$ are there $3$ power of $2$
There can be either $2$ or $3$ powers of $2$ between any two consecutive powers of $5$ $5^n$ and $5^{n+1}$
The first power of $2$ is between $5^n$ and $2 \cdot 5^n$
The second power of $2$ is between $2 \cdot 5^n$ and $4 \cdot 5^n$
The third power of $2$ is between $4 \cdot 5^n$ and $8 \cdot 5^n$ , meaning that it can be between $5^n$ and $5^{n+1}$ or not.
If there are only $2$ power of $2$ s between every consecutive powers of $5$ up to $5^{867}$ , there would be $867\cdot 2 = 1734$ power of $2$ s. However, there are $2013$ powers of $2$ before $5^{867}$ , meaning the answer is $2013 - 1734 = \boxed{279}$


Answer: 3.0

problem_id
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
9f71a48715f97d1d12960b17799f7db2    Walking down Jane Street, Ralph passed four ho...
Name: Text, dtype: object

Solution:
Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\text{B}-\text{O}-\text{R}-\text{Y}$
$\text{O}-\text{B}-\text{R}-\text{Y}$
The answer is $1+2=\boxed{3}$

Solution:
There are $4!=24$ arrangements without restrictions. There are $3!\cdot2!=12$ arrangements such that the blue house neighboring the yellow house (calculating the arrangments of [ $\text{BY}$ ], $\text{O}$ , and $\text{R}$ ). Hence, there are $24-12=12$ arrangements with the blue and yellow houses non-adjacent.
By symmetry, exactly half of the $12$ arrangements have the blue house before the yellow house, and exactly half of those $6$ arrangements have the orange house before the red house, so our answer is $12\cdot\frac{1}{2}\cdot\frac{1}{2}= \boxed{3}$

Solution:
To start with, the blue house is either the first or second house.
If the blue house is the first, then the orange must follow, leading to $2$ cases: $\text{B-O-R-Y}$ and $\text{B-O-Y-R}$
If the blue house is second, then the orange house must be first and the yellow house last, leading to $1$ case: $\text{O-B-R-Y}$ .
Therefore, our answer is $\boxed{3}$


Answer: 13.0

problem_id
a45915bc778c3b967b94f7cee4faa46d    Find all integers $n \ge 3$ such that among an...
a45915bc778c3b967b94f7cee4faa46d    Find all integers $n \ge 3$ such that among an...
Name: Text, dtype: object

Solution:
Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$ $b_i \le n.$
Now three arbitrary sidelengths $x$ $y$ , and $z$ , with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).
We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$ , then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{13}$

Solution:
Outline:
1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \ge 3$
2. If the chosen $n$ is such that $F_n \le n^2$ , then choose the sequence $a_n$ such that $a_k = \sqrt{F_k}$ for $1 \le k \le n$ . It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \sqrt{F_a}, y = \sqrt{F_b}, z = \sqrt{F_c}$ with $a<b<c$ $x^2 + y^2 = F_a + F_b \le F_{b-1} + F_b = F_{b+1} \le F_c = z^2$ , x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \le n^2$ do not work.
3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \ge F_1a_1^2, a_3^2 \ge a_2^2 + a_1^2 \ge F_2a^2$ , and by induction $a_n^2 > F_na_1^2$ , a contradiction to the condition's inequality.)
4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$ . It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$ . Thus, $\boxed{13}$ is the desired solution set.


Answer: 30.0

problem_id
ae79010feec50f73241383732e6c476e    The mean of three numbers is $10$ more than th...
ae79010feec50f73241383732e6c476e    The mean of three numbers is $10$ more than th...
Name: Text, dtype: object

Solution:
Let $m$ be the mean of the three numbers. Then the least of the numbers is $m-10$ and the greatest is $m + 15$ . The middle of the three numbers is the median, $5$ . So $\dfrac{1}{3}[(m-10) + 5 + (m + 15)] = m$ , which can be solved to get $m=10$ .
Hence, the sum of the three numbers is $3\cdot 10 = \boxed{30}$

Solution:
Say the three numbers are $x$ $y$ and $z$ . When we arrange them in ascending order then we can assume $y$ is in the middle therefore $y = 5$
We can also assume that the smallest number is $x$ and the largest number of the three is $y$ .
Therefore,
\[\frac{x+y+z}{3} = x + 10 = z - 15\] \[\frac{x+5+z}{3} = x + 10 = z - 15\]
Taking up the first equation $\frac{x+5+z}{3} = x + 10$ and simplifying we obtain $z - 2x - 25 = 0$ doing so for the equation $\frac{x+5+z}{3} = z - 15$ we obtain the equation $x - 2z + 50 = 0$
\[x - 2z + 50 = 2x - 4z + 100\]
when solve the above obtained equation and $z - 2x - 25 = 0$ we obtain the values of $z = 25$ and $x = 0$
Therefore the sum of the three numbers is $25 + 5 + 0 = \boxed{30}$


Answer: 782.0

problem_id
afa106734f55c02711ecd5e8bbf4e8d3    A charity sells $140$ benefit tickets for a to...
afa106734f55c02711ecd5e8bbf4e8d3    A charity sells $140$ benefit tickets for a to...
Name: Text, dtype: object

Solution:
Let's multiply ticket costs by $2$ , then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.
Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.
Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single full price ticket costs $\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$ . Keeping in mind that $0\leq h\leq 140$ , we are looking for a divisor between $140$ and $280$ , inclusive.
The prime factorization of $4002$ is $4002=2\cdot 3\cdot 23\cdot 29$ . We can easily find out that the only divisor of $4002$ within the given range is $2\cdot 3\cdot 29 = 174$
This gives us $280-h=174$ , hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.
In our modified setting (with prices multiplied by $2$ ) the price of a half price ticket is $\frac{4002}{174} = 23$ . In the original setting this is the price of a full price ticket. Hence $23\cdot 34 = \boxed{782}$ dollars are raised by the full price tickets.

Solution:
Let the cost of the full price ticket be $x$ , the number of full-price tickets be $A$ , and the number of half-price tickets be $B$
Let's multiply both sides of the equation that naturally follows by 2. We have
\[2Ax+Bx=4002\]
And we have $A+B=140\implies B=140-A$
Plugging in, we get $\implies 2Ax+(140-A)(x)=4002$
Simplifying, we get $Ax+140x=4002$
Factoring out the $x$ , we get $x(A+140)=4002\implies x=\frac{4002}{A+140}$
We see that the fraction has to simplify to an integer (the full price is a whole dollar amount)
Thus, $A+140$ must be a factor of 4002.
Consider the prime factorization of $4002$ $2\times3\times23\times29$
$A$ must be a positive integer. So, we seek a factor of $4002$ to set equal to $A+140$ so that we get an integer solution for $A$ that is less than $140$ . By guess-and-check OR inspection, the appropriate factor is $174$ $2\times3\times29$ ), meaning that $A$ has a value of $34$ . Plug this into the above equation for $x$ to get $x = 23$
Therefore, the price of full tickets out of $2001$ is $23\times34=\boxed{782}$

Solution:
Let $f$ equal the number of full-price tickets, and let $h$ equal the number of half-price tickets. Additionally, suppose that the price of $f$ is $p$ . We are trying to solve for $f \cdot p$
Since the total number of tickets sold is $140$ , we know that \[f+h=140.\] The sales from full-price tickets ( $f \cdot p$ ) plus the sales from half-price tickets $\Big(\frac{h \cdot p}{2}$ , because each hall-price ticket costs $\frac{p}{2}$ dollars $\Big)$ equals $2001.$ Then we can write \[fx + \frac{hx}{2}=2001.\]
Substituting $h=140-f$ into the second equation, we get \[f \cdot p +\frac{(140-f)p}{2}=f \cdot p+\frac{140p-f\cdot p}{2}=\frac{f\cdot p+140p}{2}=2001.\]
Multiplying by $2$ and subtracting $140p$ gives us \[f\cdot p=4002-140p.\]
Since the problem states that $x$ is a whole number, $140p$ will be some integer multiple of $140$ that ends in a $0$ . Thus, $4002-140p$ will end in a $2$ . Looking at the answer choices, only $\boxed{782}$ satisfies that condition.


Answer: 30.0

problem_id
b36e53894fd42003fe068f1aec8a38d0    Two farmers agree that pigs are worth $300$ do...
b36e53894fd42003fe068f1aec8a38d0    Let us simplify this problem.  Dividing by $30...
Name: Text, dtype: object

Solution: The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\boxed{30}.$

Solution: Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly.  Since our supposition can be achieved, the answer is $\boxed{30}$


Answer: 219.95

problem_id
bdf1157d287c23856113e8dc3c9bdb32    A customer who intends to purchase an applianc...
bdf1157d287c23856113e8dc3c9bdb32    A customer who intends to purchase an applianc...
Name: Text, dtype: object

Solution:
Let the listed price be $x$ . Since all the answer choices are above $\textdollar100$ , we can assume $x > 100$ . Thus the discounts after the coupons are used will be as follows:
Coupon 1: $x\times10\%=.1x$
Coupon 2: $20$
Coupon 3: $18\%\times(x-100)=.18x-18$
For coupon $1$ to give a greater price reduction than the other coupons, we must have $.1x>20\implies x>200$ and $.1x>.18x-18\implies.08x<18\implies x<225$
The only choice that satisfies such conditions is $\boxed{219.95}$

Solution: For coupon $1$ to be the most effective, we want 10% of the price to be greater than 20. This clearly occurs if the value is over 200. For coupon 1 to be more effective than coupon 3, we want to minimize the value over 200, so $\boxed{219.95}$ is the smallest number over 200.


Answer: 56.0

problem_id
c1c2900151c908ac390988a490c7e35c    The plane is tiled by congruent squares and co...
c1c2900151c908ac390988a490c7e35c    The plane is tiled by congruent squares and co...
Name: Text, dtype: object

Solution:
Consider any single tile:

If the side of the small square is $a$ , then the area of the tile is $9a^2$ , with $4a^2$ covered by squares and $5a^2$ by pentagons.
Hence exactly $5/9$ of any tile are covered by pentagons, and therefore pentagons cover $5/9$ of the plane. When expressed as a percentage, this is $55.\overline{5}\%$ , and the closest integer to this value is $\boxed{56}$


Answer: 4.0

problem_id
c84575ab947ea92b2fa92f55386966ac    Also refer to the 2007 AMC 10B #25 (same problem)
c84575ab947ea92b2fa92f55386966ac    How many pairs of positive integers $(a,b)$ ar...
Name: Text, dtype: object

Solution:
Rewrite the equation\[ $\frac{a}{b}+\frac{14b}{9a}=k$ \]in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain\[ $\frac{14b^2}{9a}=bk-a.$ \]Because $a$ $b$ , and $k$ are integers, $14b^2$ must be a multiple of $a$ , and because $a$ and $b$ have no common factors greater than 1, it follows that 14 is divisible by $a$ . Next, multiply both sides of the original equation by $9a$ and subtract $14b$ to obtain\[ $\frac{9a^2}{b}=9ak-14b.$ \]This shows that $9a^2$ is a multiple of $b$ , so 9 must be divisible by $b$ . Thus if $(a,b)$ is a solution, then $b=1$ $3$ , or $9$ , and $a=1$ , 2, 7, or 14. This gives a total of twelve possible solutions $(a,b)$ , each of which can be checked quickly. $\[
\frac{a}{b}+\frac{14b}{9a}\]$ (Error compiling LaTeX. Unknown error_msg) will only be an integer when $(a,b) \in \{(1,3),(2,3), (7,3), (14,3)\},$ for a total of $\boxed{4}$ pairs.


Answer: 210.0

problem_id
d52fac580a189f0980006e4b2e9b254b    David drives from his home to the airport to c...
d52fac580a189f0980006e4b2e9b254b    David drives from his home to the airport to c...
Name: Text, dtype: object

Solution:
Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.
Let $d$ be the distance still needed to travel after $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$ , where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.
Simplifying gives $7d+525=10d$ , or $d=175$
Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{210}$ miles.

Solution:
Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.
Quickly checking, we know that neither choice $\textbf{(A)}$ or choice $\textbf{(B)}$ work, but $\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \text{ mph}$ , David has $175$ miles left. This then takes him $3.5$ hours at $50 \text{ mph}$ . But $210/35 = 6 \text{ hours}$ . Since $1+3.5 = 4.5 \text{ hours}$ is $1.5$ hours less than $6$ , our answer is $\boxed{210}$

Solution: Let the total distance be $d$ . Then $d=35(t+1)$ . Since 1 hour has passed, and he increased his speed by $15$ miles per hour, and he had already traveled $35$ miles, the new equation is $d-35=50(t-1-\frac{1}{2})=50(t-\frac{3}{2})$ . Solving, $d=35t+35=50t-40$ $15t=75$ $t=5$ , and $d=35(5+1)=35\cdot 6=210 \Longrightarrow \boxed{210}$


Answer: 6.0

problem_id
f25c446de036ffaadaf5676a0b0756b1    Let $k$ be a positive integer. Two players $A$...
f25c446de036ffaadaf5676a0b0756b1    Let $k$ be a positive integer. Two players $A$...
Name: Text, dtype: object

Solution: We claim that the minimum $k$ such that A cannot create a $k$ in a row is $\boxed{6}$


Answer: 41.0

problem_id
f530163b6184f697cd4b5054402e0ccf    Three dice with faces numbered $1$ through $6$...
f530163b6184f697cd4b5054402e0ccf    Three dice with faces numbered 1 through 6 are...
Name: Text, dtype: object

Solution:
The numbers on one die total $1+2+3+4+5+6 = 21$ , so the numbers
on the three dice total $63$ . Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$ .
This leaves $63 - 22 = \boxed{41}$ not seen.


Answer: 30.0

problem_id
ff8b260c6e48fc087b54f3971592eb5d    Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d    Let us simplify this problem.  Dividing by $30...
Name: Text, dtype: object

Solution: The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\boxed{30}.$

Solution: Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly.  Since our supposition can be achieved, the answer is $\boxed{30}$


Answer: 896.0

rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000.

Solution:
Denote the number of horizontal lines drawn as $x$ , and the number of vertical lines drawn as $y$ . The number of basic rectangles is $(x - 1)(y - 1)$ $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$ . Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$
FOIL this to get a quadratic, $-\frac 54x^2 + 502x - \frac{2003}4$ . Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201$ . However, this gives a non-integral answer for $y$ . The closest two values that work are $(199,253)$ and $(203,248)$
We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$ . The solution is $\boxed{896}$


Answer: 720.0

regular 12-gon is inscribed in a circle of radius 12.  The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$ $b^{}_{}$ $c^{}_{}$ , and $d^{}_{}$ are positive integers.  Find $a + b + c + d^{}_{}$

Solution:
1990 AIME-12.png
The easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles . Drawing the altitude of those triangles and then solving will yield the respective lengths.
Adding all of these up, we get $12[6(\sqrt{6} - \sqrt{2}) + 12 + 12\sqrt{2} + 12\sqrt{3} + 6(\sqrt{6}+\sqrt{2})] + 6 \cdot 24$
$= 12(12 + 12\sqrt{2} + 12\sqrt{3} + 12\sqrt{6}) + 144 = 288 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6}$ . Thus, the answer is $144 \cdot 5 = \boxed{720}$

Solution:
A second method involves drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon.  Since the distance from the center to a vertex of the 12-gon is $12$ , the Law of Cosines can be applied to this isosceles triangle, to give:
$a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta$
$a^2 = 2\cdot 12^2 - 2\cdot 12^2 \cos \theta$
$a^2 = 2\cdot 12^2 (1 -  \cos \theta)$
$a = 12\sqrt{2} \cdot \sqrt{1 - \cos \theta}$
There are six lengths of sides/diagonals, corresponding to $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$
Call these lengths $a_1, a_2, a_3, a_4, a_5, a_6$ from shortest to longest.  The total length $l$ that is asked for is
$l = 12(a_1 + a_2 + a_3 + a_4 + a_5) + 6a_6$ , noting that $a_6$ as written gives the diameter of the circle, which is the longest diagonal.
$l = 12[12\sqrt{2} (\sqrt{1 - \cos 30^{\circ}}+\sqrt{1-\cos 60^{\circ}}+\sqrt{1-\cos 90^{\circ}}+\sqrt{1-\cos 120^{\circ}}+\sqrt{1 - \cos 150^{\circ}})] + 6 \cdot 24$
$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{1}{2}} + \sqrt{1-0} + \sqrt{1 + \frac{1}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}\right) + 144$
$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{6}}{2} + \sqrt{1 + \frac{\sqrt{3}}{2}}\right) + 144$
To simplify the two nested radicals, add them, and call the sum $x$
$x = \sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}$
Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving:
$x^2 = 2 + 2\sqrt{\left(1 - \frac{\sqrt{3}}{2}\right)\left(1 + \frac{\sqrt{3}}{2}\right)}$
$x^2 = 2 + 2\sqrt{1 - \frac{3}{4}}$
$x^2 = 2 + 2\sqrt{\frac{1}{4}}$
$x^2 = 2 + 2\cdot\frac{1}{2}$
$x = \sqrt{3}.$
Plugging that sum $x$ back into the equation for $l$ , we find
$l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{6}}{2} + \sqrt{3}\right) + 144$
$l = 144 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6} + 144$
Thus, the desired quantity is $144\cdot 5 = \boxed{720}$

Solution: Begin as in solution 2, drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Apply law of cosines on $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$ to get $d^2 = 288 - 288 \cos \theta$ where $d$ is the diagonal or sidelength distance between two points on the 12-gon. Now, $d=\sqrt{288-288 \cos \theta}$ . Instead of factoring out $288$ as in solution 2, factor out $\sqrt{576}$ instead-the motivation for this is to make the expression look like the half angle identity, and the fact that $\sqrt{576}$ is an integer doesn't hurt. Now, we have that $d=24 \sin \frac{\theta}{2}$ , which simplifies things quite nicely. Continue as in solution 2, computing half-angle sines instead of nested radicals, to obtain $\boxed{720}$


Answer: 29.0

regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$

Solution:
If a point $z = r\text{cis}\,\theta$ is in $R$ , then the point $\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)$ is in $S$ (where cis denotes $\text{cis}\, \theta = \cos \theta + i \sin \theta$ ). Since $R$ is symmetric every $60^{\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \le \theta \le 30$ , and then to multiply by $6$ to account for the entire area.
We note that if the region $S_2 = \left\lbrace\frac{1}{z}|z \in R_2\right\rbrace$ , where $R_2$ is the region (in green below) outside the circle of radius $1/\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below).
The equation of the hexagon side in that region is $x = r \cos \theta = \frac{1}{2}$ , which is transformed to $\frac{1}{r} \cos (-\theta) = \frac{1}{r} \cos \theta =$ 2 . Let $r\text{cis}\,\theta = a+bi$ where $a,b \in \mathbb{R}$ ; then $r = \sqrt{a^2 + b^2}, \cos \theta = \frac{a}{\sqrt{a^2 + b^2}}$ , so the equation becomes $a^2 - 2a + b^2 = 0 \Longrightarrow (a-1)^2 + b^2 = 1$ . Hence the side is sent to an arc of the unit circle centered at $(1,0)$ , after considering the restriction that the side of the hexagon is a segment of length $1/\sqrt{3}$
Including $S_2$ , we find that $S$ is the union of six unit circles centered at $\text{cis}\, \frac{k\pi}{6}$ $k = 0,1,2,3,4,5$ , as shown below.
The area of the regular hexagon is $6 \cdot \left( \frac{\left(\sqrt{3}\right)^2 \sqrt{3}}{4} \right) = \frac{9}{2}\sqrt{3}$ . The total area of the six $120^{\circ}$ sectors is $6\left(\frac{1}{3}\pi - \frac{1}{2} \cdot \frac{1}{2} \cdot \sqrt{3}\right) = 2\pi - \frac{3}{2}\sqrt{3}$ . Their sum is $2\pi + \sqrt{27}$ , and $a+b = \boxed{029}$ .
- Th3Numb3rThr33

Solution:
We can describe the line parallel to the imaginary axis $x=\frac{1}{2}$ using polar coordinates as $r(\theta)=\dfrac{1}{2\cos{\theta}},$
which rearranges to $z=\left(\dfrac{1}{2\cos{\theta}}\right)(cis{\theta})\implies \frac{1}{z}=2\cos{\theta}cis(-\theta).$
Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral:
$[S] = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}r^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}\frac{1}{2}(2\cos\theta)^2 d\theta.$
Thankfully, this is a routine computation:
$[S] = 12\int_{0}^{\frac{\pi}{6}}2(\cos\theta)^2 d\theta = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta$
$[S] = 12\int_{0}^{\frac{\pi}{6}}(\cos{2\theta}+1)d\theta = 12\left[\frac{1}{2}\sin{2\theta}+\theta\right]_0^{\frac{\pi}{6}}=12\left(\frac{\sqrt{3}}{4}+\frac{\pi}{6}\right)=2\pi+3\sqrt{3}=2\pi + \sqrt{27}$
$a+b = \boxed{029}$


Answer: 810.0

regular icosahedron is a $20$ -faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane, and a lower pentagon of five vertices all adjacent to the bottom vertex and all in another horizontal plane. Find the number of paths from the top vertex to the bottom vertex such that each part of a path goes downward or horizontally along an edge of the icosahedron, and no vertex is repeated.
[asy] size(3cm); pair A=(0.05,0),B=(-.9,-0.6),C=(0,-0.45),D=(.9,-0.6),E=(.55,-0.85),F=(-0.55,-0.85),G=B-(0,1.1),H=F-(0,0.6),I=E-(0,0.6),J=D-(0,1.1),K=C-(0,1.4),L=C+K-A; draw(A--B--F--E--D--A--E--A--F--A^^B--G--F--K--G--L--J--K--E--J--D--J--L--K); draw(B--C--D--C--A--C--H--I--C--H--G^^H--L--I--J^^I--D^^H--B,dashed); dot(A^^B^^C^^D^^E^^F^^G^^H^^I^^J^^K^^L); [/asy]

Solution: Think about each plane independently. From the top vertex, we can go down to any of 5 different points in the second plane. From that point, there are 9 ways to go to another point within the second plane: rotate as many as 4 points clockwise or as many 4 counterclockwise, or stay in place.  Then, there are 2 paths from that point down to the third plane.  Within the third plane, there are 9 paths as well (consider the logic from the second plane).  Finally, there is only one way down to the bottom vertex.  Therefore there are $5 \cdot 9 \cdot 2 \cdot 9 \cdot 1=\boxed{810}$ paths.

Solution:
Assume an ant is on the top of this icosahedron. Note that the icosahedron has two pentagon planes and two points where the ant starts and ends. Also note that when the ant hits a vertex of the pentagon, there is only two ways to go down. When the ant ends up at the last vertex and is about to head down, there is only one way to go down.
Case $1$ : The ant move down to the first pentagon, then the next pentagon, and finally to the last point in a total of four moves; there are a total of $5 \cdot 2 \cdot 1 = 10$ ways to achieve this.
Case $2$ : The ant goes to the first pentagon and moves to another vertex in the same pentagon. The ant has a total of $8$ paths doing this since it may go through $1, 2, 3,$ or $4$ edges from either two directions when going across the edges of the pentagon (recall the ant may not repeat a move to another vertex) . Now the ant moves to the second pentagon below. The ant again has a total of $8$ moves if he wanders around the pentagon. Finally, the ant moves down after wandering to the last vertex. There is a total of $5 \cdot 8 \cdot 2 \cdot 8 = 640$ ways.
Case $3$ : Assume the ant goes to the first pentagon and wanders around. Then the ant goes the next pentagon and then heads directly to the last vertex. There are $5 \cdot 8 \cdot 2 = 80$ ways.
Case $4$ : Now let the ant go to the first pentagon and then go directly down to the next pentagon. The ant wanders around the second pentagon before heading to the last vertex. Like case $3$ above, there are $5 \cdot 2 \cdot 8 = 80$ ways.
Adding up all four cases, we get $10 + 640 + 80 + 80 =\boxed{810}$ total paths, as desired.      ~skyscraper

Solution: Go to 2020 AMC 10A #19 , and connect all of the centers of the faces on the dodecahedron to get the icosahedron. The answer is $\boxed{810}$


Answer: 1.0

regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$
[asy] // Diagram by TheMathGuyd import graph; // The Solid // To save processing time, do not use three (dimensions) // Project (roughly) to two size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05)); pair g = (-8,0); // Define Gap transform real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); // Make arrow // Time for the NET pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label("$7$",trf*DC); label("$Q$",trf*DC+DA-DB); label("$5$",trf*DC-DB); label("$3$",trf*DC-DA-DB); label("$6$",trf*CD); label("$4$",trf*CD-DA); label("$2$",trf*CD-DA-DB); label("$1$",trf*CD-2DA); [/asy]
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: We color face $6$ red and face $5$ yellow. Note that from the octahedron, face $5$ and face $?$ do not share anything in common. From the net, face $5$ shares at least one vertex with all other faces except face $1,$ which is shown in green:  Therefore, the answer is $\boxed{1}.$

Solution: We label the octohedron going triangle by triangle until we reach the $?$ triangle. The triangle to the left of the $Q$ should be labeled with a $6$ . Underneath triangle $6$ is triangle $5$ . The triangle to the right of triangle $5$ is triangle $4$ and further to the right is triangle $3$ . Finally, the side of triangle $3$ under triangle $Q$ is $2$ , so the triangle to the right of $Q$ is $\boxed{1}$

Solution: Notice that the triangles labeled $2, 3, 4,$ and $5$ make the bottom half of the octahedron, as shown below:  Therefore, $\textbf{(B)}, \textbf{(C)}, \textbf{(D)},$ and $\textbf{(E)}$ are clearly not the correct answer. Thus, the only choice left is $\boxed{1}$

Solution: The first half of the octahedron will need $4$ triangles connected to one another to form it. We can choose the triangles $4$ $5$ $6$ , and $7$ and form the half around the vertex they all share. That leaves triangles $1$ $3$ $2$ , and $Q$ to form the second half. Triangle $3$ will definitely share its sides with triangles $1$ and $2$ , leaving them to share their second side with triangle $Q$ . Since triangle $Q$ will certainly share its left side with triangle $2$ , the only triangle left to share its right side is triangle $\boxed{1}$


Answer: 625.0

right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125$ , and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}.$ Find the least distance that the fly could have crawled.

Solution:
The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$ -axis and the angle $\theta$ going counterclockwise. The circumference of the base is $C=1200\pi$ . The sector's radius (cone's sweep) is $R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800$ . Setting $\theta R=C\implies 800\theta=1200\pi\implies\theta=\frac{3\pi}{2}$
If the starting point $A$ is on the positive $x$ -axis at $(125,0)$ then we can take the end point $B$ on $\theta$ 's bisector at $\frac{3\pi}{4}$ radians along the $y=-x$ line in the second quadrant. Using the distance from the vertex puts $B$ at $(-375,-375)$ . Thus the shortest distance for the fly to travel is along segment $AB$ in the sector, which gives a distance $\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}$


Answer: 544.0

semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$

Solution: We can just look at a quarter circle inscribed in a $45-45-90$ right triangle. We can then extend a radius, $r$ to one of the sides creating an $r,r, r\sqrt{2}$ right triangle. This means that we have $r + r\sqrt{2} = 8\sqrt{2}$ so $r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}$ . Then the diameter is $32 - \sqrt{512}$ giving us $32 + 512 = \boxed{544}$

Solution: We proceed by finding the area of the square in 2 different ways. The square is obviously 8*8=64, but we can also find the area in terms of d. From the center of the circle, draw radii that hit the points where the square is tangent to the semicircle. Then the square's area is the area of the small square +2* the area of the trapezoids on the corners+ the area of an isoceles triangle. Adding these all up gives $64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{2})$ Simplifying gives $d-16\sqrt{2}+d\sqrt{2}=0$ . Solving gives $d=32-16\sqrt{2}=32-\sqrt{512}$ so the answer is $32+512= \boxed{544}$

Solution: It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$ , we can see $x=\sqrt{2}(8-x)$ and we get $2x=32-\sqrt{512}$ and we have the answer $\boxed{544}$ ~bluesoul


Answer: 834.0

sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.

Solution:
Define the sum as $s$ . Since $a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$ , the sum will be:
Thus $s = \frac{a_{28} + a_{30}}{2}$ , and $a_{28},\,a_{30}$ are both given; the last four digits of their sum is $3668$ , and half of that is $1834$ . Therefore, the answer is $\boxed{834}$ .−

Solution:
Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
$a_{1}\equiv 1 \pmod {1000} \\ a_{2}\equiv 1 \pmod {1000} \\ a_{3}\equiv 1 \pmod {1000} \\ a_{4}\equiv 3 \pmod {1000} \\ a_{5}\equiv 5 \pmod {1000} \\ \cdots \\ a_{25} \equiv 793 \pmod {1000} \\ a_{26} \equiv 281 \pmod {1000} \\ a_{27} \equiv 233 \pmod {1000} \\ a_{28} \equiv 307 \pmod {1000}$
Adding all the residues shows the sum is congruent to $\boxed{834}$ mod $1000$

Solution: All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$ , so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$ . From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\frac{1}{2}, 0, \frac{1}{2})$ , at least for the first few terms. From this, we have that $\sum_{k=1}^{28}{a_k} = \frac{a_{28}+a_{30}}{2} \equiv{\boxed{834}(\mod 1000)$


Answer: 224.0

sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$

Solution:
Define a function $f(n)$ on the non-negative integers, as \[f(n) = \frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n}}\] We want $\left\lfloor f(2006) \right\rfloor$
Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Dividing through by $a_{n}a_{n-1}$ , we get \[.\phantom{------------} \frac{a_{n+1}}{a_{n}} = \frac{a_{n}}{a_{n-1}} + \frac{2007}{a_{n}a_{n-1}} \phantom{------------} (1)\] and dividing through by $a_{n}a_{n+1}$ , we get \[.\phantom{------------} \frac{a_{n-1}}{a_{n}} = \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} \phantom{------------} (2)\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \[\frac{a_{n+1}}{a_{n}} + \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} = \frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}} + \frac{2007}{a_{n}a_{n-1}}\] i.e. \[f(n)+ \frac{2007}{a_{n}a_{n+1}} = f(n-1) + \frac{2007}{a_{n}a_{n-1}}\] Summing over $n=1$ to $n=2006$ , we notice that most of the terms on each side cancel against the corresponding term on the other side. We are left with \[f(2006) + \frac{2007}{a_{2006}a_{2007}} = f(0) + \frac{2007}{a_{1}a_{0}}\] We have $f(0) = 2$ , and $2007/a_0a_1 = 2007/9 = 223$ . So \[f(2006) = 2 + 223 - \frac{2007}{a_{2006}a_{2007}} = 224 + \left( 1 - \frac{2007}{a_{2006}a_{2007}}\right)\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$ , this means that the sequence $(a_n)$ is increasing. Since $a_3=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$ . It follows that $\left\lfloor f(2006) \right\rfloor = \boxed{224}$

Solution:
We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$ $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$
Add these two equations to get
This is an invariant . Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$ , the above equation means
$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$
We can thus calculate that $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$ . Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$ , notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$ . This means that
$b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225$ . It is only a tiny bit less because all the $b_i$ are greater than $1$ , so we conclude that the floor of $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}$ is $\boxed{224}$

Solution:
The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant \[\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.\] Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\alpha b_n+\beta b_{n-1}$ for $n\ge 2$ . We wish to find $\alpha$ and $\beta$ such that $a_n=b_n$ for all $n\ge 1$ . To do this, we use the following matrix form of a linear recurrence relation
\[\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
When we take determinants, this equation becomes
\[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
We want \[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=2007\] for all $n$ . Therefore, we replace the two matrices by $2007$ to find that
\[2007=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\cdot 2007\] \[1=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)=-\beta.\] Therefore, $\beta=-1$ . Computing that $a_3=672$ , and using the fact that $b_3=\alpha b_2-b_1$ , we conclude that $\alpha=225$ . Clearly, $a_1=b_1$ $a_2=b_2$ , and $a_3=b_3$ . We claim that $a_n=b_n$ for all $n\ge 1$ . We proceed by induction . If $a_k=b_k$ for all $k\le n$ , then clearly, \[b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.\] We also know by the definition of $b_{n+1}$ that
\[\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).\]
We know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$ . After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$ .  Thinking of this as a linear equation in the variable $b_{n+1}$ , we already know that this has the solution $b_{n+1}=a_{n+1}$ . Therefore, by induction, $a_n=b_n$ for all $n\ge 1$ . We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$
It's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\ge 3$ . Now
\[\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.\]
\[=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.\]
\[=225-\frac{2007}{a_{2005}a_{2006}}.\]
The sequence certainly grows fast enough such that $\frac{2007}{a_{2005}a_{2006}}<1$ . Therefore, the largest integer less than or equal to this value is $\boxed{224}$

Solution:
We will try to manipulate $\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\frac{a_1^2+a_2^2}{a_1a_2}$
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1}  = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$ $= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}$
We can keep on using this method to get that $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}$
This telescopes to $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}$
or $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}$
Finding the first few values, we notice that they increase rapidly, so $\frac{2007}{a_{2006}a_{2007}} < 1$ . Calculating the other values, $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}$
The greatest number that does not exceed this is $\boxed{224}$

Solution:
Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$ , and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$ , respectively. It will be clear why I decided to factor these expressions as I did momentarily.
Next, let's see what the expression $\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$ . For $n = 1$ , we get $\frac{1 + 224^2}{224}$ , the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}$ . It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$ , while the second-largest term in the denominator is smaller than $224^2$ . Thus, the floor of this expression will come out to be $224$ as well.
Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$ , the degrees of both the numerator and denominator increase by $2$ , because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2$ ). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\approx224:1$ ratio between the two.
For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$ , they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\geq2$ , whose $\lim_{k\to\infty}=$ $\boxed{224}$


Answer: 986.0

sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$ . What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492?

Solution:
The problem gives us a sequence defined by a recursion , so let's calculate a few values to get a feel for how it acts.  We aren't given initial values, so let $a_1 = a$ and $a_2 = b$ .  Then $a_3 = b - a$ $a_4 = (b - a) - b = -a$ $a_5 = -a - (b - a) = -b$ $a_6 = -b - (-a) = a - b$ $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$ .  Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat.  Thus, in particular $a_{j + 6} = a_j$ for all $j$ , and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$
Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \cdot 248$ is the largest multiple of 6 less than 1492, so \[\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_{6n + j}\] \[=(a + b + (b - a) + (-a)) + \sum_{n = 0}^{247}\sum_{j = 1}^6 a_j = 2b - a,\] where we can make this last step because $\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero
Similarly, since $1980 = 6 \cdot 330$ $\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$
Finally, $\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$
Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\cdot 493 = \boxed{986}$ .
Minor edit by: PlainOldNumberTheory


Answer: 173.0

sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$

Solution:
Let the sum of all of the terms in the sequence be $\mathbb{S}$ . Then for each integer $k$ $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \ldots, 100$
\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{align*}
Now, substituting for $x_{50}$ , we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$ , and the answer is $75+98=\boxed{173}$

Solution:
Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$
In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.$ Therefore, $x_{50}+50=99x_{50}-\dfrac{50}{2}$
Solving, we get $x_{50}=\dfrac{75}{98}.$ The answer is $75+98=\boxed{173}.$


Answer: 973.0

sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression , the second, third, and fourth terms are in arithmetic progression , and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than $1000$ . Find $n+a_n.$

Solution:
Let $x = a_2$ ; then solving for the next several terms, we find that $a_3 = x^2,\ a_4 = x(2x-1),\ a_5$ $= (2x-1)^2,\ a_6$ $= (2x-1)(3x-2)$ , and in general, $a_{2n} = f(n-1)f(n)$ $a_{2n+1} = f(n)^2$ , where $f(n) = nx - (n-1)$
From \[a_9 + a_{10} = f(4)^2 + f(4)f(5) = (4x-3)(9x-7) = 646 = 2\cdot 17 \cdot 19\] , we find that by either the quadratic formula or trial-and-error/modular arithmetic that $x=5$ . Thus $f(n) = 4n+1$ , and we need to find the largest $n$ such that either $f(n)^2\, \mathrm{or}\, f(n)f(n-1) < 1000$ . This happens with $f(7)f(8) = 29 \cdot 33 = 957$ , and this is the $2(8) = 16$ th term of the sequence.
The answer is $957 + 16 = \boxed{973}$

Solution: Let $x = a_2$ . It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$ , hence our answer is $957 + 16 = \boxed{973}$

Solution:
We can find the value of $a_{9}$ by its bounds using three conditions:
Rearranging conditions 1 and 2, we get:
\[\frac{646}{3} < a_{9} < \frac{646}{2}\]
trying all the terms from the third condition, it is clear that $a_9 = 289$ is the only solution.
Then we can calculate the next few terms from there since we have $a_{10}$ as well, to find that $a_{16} = 957$ and $a_{17} = 1089$ , thus we have our answer of $957 + 16 = \boxed{973}$


Answer: 1.0

sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution: Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{1}$

Solution:
Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ .  Thus, we get the following equations:
$9r=11+d\Rightarrow d=9r-11$
$9r^2=29+2d$
Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ .  If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ .  Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{1}$

Solution:
Let the three numbers be, in increasing order, $9,y,z$
Hence, we have that $9-y=y-z\implies 9+z=2y$
Also, from the second part of information given, we get that
$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2$
Plugging back in...
$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$
Simplifying, we get that $z^2-10z-551=0$
Applying the quadratic formula, we get that $z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}$
Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$
However, the problem asks for the minimum value of the third term in a geometric progression.
Hence, the answer is $-19+20=\boxed{1}$

Solution: Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1}$

Solution: The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\boxed{1}$

Solution: List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\boxed{1}$


Answer: 1.0

Solution: Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{1}$

Solution:
Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ .  Thus, we get the following equations:
$9r=11+d\Rightarrow d=9r-11$
$9r^2=29+2d$
Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ .  If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ .  Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{1}$

Solution:
Let the three numbers be, in increasing order, $9,y,z$
Hence, we have that $9-y=y-z\implies 9+z=2y$
Also, from the second part of information given, we get that
$\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2$
Plugging back in...
$9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)$
Simplifying, we get that $z^2-10z-551=0$
Applying the quadratic formula, we get that $z=\frac{10\pm \sqrt{2304}}{2}\implies \frac{10\pm48}{2}$
Obviously, in order to minimize the value of $z$ , we have to subtract. Hence, $z=-19$
However, the problem asks for the minimum value of the third term in a geometric progression.
Hence, the answer is $-19+20=\boxed{1}$

Solution: Let the arithmetic sequence be $9,9+x,9+2x$ and let the geometric sequence be $9,11+x,29+2x$ . Now, we just try all the solutions. If the last term is $1$ , then $x=-14$ . This gives the geometric sequence $9,-3,1$ which indeed works. The answer is $\boxed{1}$

Solution: The terms of the arithmetic progression are 9, $9+d$ , and $9+2d$ for some real number $d$ . The terms of the geometric progression are 9, $11+d$ , and $29+2d$ . Therefore $(11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0.$ Thus $d=10$ or $d=-14$ . The corresponding geometric progressions are $9, 21, 49$ and $9, -3, 1,$ so the smallest possible value for the third term of the geometric progression is $\boxed{1}$

Solution: List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is $\boxed{1}$


Answer: 512.0

solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum -shaped solid $F,$ in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k$ . Given that $k=\frac m n,$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$

Solution:
Our original solid has volume equal to $V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi$ and has surface area $A = \pi r^2 + \pi r \ell$ , where $\ell$ is the slant height of the cone.  Using the Pythagorean Theorem , we get $\ell = 5$ and $A = 24\pi$
Let $x$ denote the radius of the small cone. Let $A_c$ and $A_f$ denote the area of the painted surface on cone $C$ and frustum $F$ , respectively, and let $V_c$ and $V_f$ denote the volume of cone $C$ and frustum $F$ , respectively.  Because the plane cut is parallel to the base of our solid, $C$ is similar to the uncut solid and so the height and slant height of cone $C$ are $\frac{4}{3}x$ and $\frac{5}{3}x$ , respectively. Using the formula for lateral surface area of a cone, we find that $A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2$ . By subtracting $A_c$ from the surface area of the original solid, we find that $A_f=24\pi - \frac{5}{3}\pi x^2$
Next, we can calculate $V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3$ .  Finally, we subtract $V_c$ from the volume of the original cone to find that $V_f=12\pi - \frac{4}{9}\pi x^3$ . We know that $\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.$ Plugging in our values for $A_c$ $A_f$ $V_c$ , and $V_f$ , we obtain the equation $\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}$ .  We can take reciprocals of both sides to simplify this equation to $\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1$ and so $x = \frac{15}{8}$ .  Then $k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}= \frac{125}{387} = \frac mn$ so the answer is $m+n=125+387=\boxed{512}$

Solution:
Our original solid $V$ has surface area $A_v = \pi r^2 + \pi r \ell$ , where $\ell$ is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain $\ell = 5$ and lateral area $A_\ell = 15\pi$ . The area of the base is $A_B = 3^2\pi = 9\pi$
$V$ and $C$ are similar cones, because the plane that cut out $C$ was parallel to the base of $V$ . Let $x$ be the scale factor between the original cone and the small cone $C$ in one dimension. Because the scale factor is uniform in all dimensions, $x^2$ relates corresponding areas of $C$ and $V$ , and $x^3$ relates corresponding volumes. Then, the ratio of the painted areas $\frac{A_c}{A_f}$ is $\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8 - 5 x^2} = k$ and the ratio of the volumes $\frac{V_c}{V_f}$ is $\frac{x^3}{1 - x^3} = k$ . Since both ratios are equal to $k$ , they are equal to each other. Therefore, $\frac{5 x^2}{8 - 5 x^2} = \frac{x^3}{1 - x^3}$
Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives $5 x^2(1 - x^3) = x^3(8 - 5 x^2)$ . Dividing both sides by $x^2$ and distributing the $x$ on the right, we have $5 - 5 x^3 = 8 x - 5 x^3$ , and so $8 x = 5$ and $x = \frac{5}{8}$ . Substituting back into the easier ratio, we have $\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}$ . And so we have $m + n = 125 + 387 = \boxed{512}$


Answer: 8.0

sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$

Solution:
We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.
Let $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length $s\sqrt{3}$ where $s$ is a side of the cube, the ratio of a side of the inner square to a side of the outer square is $\frac{1}{\sqrt{3}}$ (since the side of the outer square = the diagonal of the inner square). So we have $\frac{S}{24} = \left(\frac{1}{\sqrt{3}}\right)^2$ . Thus $S = 8\Rightarrow \mathrm{\boxed{8}$

Solution: First of all, it is pretty easy to see the length of each edge of the bigger cube is $2$ so the radius of the sphere is $1$ . We know that when a cube is inscribed in a sphere, assuming the edge length of the square is $x$ the radius can be presented as $\frac {\sqrt3x}{2} = 1$ , we can solve that $x=\frac {2\sqrt3}{3}$ and now we only need to apply the basic formula to find the surface and we got our final answer as $\frac {2\sqrt3}{3}*\frac {2\sqrt3}{3}*6=8$ and the final answer is $\boxed{8}$


Answer: 8.0

Solution: Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives \[l^2 + l^2 + l^2 = 2^2 = 4.\] Hence each face has surface area \[l^2 = \frac{4}{3} \ \text{square meters}.\] So the surface area of the inscribed cube is $6\cdot \frac43 = \boxed{8}$ square meters.