Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the minimum value of $f(x)=\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$
$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$

Solution: By the triangle inequality, $|x-1|+|2x-1|+|3x-1|+\cdots + |119x-1| \geq |(x-1)+(2x-1)+\cdots+(119x)-1|.$ However, we may change signs of some of these terms to cancel out the $x$ 's.
Since the minimum exists, we want all the $x$ s to cancel out. Thus, we want to find some $n$ such that \[1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119\] \[\frac{n(n+1)}{2}=\frac{119\cdot120}{2}-\frac{n(n+1)}{2}\] \[n^2+n-7140=0\] \[n=84\]
Then, $x=\frac{1}{n}= \frac{1}{84}$ . The answer(expression's value) is then $84*1+(119-85+1)*(-1)$ , which becomes $84-35=\boxed{49}$

Answer:  49
Problem: Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$

Solution: The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to $0$
First, let us find the number of zeros of the inside of the logarithm.
\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ \sin(\pi x) &= 0\\ x &= 0, 1\\ \sin(2 \pi x) &= 0\\ x &= 0, \frac{1}{2}, 1\\ \sin(3 \pi x) &= 0\\ x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ \sin(4 \pi x) &= 0\\ x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ &\cdots\end{align*}
After counting up the number of zeros for each factor and eliminating the excess cases we get $23$ zeros and $22$ intervals.
In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.
The first interval $\left(0, \frac{1}{8}\right)$ is obviously positive. This means the next interval $\left(\frac{1}{8}, \frac{1}{7}\right)$ is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals from $0$ to $\frac{1}{2}$ . Since the function is symmetric, we know that there are also $5$ negative intervals from $\frac{1}{2}$ to $1$
And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12}$

Answer: B 12
Problem: Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$ . The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$

Solution: Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Let $x_1, x_2, x_3, ..., x_8$ temporarily be the dependent variables of the functions $y_1 = \sin(\pi x_1), y_2 = \sin(2\pi x_2), ..., y_8 = \sin(8\pi x_8)$ . It is easy to see that for $y_i$ to be positive for $1\leq i\leq8$ $\lfloor i x_i \rfloor$ must be even for $1 \leq i\leq8$ . Since an even number of positives times an even number of negatives equals a positive, there can be $2, 4, 6,$ or $8$ positive values of $y_i$ for $1 \leq i\leq 8$ for a given value of $x$ . (since $y_1$ is always positive on the range $[0, 1]$ )
Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from $0$ to $1$ . (I recommend increments of at most $0.1$ .) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for $y_1, y_2, ..., y_8$ , respectively. Draw a large bar (label it with $+$ so you know it's positive) for all values of $x_i$ such that $\lfloor i x_i \rfloor$ is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at $0$ , and slowly work your way to $1$ , marking all disjoint intervals in which your straightedge touches $2, 4, 6,$ or $8$ boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. )
If done correctly, you should obtain $\boxed{12}$ as your answer.

Answer: B 12
Problem: Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$

Solution: As with solution $1$ we would  like to note that given any quadrilateral we can change its angles to make a cyclic one.
Let $a \ge b \ge c\ge d$ be the sides of the quadrilateral.
There are $\binom{31}{3}$ ways to partition $32$ . However, some of these will not be quadrilaterals since they would have one side bigger than the sum of the other three.  This occurs when $a \ge 16$ .  For $a=16$ $b+c+d=16$ . There are $\binom{15}{2}$ ways to partition $16$ . Since $a$ could be any of the four sides, we have counted $4\binom{15}{2}$ degenerate quadrilaterals. Similarly, there are $4\binom{14}{2}$ $4\binom{13}{2} \cdots 4\binom{2}{2}$ for other values of $a$ .  Thus, there are $\binom{31}{3} - 4\left(\binom{15}{2}+\binom{14}{2}+\cdots+\binom{2}{2}\right) = \binom{31}{3} - 4\binom{16}{3} = 2255$ non-degenerate partitions of $32$ by the hockey stick theorem.  We then account for symmetry. If all sides are congruent (meaning the quadrilateral is a square), the quadrilateral will be counted once. If the quadrilateral is a rectangle (and not a square), it will be counted twice. In all other cases, it will be counted 4 times. Since there is $1$ square case, and $7$ rectangle cases, there are $2255-1-2\cdot7=2240$ quadrilaterals counted 4 times. Thus there are $1+7+\frac{2240}{4} = \boxed{568}$ total quadrilaterals.

Answer: C 568
Problem: Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?
$\textbf{(A)}\ 560 \qquad \textbf{(B)}\ 564 \qquad \textbf{(C)}\ 568 \qquad \textbf{(D)}\ 1498 \qquad \textbf{(E)}\ 2255$

Solution: As with solution $1$ we find that there are $2255$ ways to form a quadrilateral if we don't account for rotations. We now apply Burnside's_Lemma . There are four types of actions in the group acting on the set of quadrilaterals. We will consider each individually:
Identity: maps a quadrilateral with sides $a,b,c,d$ in that order to $a,b,c,d$ . Obviously all members of the set of quadrilaterals are fixed points, for a total of $2255$ .
Rotation by one: maps a quadrilateral from $a,b,c,d$ to $b,c,d,a$ . For this to have a fixed point we need $a=b,b=c,c=d,d=a$ , so the only quadrilateral that is a fixed point is the square with side length $8$ , for a total of $1$ .
Rotation by two: maps a quadrilateral from $a,b,c,d$ to $c,d,a,b$ . For this to be a fixed point we need $a=c$ and $b=d$ . Thus the quadrilateral is of the form $x,y,x,y$ —a rectangle. We can count that there are $15$ rectangle cases, namely $(a, b, c, d) = \{ (1, 15, 1, 15), (2, 14, 2, 14), \cdots, (15, 1, 15, 1)  \}$ .
Roration by three: maps a quadrilateral from $a,b,c,d$ to $d,a,b,c$ . Similarly to the rotation by one case, there is one fixed point here.
Summing up, we get that the total number of groups is $2255+1+15+1=2272$ . Since there are $4$ members of the group our final answer is $\frac{2272}{4}=\boxed{568}$

Answer: C 568
Problem: Let $n$ be the smallest positive integer such that $n$ is divisible by $20$ $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution: We know that $n^2 = k^3$ and $n^3 = m^2$ . Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$ . Let $a = n^6$ . Now we know $a$ must be a perfect $36$ -th power because $lcm(9,4) = 36$ , which means that $n$ must be a perfect $6$ -th power. The smallest number whose sixth power is a multiple of $20$ is $10$ , because the only prime factors of $20$ are $2$ and $5$ , and $10 = 2 \times 5$ . Therefore our is equal to number $10^6 = 1000000$ , with $7$ digits $\Rightarrow \boxed{7}$

Answer: E 7
Problem: For what value of $x$ does
\[\log_{\sqrt{2}}\sqrt{x}+\log_{2}{x}+\log_{4}{x^2}+\log_{8}{x^3}+\log_{16}{x^4}=40?\]
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 256 \qquad \textbf{(E)}\ 1024$

Solution: Using the fact that $\log_{x^n}{y^n} = \log_{x}{y}$ , we see that the equation becomes $\log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} + \log_{2}{x} = 40$ . Thus, $5\log_{2}{x} = 40$ and $\log_{2}{x} = 8$ , so $x = 2^8 = 256$ , or $\boxed{256}$

Answer: D 256
Problem: Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$

Solution: We want to try make $a+b$ $b+c$ $c+d$ , and $d+e$ as close as possible so that $M$ , the maximum of these, is smallest.
Notice that $2010=670+670+670$ . In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): $2010=670+1+670+1+668$ or $2010=670+1+669+1+669$ . We see that in both cases, the value of $M$ is $671$ , so the answer is $671 \Rightarrow \boxed{671}$

Answer: B 671
Problem: Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$

Solution: Since $a + b \le M$ $d + e \le M$ , and $c < b + c \le M$ , we have that $2010 = a + b + c + d + e < 3M$ . Hence, $M > 670$ , or $M \ge 671$
For the values $(a,b,c,d,e) = (669,1,670,1,669)$ $M = 671$ , so the smallest possible value of $M$ is $\boxed{671}$ . The answer is (B).

Answer:  671
Problem: Let $a$ $b$ $c$ $d$ , and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$ $b+c$ $c+d$ and $d+e$ . What is the smallest possible value of $M$
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$

Solution: Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.)
Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do
1. (high, low, high, low, high)
or
2. (low, high, low, high, low)
In the extreme case that each "low" = $0$ $2010$ will be divided into either $3$ or $2$ numbers for cases 1 and 2, respectively. Obviously dividing by $3$ will yield a lower number, so we consider case 1.
Dividing $2010$ by $3$ yields $670$ , or
$670, 0, 670, 0, 670$
However, all five numbers must be positive, and the closest we can get to this is
$668, 3, 668, 3, 668$
The lowest possible sum of two adjacent numbers then becomes $671$ , or $\boxed{671}$

Answer: B 671
Problem: For how many ordered triples $(x,y,z)$ of nonnegative integers less than $20$ are there exactly two distinct elements in the set $\{i^x, (1+i)^y, z\}$ , where $i=\sqrt{-1}$
$\textbf{(A)}\ 149 \qquad \textbf{(B)}\ 205 \qquad \textbf{(C)}\ 215 \qquad \textbf{(D)}\ 225 \qquad \textbf{(E)}\ 235$

Solution: We have either $i^{x}=(1+i)^{y}\neq z$ $i^{x}=z\neq(1+i)^{y}$ , or $(1+i)^{y}=z\neq i^x$
$i^{x}=(1+i)^{y}$ only occurs when it is $1$ $(1+i)^{y}=1$ has only one solution, namely, $y=0$ $i^{x}=1$ has five solutions between 0 and 19, $x=0, x=4, x=8, x=12$ , and $x=16$ $z\neq 1$ has nineteen integer solutions between zero and nineteen. So for $i^{x}=(1+i)^{y}\neq z$ , we have $5\cdot 1\cdot 19=95$ ordered triples.
For $i^{x}=z\neq(1+i)^{y}$ , again this only occurs at $1$ $(1+i)^{y}\neq 1$ has nineteen solutions, $i^{x}=1$ has five solutions, and $z=1$ has one solution, so again we have $5\cdot 1\cdot 19=95$ ordered triples.
For $(1+i)^{y}=z\neq i^x$ , this occurs at $1$ and $16$ $(1+i)^{y}=1$ and $z=1$ both have one solution while $i^{x}\neq 1$ has fifteen solutions. $(1+i)^{y}=16$ and $z=16$ both have one solution, namely, $y=8$ and $z=16$ , while $i^{x}\neq 16$ has twenty solutions ( $i^x$ only cycles as $1, i, -1, -i$ ). So we have $15\cdot 1\cdot 1+20\cdot 1\cdot 1=35$ ordered triples.
In total we have ${95+95+35=\boxed{225}$ ordered triples

Answer: D 225
Problem: A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solution: By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$ . Since $\tan x=\frac{\sin x}{\cos x}$ , we can rewrite this as $\cos^3x=\sin^2x$
The common ratio of the sequence is $\frac{\cos x}{\sin x}$ , so we can write
\[a_1= \sin x\] \[a_2= \cos x\] \[a_3= \frac{\cos^2x}{\sin x}\] \[a_4=\frac{\cos^3x}{\sin^2x}=1\] \[a_5=\frac{\cos x}{\sin x}\] \[a_6=\frac{\cos^2x}{\sin^2x}\] \[a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}\] \[a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}\]
Since $\cos^3x=\sin^2x=1-\cos^2x$ , we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$ , which is $a_8$ , making our answer $8 \Rightarrow \boxed{8}$

Answer: E 8
Problem: A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solution: Notice that the common ratio is $r=\frac{\cos(x)}{\sin(x)}$ ; multiplying it to $\tan(x)=\frac{\sin(x)}{\cos(x)}$ gives $a_4=1$ . Then, working backwards we have $a_3=\frac{1}{r}$ $a_2=\frac{1}{r^2}$ and $a_1=\frac{1}{r^3}$ . Now notice that since $a_1=\sin(x)$ and $a_2=\cos(x)$ , we need $a_1^2+a_2^2=1$ , so $\frac{1}{r^6}+\frac{1}{r^4}=\frac{r^2+1}{r^6}=1\implies r^2+1=r^6$ . Dividing both sides by $r^2$ gives $1+\frac{1}{r^2}=r^4$ , which the left side is equal to $1+\cos(x)$ ; we see as well that the right hand side is equal to $a_8$ given $a_4=1$ , so the answer is $\boxed{8}$ . - mathleticguyyy

Answer: E 8
Problem: Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$

Solution: $P(x) = (x - a)^2 - b, Q(x) = (x - c)^2 - d$ . Notice that $P(x)$ has roots $a\pm \sqrt {b}$ , so that the roots of $P(Q(x))$ are the roots of $Q(x) = a + \sqrt {b}, a - \sqrt {b}$ . For each individual equation, the sum of the roots will be $2c$ (symmetry or Vieta's). Thus, we have $4c = - 23 - 21 - 17 - 15$ , or $c = - 19$ . Doing something similar for $Q(P(x))$ gives us $a = - 54$ .
We now have $P(x) = (x + 54)^2 - b, Q(x) = (x + 19)^2 - d$ . Since $Q$ is monic, the roots of $Q(x) = a + \sqrt {b}$ are "farther" from the axis of symmetry than the roots of $Q(x) = a - \sqrt {b}$ . Thus, we have $Q( - 23) = - 54 + \sqrt {b}, Q( -21) =- 54 - \sqrt {b}$ , or $16 - d = - 54 + \sqrt {b}, 4 - d = - 54 - \sqrt {b}$ . Adding these gives us $20 - 2d = - 108$ , or $d = 64$ . Plugging this into $16 - d = - 54 + \sqrt {b}$ , we get $b = 36$ .
The minimum value of $P(x)$ is $- b$ , and the minimum value of $Q(x)$ is $- d$ . Thus, our answer is $- (b + d) = - 100$ , or answer $\boxed{100}$

Answer: A 100
Problem: Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$

Solution: Let $P(x) = x^2 + Bx + C$ and $Q(x) = x^2 + Ex + F$
Then $P(Q(x))$ is $(x^2 + Ex + F)^2 + B(x^2 + Ex + F) + C$ , which simplifies to:
$P(Q(x)) = x^4 + 2Ex^3 + (E^2 + 2F + B)x^2 + (2EF + BE)x + (F^2 + BF + C)$
We can find $Q(P(x))$ by simply doing $B\Leftrightarrow E$ and $C \Leftrightarrow F$ to get:
$Q(P(x)) = x^4 + 2Bx^3 + (B^2 + 2C + E)x^2 + (2BC + BE)x + (C^2 + EC + F)$
The sum of the zeros of $P(Q(x))$ is $-76$ . From Vieta, the sum is $-2E$ . Therefore, $E = 38$
The sum of the zeros of $Q(P(x))$ is $-216$ . From Vieta, the sum is $-2B$ . Therefore, $B = 108$
Plugging in, we get:
$P(Q(x)) = x^4 + 76x^3 + (1552 + 2F)x^2 + (76F + 4104)x + (F^2 + 108F + C)$ $Q(P(x)) = x^4 + 216x^3 + (11702 + 2C)x^2 + (216C + 4104)x + (C^2 + 38C + F)$
Let's tackle the $x^2$ coefficients, which is the sum of the six double-products possible. Since $23 \cdot (21 + 17 + 15) + 21 \cdot (17 + 15) + 17 \cdot 15$ gives the sum of these six double products of the roots of $P(Q(x))$ , we have:
$1552 + 2F = 23 \cdot (21 + 17 + 15) + 21 \cdot (17 + 15) + 17 \cdot 15$
$1552 + 2F = 2146$
$F = 297$
Similarly with $Q(P(x))$ , we get:
$11702 + 2C = 59(57 + 51 + 49) + 57(51 + 49) + 51(49)$
$11702 + 2C = 17462$
$C = 2880$
Thus, our polynomials are $P(x) = x^2 + 108x + 2880$ and $Q(x) = x^2 + 38x + 297$
The minimum value of $P(x)$ happens at $x = -\frac{108}{2} = -54$ , and is $54^2 - 108 \cdot 54 + 2880 = 2880 - 54^2$
The minimum value of $Q(x)$ happens at $x = -\frac{38}{2} = -19$ , and is $19^2 - 38 \cdot 19 + 297 = 297 - 19^2$
The sum of these minimums is $2880 +297 - 54^2 - 19^2 = \boxed{100}$ . -srisainandan6

Answer:  100
Problem: Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$

Solution: Let $P(x) = x^2 - (a+b)x + ab$ and $Q(x) = x^2 - (c+d)x + cd$ . Notice that the roots of $P(x)$ are $a,b$ and the roots of $Q(x)$ are $c,d.$ Then we get:
\begin{align*} P(Q(x)) &= a, b \\ x^2 - (c+d)x + cd &= a, b \end{align*} The two possible equations are then $x^2 - (c+d)x + cd-a=0$ and $x^2 - (c+d)x + cd-b=0$ . The solutions are $-23, -21, -17, -15$ . From Vieta's we know that the total sum $2(c+d) = -76 \implies c+d = -38$ so the roots are paired $-23, -15$ and $-21, -17$ . Let $cd - a = 23*15$ and $cd - b = 21*17$
We can similarly get that $ab - c = 59*49$ and $ab - d = 57*51$ , and $a+b = -108$ . Add the first two equations to get \[2cd - (a+b) = 23*15 + 21*17 \implies cd = \frac{23*15+21*17 - 108}{2} = 297.\] This means $Q(x) = x^2 + 38x + 297$
Once more, we can similarly obtain \[ab = \frac{59*49 + 57*51 - 38}{2} = 2880.\] Therefore $P(x) = x^2 + 108x + 2880$
Now we can find the minimums to be \[19^2 - 19*38 + 297 = -64\] and \[54^2 - 54*108 + 2880 = -36.\] Summing, the answer is $\boxed{100}.$

Answer: A 100
Problem: Monic quadratic polynomial $P(x)$ and $Q(x)$ have the property that $P(Q(x))$ has zeros at $x=-23, -21, -17,$ and $-15$ , and $Q(P(x))$ has zeros at $x=-59,-57,-51$ and $-49$ . What is the sum of the minimum values of $P(x)$ and $Q(x)$
$\textbf{(A)}\ -100 \qquad \textbf{(B)}\ -82 \qquad \textbf{(C)}\ -73 \qquad \textbf{(D)}\ -64 \qquad \textbf{(E)}\ 0$

Solution: Let $P(x) = (x+a)(x+b)$ $Q(x) = (x+c)(x+d)$
$P(Q(x)) = (x^2 + cx + dx + cd + a)(x^2 + cx + dx + cd + b)$
$Q(P(x)) = (x^2 + ax + bx + ab + c)(x^2 + ax + bx + ab + d)$
Notice how the coefficient for $x$ has to be the same for the two quadratics that are multiplied to create $P(Q(x))$ , and $Q(P(x))$
$P(Q(x)) = (x+ 23)(x+ 21)(x+ 17)(x+ 15) = (x^2 + 38x + 345)(x^2 + 38x + 357)$
$Q(P(x)) = (x+ 59)(x+ 57)(x+ 51)(x+ 49) = (x^2 + 108x + 2891)(x^2 + 108x + 2907)$
$c + d = 38$ $cd + a = 345$ $cd + b = 357$ $a + b = 108$ $ab + c = 2891$ $ab + d = 2907$
$cd = \frac{345 + 357 - 108}{2} = 297$ $297 = 3^3 \cdot 11$ $c = 27$ $d = 11$
$a = 345 - 27*11 = 48$ $b = 357 - 27*11 = 60$
$P(x) = (x+48)(x+60) = x^2 + 108x + 2880 = (x+54)^2 - 36$
$Q(x) = (x+27)(x+11) = x^2 + 38x + 297 = (x+19)^2 -64$
$-36 -64 = \boxed{100}$

Answer: A 100
Problem: For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78$

Solution: Because 67 is the largest prime factor of 2010, it means that in the prime factorization of $\prod_{n=2}^{5300}\text{pow}(n)$ , there'll be $p_1 ^{e_1} \cdot p_2 ^{e_2} \cdot .... 67^x ...$ where $x$ is the desired value we are looking for. Thus, to find this answer, we need to look for the number of times $67$ is incorporated into the giant product.
All numbers $n=67 \cdot x$ , given $x = p_1 ^ {e_1} \cdot p_2 ^{e_2} \cdot ... \cdot p_m ^ {e_m}$ such that for any integer $x$ between $1$ and $m$ , prime $p_x$ must be less than $67$ , contributes a 67 to the product. Considering $67 \cdot 79 < 5300 < 67 \cdot 80$ , the possible values of x are $1,2,...,70,72,74,...78$ , since $x=71,73,79$ are primes that are greater than 67. However, $\text{pow}\left(67^2\right)$ contributes two $67$ s to the product, so we must count it twice. Therefore, the answer is $70 + 1 + 6 = \boxed{77}$

Answer: D 77
Problem: For every integer $n\ge2$ , let $\text{pow}(n)$ be the largest power of the largest prime that divides $n$ . For example $\text{pow}(144)=\text{pow}(2^4\cdot3^2)=3^2$ . What is the largest integer $m$ such that $2010^m$ divides
$\textbf{(A)}\ 74 \qquad \textbf{(B)}\ 75 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 77 \qquad \textbf{(E)}\ 78$

Solution: After finding the prime factorization of $2010=2\cdot3\cdot5\cdot67$ , divide $5300$ by $67$ and add $5300$ divided by $67^2$ in order to find the total number of multiples of $67$ between $2$ and $5300$ $\lfloor\frac{5300}{67}\rfloor+\lfloor\frac{5300}{67^2}\rfloor=80$ Since $71$ $73$ , and $79$ are prime numbers greater than $67$ and less than or equal to $80$ , subtract $3$ from $80$ to get the answer $80-3=\boxed{77}$

Answer: D 77
Problem: The first three terms of an arithmetic sequence are $2x - 3$ $5x - 11$ , and $3x + 1$ respectively. The $n$ th term of the sequence is $2009$ . What is $n$
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 502 \qquad \textbf{(C)}\ 1004 \qquad \textbf{(D)}\ 1506 \qquad \textbf{(E)}\ 8037$

Solution: As this is an arithmetic sequence, the difference must be constant: $(5x-11) - (2x-3) = (3x+1) - (5x-11)$ . This solves to $x=4$ . The first three terms then are $5$ $9$ , and $13$ . In general, the $n$ th term is $1+4n$ . Solving $1+4n=2009$ , we get $n=\boxed{502}$

Answer:  502