Answer: 16.0

The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$

Solution:
The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.
The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$
As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$ , for some nonnegative integer $k$
$a^2-8a=k^2$
$a(a-8)=k^2$
$((a-4)+4)((a-4)-4)=k^2$
$(a-4)^2-4^2=k^2$
$(a-4)^2=k^2+4^2$
From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$
In the case $k=0$ , the equation simplifies to $|a-4|=4$ . From this equation, we have $a=0,8$ . For both $a=0$ and $a=8$ $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")
If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$ . Hence, $a=-1,9$ . Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$ , so these two values also satisfy the original constraints.
There are a total of four possible values for $a$ $-1,0,8,$ and $9$ . Hence, the sum of all of the possible values of $a$ is $\boxed{16}$

Solution:
By the quadratic formula, the roots $r$ can be represented by \[r=\frac{a\pm\sqrt{a^2-8a}}{2}\] For $r\in\mathbb{Z}$ $a\in\mathbb{Z}$ , since $\frac{\sqrt{a^2-8a}}{2}$ and $\frac{a}{2}$ will have different mantissas (mantissae?).
Now observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$ and have two cases.
Positive $a$
$a\geq8$ and $a\leq0$ , since $1\geq a \geq7$ gives imaginary roots. Testing positive $a$ values, quickly see that $a\leq9$ . After $16$ and $36$ , the difference between the closest nonzero factor pairs of perfect squares exceeds $8$ . For $8\geq a \geq9$ $a=8,9$ . Checking both yields an integer.
Negative $a$
We can instead test with $\sqrt{-a(8-a)}$ . If $b=8-a$ , we have our original expression. Thus, for the same reasons, $b=8,9\implies 8,9=8-a$ $a=-1$ (0 does not affect the answer).
$-1+8+9=16\implies\boxed{16}$

Solution:
Let $m$ and $n$ be the roots of $x^2-ax+2a$
By Vieta's Formulas, $n+m=a$ and $mn=2a$
Substituting gets us $n+m=\frac{mn}{2}$
$2n-mn+2m=0$
Using Simon's Favorite Factoring Trick:
$n(2-m)+2m=0$
$-n(2-m)-2m=0$
$-n(2-m)-2m+4=4$
$(2-n)(2-m)=4$
This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1,0,8,$ and $9$ . Adding these up gets $\boxed{16}$

Solution:
The quadratic formula gives \[x = \frac{a \pm \sqrt{a(a-8)}}{2}\] . For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$ ; this is a quadratic in itself and the quadratic formula gives \[a = 4 \pm \sqrt{16 + b^2}\]
We want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$ , we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$ . These are the only ones; if we want to make sure then we must hand check up to $b=8$ . Indeed, for $b \geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find $\boxed{16}$ . as the answer.

Solution:
First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.
We can complete the square and rearrange to get $(a-4)^2-b^2=16$ . Let's define $m=a-4$ , just to make things a little easier to write, so now we have $(m+b)(m-b)=16$ . We can now list out the integer factor pairs of 16 and the resulting values of $m$ and $b$ . (Note that $m$ and $b$ must both be integers)
$(m+b)(m-b)=16$
$16*1=16$ $\Rightarrow$ Doesn't work
$8*2=16$ $\Rightarrow$ $m=5, b=3$
$4*4=16$ $\Rightarrow$ $m=4, b=0$
$2*8=16$ $\Rightarrow$ $m=5, b=-3$
$1*16=16$ $\Rightarrow$ Doesn't work
$-16*-1=16$ $\Rightarrow$ Doesn't work
$-8*-2=16$ $\Rightarrow$ $m=-5, b=-3$
$-4*-4=16$ $\Rightarrow$ $m=-4, b=0$
$-2*-8=16$ $\Rightarrow$ $m=-5, b=-3$
$-1*-16=16$ $\Rightarrow$ Doesn't work
We want the possible values of $m$ , which are $-5,-4,4,$ and $5$ . As $m+4=a$ $a$ can equal $-1,0,8,$ or $9.$ Adding all of that up gets us our answer, $\boxed{16}$


Answer: 3.0

There are $10$ horses, named Horse $1$ , Horse $2$ , . . . , Horse $10$ . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time $0$ all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520$ . Let $T > 0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T?$
$\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6$

Solution: If we have horses, $a_1, a_2, \ldots, a_n$ , then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that $\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12$ . Finally, $1+2 = \boxed{3}$

Solution: We are trying to find the smallest number that has $5$ one-digit divisors. Therefore we try to find the LCM for smaller digits, such as $1$ $2$ $3$ , or $4$ . We quickly consider $12$ since it is the smallest number that is the LCM of $1$ $2$ $3$ and $4$ . Since $12$ has $5$ single-digit divisors, namely $1$ $2$ $3$ $4$ , and $6$ , our answer is $1+2 = \boxed{3}$

Solution: First, for 5 horses to simultaneously pass the starting line after $T$ seconds, $T$ must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of $T$ , and therefore meaning $T$ must have at least $5$ $1$ -digit divisors. Since we want to minimize $T$ , we will start by guessing the lowest natural number, $1$ $1$ has only $1$ factor, so it does not work, we now repeat the process for the numbers between $2$ and $12$ (This should not take more than a minute) to get that $12$ is the first number to have $5$ or more single-digit divisors ( $1, 2, 3, 4, 6$ ). The sum of the digits of $12$ is $1+2 = \boxed{3}$ , which is our answer.

Solution: By inspection, $(1, 2, 3, 4, 6)$ yields the lowest answer of $12$ and the sum of the digits is $1+2 \Longrightarrow \boxed{3}$


Answer: 3.0

There are $10$ horses, named Horse 1, Horse 2, $\ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution:
We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$ . Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers $\{1,2,\cdots,10\}$ divide $n$ . Thus, $n$ has at least $5$ positive integer divisors.
We quickly see that $12$ is the smallest number with at least $5$ positive integer divisors and that $1,2,3,4,6$ are each numbers of horses. Thus, our answer is $1+2=\boxed{3}$

Solution: In order for at least $5$ horses to finish simultaneously, the current time needs to have at least $5$ divisors. Thus the number must have a form of either $p^4$ or $p^2*q$ , which have $5$ and $6$ factors respectively. The smallest number of the first form is $16$ , and the smallest number of the second form is $12.$ Thus, our answer is $1+2=\boxed{3}$


Answer: 13.0

There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\  13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$

Solution:
Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.
Case 1: $0$ diameters
There are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1$ $2$ pairs with $3$ , etc.
Case 2: $1$ diameter
There are $5$ possible diameters to draw (everyone else pairs with the person next to them).
Note that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.
Case 3: $3$ diameters
There are $5$ possible sets of $3$ diameters to draw.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5*2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5$
Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Case 4: $5$ diameters
There is only $1$ way to do this.
Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.
- Minor edits by Pearl2008

Solution: If each person knows exactly $3$ people, that means we form " $4$ -person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\dbinom{4}{2}=6$ . The $2$ nd pair is just $\dbinom{2}{2} =1$ . We need to multiply these together since these are $1$ group. The $3$ rd pair would be $\dbinom{4}{2}=6$ . The $4$ th pair is $\dbinom{2}{2}=1$ . We multiply these $2$ together and get $6$ . The final group would be $\dbinom{2}{2}=1$ . So we add these up and we have $6 + 6 + 1 = \boxed{13}$ possible ways.


Answer: 13.0

Solution:
Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.
Case 1: $0$ diameters
There are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1$ $2$ pairs with $3$ , etc.
Case 2: $1$ diameter
There are $5$ possible diameters to draw (everyone else pairs with the person next to them).
Note that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.
Case 3: $3$ diameters
There are $5$ possible sets of $3$ diameters to draw.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5*2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5$
Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Case 4: $5$ diameters
There is only $1$ way to do this.
Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.
- Minor edits by Pearl2008

Solution: If each person knows exactly $3$ people, that means we form " $4$ -person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\dbinom{4}{2}=6$ . The $2$ nd pair is just $\dbinom{2}{2} =1$ . We need to multiply these together since these are $1$ group. The $3$ rd pair would be $\dbinom{4}{2}=6$ . The $4$ th pair is $\dbinom{2}{2}=1$ . We multiply these $2$ together and get $6$ . The final group would be $\dbinom{2}{2}=1$ . So we add these up and we have $6 + 6 + 1 = \boxed{13}$ possible ways.


Answer: 40.0

There are $120$ seats in a row.  What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$

Solution:
Let $p$ be a person seated and $o$ is an empty seat
The pattern of seating that results in the fewest occupied seats is $\text{opoopoopoo...po}$ .
We can group the seats in 3s like this: $\text{opo opo opo ... opo}.$
There are a total of $40=\boxed{40}$ groups


Answer: 95000.0

Solution: We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \cdot 20 = \boxed{95000}$


Answer: 3.0

There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution:
By PIE (Property of Inclusion/Exclusion), we have
$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.$ Number of people in at least two sets is $\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.$ So, $20 = (10 + 13 + 9) - (9 + 2x) + x,$ which gives $x = \boxed{3}.$

Solution: The total number of classes taken among the 20 students is $10 + 13 + 9 = 32$ . Each student is taking at least one class so let's subtract the $20$ classes ( $1$ per each of the $20$ students) from $32$ classes to get $12$ $12$ classes is the total number of extra classes taken by the students who take $2$ or $3$ classes. Since we know that there are $9$ students taking at least $2$ classes, there must be $12 - 9 = \boxed{3}$ students that are taking all $3$ classes.

Solution: Total class count is 32. Assume there are $a$ students taking one class, $b$ students taking two classes, ad $c$ students taking three classes. Because there are $20$ students total, $a+b+c = 20$ . Because each student taking two classes is counted twice, and each student taking three classes is counted thrice in the total class count, $a+2b+3c = 10+13+9 = 32$ . There are $9$ students taking two or three classes, so $b+c = 9$ . Solving this system of equations gives us $c=\boxed{3}$

Solution:
Let us assign the following variables and put them in our Venn Diagram [1] $a$ which designates the number of people taking exactly Bridge and Yoga. $b$ which designates the number of people taking exactly Bridge and Painting. $c$ which designates the number of people that took all $3$ classes or what we want to find. $d$ which designates the number of people taking exactly Yoga and Painting.
Let's now recall what information we have given: There are exactly $9$ people that are taking at least $2$ classes meaning in other words, $9$ people total are taking strictly $2$ classes or strictly all the available classes meaning that $a+b+c+d=9$
Let's now start filling out the Venn Diagram:
Strictly taking Bridge, no other classes: We know in total, the number is $13$ , however this includes the people taking other classes too meaning we'd need to do some subtraction. From our Venn Diagram we see that we'd need to subtract the following variables to get our wanted outcome here, $a, b, c$ . Giving our answer as $13-(a+b+c)$
However, this equation seems complicated as it has $3$ different variables, so to make this look a lot less complicated we can use our earlier equation: $a+b+c+d=9$ to see that $a+b+c=9-d$ . This means that this can also be written as $13-(9-d)=4+d$
Strictly taking Yoga Only: The total number of people is $10$ , but this would also count people taking other classes too along with it, so we need to subtract this overcount which is visible in the Venn Diagram giving us: $10-(a+c+d)$
Again, we can use substitution to see that $a+c+d=9-b$ . This simplifies our equation to $10-(9-b)=1+b$
Strictly taking Painting Only: We know again, in total this number is $9$ , which also accounts for the people taking other classes too. From our Venn Diagram it is visible that we need to subtract: $b, c, d$ giving $9-(b+c+d)$
Again, through substitution of our first equation we see that $b+c+d=9-a$ meaning we can simplify this equation to $9-(9-a)=a$
If we add these newly made equations of strictly taking one class, we get the total number of people taking exactly one class as these equations each were a subcase for it. We can also find the exact number for this because we are given that there are exactly $20$ students in total, and $9$ students are taking exactly $2$ or $3$ classes, meaning that if we do $20-9$ we get our answer for the number of students taking exactly $1$ class because those taking exactly one class have no overlap with those taking exactly $2$ or exactly $3$ classes as shown in our Venn Diagram and because $1$ $2$ , and $3$ classes are subcases for finding the total number of students as we know that each student is in exactly $1, 2$ , or $3$ classes. This means that exactly $11$ students took strictly $1$ class.
We can add up our equations we found to equal $11$ because those equations were for subcases of having exactly $1$ class giving: $(4+d)+(1+b)+a=11$ $=5+a+b+d=11$
From our equation $a+b+c+d=9$ , we can substitute $9-c$ for $a+b+d$ giving us: $5+9-c=11$
This gives $c=3$ . We assigned $c$ for the number of people taking exactly $3$ classes meaning that when we find $c$ , we find the answer. This means our answer is $\boxed{3}$

Solution:
We are told that there are $20$ students in all, and $10$ take yoga, $13$ take bridge, and $9$ take painting.
Representing each student as a number from $1$ to $20$ , we can then make a list of which classes they are taking. Students 1-10 take yoga, and students $11$ to $20$ take bridge. However, this means that only $10$ students are taking bridge. To make up for it, we go back to the top of the list and start over from student $1$ . Students $1$ $2$ , and $3$ will also take bridge giving the desired count of $13$ total bridge students.
Now, all that is left are the students who take painting. There are $9$ students who take painting, so students $4$ through students $9$ take painting. Note that because $9$ people take at least two classes, students $10$ through $20$ are unable to take more than one class. This means that we must once more start over from the top. Already $6$ painting slots have been filled, so students $1$ $2$ , and $3$ will also take painting. This gives a total of $3$ students (students $1$ $2$ , and $3$ ) who take all three classes. Therefore, our answer is $\boxed{3}$


Answer: 12.0

There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

Solution:
Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$ . So
$z^6=e^{\frac{ni\pi}{2}}$
This is real if $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{12}$

Solution:
$z = \sqrt[24]{1} = 1^{\frac{1}{24}}$
By Euler's identity $1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+2k\pi)$ , where $k$ is an integer.
Using De Moivre's Theorem , we have $z = 1^{\frac{1}{24}} = {\cos (\frac{k\pi}{12}) + i \sin (\frac{k\pi}{12})}$ , where $0 \leq k<24$ that produce $24$ unique results.
Using De Moivre's Theorem again, we have $z^6 = {\cos (\frac{k\pi}{2}) + i \sin (\frac{k\pi}{2})}$
For $z^6$ to be real, $\sin(\frac{k\pi}{2})$ has to equal $0$ to negate the imaginary component. This occurs whenever $\frac{k\pi}{2}$ is an integer multiple of $\pi$ , requiring that $k$ is even. There are exactly $\boxed{12}$

Solution: From the start, recall from the Fundamental Theorem of Algebra that $z^{24} = 1$ must have $24$ solutions (and these must be distinct since the equation factors into $0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)$ ), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be $24$ . Notice that $1 = z^{24} = (z^6)^4$ , so for any solution $z$ $z^6$ will be one of the 4th roots of unity ( $1$ $i$ $-1$ , or $-i$ ). Then $6$ solutions $z$ will satisfy $z^6 = 1$ $6$ will satisfy $z^6 = -1$ (and this is further justified by knowledge of the 6th roots of unity), so there must be $\boxed{12}$ such $z$

Solution: Let $a\in\mathbb{R}$ and $a = z^6.$ We have \[a^4 = 1 \implies a = 1,-1.\] $z^6 = \pm 1$ has 6 solutions for $1$ and $-1$ respectively, so $6+6=\boxed{12}.$ \[\] -svyn


Answer: 548.0

There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution:
Let $a_{n}$ be the number of ways to form $n$ -letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$ -letter string, there are $3$ possibilities for the last letter chain: it must be either $1$ $2$ , or $3$ letters long. Removing this last chain will make a new string that is $n-1$ $n-2$ , or $n-3$ letters long, respectively.
Therefore we can deduce that $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$
We can see that \[a_{1}=2\] \[a_{2}=2^{2}=4\] \[a_{3}=2^{3}=8\] so using our recursive relation we find \[a_{4}=14\] \[a_{5}=26\] \[a_{6}=48\] \[a_{7}=88\] \[a_{8}=162\] \[a_{9}=298\] \[a_{10}=\boxed{548}\]

Solution:
The solution is a simple recursion:
We have three cases for the ending of a string: three in a row, two in a row, and a single:
...AAA $(1)$ ...BAA $(2)$ ...BBA or ...ABA $(3)$
(Here, WLOG each string ends with A. This won't be the case when we actually solve for values in recursion.)
For case $(1)$ , we could only add a B to the end, making it a case $(3)$ .
For case $(2)$ , we could add an A or a B to the end, making it a case $(1)$ if you add an A, or a case $(3)$ if you add a B.
For case $(3)$ , we could add an A or a B to the end, making it a case $(2)$ or a case $(3)$
Let us create three series to represent the number of permutations for each case: $\{a\}$ $\{b\}$ , and $\{c\}$ representing case $(1)$ $(2)$ , and $(3)$ respectively.
The series have the following relationship:
$a_n=b_{n-1}$ $b_n=c_{n-1}$ $c_n=c_{n-1}+a_{n-1}+b_{n-1}$
For $n=3$ $a_3$ and $b_3$ both equal $2$ $c_3=4$ . With some simple math, we have: $a_{10}=88$ $b_{10}=162$ , and $c_{10}=298$ .
Summing the three up we have our solution: $88+162+298=\boxed{548}$

Solution:
This is a recursion problem. Let $a_n$ be the number of valid strings of $n$ letters, where the first letter is $A$ .  Similarly, let $b_n$ be the number of valid strings of $n$ letters, where the first letter is $B$
Note that $a_n=b_{n-1}+b_{n-2}+b_{n-3}$ for all $n\ge4$
Similarly, we have $b_n=a_{n-1}+a_{n-2}+a_{n-3}$ for all $n\ge4$
Here is why:  every valid strings of $n$ letters $(n\ge4)$ where the first letter is $A$ must begin with one of the following:
$AAAB$ - and the number of valid ways is $b_{n-3}$
$AAB$ - and the number of valid ways is $b_{n-2}$
$AB$ - and there are $b_{n-1}$ ways.
We know that $a_1=1$ $a_2=2$ , and $a_3=4$ .  Similarly, we have $b_1=1$ $b_2=2$ , and $b_3=4$ .  We can quickly check our recursion to see if our recursive formula works.  By the formula, $a_4=b_3+b_2+b_1=7$ , and listing out all $a_4$ , we can quickly verify our formula.
Therefore, we have the following:
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\\b & 1 & 2 & 4 & 7 & 13 & 24 & 44 & 81 & 149 & 274\end{tabular}$
The total number of valid $10$ letter strings is equal to $a_{10}+b_{10}=274+274=\boxed{548}$

Solution:
Playing around with strings gives this approach: We have a certain number of As, then Bs, and so on. Therefore, what if we denoted each solution with numbers like $(3,3,3,1)$ to denote AAABBBAAAB or vice versa (starting with Bs)? Every string can be represented like this!
We can have from 4 to 10 numbers in our parentheses. For each case, we will start with the largest number possible, usually a bunch of 3s, then go down systematically. Realize also that if we are left with just 2s and 1s, there is only one number of 2s and 1s that adds up to the leftover amount. Our final answer is the sum of all of these parenthetical sets [each set multiplied by its permutations, as order matters] multiplied by two [starting with either A or B, and alternating as we go along].
$4 \rightarrow (3,3,3,1) = 4, (3,3,2,2) = 6$
$5 \rightarrow (3, 3, 2, 1, 1) = 30, (3, 2, 2, 2, 1) = 20, (2,2,2,2,2)=1$
$6 \rightarrow (3,3,1,1,1,1) = 15, (3,2,2,1,1,1) = 60, (2,2,2,2,1,1) = 15$
$7 \rightarrow (3,2,1,1,1,1,1) = 42, (2,2,2,1,1,1,1) = 35$
$8 \rightarrow (3,1...1) = 8, (2,2,1...1) = 28$
$9 \rightarrow (2,1...1) = 9$
$10 \rightarrow (1,1....1) =1$
Adding them all up gives you 274; multiplying by 2 gives $\boxed{548}$

Solution:
We are going to build the string, 1 character at a time. And, we are going to only care about the streak of letters at the end of the string.
Let $a_n$ be the number of strings of length n that satisfy the problem statement and also has a "streak" of length 1 at the end. ABABBA has a streak of length 1.
Let $b_n$ be the number of strings of length n that satisfy the problem statement and also has of length 2 at the end. ABABBAA has a streak of length 2. There are 2 "A" s at the end of the string.
Let $c_n$ be the number of string of length n that that satisfy the problem statement and also has a "streak" of length 3 at the end. ABABBAAA has a streak of length 3. There are 3 "A" s at the end of the string.
Let's establish a recursive relationship. $a_{n+1} = a_n+b_n+c_n$ , since you can simply break the streak. $b_{n+1} = a_n$ , and $c_{n+1} = b_n$ Since you can just add to the streak.
We can log everything using a table.
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c}Value of n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\\hline a & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 & 298\\b & 0 & 2 & 2 & 4 & 8 & 14 & 26 & 48 & 88 & 162 \\c & 0 & 0 & 2 & 2 &4 & 8 & 14 & 26 & 48 & 88\end{tabular}$
Adding $a_{10}$ $b_{10}$ $c_{10}$ gets the total number of numbers that doesn't have more than 3 concecutive letters.
That gets a total of $298+162+88 = \boxed{548}$

Solution:
Let $S_n$ be the number of n-letter strings satisfying the problem criteria. Then we can easily see $S_1 = 2, S_2 = 4, S_3 = 8, S_4 = 16-2 =14$ . For $n \ge 4$ consider the last three elements of the list. For example to get $S_5$ we could try to take $S_4$ and multiply by 2 since for each $S_4$ string we have two choices for the fifth character. However we have to be careful if the last three characters of $S_4$ are all the same, as in that case we would be overcounting, (i.e if we have a string $\dots aaa$ , we can only add $b$ to the end so that the problem criteria is satisfied. Additionally strings such as $\dots aaa$ or $\dots bbb$ should only be counted once as we only have one choice for the $n$ th character to add ( $b$ and $a$ respectively).
Thus to compute $S_n$ we start with $S_{n-1}$ then take out the strings ending in $\dots bbb$ or $\dots aaa$ . There are $S_{n-4}$ remaining valid strings (i.e if we pick any of the $S_{n-4}$ valid strings, examine the last character, if it is a $b$ then we append $aaa$ to it, and if if it an $a$ we append $bbb$ to it, hence the number of $S_{n-1}$ strings are in one-to-one correspondence with the number of strings in $S_{n-4}$ . Thus we have the recursion $S_n = 2(S_{n-1} - S_{n-4}) + S_{n-4}$ (where we are first taking away $S_{n-4}$ , doubling the result, then adding $S_{n-4}$ back in signifying that we only count it once, as described above).
The recursion simplifies to $S_n = 2S_{n-1} - S_{n-4}$ and we can now quickly compute the remaining values: \[S_5 = 2S_4 - S_1 = 26, S_6 = 2(26)-4 = 48, S_7 = 2(48) - 8 = 88, S_8 = 2(88) - 14 = 162, S_9 = 2(162) - 26 = 298, S_{10} = 2S_9 - 48 = \boxed{548}\]

Solution:
Let us define a "run" as a set of consecutive letters that are all the same and let a "maximal run" be a run that is not the proper subset of any other run, in other words, it cannot be expanded.
From now on, we consider all runs to be maximal.
Note that the minimum number of runs in the $10$ -letter string is $4$ (or else the Pigeonhole Principle tells us that at least $1$ run has $4$ or more letters in it, contradiction), and the maximum number of runs is clearly $10.$
Since an arbitrary run can have anywhere from $1$ to $3$ letters in it, it follows that the number of $10$ -letter strings with $n$ runs (where $4 \leq n \leq 10$ ) is \[2[x^{10}] (x+x^2+x^3)^n,\] i.e. twice the coefficient of $x^{10}$ in the expansion of $(x+x^2+x^3)^n.$ (Note that we multiplied by $2$ because there are two choices for which letter the first run has and then the rest are fixed).
Hence, we wish to find \[2\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n.\] First, we can rewrite this as
\begin{align*}
\sum_{n=4}^{10} [x^{10}] (x+x^2+x^3)^n &= \sum_{n=4}^{10} [x^{10}] \ x^n \cdot (1+x+x^2)^n \\
&= \sum_{n=4}^{10} [x^{10-n}] \left(\frac{1-x^3}{1-x}\right)^n \\
&= \sum_{n=4}^{10} [x^{10-n}] \frac{(1-x^3)^n}{(1-x)^n}. \\
\end{align*}
Now, we proceed case by case, utilizing the Binomial Theorem for the numerator in all of the cases:
For $n=4,$ we have
\begin{align*}
[x^6] \frac{(1-x^3)^4}{(1-x)^4} &= \binom{4}{0} [x^6] (1-x)^{-4} - \binom{4}{1} [x^3] (1-x)^{-4} + \binom{4}{2} [x^0] (1-x)^{-4} \\
&= 1\cdot\binom{9}{3} - 4\cdot\binom{6}{3} + 6\cdot\binom{3}{3} \\
&= 1\cdot84-4\cdot20+6\cdot1 \\
&= 84-80+6 \\
&= \underline{10}.
\end{align*}
For $n=5,$ we have
\begin{align*}
[x^5] \frac{(1-x^3)^5}{(1-x)^5} &= \binom{5}{0} [x^5] (1-x)^{-5} - \binom{5}{1}[x^2] (1-x)^{-5} \\
&= 1\cdot\binom{9}{4}-5\cdot\binom{6}{4} \\
&= 1\cdot126-5\cdot15 \\
&= 126-75 \\
&= \underline{51}.
\end{align*}For $n=6,$ we have
\begin{align*}
[x^4] \frac{(1-x^3)^6}{(1-x)^6} &= \binom{6}{0} [x^4](1-x)^{-6} - \binom{6}{1}[x^1](1-x)^{-6} \\
&= 1 \cdot \binom{9}{5} - 6 \cdot \binom{6}{5} \\
&= 1 \cdot 126 - 6 \cdot 6 \\
&= 126-36 \\
&= \underline{90}.
\end{align*}For $n=7,$ we have
\begin{align*}
[x^3] \frac{(1-x^3)^7}{(1-x)^7} &= \binom{7}{0}[x^3](1-x)^{-7} - \binom{7}{1}[x^0](1-x)^{-7} \\
&= 1\cdot \binom{9}{6} - 7\cdot\binom{6}{6} \\
&= 1\cdot84-7\cdot1\\
&= 84-7 \\
&= \underline{77}.
\end{align*}For $n=8,$ we have
\begin{align*}
[x^2] \frac{(1-x^3)^8}{(1-x)^8} &=\binom{8}{0} [x^2](1-x)^{-8} \\
&= 1\cdot\binom{9}{7} \\
&= 1\cdot36 \\
&= \underline{36}.
\end{align*}For $n=9,$ we have
\begin{align*}
[x^1] \frac{(1-x^3)^9}{(1-x)^9} &= \binom{9}{0} [x^1] (1-x)^{-9}\\
&= 1\cdot\binom{9}{8} \\
&= 1\cdot9 \\
&= \underline{9}.
\end{align*}For $n=10,$ we have
\begin{align*}
[x^0] \frac{(1-x^3)^{10}}{(1-x)^{10}} &= [x^0] (1-x)^{-10} \\
&= 1\cdot\binom{9}{9} \\
&= 1\cdot1 \\
&= \underline{1}.
\end{align*}Hence, the answer is
\begin{align*}
2\sum_{n=4}^{10} [x^{10-n}]  \frac{(1-x^3)^n}{(1-x)^n} &= 2(10+51+90+77+36+9+1) \\
&= 2\cdot274 \\
&= \boxed{548}.
\end{align*}


Answer: 840.0

There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$

Solution:
$z$ can be written in the form $\text{cis\,}\theta$ . Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$
Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta =  -\frac{1}{2}+\frac {\sqrt{3}}{2}i$ , or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta =  -\frac{1}{2}- \frac{\sqrt{3}}{2}i$
Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$ , we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$ . We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
Again setting up equations ( $Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$ ) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$
Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed{840}$ degrees. We only want the sum of a certain number of theta, not all of it.


Answer: 5.0

There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$

Solution: Pretend you have $52$ people you want to place  in $12$ boxes, because there are $12$ months in a year. By the Pigeonhole Principle , one box must have at least $\left\lceil \frac{52}{12} \right\rceil$ people $\longrightarrow \boxed{5}$


Answer: 440.0

There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$

Solution:
Be wary of "position" versus "number" in the solution!
Each POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$ . The $n$ th position is defined by this ordered triple where $i$ is $n \mod 2$ $j$ is $n \mod 3$ , and $k$ is $n \mod 5$ . There are 2 choices for $i$ , 3 for $j$ , and 5 for $k$ , yielding $2 \cdot 3 \cdot 5=30$ possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if $i$ is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If $j$ is the same, then the two numbers must be equivalent $\mod 3$ , and if $k$ is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have $1 \mod 2$ ! It's that the POSITION which NUMBER 1 occupies has $1 \mod 2$
The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement.
The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed.
Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of $1\cdot 1 \cdot 3=3$ choices for the placement of 3.
As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of $1 \cdot 1 \cdot 2=2$ possible placements for 4.
5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5.
All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed.
Thus, $N = 30 \cdot 8 \cdot 3 \cdot 2=30 \cdot 48=1440$ . Thus, the remainder when $N$ is divided by $1000$ is $\boxed{440}.$

Solution: Note that $30=2\cdot 3\cdot 5$ . Since $\gcd(2, 3, 5)=1$ , by CRT, for each value $k=0\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\equiv a\pmod{2}$ $k\equiv b\pmod{3}$ , and $k\equiv c\pmod{5}$ . Therefore, we can independently assign the residues modulo $2, 3, 5$ , so $N=2!\cdot 3!\cdot 5!=1440$ , and the answer is $\boxed{440}$

Solution:
First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\equiv 0 \pmod 2$ or $\equiv 1 \pmod 2$ . Another way to say this is each element in the sequence $a_1,a_3,a_5,a_7...$ would be the same mod 2, and similarly for the other sequence. There are 2! = 2 ways to assign the mods to the sequences.
Now when $m$ is equal to 3, the sequences are $T_1 = a_1, a_4,a_7..$ $T_2 = a_2,a_5,a_8,a_11...$ , and $T_3 = a_3,a_6,a_9...$ . Again, for each sequence, all of its elements are congruent either $0$ $1$ , or $2$ mod $3$ . There are $3! = 6$ ways to assign the mods to the sequences.
Finally do the same thing for $m = 5$ . There are $5!$ ways. In total there are $2 * 6*120 = 1440$ and the answer is $\boxed{440}$


Answer: 332.0

There are $n$ mathematicians seated around a circular table with $n$ seats numbered $1,$ $2,$ $3,$ $...,$ $n$ in clockwise order. After a break they again sit around the table. The mathematicians note that there is a positive integer $a$ such that
Find the number of possible values of $n$ with $1 < n < 1000.$

Solution:
It is a well-known fact that the set $0, a, 2a, ... (n-1)a$ forms a complete set of residues if and only if $a$ is relatively prime to $n$
Thus, we have $a$ is relatively prime to $n$ . In addition, for any seats $p$ and $q$ , we must have $ap - aq$ not be equivalent to either $p - q$ or $q - p$ modulo $n$ to satisfy our conditions. These simplify to $(a-1)p \not\equiv (a-1)q$ and $(a+1)p \not\equiv (a+1)q$ modulo $n$ , so multiplication by both $a-1$ and $a+1$ must form a complete set of residues mod $n$ as well.
Thus, we have $a-1$ $a$ , and $a+1$ are relatively prime to $n$ . We must find all $n$ for which such an $a$ exists. $n$ obviously cannot be a multiple of $2$ or $3$ , but for any other $n$ , we can set $a = n-2$ , and then $a-1 = n-3$ and $a+1 = n-1$ . All three of these will be relatively prime to $n$ , since two numbers $x$ and $y$ are relatively prime if and only if $x-y$ is relatively prime to $x$ . In this case, $1$ $2$ , and $3$ are all relatively prime to $n$ , so $a = n-2$ works.
Now we simply count all $n$ that are not multiples of $2$ or $3$ , which is easy using inclusion-exclusion. We get a final answer of $998 - (499 + 333 - 166) = \boxed{332}$


Answer: 7425.0

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
$\textbf{(A)}\ 5724 \qquad \textbf{(B)}\ 7245 \qquad \textbf{(C)}\ 7254 \qquad \textbf{(D)}\ 7425 \qquad \textbf{(E)}\ 7542$

Solution: We begin by narrowing down the possibilities. If the larger number were twice the smaller number, then the smallest possibility for the larger number is $2457\times2=4914$ , since $2457$ is the smallest number in the set. The largest possibility would have to be twice the largest number in the set such that when it is multiplied by $2$ , it is less than or equal to $7542$ , the largest number in the set. This happens to be $2754\times2=5508$ . Therefore, the number would have to be between $4914$ and $5508$ , and also even. The only even numbers in the set and in this range are $5472$ and $5274$ . A quick check reveals that neither of these numbers is twice a number in the set. The number can't be quadruple or more another number in the set since $2457\times4=9828$ , well past the range of the set. Therefore, the number must be triple another number in the set. The least possibility is $2457\times3=7371$ and the greatest is $2475\times3=7425$ , since any higher number in the set multiplied by $3$ would be out of the range of the set. Reviewing, we find that the upper bound does in fact work, so the multiple is $2475\times3=7425, \boxed{7425}$

Solution:
There are only 5 options for the problem so we can just try them. It is easy since that we only need try to use $2$ $3$ to divide them. Even $5$ will leads to a solution start with $1$ which we don't need.
$5724/2=2862$ $5724/3=1908$ $7245/3=2415$ $7254/2=3612$ $7254/3=2418$ $7425/3=2475$ . The answer is $\boxed{7425}$ . You can obtain the answer in only 6 calculations.


Answer: 3.0

There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$

Solution:
Notice that we don't need to find what $A, B,$ and $C$ actually are, just their average. In other words, if we can find $A+B+C$ , we will be done.
Adding up the equations gives $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$ . Our answer is $\boxed{3}$

Solution: Start by isolating $B$ and $C$ in both of the equations, in order to represent the variables $C$ and $B$ in terms of A. Ending up with the two equations $C = 2A +4$ and $B = -3A + 5$ , we have to calculate the value of the expression $\frac{A+B+C}{3}$ . Plugging in $2A + 4$ for $C$ and $-3A + 5$ for $B$ , we add them up and end up with a value of 9. Dividing 9 by 3 to compute the average, we get our answer of $\boxed{3}$


Answer: 27720.0

There are exactly $77,000$ ordered quadruplets $(a, b, c, d)$ such that $\gcd(a, b, c, d) = 77$ and $\operatorname{lcm}(a, b, c, d) = n$ . What is the smallest possible value for $n$
$\textbf{(A)}\ 13,860\qquad\textbf{(B)}\ 20,790\qquad\textbf{(C)}\ 21,560 \qquad\textbf{(D)}\ 27,720 \qquad\textbf{(E)}\ 41,580$

Solution:
Let $A=\frac{a}{77},\ B=\frac{b}{77}$ , etc., so that $\gcd(A,B,C,D)=1$ . Then for each prime power $p^k$ in the prime factorization of $N=\frac{n}{77}$ , at least one of the prime factorizations of $(A,B,C,D)$ has $p^k$ , at least one has $p^0$ , and all must have $p^m$ with $0\le m\le k$
Let $f(k)$ be the number of ordered quadruplets of integers $(m_1,m_2,m_3,m_4)$ such that $0\le m_i\le k$ for all $i$ , the largest is $k$ , and the smallest is $0$ . Then for the prime factorization $N=2^{k_2}3^{k_3}5^{k_5}\ldots$ we must have $77000=f(k_2)f(k_3)f(k_5)\ldots$ So let's take a look at the function $f(k)$ by counting the quadruplets we just mentioned.
There are $14$ quadruplets which consist only of $0$ and $k$ . Then there are $36(k-1)$ quadruplets which include three different values, and $12(k-1)(k-2)$ with four. Thus $f(k)=14+12(k-1)(3+k-2)=14+12(k^2-1)$ and the first few values from $k=1$ onwards are \[14,50,110,194,302,434,590,770,\ldots\] Straight away we notice that $14\cdot 50\cdot 110=77000$ , so the prime factorization of $N$ can use the exponents $1,2,3$ . To make it as small as possible, assign the larger exponents to smaller primes. The result is $N=2^33^25^1=360$ , so $n=360\cdot 77=27720$ which is answer $\boxed{27,720}$


Answer: 78.0

There are exactly $N$ distinct rational numbers $k$ such that $|k|<200$ and \[5x^2+kx+12=0\] has at least one integer solution for $x$ .  What is $N$
$\textbf{(A) }6\qquad \textbf{(B) }12\qquad \textbf{(C) }24\qquad \textbf{(D) }48\qquad \textbf{(E) }78\qquad$

Solution: Factor the quadratic into \[\left(5x + \frac{12}{n}\right)\left(x + n\right) = 0\] where $-n$ is our integer solution. Then, \[k = \frac{12}{n} + 5n,\] which takes rational values between $-200$ and $200$ when $|n| \leq 39$ , excluding $n = 0$ . This leads to an answer of $2 \cdot 39 = \boxed{78}$

Solution: Solve for $k$ so \[k=-\frac{12}{x}-5x.\] Note that $x$ can be any integer in the range $[-39,0)\cup(0,39]$ so $k$ is rational with $\lvert k\rvert<200$ .  Hence, there are $39+39=\boxed{78}.$

Solution: Plug in $k=200$ to find the upper limit. You will find the limit to be a number from $0<x<-1$ and one that is just below $-39.$ All the integer values from $-1$ to $-39$ can be attainable through some value of $k$ . Since the question asks for the absolute value of $k$ , we see that the answer is $39\cdot2 = \boxed{78.}$


Answer: 3.0

There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution:
$\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$
The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
$a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$ $\frac{7}{12}$ is the result when $a, b,$ and $c$ are $3, 2,$ and $2$
$b$ being $3$ will make the fraction $\frac{2}{3}$ which is close to $\frac{25}{36}$
Finally, with $c$ being $6$ , the fraction becomes $\frac{25}{36}$ . In this case $a, b,$ and $c$ work, which means that $b$ must equal $\boxed{3.}$ ~lopkiloinm

Solution:
As above, notice that you get $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
Now, combine the fractions to get $\frac{bc+c+1}{abc}=\frac{25}{36}$
Let us assume $bc+c+1=25$ and $abc=36$ as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
From the first equation, we get $c(b+1)=24$ . Note also that from the second equation, $b$ and $c$ must be factors of 36.
After listing out the factors of 36 and utilising trial and error, we find that $c=6$ and $b=3$ works, with $a=2$ . So our answer is $\boxed{3.}$

Solution: Collapsed, $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1}}$ . Comparing this to $\sqrt[36]{N^{25}}$ , observe that $bc+c+1=25$ and $abc=36$ . The first can be rewritten as $c(b+1)=24$ . Then, $b+1$ has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows $36=2^2 3^2$ and $24=2^33$ . Then, $b=\boxed{3}$ , as only 4 and 3 factor into 36 and 24 while being 1 apart.


Answer: 119.0

There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10.  The sum of these two multiples of 7 is
$\text{(A)}\ 119 \qquad \text{(B)}\ 126 \qquad \text{(C)}\ 140 \qquad \text{(D)}\ 175 \qquad \text{(E)}\ 189$

Solution: Writing out all two digit numbers that have a digital sum of $10$ , you get $19, 28, 37, 46, 55, 64, 73, 82,$ and $91$ .  The two numbers on that list that are divisible by $7$ are $28$ and $91$ .  Their sum is $28+91=119$ , choice $\boxed{119}$

Solution: Writing out all the two digit multiples of $7$ , you get $14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84,$ and $91$ .  Again you find $28$ and $91$ have a digital sum of $10$ , giving answer $\boxed{119}$


Answer: 80.0

There are nonzero integers $a$ $b$ $r$ , and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$ . For each possible combination of $a$ and $b$ , let ${p}_{a,b}$ be the sum of the zeros of $P(x)$ . Find the sum of the ${p}_{a,b}$ 's for all possible combinations of $a$ and $b$

Solution:
Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have
$(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$
Applying difference of squares, and regrouping, we have
$(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$
So matching coefficients, we obtain
$q(r^2 + s^2) = 65$
$b = r^2 + s^2 + 2rq$
$a = q + 2r$
By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair.
We proceed by determining possible values for $q$ $r$ , and $s$ and using these to determine $a$ and $b$
If $q = 1$ $r^2 + s^2 = 65$ so (r, s) = $(\pm1, \pm 8), (\pm8, \pm 1), (\pm4, \pm 7), (\pm7, \pm 4)$
Similarly, for $q = 5$ $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\pm2, \pm 3), (\pm3, \pm 2)$
For $q = 13$ $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\pm2, \pm 1), (\pm1, \pm 2)$
Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases.
The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$ ).
Our answer is then $(1)(8) + (5)(4) + (13)(4) = \boxed{080}$


Answer: 880.0

There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$

Solution:
Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$ .
Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$ .
Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$ , along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$ .
This can easily be simplified to $\log(xy)=210$ , or $xy = 10^{210}$
$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$ , and $m+n$ equals to the sum of the exponents of $2$ and $5$ , which is $210+210 = 420$ .
Multiply by two to get $2m +2n$ , which is $840$ .
Then, use the first equation ( $\log x + 2\log(\gcd(x,y)) = 60$ ) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$ , which is a contradiction to the restrains we set before). Therefore, $\gcd(x,y)=x$ . Then, turn the equation into $3\log x = 60$ , which yields $\log x = 20$ , or $x = 10^{20}$ .
Factor this into $2^{20} \cdot 5^{20}$ , and add the two 20's, resulting in $m$ , which is $40$ .
Add $m$ to $2m + 2n$ (which is $840$ ) to get $40+840 = \boxed{880}$

Solution:
First simplifying the first and second equations, we get that
\[\log_{10}(x\cdot\text{gcd}(x,y)^2)=60\] \[\log_{10}(y\cdot\text{lcm}(x,y)^2)=570\]
Thus, when the two equations are added, we have that \[\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630\] When simplified, this equals \[\log_{10}(x^3y^3)=630\] so this means that \[x^3y^3=10^{630}\] so \[xy=10^{210}.\]
Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x<y$ and $x$ and $y$ are both powers of $10$ . This means the first equation would simplify to \[x^3=10^{60}\] and \[y^3=10^{570}.\] Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so \[3\cdot 40 + 2\cdot 380 = \boxed{880}.\]

Solution: Let $x=10^a$ and $y=10^b$ and $a<b$ . Then the given equations become $3a=60$ and $3b=570$ . Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$ . Our answer is $3(20+20)+2(190+190)=\boxed{880}$

Solution:
We will use the notation $(a, b)$ for $\gcd(a, b)$ and $[a, b]$ as $\text{lcm}(a, b)$ .
We can start with a similar way to Solution 1. We have, by logarithm properties, $\log_{10}{x}+\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$ . We can do something similar to the second equation and our two equations become \[x(x, y)^2=10^{60}\] \[y[x, y]^2=10^{570}\] Adding the two equations gives us $xy(x, y)^2[x, y]^2=10^{630}$ . Since we know that $(a, b)\cdot[a, b]=ab$ $x^3y^3=10^{630}$ , or $xy=10^{210}$ . We can express $x$ as $2^a5^b$ and $y$ as $2^c5^d$ . Another way to express $(x, y)$ is now $2^{min(a, c)}5^{min(b, d)}$ , and $[x, y]$ is now $2^{max(a, c)}5^{max(b, d)}$ . We know that $x<y$ , and thus, $a<c$ , and $b<d$ . Our equations for $lcm$ and $gcd$ now become \[2^a5^b(2^a5^a)^2=10^{60}\] or $a=b=20$ . Doing the same for the $lcm$ equation, we have $c=d=190$ , and $190+20=210$ , which satisfies $xy=210$ . Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$ .
~awsomek

Solution: Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$ . Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}$ . Notice that since $\alpha, \beta$ are coprime, and $\alpha < 5^{90}$ (Prove it yourself !) , $\alpha=1, \beta = 10^{170}$ . Hence, $x=10^{20}, y=10^{190}$ giving the answer $\boxed{880}$

Solution: Add the two equations and use the fact that $\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy$ to find that $xy=10^{210}$ . So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\leq a,b\leq210$ . If $a\geq105$ then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2=10^{60}$ is $a+2\left(210-a\right)=420-a$ , so $a=360$ , contradiction. So $a<105$ . Then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2$ is $a+2a=3a$ , so $a=20$ . Similarly, $b=20$ . Then $3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}$ as desired.


Answer: 36.0

There are positive integers that have these properties:
The product of the digits of the largest integer with both properties is
$\text{(A)}\ 7 \qquad \text{(B)}\ 25 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 48 \qquad \text{(E)}\ 60$

Solution:
Five-digit numbers will have a minimum of $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$ as the sum of their squares if the five digits are distinct and non-zero.  If there is a zero, it will be forced to the left by rule #2.
No digit will be greater than $7$ , as $8^2 = 64$
Trying four digit numbers $WXYZ$ , we have $w^2 + x^2 + y^2 + z^2 = 50$ with $0 < w < x < y < z < 8$
$z=7$ will not work, since the other digits must be at least $1^2 + 2^2 + 3^2 = 14$ , and the sum of the squares would be over $50$
$z=6$ will give $w^2 + x^2 + y^2 = 14$ $(w,x,y) = (1,2,3)$ will work, giving the number $1236$ .  No other number with $z=6$ will work, as $w, x,$ and $y$ would have to be greater.
$z=5$ will give $w^2 + x^2 + y^2 = 25$ $y=4$ forces $x=3$ and $w=0$ , which has a leading zero, and then we have 345 which is a 3-digit number.
$z=4$ can only give the number $1234$ , which does not satisfy the condition of the problem.
Thus, the number in question is $1236$ , and the product of the digits is $36$ , giving $\boxed{36}$ as the answer.


Answer: 330.0

There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$

Solution:
By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\sqrt{n}\cdot i$ and $\overline{z}=m-\sqrt{n}\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\overline{z}.$
We know that \begin{align*} z+\overline{z}&=2m, & (1) \\ z\overline{z}&=m^2+n. & (2) \end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$
Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\overline{z}+z\overline{z}=0,$ or $-21(z+\overline{z})+z\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$
Finally, the answer is $m+n=\boxed{330}.$

Solution:
$(-20)^{3} + (-20)a + b = 0$ , hence $-20a + b = 8000$
Also, $(-21)^{3} + c(-21)^{2} + d = 0$ , hence $441c + d = 9261$
$m + i \sqrt{n}$ satisfies both $\Rightarrow$ we can put it in both equations and equate to 0.
In the first equation, we get $(m + i \sqrt{n})^{3} + a(m + i \sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0$
Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 \rightarrow (1)$
In the second equation, we get $(m + i \sqrt{n})^{3} + c(m + i \sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0$
Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)$
Comparing (1) and (2),
$a = 2mc$ and $am + b = m^{2}c - nc + d \rightarrow (3)$
$b = 8000 + 20a \Rightarrow b = 40mc + 8000$ $d = 9261 - 441c$
Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$
This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261$
Hence, $c|1261 \Rightarrow c \in {1,13,97,1261}$
Consider case of $c = 1$
$c = 1 \Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$
$am + b = m^{2} - n + 8820$ (because c = 1)
Also, $m^{2} + n + 40m = 820 \rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \rightarrow (5)$
Solving (4) and (5) simultaneously gives $m = 10, n = 320$
[AIME can not have more than one answer, so we can stop here ... Not suitable for Subjective exam]
Hence, $m + n = 10 + 320 = \boxed{330}$

Solution: start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$ . After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$ , respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore $(-20)^3-20(a)+b=0$ and $(-21)^3+c*(-21)^2+d=0$ and you can set these two equations equal to each other while also substituting the values of $a$ $b$ $c$ , and $d$ above to give you $21m^2+21n-1682m+8000=0$ , then you can rearrange the equation into $21n = -21m^2+1682m-8000$ . With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$ therefore that means $1682m-8000=0(mod 21)$ which results in $2m-20=0(mod 21)$ which finally gives us m=10 mod 21. We can test the first obvious value of $m$ which is $10$ and we see that this works  as we get $m=10$ and $n=320$ . That means your answer will be $m + n = 10 + 320 = \boxed{330}$

Solution:
We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$
Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$ , and $Q(x) = x^2 + (c-21)x + (441 - 21c)$
By the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$
Therefore, $c-21 = -20$ and $441-21c = 400 + a$ . Solving the system of equations, we get $a = 20$ and $c = 1$ , so $P(x) = Q(x) = x^2 - 20x + 420$
Finally, by the quadratic formula, we have roots of $\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i$ , so our final answer is $10 + 320 = \boxed{330}$

Solution:
We plug -20 into the equation obtaining $(-20)^3-20a+b$ , likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$
Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$
By Vieta's, the sum of the roots in the first equation is $0$ , so $m$ must be $10$
Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$ . However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$
Given that $m$ is $10$ , then $n$ is equal to $320$
Therefore, the answer to the equation is $\boxed{330}$

Solution:
Since $m+i\sqrt{n}$ is a common root and all the coefficients are real, $m-i\sqrt{n}$ must be a common root, too.
Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients).
First, however, the product of the two common roots is: \begin{align*} &&&(x-m-i\sqrt{n})(x-m+i\sqrt{n})\\ &=&&x^2-x(m+i\sqrt{n}+m-i\sqrt{n})+(m+i\sqrt{n})(m-i\sqrt{n})\\ &=&&x^2-2xm+(m^2-i^2n)\\ &=&&x^2-2xm+m^2+n \end{align*}
Now, let's equate the two forms of both the polynomials: \[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\] \[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\] Now we can match the zero coefficients. \[-2m+20=0\to m=10\text{ and}\] \[-42m+m^2+n=0\to-420+100+n=0\to n=320\text{.}\] Thus, $m+n=10+320=\boxed{330}$


Answer: 4.0

There are several sets of three different numbers whose sum is $15$ which can be chosen from $\{ 1,2,3,4,5,6,7,8,9 \}$ .  How many of these sets contain a $5$
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution:
Let the three-element set be $\{ a,b,c \}$ and suppose that $a=5$
We need $b+c=10$ and $b\neq c$ .  This gives us four solutions, so there are $4$ sets with a $5$ also with the desired properties $\rightarrow \boxed{4}$


Answer: 5724.0

There are twenty-four $4$ -digit numbers that use each of the four digits $2$ $4$ $5$ , and $7$ exactly once.  Listed in numerical order from smallest to largest, the number in the $17\text{th}$ position in the list is
$\text{(A)}\ 4527 \qquad \text{(B)}\ 5724 \qquad \text{(C)}\ 5742 \qquad \text{(D)}\ 7245 \qquad \text{(E)}\ 7524$

Solution:
For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit.  Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.
We can just list from here: $5247,5274,5427,5472,5724 \rightarrow \boxed{5724}$


Answer: 23.0

There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$

Solution: There are three red faces, and two are on the card that is completely red, so our answer is $\frac{2}{3}$ , which is $\boxed{23}$


Answer: 392.0

There are two natural ways to inscribe a square in a given isosceles right triangle.
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$ .
What is the area (in $\text{cm}^2$ ) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?
[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (-25,10), W); label("B", (-25,0), W); label("C", (-15,0), E); label("Figure 1", (-20, -5)); label("Figure 2", (5, -5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); [/asy]
$\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$

Solution:
We are given that the area of the inscribed square is $441$ , so the side length of that square is $21$ . Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$ , then the legs of the larger isosceles right triangle ( $BC$ and $AB$ ) are equal to $42$
We now have that $3S=42\sqrt{2}$ , so $S=14\sqrt{2}$ . But we want the area of the square which is $S^2=(14\sqrt{2})^2= \boxed{392}$


Answer: 16.0

There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$ . What is the sum of those values of $a$
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$

Solution:
quadratic equation has exactly one root if and only if it is a perfect square .  So set
$4x^2 + ax + 8x + 9 = (mx + n)^2$
$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$
Two polynomials are equal only if their coefficients are equal, so we must have
$m^2 = 4, n^2 = 9$
$m = \pm 2, n = \pm 3$
$a + 8= 2mn = \pm 2\cdot 2\cdot 3 = \pm 12$
$a = 4$ or $a = -20$
So the desired sum is $(4)+(-20)=\boxed{16}$

Solution: Since this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have \[(a+8)^2 - 4(4)(9) = 0 \implies a^2 + 16a - 80.\] We can use the quadratic formula to solve for its roots (we can ignore the expression in the radical sign as it will cancel out due to the $\pm$ sign when added). So we must have \[\frac{-16 + \sqrt{\text{something}}}{2} + \frac{-16 - \sqrt{\text{something}}}{2}.\] $\frac{-32}{2} = \boxed{16}$

Solution: There is only one positive value for $k$ such that the quadratic equation would have only one solution. $k-8$ and $-k-8$ are the values of $a$ $-8-8=-16$ , so the answer is $\boxed{16}$


Answer: 9.0

Solution:
Multiply out the $7!$ to get
\[5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\]
By Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}$ , so $a_7 = 2$ . Then we move $a_7$ to the left and divide through by $7$ to obtain
\[\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.\]
We then repeat this procedure $\pmod{6}$ , from which it follows that $a_6 \equiv 514 \equiv 4 \pmod{6}$ , and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem ). The answer is $9 \Longrightarrow \boxed{9}$

Solution: We start by multiplying both sides by $7!$ , and we get: \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] After doing some guess and check, we find that the answer is $\boxed{9}$

Solution:
Let's clear up the fractions: \[\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}\] \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] Notice that if we divide everything by $7$ then we would have: \[514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7\] Since $0 \le a_7<7$ and $a_7$ must be an integer, then we have $\frac{2}{7}=\frac{1}{7}a_7$ , so $a_7=2$
Similarly, if we divide everything by $6$ , then we would have: \[85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6\] Again, since $0 \le a_6<6$ and $a_6$ must be an integer, we have $\frac{4}{6}=\frac{1}{6}a_6$ , so $a_6=4$
The pattern repeats itself, so in the end we have $a_2=1$ $a_3=1$ $a_4=1$ $a_5=0$ $a_6=4$ $a_7=2$ . So $a_2+a_3+a_4+a_5+a_6+a_7=\boxed{9}$

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


Answer: 21.0

There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$

Solution:
In base $3$ , we find that $\overline{2008}_{10} = \overline{2202101}_{3}$ . In other words,
In order to rewrite as a sum of perfect powers of $3$ , we can use the fact that $2 \cdot 3^k = 3^{k+1} - 3^k$
The answer is $7+5+4+3+2+0 = \boxed{021}$


Answer: 80.0

There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$

Solution:
Completing the square $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$ . Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares
Since $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \cdot 61$ , the factors must be $2$ and $122$ . Since $x,y > 0$ , we have $y - x - 42 = 2$ and $y + x + 42 = 122$ ; the latter equation implies that $x + y = \boxed{080}$

Solution:
We complete the square like in the first solution: $y^2 = (x+42)^2 + 244$ . Since consecutive squares differ by the consecutive odd numbers, we note that $y$ and $x+42$ must differ by an even number. We can use casework with the even numbers, starting with $y-(x+42)=2$ \begin{align*}2(x+42)+1+2(x+42)+3&=244\\ \Leftrightarrow x&=18\end{align*}
Thus, $y=62$ and the answer is $\boxed{080}$

Solution:
We see that $y^2 \equiv x^2 + 4 \pmod{6}$ . By quadratic residues , we find that either $x \equiv 0, 3 \pmod{6}$ . Also, $y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}$ , so $x \equiv 0, 2 \mod{4}$ . Combining, we see that $x \equiv 0 \mod{6}$
Testing $x = 6$ and other multiples of $6$ , we quickly find that $x = 18, y = 62$ is the solution. $18+62=\boxed{080}$

Solution:
We solve for x: $x^2 + 84x + 2008-y^2 = 0$
$x=\dfrac{-84+\sqrt{84^2-4\cdot 2008+4y^2}}{2}=-42+\sqrt{y^2-244}$
So $y^2-244$ is a perfect square. Since 244 is even, the difference $\sqrt{y^2-244} -y^2$ is even, so we try $y^2-244=(y-2)^2$ $-244=-4y+4$ $y=62$
Plugging into our equation, we find that $x=18$ , and $(x,y)=(18,62)$ indeed satisfies the original equation. $x+y=\boxed{080}$

Solution:
Let $y=x+d$ for some $d>0$ , substitute into the original equation to get $84x + 2008 = 2xd + d^2$
All terms except for the last one are even, hence $d^2$ must be even, hence let $d=2e$ . We obtain $21x + 502 = xe + e^2$ . Rearrange to $502-e^2 = x(e-21)$
Obviously for $0<e<21$ the right hand side is negative and the left hand side is positive. Hence $e\geq 21$ . Let $e=21+f$ , then $f\geq 0$
We have $502 - (21+f)^2 = xf$ . Left hand side simplifies to $61 - 42f + f^2$ . As $x$ must be an integer, $f$ must divide the left hand side. But $61$ is a prime, which only leaves two options: $f=1$ and $f=61$
Option $f=61$ gives us a negative $x$ . Option $f=1$ gives us $x=61/f - 42 + f = 18$ , and $y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62$ , hence $x+y=\boxed{080}$

Solution: First complete the square to get $y^2 = (x+42)^2 + 244$ . Remember that squares are the sums of consecutive odd integers, so when the difference between the two squares is 244, the two squares must be an even number of odd integers apart. However, there is only one distinct solution, as the problem states, and very quickly you will realize that only two odd integers work. When there are four, then the numbers are not odd, and when it is any other even integer it does not divide. So we need two consecutive odd integers that sum to 244. Easily we find 121 and 123. 121 is the 61st odd integer and 123 is the 62nd odd integer, so $(x+42)^2$ is the sum of the first 60 odd integers, or $(60)^2$ , while $y^2$ is $62^2$ for the same reasons. That way we get $x=12$ $y=62$ , hence $x+y=\boxed{080}$


Answer: 944.0

There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution:
Define $\left\{ x \right\} = x - \left\lfloor x \right\rfloor$
First, we bound $U$
We establish an upper bound of $U$ . We have \begin{align*} U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\ & = \frac{1}{5} \sum_{n=1}^{2023} n^2 - \frac{a}{5} \sum_{n=1}^{2023} n \\ & = \frac{1012 \cdot 2023}{5} \left( 1349 - a \right) \\ & \triangleq UB . \end{align*}
We establish a lower bound of $U$ . We have \begin{align*} U & =  \sum_{n=1}^{2023} \left(  \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\}  \right) \\ & = \sum_{n=1}^{2023} \frac{n^2 - na}{5} - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = UB -  \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & \geq UB -  \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} . \end{align*}
We notice that if $5 | n$ , then $\frac{n^2 - na}{5} \in \Bbb Z$ .
Thus, \begin{align*} U & \geq UB -  \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\ & \geq UB -  \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\ & = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\ & = UB - 1619 \\ & \triangleq LB . \end{align*}
Because $U \in \left[ - 1000, 1000 \right]$ and $UB - LB = 1619 < \left( 1000 - \left( - 1000 \right) \right)$ , we must have either $UB \in \left[ - 1000, 1000 \right]$ or $LB \in \left[ - 1000, 1000 \right]$
For $UB \in \left[ - 1000, 1000 \right]$ , we get a unique $a = 1349$ .
For $LB \in \left[ - 1000, 1000 \right]$ , there is no feasible $a$
Therefore, $a = 1349$ . Thus $UB = 0$
Next, we compute $U$
Let $n = 5 q + r$ , where $r = {\rm Rem} \ \left( n, 5 \right)$
We have \begin{align*} \left\{ \frac{n^2 - na}{5} \right\} & = \left\{ \frac{\left( 5 q + r \right)^2 - \left( 5 q + r \right)\left( 1350 - 1 \right)}{5} \right\} \\ & = \left\{ 5 q^2 + 2 q r - \left( 5 q + r \right) 270 + q + \frac{r^2 + r}{5}  \right\} \\ & = \left\{\frac{r^2 + r}{5}  \right\} \\ & = \left\{ \begin{array}{ll} 0 & \mbox{ if } r = 0, 4 \\ \frac{2}{5} & \mbox{ if } r = 1, 3 \\ \frac{1}{5} & \mbox{ if } r = 2 \end{array} \right. . \end{align*}
Therefore, \begin{align*} U & =  \sum_{n=1}^{2023} \left(  \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\}  \right) \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{q=0}^{404} \sum_{r=0}^4 \left\{\frac{r^2 + r}{5}  \right\} + \left\{ \frac{0^2 - 0 \cdot a}{5} \right\} + \left\{ \frac{2024^2 - 2024a}{5} \right\} \\ & = - \sum_{q=0}^{404} \left( 0 + 0 + \frac{2}{5} + \frac{2}{5} + \frac{1}{5} \right) + 0 + 0 \\ & = - 405 . \end{align*}
Therefore, $a + U = 1349 - 405 = \boxed{944}$

Solution:
We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$ . Since for any real number $x$ $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$ , we have $U \le U' \le U + 2023$ . Now, since $-1000 \le U \le 1000$ , we have $-1000 \le U' \le 3023$
Now, we can solve for $U'$ in terms of $a$ . We have: \begin{align*} U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\ &= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\ &= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\ &= \frac{2023(2024)(4047-3a)}{30} \\ \end{align*} So, we have $U' = \frac{2023(2024)(4047-3a)}{30}$ , and $-1000 \le U' \le 3023$ , so we have $-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023$ , or $-30000 \le 2023(2024)(4047-3a) \le 90690$ . Now, $2023 \cdot 2024$ is much bigger than $90690$ or $30000$ , and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$ , or $a = 1349$ , and $U' = 0$
Now, we can find $U - U'$ . We have: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\ &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \end{align*}. Now, if $n^2-1349n \equiv 0 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0$ , and if $n^2-1349n \equiv 1 \text{ (mod 5)}$ , then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}$ , and so on. Testing with $n \equiv 0,1,2,3,4, \text{ (mod 5)}$ , we get $n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$ , we get: \begin{align*} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\ &= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot t0 \\ &= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\ &= -405 \end{align*} Since $U' = 0$ , this gives $U = -405$ , and we have $a + U = 1349-405 = \boxed{944}$

Solution:
We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$ ) from the numerator of $\frac{n^2-na}{5}$ . For example, $\left\lfloor \frac{7}{5} \right\rfloor = \frac{7-2}{5} = 1$
Note that the congruence of $n^2 - na$ (mod $5$ ) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$ ) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$ ).
For example, if $a \equiv 1$ (mod $5$ ), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \equiv 1,2,3,4,0$ (mod $5$ ). Let the sum of these elements be $s$ . Note that for each “loop” of the numerator (mod $5$ ), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$ ) throughout the sum, as $5 \cdot 404=2020$ . Then
$U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2}  -404s \right)}{5}$
Where $n=2023$ . Note that since $5 \cdot 404=2020$ , this is an approximation for $U$ because the equation disregards the remainder (mod $5$ ) when $n=2021, 2022$ , and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$ , so we can proceed with this approximation to solve for $a$
Factoring our approximation gives $U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$
We set $a= \frac{(2n+1)}{3} = 1349$ to make $\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$ , accordingly minimizing $|U|$ , yielding $U \approx  \frac{-404s}{5}$
If $a$ increases or decreases by $1$ , then $U$ changes by $\frac {(n)(n+1)}{2 \cdot 5} = \frac {2023 \cdot 2024}{10}$ which clearly breaks the inequality on $U$ . Therefore $a=1349 \equiv 4$ (mod $5$ ) giving the set of remainders $(2,1,2,0,0)$ , so $s=5$ and our approximation yields $U \approx -404$ . However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \frac{(2+1+2)}{5} = - 405$ , giving an answer of $1349-405= \boxed{944}$

Solution: Consider the integral \[\int_{0}^{2023} \dfrac{n^2-na}{5} \, dn.\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated(if you know calculus) to be \[\dfrac{1}{15}2023^3-\dfrac{a}{10}2023^2=2023^2\left(\dfrac{2023}{15}-\dfrac{a}{10}\right).\] Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set \[U'=\sum_{n=1}^{2023}\dfrac{n^2-1349n}{5}\] and \[U''=\sum_{n=1}^{2023}\{\dfrac{n^2-1349n}{5}\}.\] Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is \[1349+(-405)=\boxed{944}.\] BS2012


Answer: 137.0

There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$

Solution:
First, substitute $2^{17}$ with $x$ .
Then, the given equation becomes $\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.
Now consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \cdot (2^{17}-1)$ .
Note that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by difference of powers factorization (or by considering the expansion of $2^{17}=2^{16}+2^{15}+...+2+2$ ).
Thus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2.
Applying the same method to each of $x^{14}-x^{13}$ $x^{12}-x^{11}$ , ... , $x^{2}-x^{1}$ , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \cdot 8=136$ .
But we must count also the $x^0$ term.
Thus, Our answer is $136+1=\boxed{137}$

Solution:
Multiply both sides by $2^{17}+1$ to get \[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\]
Notice that $a_1 = 0$ , since there is a $1$ on the LHS. However, now we have an extra term of $2^{18}$ on the right from $2^{a_1+17}$ . To cancel it, we let $a_2 = 18$ . The two $2^{18}$ 's now combine into a term of $2^{19}$ , so we let $a_3 = 19$ . And so on, until we get to $a_{18} = 34$ . Now everything we don't want telescopes into $2^{35}$ . We already have that term since we let $a_2 = 18 \implies a_2+17 = 35$ . Everything from now on will automatically telescope to $2^{52}$ . So we let $a_{19}$ be $52$
As you can see, we will have to add $17$ $a_n$ 's at a time, then "wait" for the sum to automatically telescope for the next $17$ numbers, etc, until we get to $2^{289}$ . We only need to add $a_n$ 's between odd multiples of $17$ and even multiples. The largest even multiple of $17$ below $289$ is $17\cdot16$ , so we will have to add a total of $17\cdot 8$ $a_n$ 's. However, we must not forget we let $a_1=0$ at the beginning, so our answer is $17\cdot8+1 = \boxed{137}$

Solution: In order to shorten expressions, $\#$ will represent $16$ consecutive $0$ s when expressing numbers. Think of the problem in binary. We have $\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}$ Note that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)$ $= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$ and $(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)$ $= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$ Since $\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$ $-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$ $= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2$ this means that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}$ so $\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})$ $= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})$ Expressing each of the pairs of the form $2^{n + 17} - 2^n$ in binary, we have $\phantom{=\ } 1000000000000000000 \cdots 0_2$ $-\ \phantom{10000000000000000} 10 \cdots 0_2$ $= \phantom{1} 111111111111111110 \cdots 0_2$ or $2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}$ This means that each pair has $17$ terms of the form $2^n$ Since there are $8$ of these pairs, there are a total of $8 \cdot 17 = 136$ terms. Accounting for the $2^0$ term, which was not in the pair, we have a total of $136 + 1 = \boxed{137}$ terms. ~ emerald_block


Answer: 137.0

Solution: Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\boxed{137}$


Answer: 137.0

Solution:
First, substitute $2^{17}$ with $x$ .
Then, the given equation becomes $\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.
Now consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \cdot (2^{17}-1)$ .
Note that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by difference of powers factorization (or by considering the expansion of $2^{17}=2^{16}+2^{15}+...+2+2$ ).
Thus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2.
Applying the same method to each of $x^{14}-x^{13}$ $x^{12}-x^{11}$ , ... , $x^{2}-x^{1}$ , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \cdot 8=136$ .
But we must count also the $x^0$ term.
Thus, Our answer is $136+1=\boxed{137}$

Solution:
Multiply both sides by $2^{17}+1$ to get \[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\]
Notice that $a_1 = 0$ , since there is a $1$ on the LHS. However, now we have an extra term of $2^{18}$ on the right from $2^{a_1+17}$ . To cancel it, we let $a_2 = 18$ . The two $2^{18}$ 's now combine into a term of $2^{19}$ , so we let $a_3 = 19$ . And so on, until we get to $a_{18} = 34$ . Now everything we don't want telescopes into $2^{35}$ . We already have that term since we let $a_2 = 18 \implies a_2+17 = 35$ . Everything from now on will automatically telescope to $2^{52}$ . So we let $a_{19}$ be $52$
As you can see, we will have to add $17$ $a_n$ 's at a time, then "wait" for the sum to automatically telescope for the next $17$ numbers, etc, until we get to $2^{289}$ . We only need to add $a_n$ 's between odd multiples of $17$ and even multiples. The largest even multiple of $17$ below $289$ is $17\cdot16$ , so we will have to add a total of $17\cdot 8$ $a_n$ 's. However, we must not forget we let $a_1=0$ at the beginning, so our answer is $17\cdot8+1 = \boxed{137}$

Solution: In order to shorten expressions, $\#$ will represent $16$ consecutive $0$ s when expressing numbers. Think of the problem in binary. We have $\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}$ Note that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)$ $= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$ and $(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)$ $= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$ Since $\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$ $-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$ $= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2$ this means that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}$ so $\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})$ $= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})$ Expressing each of the pairs of the form $2^{n + 17} - 2^n$ in binary, we have $\phantom{=\ } 1000000000000000000 \cdots 0_2$ $-\ \phantom{10000000000000000} 10 \cdots 0_2$ $= \phantom{1} 111111111111111110 \cdots 0_2$ or $2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}$ This means that each pair has $17$ terms of the form $2^n$ Since there are $8$ of these pairs, there are a total of $8 \cdot 17 = 136$ terms. Accounting for the $2^0$ term, which was not in the pair, we have a total of $136 + 1 = \boxed{137}$ terms. ~ emerald_block


Answer: 137.0

Solution: Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\boxed{137}$


Answer: 107.0

There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a+b$

Solution: Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{4}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$ . Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}$ . Therefore, the probability that rains on at least one of the days is $1-\dfrac{33}{70}=\dfrac{37}{70}$ , so adding up the $2$ numbers, we have $37+70=\boxed{107}$


Answer: 902.0

There is a collection of $25$ indistinguishable white chips and $25$ indistinguishable black chips. Find the number of ways to place some of these chips in the $25$ unit cells of a $5\times5$ grid such that:

Solution:
The problem says "some", so not all cells must be occupied.
We start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is $1$ way to select $5$ cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- $1$ way . There are $5$ ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining $4$ cells have $2^4-1$ different ways( $-1$ comes from all blank). This gives us $75$ ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is $10*15$ $10*15$ $5*15$ . Adding these up, we have $1+75+150+150+75 = 451$ . Multiplying this by 2, we get $\boxed{902}$ .
~westwoodmonster

Solution:
Note that the answer is equivalent to the number of ways to choose rows and columns that the white chips occupy, as once those are chosen, there is only one way to place the chips, and every way to place the chips corresponds to a set of rows and columns occupied by the white pieces.
If the white pieces occupy none of the rows, then because they don't appear on the board, they will not occupy any of the columns. Similar logic can be applied to show that if white pieces occupy all of the rows, they will also occupy all of the columns.
The number of sets of rows and columns that white can occupy are $2^{5} - 2 = 30$ each, accounting for the empty and full set.
So, including the board with 25 white pieces and the board with 25 black pieces, the answer is $30^{2}+2 = \boxed{902}$

Solution:
Case 1: All chips on the grid have the same color.
In this case, all cells are occupied with chips with the same color.
Therefore, the number of configurations in this case is 2.
Case 2: Both black and white chips are on the grid.
Observation 1: Each colored chips must occupy at least one column and one row.
This is because, for each given color, there must be at least one chip. Therefore, all chips placed in the cells that are in the same row or the same column with this given chip must have the same color with this chip.
Observation 2: Each colored chips occupy at most 4 rows and 4 columns.
This directly follows from Observation 1.
Observation 3: For each color, if all chips in this color occupy columns with $x$ -coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$ -coordinates $\left\{ y_1, \cdots, y_n \right\}$ , then every cell $\left( x, y \right)$ with $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$ is occupied by a chip with the same color.
This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.
If there is any cell in this region that is empty, then it violates Condition 3.
Observation 4: For each color, if all chips in this color occupy columns with $x$ -coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$ -coordinates $\left\{ y_1, \cdots, y_n \right\}$ , then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \in \left\{ y_1, \cdots , y_n \right\}$ , or $x \in \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is empty.
This is because, if there is any cell in this region occupied by a chip with a different color, it violates Condition 2.
Observation 5: For each color, if all chips in this color occupy columns with $x$ -coordinates $\left\{ x_1, \cdots, x_m \right\}$ and rows with $y$ -coordinates $\left\{ y_1, \cdots, y_n \right\}$ , then every cell $\left( x, y \right)$ with $x \notin \left\{ x_1, \cdots , x_m \right\}$ and $y \notin \left\{ y_1, \cdots , y_n \right\}$ is occupied by chips with the different color.
This follows from Condition 3.
By using the above observations, the number of feasible configurations in this case is given by
\begin{align*}
\sum_{n=1}^4 \sum_{m=1}^4 \binom{5}{n} \binom{5}{m}
& = \left( \sum_{n=1}^4 \binom{5}{n} \right)
\left( \sum_{m=1}^4 \binom{5}{m} \right) \\
& = \left( 2^5 - 2 \right)^2 \\
& = 900 .
\end{align*}
Putting all cases together, the total number of feasible configurations is $2 + 900 =  \boxed{902}$