Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$

Solution: Let $S_n$ be the number of possible seating arrangements with $n$ women. Consider $n \ge 3,$ and focus on the rightmost woman. If she returns back to her seat, then there are $S_{n-1}$ ways to seat the remaining $n-1$ women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us $S_{n-2}$ ways to seat the other $n-2$ women, so we obtain the recursion \[S_n = S_{n-1}+S_{n-2}.\]
Starting with $S_1=1$ and $S_2=2,$ we can calculate $S_{10}=\boxed{89}.$

Answer:  89
Problem: Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$

Solution: The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$
In our setting where $A=(0,0)$ $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$
In other words, we need to count the triples of integers $(k,s,t)$ where $k>1$ $s,t>0$ and $(k-1)st = 1,\!000,\!000 = 2^6 5^6$
These can be counted as follows: We have $6$ identical red balls (representing powers of $2$ ), $6$ blue balls (representing powers of $5$ ), and three labeled urns (representing the factors $k-1$ $s$ , and $t$ ). The red balls can be distributed in ${8\choose 2} = 28$ ways, and for each of these ways, the blue balls can then also be distributed in $28$ ways. (See Distinguishability for a more detailed explanation.)
Thus there are exactly $28^2 = 784$ ways how to break $1,\!000,\!000$ into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is $784 \longrightarrow \boxed{784}$

Answer: C 784
Problem: Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$

Solution: We know that $A$ is $(0, 0)$ . Since $B$ is on the line $y=x$ , let it be represented by the point $(b, b)$ . Similarly, let $D$ be $(d, kd)$ . Since this is a parallelogram, sides $\overline{AD}$ and $\overline{BC}$ are parallel. Therefore, the distance and relative position of $D$ to $A$ is equivalent to that of $C$ to $B$ (if we take the translation of $A$ to $D$ and apply it to $B$ , we will get the coordinates of $C$ ). This yields $C (b+d, b+kd)$ . Using the Shoelace Theorem we get
$1,000,000 = \frac{1}{2}|\left(b(b+kd) + (b+d)(kd)\right) - \left(b(b+d) + (b+kd)d\right)|$
$\Rightarrow 2,000,000 = |2kbd - 2bd|$
$\Rightarrow 1,000,000 = |kbd - bd|$
Since $k > 1, kbd > bd$ . The equation becomes
$1,000,000 = (k-1)bd$
$\Rightarrow \frac{1,000,000}{k-1} = bd$
Since $k$ must be a positive integer greater than $1$ , we know $k-1$ will be a positive integer. We also know that $bd$ is an integer, so $k-1$ must be a factor of $1,000,000$ . Therefore $bd$ will also be a factor of $1,000,000$
Notice that $1,000,000 = 10^6 = 2^6*5^6$
Let $b$ be $2^x5^y$ such that $x, y$ are integers on the interval $[0, 6]$
Let $d$ be $2^w5^z$ such that $w, z$ are integers, $x+w \le 6$ , and $y+z \le 6$
For a pair $(x, y)$ , there are $7-x$ possibilities for $w$ and $7-y$ possibilites for $z$ $d$ doesn't have to be the co-factor of $1,000,000$ , it just can't be big enough such that $bd > 1,000,000$ ), for a total of $(7-x)(7-y)$ possibilities. So we want
$\sum_{k=0, i=0}^6 (7-k)(7-i)$
$\left(\text{the sum of the number of possible pairs}(w, z) \text{ for all pairs}(k , i)\text{ for } k[0, 6]\text{ and } i[0, 6]\right)$
Notice that if we "fix" the value of $k$ , at, say $6$ , then run through all of the values of $i$ , change the value of $k$ to $5$ , and run through all of the values of $i$ again, and so on until we exhaust all $49$ combinations of $(k, i)$ we get something like this:
$1*1 + 1*2 + ... + 1*6 + 1*7 + 2*1 + 2*2 + ... + 2*6 + 2*7 + ..... + 7*1 + 7*2 + ... + 7*6 + 7*7$
which can be rewritten
$1(1+2+...+7)+2(1+2+...+7)+.....+7(1+2+...+7)$
$\Rightarrow (1+2+...+7)(1+2+...+7)$
$\Rightarrow 28^2$
$\Rightarrow 784$
So there are $784$ possible sets of coordinates $B,$ $C$ , and $D \Rightarrow \boxed{784}$

Answer: C 784
Problem: A region $S$ in the complex plane is defined by \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] A complex number $z = x + iy$ is chosen uniformly at random from $S$ . What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78$

Solution: We multiply $z$ and $(\frac{3}{4}+\frac{3}{4}i)$ to get \[(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).\] Since we want to find the probability that this number is in $S$ , we need the real and complex coefficients of this number to be less than or equal to $1$ or greater than or equal to $-1.$ This gives us the equations \[-1\le \frac{3}{4}x-\frac{3}{4}y \le 1\] and \[-1\le \frac{3}{4}x+\frac{3}{4}y\le 1.\] Now, we see that we can solve this by graphing. We can graph our barriers $-1\le x\le 1$ and $-1\le y\le 1$ to form a $2$ by $2$ square centered at the origin. Graphing our two equations gives us the four lines \[x-y=\frac{4}{3},\] \[x-y=-\frac{4}{3},\] \[x+y=\frac{4}{3},\] \[x+y=-\frac{4}{3}.\] The square that is formed is the region that satisfies these four equations. Now, the barriers and this square gives us an octagon as the desired region. The area of this octagon is the total area of the square minus the 4 small triangles on each corner, each with $\frac{2}{9}$ area. Therefore, the octagon has area of $\frac{28}{9}.$ Finally, to find the probability of it working, we find the area of the octagon divided by the area of the entire square which is $\frac{\frac{28}{9}}{4}=\frac{7}{9}$ or $\boxed{79}.$

Answer: D 79
Problem: For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$

Solution: First of all, we have to agree on the range of $\sin^{-1}$ and $\cos^{-1}$ . This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: $\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ and $\forall x: 0\leq \cos^{-1}(x) \leq \pi$
Hence we get that $\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x$ , thus our equation simplifies to $\sin^{ - 1}(\sin 6x) = x$
Consider the function $f(x) = \sin^{ - 1}(\sin 6x)  - x$ . We are looking for roots of $f$ on $[0,\pi]$
By analyzing properties of $\sin$ and $\sin^{-1}$ (or by computing the derivative of $f$ ) one can discover the following properties of $f$
For $x=\pi/6$ we have $f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$ . Hence $f$ has exactly one root on $(0,\pi/6)$
For $x=2\pi/6$ we have $f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$ . Hence $f$ is negative on the entire interval $[\pi/6,2\pi/6]$
Now note that $\forall t: \sin^{-1}(t) \leq \pi/2$ . Hence for $x > 3\pi/6$ we have $f(x) < 0$ , and we can easily check that $f(3\pi/6)<0$ as well.
Thus the only unknown part of $f$ is the interval $(2\pi/6,3\pi/6)$ . On this interval, $f$ is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.
To prove that there are two roots, it is enough to find any $x$ from this interval such that $f(x)>0$
A good guess is its midpoint, $x=5\pi/12$ , where the function $\sin^{-1}(\sin 6x)$ has its local maximum. We can evaluate: $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$
Summary: The function $f$ has $\boxed{4}$ roots on $[0,\pi]$ : the first one is $0$ , the second one is in $(0,\pi/6)$ , and the last two are in $(2\pi/6,3\pi/6)$

Answer: B 4
Problem: For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$

Solution: Since $0\leq \cos^{-1}(a) \leq \pi$ for all $a$ , the equation reduces to $\sin^{-1}(\sin(6x)) = x$ . Since $-\pi/2 \leq \sin^{-1}(a) \leq \pi/2$ for all $a$ $0 \leq x \leq \pi/2$ . To make the problem easier, we will measure angles in degrees. We will consider each sixth of the interval $[0, 90]$
For $0 \leq x \leq 15$ $6x$ is in the first quadrant. Thus, $\sin^{-1}(\sin(6x)) = 6x$ . Setting this equal to $x$ yields the solution $x = 0$
For $15 \leq x \leq 30$ $6x$ is in the second quadrant. Thus, $\sin^{-1}(\sin(6x)) = 180 - 6x$ . This yields the solution $x = \frac{180}7$
For $30 \leq x \leq 45$ $6x$ is in the third quadrant. Thus, $\sin^{-1}(\sin(6x)) = 180 - 6x$ . As $\frac{180}{7}$ is not on the interval $30 \leq x \leq 45$ , this yields no solution.
For $45 \leq x \leq 60$ $6x$ is in the fourth quadrant. Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$ . As $72$ is not on the interval $45 \leq x \leq 60$ , this yields no solution.
For $60 \leq x \leq 75$ $6x$ is in the first quadrant plus a full revolution. Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$ . This yields the solution $x = 72$
For $75 \leq x \leq 90$ $6x$ is in the second quadrant plus a full revolution. Thus $\sin^{-1}(\sin(6x)) = 540 - 6x$ . This yields the solution $x = \frac{540}7$
There are $\boxed{4}$ solutions, $x=0$ $x=\pi/7$ $x=2\pi/5$ , and $x=3\pi/7$

Answer: B 4
Problem: For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$

Solution: Algebraically, the inverse function of a function should just cancel out, leaving $6x=x$ . However, upon inspection we find that the graphs of these "inverse function of the function" equations are also periodic, like their normal trig function counterparts, due to the fact that inverse trig functions will never return any angle value higher than $2\pi$ . But instead of a smooth wave, these graphs are made up of zigzags with slope $1$ and $-1$ . Trying a few values, we see that
$y = \arcsin{\sin{x}}$
peaks at $\frac{\pi}{2}$ and
$y = \arccos{\cos{x}}$
peaks at $\pi$
But we want the graph of $y = \arcsin{(\sin6{x})}$ , which has a period of $\frac{\pi}{3}$ instead of $2\pi$ . So this means the interval $[0, \pi]$ will show $3$ periods instead of $\frac{1}{2}$ of a period. Visually it would be lines with slopes $6$ and $-6$ . Using the graph paper given to us, we plot out the two equations according to the above and we see that they intersect $4$ times $\Rightarrow \boxed{4}$

Answer: B 4
Problem: For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$

Solution: The most conventional domain and range for $\cos^{-1} x$ is $x \in [-1, 1]$ and $\cos^{-1} x \in [0, \pi]$ . For $\cos x$ $x \in [0, \pi]$ and $\cos x \in [-1, 1]$ . Since the domain of $\cos x$ is equal to the range of $\cos^{-1} x$ , and the range of $\cos x$ is equal to the domain of $\cos^{-1} x$ $\cos^{-1} (\cos x) = x$
Meaning that $\sin^{ - 1}(\sin 6x) = x$ $\sin x = \sin (6x)$
By graphing $\sin x$ and $\sin 6x$ together, it can be seen that there are 7 intersections for $x \in [0, \pi]$
However, the range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . Therefore $x \in [0, \frac{\pi}{2}]$ for $x = \sin^{ - 1}(\sin 6x)$ . There are $\boxed{4}]$

Answer: B 4
Problem: Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?
[asy] unitsize(.8cm);  pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label("2",(-0.5,0.5)); draw(unitsquare,p); label("32",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label("16",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label("4",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label("8",(0.5,-0.5)); [/asy]
$\mathrm{(A)}\ 154\qquad\mathrm{(B)}\ 159\qquad\mathrm{(C)}\ 164\qquad\mathrm{(D)}\ 167\qquad\mathrm{(E)}\ 189$

Solution: Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$ , 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexposed face. $2(51)+32+16+8+4+2=164 \implies \boxed{164}$

Answer: C 164
Problem: The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$

Solution: Let $A = \log(a)$ and $B = \log(b)$
The first three terms of the arithmetic sequence are $3A + 7B$ $5A + 12B$ , and $8A + 15B$ , and the $12^\text{th}$ term is $nB$
Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$
Since the first three terms in the sequence are $13B$ $22B$ , and $31B$ , the $k$ th term is $(9k + 4)B$
Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{112}$

Answer: D 112
Problem: The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$

Solution: If $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are in arithmetic progression , then $a^3b^7$ $a^5b^{12}$ , and $a^8b^{15}$ are in geometric progression . Therefore,
\[a^2b^5=a^3b^3 \Rightarrow a=b^2\]
Therefore, $a^3b^7=b^{13}$ $a^5b^{12}=b^{22}$ , therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} \Rightarrow \boxed{112}$

Answer: D 112
Problem: The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$

Solution: Given the first three terms form an arithmetic progression, we have: \[a = \log(a^3b^7)\] \[a+d = \log(a^5b^{12})\] \[a+2d = \log(a^8b^{15}).\] Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for $d$ \[d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5\] \[d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.\] The desired $12$ th term in the sequence is $a+11d$ , so we can substitute our values for $a$ and $d$ (using either one of our two expressions for $d$ ): \[a+11d = \log a^3b^7 + 11\log(a^2b^5)\] \[= \log a^3b^7 + \log(a^{22}b^{55})\] \[= \log a^{25}b^{62}.\] The answer must be expressed as $\log(b^n)$ , however. We're in luck: the two different yet equal expressions for $d$ allow us to express $a$ and $b$ in terms of each other: \[\log a^2b^5 = \log a^3b^3\] \[a^2b^5 = a^3b^3\] \[a=b^2.\] Plugging in $a=b^2$ , we have: \[a+11d = \log b^{50}b^{62}\] \[= \log b^{112} \Rightarrow \boxed{112}.\] ~ Jingwei325 $\smiley$

Answer: D 112
Problem: Triangle $ABC$ , with sides of length $5$ $6$ , and $7$ , has one vertex on the positive $x$ -axis, one on the positive $y$ -axis, and one on the positive $z$ -axis. Let $O$ be the origin . What is the volume of tetrahedron $OABC$
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}$

Solution: Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7.  Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively.  Then by the Pythagorean Theorem \begin{align*} a^2+b^2 &=5^2 , \\  b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*} so $a^2 = (5^2+7^2-6^2)/2 = 19$ ; similarly, $b^2 = 6$ and $c^2 = 30$ . Since $OA$ $OB$ , and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice $\boxed{95}$

Answer: C 95
Problem: In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$

Solution: Let $A = \left(1 + x + x^2 + \cdots + x^{14}\right)$ and $B = \left(1 + x + x^2 + \cdots + x^{27}\right)$ . We are expanding $A \cdot A \cdot B$
Since there are $15$ terms in $A$ , there are $15^2 = 225$ ways to choose one term from each $A$ . The product of the selected terms is $x^n$ for some integer $n$ between $0$ and $28$ inclusive. For each $n \neq 0$ , there is one and only one $x^{28 - n}$ in $B$ . For example, if I choose $x^2$ from $A$ , then there is exactly one power of $x$ in $B$ that I can choose; in this case, it would be $x^{24}$ . Since there is only one way to choose one term from each $A$ to get a product of $x^0$ , there are $225 - 1 = 224$ ways to choose one term from each $A$ and one term from $B$ to get a product of $x^{28}$ . Thus the coefficient of the $x^{28}$ term is $224 \Rightarrow \boxed{224}$

Answer: C 224
Problem: In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$

Solution: Let $P(x) = \left(1 + x + x^2 + \cdots + x^{14}\right)^2 = a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28}$ . Then the $x^{28}$ term from the product in question $\left(1 + x + x^2 + \cdots + x^{27}\right)(a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28})$ is
$1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + \cdots + x^{27}a_1x = \left(a_1 + a_2 + \cdots a_{28}\right)x^{28}$
So we are trying to find the sum of the coefficients of $P(x)$ minus $a_0$ . Since the constant term $a_0$ in $P(x)$ (when expanded) is $1$ , and the sum of the coefficients of $P(x)$ is $P(1)$ , we find the answer to be $P(1) - a_0 = \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1 = 15^2 - 1 = 224 \Rightarrow \boxed{224}$

Answer: C 224
Problem: In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$

Solution: We expand $(1 + x + x^2 + x^3 + \cdots + x^{14})^2$ to $(1 + x + x^2 + x^3 + \cdots + x^{14}) * (1 + x + x^2 + x^3 + \cdots + x^{14})$ and use FOIL to multiply. It expands out to:
$1 + x + x^2 + x^3 + x^4 + \cdots + x^{14} +$
$\qquad x + x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} +$
$\qquad \qquad x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} + x^{16} + \cdots$
It becomes apparent that
$(1 + x + x^2 + x^3 + \cdots + x^{14})^2 = 1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}$
Now we have to find the coefficient of $x^{28}$ in the product:
$(1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}) \cdot (1 + x + x^2 + x^3 + \cdots + x^{27})$
We quickly see that the we get $x^{28}$ terms from $x^{27} \cdot 2x$ $x^{26} \cdot 3x^2$ $x^{25} \cdot 4x^3$ , ... $15x^{14} \cdot x^{14}$ , ... $x^{28} \cdot 1$ . The coefficient of $x^{28}$ is just the sum of the coefficients of all these terms. $1 + 2 + 3 + 4 + \cdots + 15 + 14 + 13 + \cdots + 4 + 3 + 2 = 224$ , so the answer is $\boxed{224}$

Answer: C 224
Problem: In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$

Solution: Rewrite the product as $\frac{(x^{28} - 1)(x^{15} - 1)(x^{15} - 1)}{(x - 1)^3}$ . It is known that
\[\frac{1}{(1 - x)^n} = \binom{n - 1}{n - 1} + \binom{n}{n - 1}x + \binom{n + 1}{n - 1}x^2 + \binom{n + 2}{n - 1}x^3 + \cdots + \binom{n - 1 + k}{n - 1}x^k + \cdots .\]
Thus, our product becomes
\[-\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( \binom{2}{2} + \binom{3}{2}x + \binom{4}{2}x^2 + \cdots \right).\]
\[= -\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( 1 + 3x + 6x^2 + \cdots \right).\]
We determine the $x^{28}$ coefficient by doing casework on the first three terms in our product. We can obtain an $x^{28}$ term by choosing $x^{28}$ in the first term, $-1$ in the second and third terms, and $1$ in the fourth term. We can get two $x^{28}$ terms by choosing $x^{15}$ in either the second or third term, $-1$ in the first term, $-1$ in the second or third term from which $x^{15}$ has not been chosen, and the $\binom{15}{2}x^{13}$ in the fourth term. We get $\binom{15}{2} * 2 = 210$ $x^{28}$ terms this way. (We multiply by $2$ because the $x^{15}$ term could have been chosen from the second term or the third term). Lastly, we can get an $x^{28}$ term by choosing $-1$ in the first three terms and a $\binom{30}{2}x^{28}$ from the fourth term. We have a total of $1 + 210 - 435 = -224$ for the $x^{28}$ coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of $224 \Rightarrow \boxed{224}$

Answer: C 224