Answer: 78.0

Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$

Solution: When $m=1$ $a_{n+1}=1+a_n+n$ . Hence, \[a_{2}=1+a_1+1\] \[a_{3}=1+a_2+2\] \[a_{4}=1+a_3+3\] \[\dots\] \[a_{12}=1+a_{11}+11\] Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=\boxed{78}$

Solution:
Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$ $a_{m+1}=a_m+a_{1}+m$
Since $a_1 = 1$ $a_{m+1}=a_m+m+1$
Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$ , and so on until $a_2 = a_1 + 2$
Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2$
Adding the Right Hand Sides of these equations gives
$(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$
These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ and $a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)$
Substituting $a_1 = 1$ $a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}$
Thus we have a general formula for $a_m$ and substituting $m=12$ $a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{78}$

Solution: We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since $a_{m+n} = a_m+a_n +mn$ , we know $a_2=a_1+a_1+1\cdot1=1+1+1=3$ . After this, we can use $a_2$ to find $a_4$ $a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10$ . Now, we can use $a_2$ and $a_4$ to find $a_6$ , or $a_6=a_4+a_2+4\cdot 2 = 10+3+8=21$ . Lastly, we can use $a_6$ to find $a_{12}$ $a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{78}$

Solution:
We can set $n$ equal to $m$ , so we can say that \[a_{m + m} = a_m + a_m + m*m\] \[a_{2m} = 2a_m + m^2\]
We set $2m = 12$ , we get $m = 6$ \[a_{12} = 2a_6 + 36\]
We set $2m = 6$ m, we get $m = 3$ \[a_6 = 2a_3 + 9\]
Solving for $a_3$ is easy, just direct substitution. \[a_2 = 1 + 1 + 1 = 3\] \[a_3 = a_{2 + 1} = 3 + 1 + 2 = 6\]
Substituting, we get \[a_6 = 2(6) + 9 = 21\] \[a_{12} = 2(21) + 36 = 78\]
Thus, the answer is $\boxed{78}$

Solution:
Note that the sequence of triangular numbers $T_n=1+2+3+...+n$ satisfies these conditions. It is immediately obvious that it satisfies $a_1=1$ , and $a_{m+n}=a_m+a_n+mn$ can be visually proven with the diagram below.

This means that we can use the triangular number formula $T_n = \frac{n(n+1)}{2}$ , so the answer is $T_{12} = \frac{12(12+1)}{2} = \boxed{78}$


Answer: 1341.0

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$ , and in general,
\[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}\]
Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$ .  What is the sum of all integers $k$ $1\le k \le 2011$ , such that $a_k=b_k?$
$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$

Solution:
First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$ (decreasing exponential function), and $g(x) = x^k$ for $k > 0$ (increasing power function for positive $x$ ). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.
We will now examine the first few terms.
Comparing $a_1$ and $a_2$ $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$
Therefore, $0 < a_1 < a_2 < 1$
Comparing $a_2$ and $a_3$ $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$
Comparing $a_1$ and $a_3$ $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$
Therefore, $0 < a_1 < a_3 < a_2 < 1$
Comparing $a_3$ and $a_4$ $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$
Comparing $a_2$ and $a_4$ $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$
Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$
Continuing in this manner, it is easy to see a pattern(see Note 1).
Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$ . Solving gives $\boxed{1341}$

Solution:
Start by looking at the first few terms and comparing them to each other. We can see that $a_1 < a_2$ , and that $a_1 < a_3 < a_2$ , and that $a_3 < a_4 < a_2$ , and that $a_3 < a_5 < a_4$ ...
From this, we find the pattern that $a_{k-1} < a_{k+1} < a_k$
Examining this relationship, we see that every new number $a_k$ will be between the previous two terms, $a_{k-1}$ and $a_{k-2}$ . Therefore, we can see that $a_1$ is the smallest number, $a_2$ is the largest number, and that all odd terms are less than even terms. Furthermore, we can see that for every odd k, $a_k < a_{k+2}$ , and for every even k, $a_k > a_{k+2}$
This means that rearranging the terms in descending order will first have all the even terms from $a_2$ to $a_{2012}$ , in that order, and then all odd terms from $a_{2011}$ to $a_1$ , in that order (so $\{b_k\}_{k=1}^{2011} = {a_2, a_4, a_6, ... a_{2008}, a_{2010}, a_{2011}, a_{2009}, ... a_5, a_3, a_1}$ ).
We can clearly see that there will be no solution k where k is even, as the $k$ th term in $\{a_k\}_{k=1}^{2011}$ will appear in the same position in its sequence as the $2k$ th term does in $\{a_k\}_{k=1}^{2011}$ , where k is even. Therefore, we only have to look at the odd terms of $a_k$ , which occur in the latter part of $b_k$
Looking at the back of both sequences, we see that term k in $\{a_k\}_{k=1}^{2011}$ progresses backwards in the equation $2012 - k$ , and that term k in $\{a_k\}_{k=1}^{2011}$ progresses backwards in the equation $2k - 1$ . Setting these two expressions equal to each other, we get $671$
However, remember that we started counting from the back of both sequences. So, plugging $671$ back into either side of the equation from earlier, we get our answer of $\boxed{1341}$


Answer: 1.0

Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term.  The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ .  What is $r_1 + r_2$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$

Solution:
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$
Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$

Solution:
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$
Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ .
This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$
Which can be further rewritten as $r_1-r_1^2 = r_2-r_2^2$ .
Rearranging the equation we get $r_1-r_2 = r_1^2-r_2^2$ .
Expressing this as a difference of squares we get $r_1-r_2 = (r_1-r_2)(r_1+r_2)$
Dividing by like terms we finally get $r_1+r_2 = \boxed{1}$ as desired.


Answer: 251.0

Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer.

Solution:
By the product-to-sum identities , we have that $2\cos a \sin b = \sin (a+b) - \sin (a-b)$ . Therefore, this reduces to a telescope series: \begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\ &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) \end{align*}
Thus, we need $\sin \left(\frac{n(n+1)\pi}{2008}\right)$ to be an integer; this can be only $\{-1,0,1\}$ , which occur when $2 \cdot \frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1$ . We know that $n$ cannot be $250$ as $250$ isn't divisible by $4$ , so 1004 doesn't divide $n(n+1) = 250 \cdot 251$ . Therefore, it is clear that $n = \boxed{251}$ is the smallest such integer.

Solution:
We proceed with complex trigonometry. We know that for all $\theta$ , we have $\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)$ and $\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)$ for some complex number $z$ on the unit circle. Similarly, we have $\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)$ and $\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)$ . Thus, we have $\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)$
$= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)$
$= \dfrac{1}{2} \left( \dfrac{1}{2i} \left(z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} \right) - \dfrac{1}{2i} \left(z^{n^2 - n} - \dfrac{1}{z^{n^2 - n}} \right) \right)$
$= \dfrac{1}{2} \left( \sin ((n^2 + n)a) - \sin ((n^2 - n)a) \right)$
$= \dfrac{1}{2} \left( \sin(((n+1)^2 - (n+1))a) - \sin((n^2 - n)a) \right)$
which clearly telescopes! Since the $2$ outside the brackets cancels with the $\dfrac{1}{2}$ inside, we see that the sum up to $n$ terms is
$\sin ((2^2 - 2)a) - \sin ((1^2 - 1)a) + \sin ((3^3 - 3)a) - \sin ((2^2 - 2)a) \cdots + \sin (((n+1)^2 - (n+1))a) - \sin ((n^2 - n)a)$
$= \sin (((n+1)^2 - (n+1))a) - \sin(0)$
$= \sin ((n^2 + n)a) - 0$
$= \sin \left( \dfrac{n(n+1) \pi}{2008} \right)$
This expression takes on an integer value iff $\dfrac{2n(n+1)}{2008} = \dfrac{n(n+1)}{1004}$ is an integer; that is, $1004 \mid n(n+1)$ . Clearly, $1004 = 2^2 \cdot 251$ , implying that $251 \mid n(n+1)$ . Since we want the smallest possible value of $n$ , we see that we must have ${n,n+1} = 251$ . If $n+1 = 251 \rightarrow n=250$ , then we have $n(n+1) = 250(251)$ , which is clearly not divisible by $1004$ . However, if $n = 251$ , then $1004 \mid n(n+1)$ , so our answer is $\boxed{251}$


Answer: 315.0

Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that
What is the smallest possible value of $a$
$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$

Solution:
We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$
Thus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$
Then, plugging in values of $2,4,6,8,$ we get
\[P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a\] \[P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a\] \[P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a\] \[P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a\]
$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $\text{lcm}(15,9,15,105)$ .
Solving, we receive $315$ , so our answer is $\boxed{315}$

Solution:
The evenly-spaced data suggests using discrete derivatives to tackle this problem.  First, note that any polynomial of degree $n$
can also be written as
Moreover, the coefficients $a_i$ are integers for $i=1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=1, 2, \ldots n$ .  This latter form is convenient for calculating discrete derivatives of $P(x)$
The discrete derivative of a function $f(x)$ is the related function $\Delta f(x)$ defined as
With this definition, it's easy to see that for any positive integer $k$ we have
This in turn allows us to use successive discrete derivatives evaluated at $x=1$ to calculate all of the coefficients $b_i$ using
We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
Thus we can read down the column for $x=1$ to find that $k! b_k = (-2)^k a$ for $k = 0, 1, \ldots, 7$ .  Interestingly, even if we choose $P(x)$ to have degree greater than $7$ , the $8$ coefficients of lowest order always satisfy these conditions.  Moreover, it's straightforward to show that the $P(x)$ of degree $7$ with $b_k$ satisfying these conditions will fit the data given in the problem statement.  Thus, to find minimal necessary and sufficient conditions on the value of $a$ , we need only consider these $8$ equations.  As a result, $P(x)$ with integer coefficients fitting the given data exists iff $k!$ divides $2^k a$ for $k = 0, 1, \ldots, 7$ .  In other words, it's necessary and sufficient that
The last condition holds if $7 \cdot 3 \cdot 5 \cdot 3 = 315$ divides evenly into $a$ .  Since such $a$ will also satisfy the first $7$ conditions, our answer is $\boxed{315}$


Answer: 315.0

Solution:
We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$
Thus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$
Then, plugging in values of $2,4,6,8,$ we get
\[P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a\] \[P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a\] \[P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a\] \[P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a\]
$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $\text{lcm}(15,9,15,105)$ .
Solving, we receive $315$ , so our answer is $\boxed{315}$

Solution:
The evenly-spaced data suggests using discrete derivatives to tackle this problem.  First, note that any polynomial of degree $n$
can also be written as
Moreover, the coefficients $a_i$ are integers for $i=1, 2, \ldots n$ iff the coefficients $b_i$ are integers for $i=1, 2, \ldots n$ .  This latter form is convenient for calculating discrete derivatives of $P(x)$
The discrete derivative of a function $f(x)$ is the related function $\Delta f(x)$ defined as
With this definition, it's easy to see that for any positive integer $k$ we have
This in turn allows us to use successive discrete derivatives evaluated at $x=1$ to calculate all of the coefficients $b_i$ using
We can also calculate the following table of discrete derivatives based on the data points given in the problem statement:
Thus we can read down the column for $x=1$ to find that $k! b_k = (-2)^k a$ for $k = 0, 1, \ldots, 7$ .  Interestingly, even if we choose $P(x)$ to have degree greater than $7$ , the $8$ coefficients of lowest order always satisfy these conditions.  Moreover, it's straightforward to show that the $P(x)$ of degree $7$ with $b_k$ satisfying these conditions will fit the data given in the problem statement.  Thus, to find minimal necessary and sufficient conditions on the value of $a$ , we need only consider these $8$ equations.  As a result, $P(x)$ with integer coefficients fitting the given data exists iff $k!$ divides $2^k a$ for $k = 0, 1, \ldots, 7$ .  In other words, it's necessary and sufficient that
The last condition holds if $7 \cdot 3 \cdot 5 \cdot 3 = 315$ divides evenly into $a$ .  Since such $a$ will also satisfy the first $7$ conditions, our answer is $\boxed{315}$


Answer: 896.0

Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$ . Find the remainder when $x$ is divided by $1000$

Solution:
The first condition implies
\[a^{128} = \log_a\log_a 2 + \log_a 24 - 128\]
\[128+a^{128} = \log_a\log_a 2^{24}\]
\[a^{a^{128}a^{a^{128}}} = 2^{24}\]
\[\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8\]
So $a^{a^{128}} = 8$
Putting each side to the power of $128$
\[\left(a^{128}\right)^{\left(a^{128}\right)} = 8^{128} = 64^{64},\]
so $a^{128} = 64 \implies a = 2^{\frac{3}{64}}$ . Specifically,
\[\log_a(x) = \frac{\log_2(x)}{\log_2(a)} = \frac{64}{3}\log_2(x)\]
so we have that
\[256 = \log_a(\log_a(x)) = \frac{64}{3}\log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[12 = \log_2\left(\frac{64}{3}\log_2(x)\right)\]
\[2^{12} = \frac{64}{3}\log_2(x)\]
\[192 = \log_2(x)\]
\[x = 2^{192}\]
We only wish to find $x\bmod 1000$ . To do this, we note that $x\equiv 0\bmod 8$ and now, by the Chinese Remainder Theorem, wish only to find $x\bmod 125$ . By Euler's Totient Theorem
\[2^{\phi(125)} = 2^{100} \equiv 1\bmod 125\]
so
\[2^{192} \equiv \frac{1}{2^8} \equiv \frac{1}{256} \equiv \frac{1}{6} \bmod 125\]
so we only need to find the inverse of $6 \bmod 125$ . It is easy to realize that $6\cdot 21 = 126 \equiv 1\bmod 125$ , so
\[x\equiv 21\bmod 125, x\equiv 0\bmod 8.\]
Using Chinese Remainder Theorem , we get that $x\equiv \boxed{896}\bmod 1000$ , finishing the solution.

Solution:
If you've found $x$ but you don't know that much number theory.
Note $192 = 3 * 2^6$ , so what we can do is take $2^3$ and keep squaring it (mod 1000).
\[2^3 = 8\] \[2^6 = 8*8 = 64\] \[2^{12} = 64*64 \equiv 96\bmod 1000\] \[2^{24} \equiv 96*96 \equiv 216\bmod 1000\] \[2^{48} \equiv 216*216 \equiv 656\bmod 1000\] \[2^{96} \equiv 656*656 \equiv 336\bmod 1000\] \[2^{192} \equiv 336*336 \equiv \boxed{896}\bmod 1000\]


Answer: 671.0

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

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

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


Answer: 671.0

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


Answer: 18.0

Let $a$ $b$ $c$ , and $d$ be real numbers with $|a-b|=2$ $|b-c|=3$ , and $|c-d|=4$ . What is the sum of all possible values of $|a-d|$
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$

Solution:
From $|a-b|=2$ we get that $a=b\pm 2$
Similarly, $b=c\pm3$ and $c=d\pm4$
Substitution gives $a=d\pm 4\pm 3\pm 2$ . This gives $|a-d|=|\pm 4\pm 3\pm 2|$ . There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm2$
$4+3+2=9$
$4+3-2=5$
$4-3+2=3$
$-4+3+2=1$
$4-3-2=-1$
$-4+3-2=-3$
$-4-3+2=-5$
$-4-3-2=-9$
Therefore, the only possible values of $|a-d|$ are $9$ $5$ $3$ , and $1$ . Their sum is $\boxed{18}$

Solution:
If we add the same constant to all of $a$ $b$ $c$ , and $d$ , we will not change any of the differences. Hence we can assume that $a=0$
From $|a-b|=2$ we get that $|b|=2$ , hence $b\in\{-2,2\}$
If we multiply all four numbers by $-1$ , we will not change any of the differences. (This is due to the fact that we are calculating $|d|$ at the end ~Williamgolly) WLOG we can assume that $b=2$
From $|b-c|=3$ we get that $c\in\{-1,5\}$
From $|c-d|=4$ we get that $d\in\{-5,1,3,9\}$
Hence $|a-d|=|d|\in\{1,3,5,9\}$ , and the sum of possible values is $1+3+5+9 = \boxed{18}$

Solution:
Let \[\begin{cases} |a-b| = 2, \\ |b-c| = 3, \\ |c-d| = 4, \\ |d-a| = X. \\ \end{cases}\] Note that we have \[\begin{cases} a-b = \pm 2, \\ b-c = \pm 3, \\ c-d = \pm 4 \\ d-a = \pm X, \\ \end{cases} \ \ \implies\] \[\pm X \pm 2 \pm 3 \pm 4 = 0, \ \ \implies X \pm 2 \pm 3 \pm 4 = 0,\] from which it follows that \[X = \pm 4 \pm 3 \pm 2.\]
Note that $X = |a-d|$ must be positive however, the only arrangements of $+$ and $-$ signs on the RHS which make $X$ positive are \[(4, 3, 2) \ \ \implies \ \ X = 9\] \[(4, 3, -2) \ \ \implies \ \ X = 5\] \[(4, -3, 2) \ \ \implies \ \ X = 3\] \[(-4, 3, 2) \ \ \implies \ \ X = 1.\] (There are no cases with $2$ or more negative as $4-3-2<0.$
Thus, the answer is \[1+3+5+9 = \boxed{18}.\]

Solution:
Let $a=0$
Then $\begin{cases} b=2 \\ b=-2 \\ \end{cases}$
Thus $\begin{cases} c=5 \\ c=-1 \\ c=1 \\ c=-5 \\ \end{cases}$
Therefore $\begin{cases} d=9 \\ d=1 \\ d=3 \\ d=-5 \\ d=5 \\ d=-3 \\ d=-1 \\ d=-9 \\ \end{cases}$
So $|a-d|=1, 9, 3, 5$ and the sum is $\boxed{18}$


Answer: 994.0

Let $a$ $b$ $c$ be the three sides of a triangle , and let $\alpha$ $\beta$ $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

Solution:
We draw the altitude $h$ to $c$ , to get two right triangles
Then $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$ , from the definition of the cotangent
Let $K$ be the area of $\triangle ABC.$ Then $h=\frac{2K}{c}$ , so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2K}$
By identical logic, we can find similar expressions for the sums of the other two cotangents: \begin{align*} \cot \alpha + \cot \beta &= \frac{c^2}{2K} \\ \cot \beta + \cot \gamma &= \frac{a^2}{2K} \\ \cot \gamma + \cot \alpha &= \frac{b^2}{2K}. \end{align*} Adding the last two equations, subtracting the first, and dividing by 2, we get \[\cot \gamma = \frac{a^2 + b^2 - c^2}{4K}.\] Therefore \begin{align*} \frac{\cot \gamma}{\cot \alpha + \cot \beta} &= \frac{(a^2 + b^2 - c^2)/(4K)}{c^2/(2K)} \\ &= \frac{a^2 + b^2 - c^2}{2c^2} \\ &= \frac{1989 c^2 - c^2}{2c^2} \\ &= \frac{1988}{2} = \boxed{994}

Solution:
We start as in solution 1, though we'll write $A$ instead of $K$ for the area.  Now we evaluate the numerator:
\[\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}\]
From the Law of Cosines and the sine area formula,
\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\ \sin{\gamma}&= \frac{2A}{ab}\\ \cot{\gamma}&= \frac{\cos \gamma}{\sin \gamma} = \frac{1988c^2}{4A} \end{align*}
Then $\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}$

Solution:
\begin{align*} \cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac {\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}\\ &= \frac {\sin{(\alpha + \beta)}}{\sin{\alpha}\sin{\beta}} = \frac {\sin{\gamma}}{\sin{\alpha}\sin{\beta}} \end{align*}
By the Law of Cosines,
\[a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2\]
Now
\begin{align*}\frac {\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac {\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma}} = \frac {\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}} = \frac {ab}{c^2}\cos{\gamma} = \frac {ab}{c^2} \cdot \frac {994c^2}{ab}\\ &= \boxed{994}

Solution:
Use Law of cosines to give us $c^2=a^2+b^2-2ab\cos(\gamma)$ or therefore $\cos(\gamma)=\frac{994c^2}{ab}$ .  Next, we are going to put all the sin's in term of $\sin(a)$ .  We get $\sin(\gamma)=\frac{c\sin(a)}{a}$ .  Therefore, we get $\cot(\gamma)=\frac{994c}{b\sin a}$
Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\cos(\beta)$ .  Therefore, $\cos(\beta)=\frac{a^2-994c^2}{ac}$ .  Also, $\sin(\beta)=\frac{b\sin(a)}{a}$ .  Hence, $\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}$
Lastly, $\cos(\alpha)=\frac{b^2-994c^2}{bc}$ . Therefore, we get $\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}$
Now, $\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}$ .  After using $a^2+b^2=1989c^2$ , we get $\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}$

Solution:
Let $\gamma$ be $(180-\alpha-\beta)$
$\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}$
WLOG, assume that $a$ and $c$ are legs of right triangle $abc$ with $\beta = 90^o$ and $c=1$
By the Pythagorean theorem, we have $b^2=a^2+1$ , and the given $a^2+b^2=1989$ . Solving the equations gives us $a=\sqrt{994}$ and $b=\sqrt{995}$ . We see that $\tan \beta = \infty$ , and $\tan \alpha = \sqrt{994}$
Our derived equation equals $\tan^2 \alpha$ as $\tan \beta$ approaches infinity.
Evaluating $\tan^2 \alpha$ , we get $\boxed{994}$

Solution:
As in Solution 1, drop an altitude $h$ to $c$ . Let $h$ meet $c$ at $P$ , and let $AP = x, BP = y$
Then, $\cot{\alpha} = \frac{1}{\tan{\alpha}} = \frac{x}{h}$ $\cot{\beta} = \frac{1}{\tan{\beta}} = \frac{y}{h}$ . We can calculate $\cot{\gamma}$ using the tangent addition formula , after noticing that $\cot{\gamma} = \frac{1}{\tan{\gamma}}$ . So, we find that
\begin{align*}
\cot{\gamma} &= \frac{1}{\tan{\gamma}} \\
&= \frac{1}{\frac{\frac{x}{h} + \frac{y}{h}}{1 - \frac{xy}{h^2}}} \\
&= \frac{1}{\frac{(x+y)h}{h^2 - xy}} \\
&= \frac{h^2 - xy}{(x+y)h}.
\end{align*}
So now we can simplify our original expression:
\begin{align*}
\frac{\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac{\frac{h^2 - xy}{(x+y)h}}{\frac{x + y}{h}} \\
&= \frac{h^2 - xy}{(x+y)^2}.
\end{align*}
But notice that $x+y = c$ , so this becomes \[\frac{h^2 - xy}{c^2}.\] Now note that we can use the Pythagorean theorem to calculate $h^2$ , we get that \[h^2 = \frac{a^2 - y^2 + b^2 - x^2}{2}.\] So our expression simplifies to \[\frac{1989c^2 - (x+y)^2}{2c^2}\] since $a^2 + b^2 = 1989c^2$ from the problem and that there is another $-\frac{2xy}{2}$ after the $h^2$ in our expression. Again note that $x+y = c$ , so it again simplifies to $\frac{1988c^2}{2c^2}$ , or $\boxed{994}$

Solution:
Since no additional information is given, we can assume that triangle ABC is right with the right angle at B.
We can use the Pythagorean theorem to say \[c^2+a^2=b^2\] We can now solve for $a$ in terms of $c$
\[c^2+a^2=1989c^2-a^2\] \[a^2=994c^2\] \[a=\sqrt{994}c\]
Using the definition of cotangent
\[cot(A)=\frac{c}{a}=\frac{1}{\sqrt{994}}\] \[cot(B)=cot(90)=0\] \[cot(C)=\frac{a}{c}=\sqrt{994}\] Plugging into our desired expression, we get $\boxed{994}$


Answer: 16.0

Let $a$ $b$ , and $c$ be digits with $a\ne 0$ . The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21$

Solution:
Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$ , and the distance between them is $2k+1$ . Let this be equivalent to $3d$ , so that the one-third of the distance between the squares is equivalent to $d$ . The numbers $abc$ and $acb$ are one-third and two-thirds of the way between $k^2$ and $(k+1)^2$ . Therefore, the distance between these two numbers is also one-third the distance between the squares, or $d$ . Setting these equal to each other, we have
$\frac{2k+1}{3} = 9(c-b)$
$\Rightarrow 2k+1 = 27(c-b)$
Notice that since $c$ and $b$ are digits, their difference is at most $9$ and at least $0$ . Also notice that since $acb$ is greater than $abc$ $c > b$ . Representing this as an inequality, we have
$27 \le 27(c-b) \le 243$
Substituting $2k+1$ , we have
$27 \le 2k+1 \le 243$
$\Rightarrow 13 \le k \le 121$
However, we know that $abc$ is a $3$ -digit number, and since $k^2$ is less than $abc$ $k^2$ must be at most $961$ , or $31^2$ . Therefore $k \le 31$ . Plugging this back into our inequality, we have
$13 \le k \le 31$
$\Rightarrow 27 \le 2k+1 \le63$
$\Rightarrow 27 \le 27(c-b) \le 63$
$\Rightarrow 1 \le (c-b) \le \frac{7}{3}$
But (c-b) must be an integer, so now we have
$1 \le (c-b) \le 2$
$\Rightarrow 27 \le 27(c-b) \le 54$
$\Rightarrow 27 \le 2k+1 \le 54$
$\Rightarrow 13 \le k \le\frac{53}{2}$
$k$ is also an integer, so now we have
$\Rightarrow 13 \le k \le 26$
$\Rightarrow 27 \le 2k+1 \le 53$
$\Rightarrow 27 \le 27(c-b) \le 53$
$\Rightarrow 1 \le (c-b) \le \frac{53}{27}$
Once again, $(c-b)$ must be an integer, so we have
$1 \le (c-b) \le 1$
$\Rightarrow (c-b) = 1$
$\Rightarrow 27(c-b) = 27$
$\Rightarrow 2k+1 = 27$
$\Rightarrow k = 13$
The two squares are $13^2$ and $14^2$ , or $169$ and $196$ . A third of the distance between them is $9$ , and $169 +9 = 178$ $1 + 7 + 8 = 16 \Rightarrow \boxed{16}$


Answer: 253.0

Let $a$ $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$
What is $a$
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$

Solution:
Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$
Now, this can be rearranged and factored.
$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares.
We see that the only is possibility is $14 = 9 + 4 + 1$
$(a-c)^2 = 9 \Rightarrow a-c = 3$ , since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$ , or whether $b-c = 1$ or $2$
We want to solve for $a$ , so take the two cases and solve them each for an expression in terms of $a$ . Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$ . Plug these values into one of the original equations to see if we can get an integer for $a$
$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$ , after some algebra, simplifies to $7a = 2021$ $2021$ is not divisible by $7$ , so $a$ is not an integer.
The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$ , which simplifies to $8a = 2024$ . Thus, $a = 253$ and the answer is $\boxed{253}$


Answer: 253.0

Solution:
Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$
Now, this can be rearranged and factored.
$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares.
We see that the only is possibility is $14 = 9 + 4 + 1$
$(a-c)^2 = 9 \Rightarrow a-c = 3$ , since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$ , or whether $b-c = 1$ or $2$
We want to solve for $a$ , so take the two cases and solve them each for an expression in terms of $a$ . Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$ . Plug these values into one of the original equations to see if we can get an integer for $a$
$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$ , after some algebra, simplifies to $7a = 2021$ $2021$ is not divisible by $7$ , so $a$ is not an integer.
The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$ , which simplifies to $8a = 2024$ . Thus, $a = 253$ and the answer is $\boxed{253}$


Answer: 1.0

Let $a$ $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$

Solution:
Rearranging, we get $a+8c=7b+4$ and $8a-c=7-4b$
Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.
Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$ , so $a^2-b^2+c^2=\boxed{1}$

Solution:
The easiest way is to assume a value for $a$ and then solve the system of equations. For $a = 1$ , we get the equations $-7b + 8c = 3$ and $4b - c = -1$ Multiplying the second equation by $8$ , we have $32b - 8c = -8$ Adding up the two equations yields $25b = -5$ , so $b = -\frac{1}{5}$ We obtain $c = \frac{1}{5}$ after plugging in the value for $b$ .
Therefore, $a^2-b^2+c^2 = 1-\frac{1}{25}+\frac{1}{25}=\boxed{1}$ .
This time-saving trick works only because we know that for any value of $a$ $a^2-b^2+c^2$ will always be constant (it's a contest), so any value of $a$ will work. This is also called without loss of generality or WLOG.


Answer: 1.0

Solution: Notice that the coefficients of $a$ and $c$ are pretty similar (15s for reading and noticing), so let $b=0$ gives $a+8c=4$ , and $8a-c=7$ (10s writing). Since the desired quantity simplifies to $a^2+c^2$ , the $ac$ term of the quadratics after squaring gets canceled by adding up the squares of the two equations because they have the same coefficients but opposite sign (15s mind-binom). This simplifies to $65(a^2+c^2)=16+49$ , or $a^2-b^2+c^2=\frac{65}{65}=\boxed{1}$ (15s writing and addition and fraction simplification and (B) circling and submission)


Answer: 16.5

Let $a$ $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$
$\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17$

Solution: Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$ . By Vieta's formula the sum of the roots is $\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}$ . To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$ . So let $b=9$ $a=8$ , and $c=7$ . Then the answer is $\dfrac{8+18+7}{2}=\boxed{16.5}$

Solution: Factoring out $(x-b)$ from the equation yields $(x-b)(2x-(a+c))=0 \Rightarrow (x-b)\left(x-\frac{a+c}{2}\right)=0$ . Therefore the roots are $b$ and $\frac{a+c}{2}$ . Because $b$ must be the larger root to maximize the sum of the roots, letting $a,b,$ and $c$ be $8,9,$ and $7$ respectively yields the sum $9+\frac{8+7}{2} = 9+7.5 = \boxed{16.5}$


Answer: 0.8

Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]
Find $\frac{a}{b}$
$\text{(A) }0.4 \qquad \text{(B) }0.5  \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$

Solution: For sake of speed, WLOG, let $b=1$ . This means that the ratio $\frac{a}{b}$ will simply be $a$ because $\frac{a}{b}=\frac{a}{1}=a.$ Solving for $a$ with some very simple algebra gives us a quadratic which is $5a^2 -9a +4=0.$ Factoring the quadratic gives us $(5a-4)(a-1)=0$ . Therefore, $a=1$ or $a=\frac{4}{5}=0.8.$ However, since $a$ and $b$ must be distinct, $a$ cannot be $1$ so the latter option is correct, giving us our answer of $\boxed{0.8}.$

Solution: The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by $b(b+10a)$ gives us $2ab+10a^2+10b^2=2b^2+20ab.$ Moving everything to the left-hand side and dividing by $2$ gives $5a^2-4b^2 -9ab,$ which factors into $(5a-4b)(a-b)=0.$ Because $a \neq b, 5a=4b \implies \frac{a}{b}=0.8$ giving us our answer of $\boxed{0.8}.$


Answer: 36.0

Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$ . The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution:
Let us call the quantity $\frac{a^3b^3+1}{a^3+b^3}$ as $N$ for convenience. Knowing that $a$ and $b$ are positive integers, we can legitimately rearrange the given inequality so that $a$ is by itself, which makes it easier to determine the pairs of $(a, b)$ that work. Doing so, we have \[\frac{ab+1}{a+b} < \frac{3}{2}\] \[\implies 2ab + 2 < 3a + 3b \implies 2ab - 3a < 3b - 2\] \[\implies a < \frac{3b - 2}{2b - 3}.\] Now, observe that if $b = 1$ we have that $N = \frac{a^3 + 1}{a^3 + 1} = 1$ , regardless of the value of $a$ . If $a = 1$ , we have the same result: that $N = \frac{b^3 + 1}{b^3 + 1} = 1$ , regardless of the value of $b$ . Hence, we want to find pairs of positive integers $(a, b)$ existing such that neither $a$ nor $b$ is equal to $1$ , and that the conditions given in the problem are satisfied in order to check that the maximum value for $N$ is not $1$
To avoid the possibility that $a = 1$ , we want to find values of $b$ such that $\frac{3b - 2}{2b - 3} > 2$ . If we do this, we will have that $a < \frac{3b - 2}{2b - 3} = k$ , where $k$ is greater than $2$ , and this allows us to choose values of $a$ greater than $1$ . Again, since $b$ is a positive integer, and we want $b > 1$ , we can legitimately multiply both sides of $\frac{3b - 2}{2b - 3} > 2$ by $2b - 3$ to get $3b - 2 > 4b - 6 \implies b < 4$ . For $b = 3$ , we have that $a < \frac{7}{3}$ , so the only possibility for $a$ greater than $1$ is obviously $2$ . Plugging these values into $N$ , we have that $N = \frac{8(27) + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}$ . For $b = 2$ , we have that $a < \frac{4}{1} = 4$ . Plugging $a = 3$ and $b = 2$ in for $N$ yields the same result of $\frac{31}{5}$ , but plugging $a = 2$ and $b = 2$ into $N$ yields that $N = \frac{8(8) + 1}{8 + 8} = \frac{65}{16}$ . Clearly, $\frac{31}{5}$ is the largest value we can have for $N$ , so our answer is $31 + 5 = \boxed{036}$

Solution:
\[\frac{ab + 1}{a + b} < \frac{3}{2} \rightarrow 2ab + 2 < 3a + 3b \rightarrow 4ab - 6a - 6b + 4 < 0 \rightarrow (2a - 3)(2b - 3) < 5.\]
\[2a - 3, 2b - 3 \in \{x \neq 2k, k \in Z \} \rightarrow\] $(2a-3)$ and $(2b-3)$ each cannot be even or else $a$ and $b$ will not be integers \[(2a - 3)(2b - 3) = 1, 3 \rightarrow (2a - 3, 2b - 3) = (1, 1), (1, 3), (3, 1).\] \[(a, b) = (2, 2), (2, 3), (3, 2).\] \[\frac{a^3 b^3 + 1}{a^3 + b^3} = \frac{65}{16}, \frac{31}{5}.\] \[\frac{31}{5} \rightarrow \boxed{036}.\]

Solution: Notice that for $\frac{a^3b^3+1}{a^3+b^3}$ to be maximized, $\frac{ab+1}{a+b}$ has to be maximized. We simplify as above to $2ab + 2 < 3a + 3b$ , which is $(a-\frac{3}{2})(b-\frac{3}{2}) < \frac{5}{4}$ . To maximize, $a$ has to be as close to $b$ as possible, making $a$ close to $\frac{3+\sqrt{5}}{2}$ . Because $a$ and $b$ are positive integers, $a = 3$ , and checking back gives $b = 2$ as the maximum or the other way around, which the answer is thus $\frac{216+1}{27+8} = \frac{217}{35} = \frac{31}{5} \rightarrow \boxed{036}$


Answer: 7.0

Let $a$ and $b$ be positive real numbers with $a\ge b$ . Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$ . Then $\rho^2$ can be expressed as a fraction $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Notice that the given equation implies
We have $2by \ge y^2$ , so $2ax \le a^2 \implies x \le \frac {a}{2}$
Then, notice $b^2 + x^2 = a^2 + y^2 \ge a^2$ , so $b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}$
The solution $(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \frac {1}{2}, 0\right)$ satisfies the equation, so $\rho^2 = \frac {4}{3}$ , and the answer is $3 + 4 = \boxed{007}$

Solution:
Consider the points $(a,y)$ and $(x,b)$ . They form an equilateral triangle with the origin. We let the side length be $1$ , so $a = \cos{\theta}$ and $b = \sin{\left(\theta + \frac {\pi}{3}\right)}$
Thus $f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}$ and we need to maximize this for $0 \le \theta \le \frac {\pi}{6}$
Taking the derivative shows that $-f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0$ , so the maximum is at the endpoint $\theta = 0$ . We then get
Then, $\rho^2 = \frac {4}{3}$ , and the answer is $3+4=\boxed{007}$

Solution:
Consider a cyclic quadrilateral $ABCD$ with $\angle B = \angle D = 90^{\circ}$ , and $AB = y, BC = a, CD = b, AD = x$ . Then \[AC^2 = a^2 + y^2 = b^2 + x^2\] From Ptolemy's Theorem $ax + by = AC(BD)$ , so \[AC^2 = (a - x)^2 + (b - y)^2 = a^2 + y^2 + b^2 + x^2 - 2(ax + by) = 2AC^2 - 2AC*BD\] Simplifying, we have $BD = AC/2$
Note the circumcircle of $ABCD$ has radius $r = AC/2$ , so $BD = r$ and has an arc of $60^{\circ}$ , so $\angle C = 30^{\circ}$ . Let $\angle BDC = \theta$
$\frac ab = \frac{BC}{CD} = \frac{\sin \theta}{\sin(150^{\circ} - \theta)}$ , where both $\theta$ and $150^{\circ} - \theta$ are $\leq 90^{\circ}$ since triangle $BCD$ must be acute . Since $\sin$ is an increasing function over $(0, 90^{\circ})$ $\frac{\sin \theta}{\sin(150^{\circ} - \theta)}$ is also increasing function over $(60^{\circ}, 90^{\circ})$
$\frac ab$ maximizes at $\theta = 90^{\circ} \Longrightarrow \frac ab$ maximizes at $\frac 2{\sqrt {3}}$ . This squared is $(\frac 2{\sqrt {3}})^2 = \frac4{3}$ , and $4 +  3 = \boxed{007}$

Solution: The problem is looking for an intersection in the said range between parabola $P$ $y = \tfrac{(x-a)^2 + b^2-a^2}{2b}$ and the hyperbola $H$ $y^2 = x^2 + b^2 - a^2$ . The vertex of $P$ is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the $H$ , which is $\sqrt{a^2 - b^2}$ . So for the intersection to exist with $x<a$ and $y \geq 0$ $P$ needs to cross x-axis between $\sqrt{a^2 - b^2}$ , and $a$ , meaning, \[(\sqrt{a^2 - b^2}-a)^2 + b^2-a^2 \geq 0\] Divide both side by $b^2$ \[(\sqrt{\rho^2 - 1}-\rho)^2 + 1-\rho^2 \geq 0\] which can be easily solved by moving $1-\rho^2$ to RHS and taking square roots. Final answer $\rho^2 \leq \frac{4}{3}$ $\boxed{007}$

Solution: The given system is equivalent to the points $(a,y)$ and $(x,b)$ forming an equilateral triangle with the origin. WLOG let this triangle have side length $1$ , so $x=\sqrt{1-a^2}$ . The condition $x<a$ implies $(x,b)$ lies to the left of $(a,y)$ , so $(x,b)$ is the top vertex. Now we can compute (by complex numbers, or the sine angle addition identity) that $b = \frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}$ , so $\frac{a}{b} = \frac{a}{\frac{\sqrt{3}}{2}a + \frac{1}{2}\sqrt{1-a^2}} = \frac{1}{\frac{\sqrt{3}}{2} + \frac{1}{2a}\sqrt{1-a^2}}$ . Minimizing this is equivalent to minimizing the denominator, which happens when $\sqrt{1-a^2} = 0$ and thus $a=1$ , resulting in $\rho = \frac{2}{\sqrt{3}}$ , so $\rho^2 = \frac{4}{3}$ and the answer is $\boxed{007}$


Answer: 2.0

Let $a$ and $b$ be real numbers greater than $1$ for which there exists a positive real number $c,$ different from $1$ , such that
\[2(\log_a{c} + \log_b{c}) = 9\log_{ab}{c}.\]
Find the largest possible value of $\log_a b.$
$\text{(A) }\sqrt{2} \qquad \text{(B) }\sqrt{3} \qquad \text{(C) }2 \qquad \text{(D) }\sqrt{6} \qquad \text{(E) }3$

Solution:
We may rewrite the given equation as \[2(\frac {\log c}{\log a} + \frac {\log c}{\log b}) = \frac {9\log c}{\log a + \log b}\] Since $c \neq 1$ , we have $\log c \neq 0$ , so we may divide by $\log c$ on both sides. After making the substitutions $x = \log a$ and $y = \log b$ , our equation becomes \[\frac {2}{x} + \frac {2}{y} = \frac {9}{x+y}\]
Rewriting the left-hand side gives \[\frac {2(x+y)}{xy} = \frac {9}{x+y}\]
Cross-multiplying gives $2(x+y)^2 = 9xy$ or \[2x^2 - 5xy + 2y^2 = 0\]
Factoring gives $(2x-y)(x-2y) = 0$ or $\frac {x}{y} = 2, \frac {1}{2}$
Recall that $\frac {x}{y} = \frac {\log a}{\log b} = \log_{a} b$ . Therefore, the maximum value of $\log_{a} b$ is $\boxed{2}$


Answer: 3.0

Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$

Solution: Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$ , the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$ . Looking at the answer choices, the only multiple of $3$ is $\boxed{3}$

Solution:
Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}$
Set $x = a^2 + b^2$ , and $y = ab$ .  Then $\frac{x + y}{x - 2y} = \frac{73}{3}$ .  Cross multiplying gives $3x + 3y = 73x - 146y$ , and simplifying gives $\frac{x}{y} = \frac{149}{70}$ .  Since $149$ and $70$ are relatively prime, we let $x = 149$ and $y = 70$ , giving $a^2 + b^2 = 149$ and $ab = 70$ .  Since $a>b>0$ , the only solution is $(a,b) = (10, 7)$ , which can be seen upon squaring and summing the various factor pairs of $70$
Thus, $a - b = \boxed{3}$

Solution:
The first step is the same as above which gives $\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$
Then we can subtract $3ab$ and then add $3ab$ to get $\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}$ , which gives $1+\frac{3ab}{(a-b)^2}=\frac{73}{3}$ $\frac{3ab}{(a-b)^2}=\frac{70}{3}$ .
Cross multiply $9ab=70(a-b)^2$ . Since $a>b$ , take the square root. $a-b=3\sqrt{\frac{ab}{70}}$ .
Since $a$ and $b$ are integers and relatively prime, $\sqrt{\frac{ab}{70}}$ is an integer. $ab$ is a multiple of $70$ , so $a-b$ is a multiple of $3$ .
Therefore $a=10$ and $b=7$ is a solution.
So $a-b=\boxed{3}$

Solution:
Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$
Canceling the $(a-b)$ , cross multiplying, and simplifying, we obtain that
$0=70a^2-149ab+70b^2$ .
Dividing everything by $b^2$ , we get that
$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$
Applying the quadratic formula....and following the restriction that $a>b>0$ ....
$\frac{a}{b}=\frac{10}{7}$
Hence, $7a=10b$
Since they are relatively prime, $a=10$ $b=7$
$10-7=\boxed{3}$

Solution: Note that the denominator, when simplified, gets $3.$ We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly $\boxed{3}$ ~mathboy282

Solution: Let us rewrite the expression as $\frac{(a-b)^2 + 3ab}{(a-b)^2}$ . Now letting $x = a - b$ , we simplify the expression to $\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}$ . Cross multiplying and doing a bit of simplification, we obtain that $ab = \frac{70x^2}{9}$ . Since $a$ and $b$ are both integers, we know that $\frac{70x^2}{9}$ has to be an integer. Experimenting with values of $x$ , we get that $x = 3$ which means $ab = 70$ . We could prime factor from here to figure out possible values of $a$ and $b$ , but it is quite obvious that $a = 10$ and $b=7$ , so our desired answer is $\boxed{3}$ ~triggod

Solution: Since the two numbers are integers, and both $(a-b)^3$ and $a^3-b^3$ would yield integers, for the denominator to have a factor of 3, $(a-b)$ must have a factor of 3. Only choice $\boxed{3}$ has a factor of 3. ~hh99754539


Answer: 1.0

Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle
in the complex plane centered at $0+0i$ and having radius $1$ . The sum of the reciprocals of the roots is necessarily
$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ -a \qquad \textbf{(E)}\ -b$

Solution: Let's denote the roots of the polynomial as $z_1, z_2, z_3, z_4$ . We know that the magnitudes of these 4 roots are 1 as given in the problem statement. Therefore, we have $z_1 \overline{z_1}, z_2 \overline{z_2}, z_3 \overline{z_3}, z_4 \overline{z_4} = 1$ . We want to find $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4}$ which is $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4}$ . Remember that these are the roots of polynomials. Whenever complex numbers are the roots of a polynomial with real coefficients, we know they come in complex-conjugate pairs. Therefore, $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4} = z_1 + z_2 + z_3 + z_4$ . However, by Vieta's, $z_1 + z_2 + z_3 + z_4 = -a$ . Thus, the answer is $\boxed{01}$


Answer: 13.0

Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$ $\gcd(b, c)=36$ $\gcd(c, d)=54$ , and $70<\gcd(d, a)<100$ . Which of the following must be a divisor of $a$
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$

Solution:
The GCD information tells us that $24$ divides $a$ , both $24$ and $36$ divide $b$ , both $36$ and $54$ divide $c$ , and $54$ divides $d$ . Note that we have the prime factorizations: \begin{align*} 24 &= 2^3\cdot 3,\\ 36 &= 2^2\cdot 3^2,\\ 54 &= 2\cdot 3^3. \end{align*}
Hence we have \begin{align*} a &= 2^3\cdot 3\cdot w\\ b &= 2^3\cdot 3^2\cdot x\\ c &= 2^2\cdot 3^3\cdot y\\ d &= 2\cdot 3^3\cdot z \end{align*} for some positive integers $w,x,y,z$ . Now if $3$ divides $w$ , then $\gcd(a,b)$ would be at least $2^3\cdot 3^2$ which is too large, hence $3$ does not divide $w$ . Similarly, if $2$ divides $z$ , then $\gcd(c,d)$ would be at least $2^2\cdot 3^3$ which is too large, so $2$ does not divide $z$ . Therefore, \[\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)\] where neither $2$ nor $3$ divide $\gcd(w,z)$ . In other words, $\gcd(w,z)$ is divisible only by primes that are at least $5$ . The only possible value of $\gcd(a,d)$ between $70$ and $100$ and which fits this criterion is $78=2\cdot3\cdot13$ , so the answer is $\boxed{13}$

Solution: We can say that $a$ and $b$ 'have' $2^3 \cdot 3$ , that $b$ and $c$ have $2^2 \cdot 3^2$ , and that $c$ and $d$ have $3^3 \cdot 2$ . Combining $1$ and $2$ yields $b$ has (at a minimum) $2^3 \cdot 3^2$ , and thus $a$ has $2^3 \cdot 3$ (and no more powers of $3$ because otherwise $\gcd(a,b)$ would be different). In addition, $c$ has $3^3 \cdot 2^2$ , and thus $d$ has $3^3 \cdot 2$ (similar to $a$ , we see that $d$ cannot have any other powers of $2$ ). We now assume the simplest scenario, where $a = 2^3 \cdot 3$ and $d = 3^3 \cdot 2$ . According to this base case, we have $\gcd(a, d) = 2 \cdot 3 = 6$ . We want an extra factor between the two such that this number is between $70$ and $100$ , and this new factor cannot be divisible by $2$ or $3$ . Checking through, we see that $6 \cdot 13$ is the only one that works. Therefore the answer is $\boxed{13}$

Solution:
First off, note that $24$ $36$ , and $54$ are all of the form $2^x\times3^y$ . The prime factorizations are $2^3\times 3^1$ $2^2\times 3^2$ and $2^1\times 3^3$ , respectively. Now, let $a_2$ and $a_3$ be the number of times $2$ and $3$ go into $a$ , respectively. Define $b_2$ $b_3$ $c_2$ , and $c_3$ similarly. Now, translate the $lcm$ s into the following: \[1) \min(a_2,b_2)=3\] \[2) \min(a_3,b_3)=1\] \[3) \min(b_2,c_2)=2\] \[4) \min(b_3,c_3)=2\] \[5) \min(c_2,d_2)=1\] \[6) \min(c_3,d_3)=3\]
From $4)$ , we see that $b_3 \geq 2$ , thus from $2)$ $a_3 = 1$ . Similarly, from $3)$ $c_2 \geq 2$ , thus from $5)$ $d_2 = 1$
Note also that $d_3 \geq 3$ and $a_2 \geq 3$ . Therefore \[\min(a_2, d_2) = 1\] \[\min(a_3, d_3) = 1\] Thus, $\gcd(a, d) = 2 \times 3 \times k$ for some $k$ having no factors of $2$ or $3$
Since $70 < \gcd(a, d) < 100$ , the only values for k are $12, 13, 14, 15, 16$ , but all have either factors of $2$ or $3$ , except $\boxed{13}$ . And we're done.

Solution: Notice that $\gcd (a,b,c,d)=\gcd(\gcd(a,b),\gcd(b,c),\gcd(c,d))=\gcd(24,36,54)=6$ , so $\gcd(d,a)$ must be a multiple of $6$ . The only answer choice that gives a value between $70$ and $100$ when multiplied by $6$ is $\boxed{13}$ . - mathleticguyyy + einstein

Solution: The relationship of $a$ $b$ $c$ , and $d$ is shown in the above diagram. $gcd(a,d)=2 \cdot 3 \cdot k$ $70 < 6k < 100$ $12 \le k \le 16$ $k=\boxed{13}$

Solution:
Just as in solution $1$ , we prime factorize $a, b, c$ and $d$ to observe that
$a=2^3\cdot{3}\cdot{w}$
$b=2^3\cdot{3^2}\cdot{x}$
$c=2^2\cdot{3^3}\cdot{y}$
$d=2\cdot{3^3}\cdot{z}.$
Substituting these expressions for $a$ and $d$ into the final given,
$70<\text{gcd}(2\cdot{3^3}\cdot{z}, 2^3\cdot{3}\cdot{w})<100.$
The greatest common divisor of these two numbers is already $6$ . If $k$ is what we wish to multiply $6$ by to obtain the gcd of these two numbers, then
$70<6k<100$ . Testing the answer choices, only $13$ works for $k$ (in order for the compound inequality to hold). so our gcd is $78$ , which means that $\boxed{13}$ must divide $a$


Answer: 145.0

Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$

Solution: From the fourth equation we get $d=\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14$ . Hence $(abc)^2 - 14(abc)-120 = 0$ . Solving, we get $abc = -6$ or $abc = 20$ . From the first and second equation, we get $ab + bc + ca = ab-3c = -4 \Longrightarrow ab = 3c-4$ . If $abc=-6$ , substituting we get $c(3c-4)=-6$ . If you try solving this you see that this does not have real solutions in $c$ , so $abc$ must be $20$ . So $d=\frac{3}{2}$ . Since $c(3c-4)=20$ $c=-2$ or $c=\frac{10}{3}$ . If $c=\frac{10}{3}$ , then the system $a+b=-3$ and $ab = 6$ does not give you real solutions. So $c=-2$ . Since you already know $d=\frac{3}{2}$ and $c=-2$ , so you can solve for $a$ and $b$ pretty easily and see that $a^{2}+b^{2}+c^{2}+d^{2}=\frac{141}{4}$ . So the answer is $\boxed{145}$

Solution:
Note that $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$ . Hence, $ab = 3c - 4$
Rewriting $abc+bcd+cda+dab = 14$ , we get $ab(c+d) + cd(a+b) = 14$ .
Substitute $ab = 3c - 4$ and solving, we get \[3c^{2} - 4c - 4d - 14 = 0.\] We refer to this as Equation 1.
Note that $abcd = 30$ gives $(3c-4)cd = 30$ . So, $3c^{2}d - 4cd = 30$ , which implies $d(3c^{2} - 4c) = 30$ or \[3c^{2} - 4c = \frac{30}{d}.\] We refer to this as Equation 2.
Substituting Equation 2 into Equation 1 gives, $\frac{30}{d} - 4d - 14 = 0$
Solving this quadratic yields that $d \in \left\{-5, \frac{3}{2}\right\}$
Now we just try these two cases:
For $d = \frac{3}{2}$ substituting in Equation 1 gives a quadratic in $c$ which has roots $c \in \left\{\frac{10}{3}, -2\right\}$
Again trying cases, by letting $c = -2$ , we get $ab = 3c-4$ , Hence $ab = -10$ .
We know that $a + b = -3$ , Solving these we get $a = -5, b = 2$ or $a= 2, b = -5$ (doesn't matter due to symmetry in $a,b$ ).
So, this case yields solutions $(a,b,c,d) = \left(-5, 2 , -2, \frac{3}{2}\right)$
Similarly trying other three cases, we get no more solutions, Hence this is the solution for $(a,b,c,d)$
Finally, $a^{2} + b^{2} + c^{2} + d^{2} = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4} = \frac{m}{n}$
Therefore, $m + n = 141 + 4 = \boxed{145}$

Solution:
For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.
Rearranging $(2)$ and solving for $c,$ we have \begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \begin{align*} ab\left(\frac{ab+4}{3}\right)d&=30 \\ d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6)  \end{align*} Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: \begin{align*} ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. \end{align*} Let $t=ab(ab+4),$ from which \[\frac{t}{3}-\frac{360}{t}=14.\] Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce \begin{align*} \left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\ \left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\ \underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\ ab&=6,-10. \end{align*} If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number.
If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\{a,b\}=\{-5,2\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\frac32,$ respectively. Together, we have \[a^2+b^2+c^2+d^2=\frac{141}{4},\] so the answer is $141+4=\boxed{145}.$

Solution:
Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: \[abc+d(ab+bc+ca)=14\] Then we plug in equation 2 to receive $abc-4d=14$ . By equation 4 we get $abc=\frac{30}{d}$ . Plugging in, we get $\frac{30}{d}-4d=14$ . Multiply by $d$ on both sides to get the quadratic equation $4d^2+14d-30=0$ . Solving using the quadratic equation, we receive $d=\frac{3}{2},d=-5$ . So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get \[ab+c(a+b)=-4\] After plugging in equation 1, we get $ab-3c=-4$ . Now we convert it into a quadratic to receive $3c^2-4c-abc=0$ . The value of $abc$ will depend on $d$ . So we obtain the discriminant $16+12abc$ .
Let d = -5.
Then $abc = \frac{30}{-5}$ , so $abc=-6$ , discriminant is $16-72$ , which makes this a dead end. Thus $d=\frac{3}{2}$ For $d=\frac{3}{2}$ , making $abc=20$ . This means the discriminant is just $256$ , so we obtain two values for $c$ as well. We get either $c=\frac{10}{3}$ or $c=-2$ . So, we must AGAIN test which one is correct.
We know $ab=3c-4$ , and $a+b=-3$ , so we use these values for testing.
Let $c=\frac{10}{3}$ .
Then $ab=6$ , so $a=\frac{6}{b}$ . We thus get $\frac{6}{b}+b=-3$ , which leads to the quadratic $b^2+3b+6$ . The discriminant for this is $9-24$ . That means this value of $c$ is wrong, so $c=-2$ . Thus we get polynomial $b^2+3b-10$ . The discriminant this time is $49$ , so we get two values for $b$ . Through simple inspection, you may see they are interchangeable, as if you take the value $b=2$ , you get $a=-5$ . If you take the value $b=-5$ , you get $a=2$ . So it doesn't matter. That means the sum of all their squares is \[\frac{9}{4}+4+4+25=\frac{141}{4}\] so the answer is $141+4=\boxed{145}.$ ~amcrunner

Solution:
Let the four equations from top to bottom be listed $(1)$ through $(4)$ respectively. Multiplying both sides of $(3)$ by $d$ and factoring some terms gives us $abcd + d^2(ab + ac + bc) = 14d$ . Substituting using equations $(4)$ and $(2)$ gives us $30 -4 d^2 = 14d$ , and solving gives us $d = -5$ or $d = \frac{3}{2}$ . Plugging this back into $(3)$ gives us $abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14$ , or using the other solution for $d$ gives us $abc - 6 = 14$ . Solving both of these equations gives us $abc = -6$ when $d = -5$ and $abc = 14$ when $d = \frac{3}{2}$
Multiplying both sides of $(2)$ by $c$ and factoring some terms gives us $abc + c^2 (a + b) = abc -3c^2 = -4c$ . Testing $abc = -6$ will give us an imaginary solution for $c$ , so therefore $abc = 14$ and $d = \frac{3}{2}$ . This gets us $14 - 3c^2 = -4c$ . Solving for $c$ gives us $c = \frac{3}{10}$ or $c = -2$ . With a bit of testing, we can see that the correct value of $c$ is $c=-2$ . Now we know $a+b = -3$ and $ab + bc + ca = ab + c(a+b) = ab + 6 = -4$ $ab = -10$ , and it is obvious that $a = -5$ and $b = 2$ or the other way around, and therefore, $a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}$ , giving us the answer $141 + 4 = \boxed{145}$ ~hihitherethere


Answer: 23.0

Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$

Solution:
Denote $P = \left( x , y \right)$
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$ $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the major axis of this ellipse, $2a$
Because $\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1$ $P$ is on an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$
Hence, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the major axis of this ellipse, $2b$
Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses.
To make this minimized, $P$ is the intersection point of the line that passes through $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ , and the line that passes through $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$
The distance between $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ is $\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26$
The distance between $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ is $\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20$
Hence, $2 a + 2 b = 26 + 20 = 46$
Therefore, $a + b = \boxed{023}.$


Answer: 16.0

Let $a,b,c,d$ be real numbers such that $b-d \ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.

Solution: Using the hint we turn the equation into $\prod_{k=1} ^4 (x_k-i)(x_k+i) \implies P(i)P(-i) \implies (b-d-1)^2 + (a-c)^2 \implies \boxed{16}$ . This minimum is achieved when all the $x_i$ are equal to $1$


Answer: 34.0

Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30     \qquad \mathrm{(B)}\ 32     \qquad \mathrm{(C)}\ 34     \qquad \mathrm{(D)}\ 40     \qquad \mathrm{(E)}\ 50$

Solution: The sum of the set is $-7-5-3-2+2+4+6+13=8$ , so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$ . Considering the set of four terms containing $13$ , this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $13-7-5-3=-2$ , and with two odd terms then its minimum value is $13-7+2-2=6$ , so we cannot achieve two sums of $4$ . The closest we could have to $4$ and $4$ is $3$ and $5$ , which can be achieved through $13-7-5+2$ and $6-3-2+4$ . So the minimum possible value is $3^2+5^2=34\Rightarrow\boxed{34}$

Solution:
Trying out the values for a bit leads to us getting $34$ with $(-3, -2, 2, 6)$ in 1 set for a total of 9 and the other numbers giving a total of 25.
We start off with trying to get a $2(4^2) = 32$ solution, so we only need to find one set with a sum of 4 (the other will automatically have a sum of 4).
We can add 7 to each number. We are now trying to get a set with 32.
\[(0, 2, 4, 5, 9, 11, 13, 20)\]
Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers $a,b,c,d$ , the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26).
If $a,b,c,d$ has 2 odd numbers and 2 even numbers we can divide them into two cases: whether $a,b,c,d$ has 20 or doesn't have 20.
Case 1: $a,b,c,d$ doesn't have 20.
The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need.
Case 2: $a,b,c,d$ has 20.
The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need.
We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest value is $\boxed{34}$


Answer: 289.0

Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \cdot e \cdot f = (a \cdot b \cdot c) \cdot (a \cdot b \cdot c - 1) \geq 6! = 720,$ so $a \cdot b \cdot c \geq 28.$ It follows that $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are $36$ and $35,$ respectively. So, we have $\{a,b,c\} = \{2,3,6\}, \{d,e,f\} = \{1,5,7\},$ and $\{g,h,i\} = \{4,8,9\},$ from which $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} = \frac{1}{288}.$
If we do not minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f > 1.$ Note that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{2}{7\cdot8\cdot9} > \frac{1}{288}.$
Together, we conclude that the minimum possible positive value of $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}$ is $\frac{1}{288}.$ Therefore, the answer is $1+288=\boxed{289}.$

Solution:
Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator.
To bash efficiently, we can start out with $7\cdot8\cdot9$ as our denominator. This, however, leaves us with the numbers $1, 2, 3, 4, 5,$ and $6$ left. The smallest we can make out of this is $1\cdot5\cdot6 - 2\cdot3\cdot4 = 30 - 24 = 6$ . When simplified, it gives us $\frac{1}{84}$ , which gives a small answer of $85$ . Obviously there are larger answers than this.
After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple $(a,b,c)=(2,3,6)$ and $(d,e,f)=(1,5,7)$
This gives us a numerator of $36-35=1$ , which is by far the smallest yet. With the remaining numbers $4, 8,$ and $9$ , we get $\frac{36-35}{4\cdot8\cdot9}=\frac{1}{288}$
Finally, we add up our numerator and denominator: The answer is $1+288=\boxed{289}$

Solution: To minimize the numerator, we must have $abc - def = 1$ . Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value $(d, e, f) = (1, 3, 5)$ cannot result in a difference of $1$ , and the next smallest product, $(d, e, f) = (1, 3, 7)$ cannot either, but $(d, e, f) = (1, 5, 7)$ can if $(a, b, c) = (2, 3, 6)$ . Thus, the denominator must be $(g, h, i) = (4, 8, 9)$ , and the smallest fraction possible is $\dfrac{36 - 35}{288} = \dfrac{1}{288}$ , making the answer $1 + 288 = \boxed{289}$


Answer: 3.0

Let $a_1 , a_2 , ...$ be a sequence for which $a_1=2$ $a_2=3$ , and $a_n=\frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \ge 3$ . What is $a_{2006}$
$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3$

Solution:
Looking at the first few terms of the sequence:
$a_1=2 , a_2=3 , a_3=\frac{3}{2}, a_4=\frac{1}{2} , a_5=\frac{1}{3} , a_6=\frac{2}{3} , a_7=2 , a_8=3 , ....$
Clearly, the sequence repeats every 6 terms.
Since $2006 \equiv 2\bmod{6}$
$a_{2006} = a_2 = \boxed{3}$

Solution: $a_n = \frac{a_{n-1}}{a_{n-2}} = \frac{\frac{a_{n-2}}{a_{n-3}}}{a_{n-2}} = \frac{1}{a_{n-3}}$ , so $a_n = a_{n-6}$ and because $2006 = 2 + 334 \times 6$ , so $a_{2006} = a_2 = \boxed{3}$


Answer: 10000.0

Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be arithmetic progressions such that $a_1 = 25, b_1 = 75$ , and $a_{100} + b_{100} = 100$ .
Find the sum of the first hundred terms of the progression $a_1 + b_1, a_2 + b_2, \ldots$
$\textbf{(A)}\ 0 \qquad  \textbf{(B)}\ 100 \qquad  \textbf{(C)}\ 10,000 \qquad  \textbf{(D)}\ 505,000 \qquad \\ \textbf{(E)}\ \text{not enough information given to solve the problem}$

Solution: Notice that $a_{100}$ and $b_{100}$ are $25+99k_1$ and $75+99k_2$ , respectively. Therefore $k_2 = -k_1$ . Now notice that $a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100$ . The sum of the first $100$ terms is $100\cdot100 = \boxed{10,000}$


Answer: 4.0

Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution:
Verify that $a^3 \equiv a \pmod{6}$ manually for all $a\in \mathbb{Z}/6\mathbb{Z}$ . We check: $0^3 \equiv 0 \pmod{6}$ $1^3 \equiv 1 \pmod{6}$ $2^3 \equiv 8 \equiv 2 \pmod{6}$ $3^3 \equiv 27 \equiv 3 \pmod{6}$ $4^3 \equiv 64 \equiv 4 \pmod{6}$ , and $5^3 \equiv 125 \equiv 5 \pmod{6}$ . We conclude that $a^3 \equiv a \pmod{6}$
Therefore, \[a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}^3 \pmod{6}.\]
Thus the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{4}$ alternates with $2$ and $4$ when $n$ increases.

Solution:
Note that $\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2\\ +\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k$
Note that $a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018}) \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6$ Therefore, $-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}$
Thus, $a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3$ . However, since cubing preserves parity, and the sum of the individual terms is even, the sum of the cubes is also even, and our answer is $\boxed{4}$

Solution:
First, we can assume that the problem will have a consistent answer for all possible values of $a_1$ . For the purpose of this solution, we will assume that $a_1 = 1$
We first note that $1^3+2^3+...+n^3 = (1+2+...+n)^2$ . So what we are trying to find is what $\left(2018^{2018}\right)^2=\left(2018^{4036}\right)$ mod $6$ . We start by noting that $2018$ is congruent to $2 \pmod{6}$ . So we are trying to find $\left(2^{4036}\right) \pmod{6}$ . Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of $2$ and see that $2^1$ is $2$ mod $6$ $2^2$ is $4$ mod $6$ $2^3$ is $2$ mod $6$ $2^4$ is $4$ mod $6$ , and so on... So we see that since $\left(2^{4036}\right)$ has an even power, it must be congruent to $4 \pmod{6}$ , thus giving our answer $\boxed{4}$ . You can prove this pattern using mods. But I thought this was easier.

Solution: First, we can assume that the problem will have a consistent answer for all possible values of $a_1$ . For the purpose of this solution, assume $a_1, a_2, ... a_{2017}$ are multiples of 6 and find $2018^{2018} \pmod{6}$ (which happens to be $4$ ). Then ${a_1}^3 + ... + {a_{2018}}^3$ is congruent to $64 \pmod{6}$ or just $\boxed{4}$

Solution: Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just $2018^{2018} \pmod{6}$ , which is $\boxed{4}$


Answer: 1700.0

Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$ , define $b_m=\min\{n: a_n \ge m\}$ , that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$ . If $a_{19}=85$ , determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$

Solution:
We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the $i$ th column we fill the first $a_i$ cells with one dot each. Then the $19$ th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row $j$ , we see that the number of blank cells is equal to $b_j-1$ . Therefore the number of filled cells in the first 19 columns of row $j$ is equal to $20-b_j$
We now count the number of cells in the first 19 columns of our array, but we do it in two different ways. First, we can sum the number of dots in each column: this is simply $a_1+\cdots+a_{19}$ . Alternatively, we can sum the number of dots in each row: this is $(20-b_1)+\cdots +(20-b_{85})$ . Since we have counted the same number in two different ways, these two sums must be equal. Therefore \[a_1+\cdots +a_{19}+b_1+\cdots +b_{85}=20\cdot 85=\boxed{1700}.\] Note that this shows that the value of the desired sum is constant.


Answer: 35.0

Let $a_n=6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$

Solution:
Firstly, we try to find a relationship between the numbers we're provided with and $49$ . We notice that $49=7^2$ , and both $6$ and $8$ are greater or less than $7$ by $1$
Thus, expressing the numbers in terms of $7$ , we get $a_{83} = (7-1)^{83}+(7+1)^{83}$
Applying the Binomial Theorem , half of our terms cancel out and we are left with $2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)$ . We realize that all of these terms are divisible by $49$ except the final term.
After some quick division, our answer is $\boxed{035}$

Solution: As the value of $a$ is obviously $6^{83}+8^{83}$ we look for a pattern with others. With a bit of digging, we discover that $6^n+6^m$ where $m$ and $n$ are odd is equal to $\boxed{35}\:49$


Answer: 112.0

Solution:
When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.
After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$
$a_{0} = 2$ $a_{1} = 5$ $a_{2} = 8$ $a_{3} = 5$ $a_{4} = 6$ $a_{5} = 10$ $a_{6} = 7$ $a_{7} = 4$ $a_{8} = 7$ $a_{9} = 6$ $a_{10} = 2$ $a_{11} = 5$ $a_{12} = 8$ $a_{13} = 5$
We can simplify the expression we need to solve to $a_{8}\cdot a_{10} \cdot a_{2}$
Our answer is $7 \cdot 2 \cdot 8$ $= \boxed{112}$

Solution:
Notice that the characteristic polynomial of this is $x^3-4x^2-4x-4\equiv 0\pmod{11}$
Then since $x\equiv1$ is a root, using Vieta's formula, the other two roots $r,s$ satisfy $r+s\equiv3$ and $rs\equiv4$
Let $r=7+d$ and $s=7-d$
We have $49-d^2\equiv4$ so $d\equiv1$ . We found that the three roots of the characteristic polynomial are $1,6,8$
Now we want to express $a_n$ in an explicit form as $a(1^n)+b(6^n)+c(8^n)\pmod{11}$
Plugging in $n=0,1,2$ we get
$(*)$ $a+b+c\equiv2,$
$(**)$ $a+6b+8c\equiv5,$
$(***)$ $a+3b+9c\equiv8$
$\frac{(***)-(*)}{2}$ $\implies b+4c\equiv3$ and $(***)-(**)$ $\implies -3b+c\equiv3$
so $a\equiv6,$ $b\equiv1,$ and $c\equiv6$
Hence, $a_n\equiv 6+(6^n)+6(8^n)\equiv(2)^{-n\pmod{10}}+(2)^{3n-1\pmod{10}}-5\pmod{11}$
Therefore
$a_{2018}\equiv4+8-5=7$
$a_{2020}\equiv1+6-5=2$
$a_{2022}\equiv3+10-5=8$
And the answer is $7\times2\times8=\boxed{112}$


Answer: 45.0

Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$

Solution: Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives \[a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10\] Which simplifies to \[a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)\] Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$ , we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$ , we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{045}$

Solution:
\[a_n \equiv a_{n-1} + n \pmod {99}\] By looking at the first few terms, we can see that \[a_n \equiv 10+11+12+ \dots + n \pmod {99}\] This implies \[a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}\] Since $a_n \equiv 0 \pmod {99}$ , we can rewrite the equivalence, and simplify \[0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}\] \[0 \equiv n(n+1) - 90 \pmod {99}\] \[0 \equiv 4n^2+4n+36 \pmod {99}\] \[0 \equiv (2n+1)^2+35 \pmod {99}\] \[64 \equiv (2n+1)^2 \pmod {99}\] The only squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$ , so \[2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}\] $2n+1 \equiv -8 \pmod {99}$ yields $n=45$ as the smallest integer solution.
$2n+1 \equiv 8 \pmod {99}$ yields $n=53$ as the smallest integer solution.
$2n+1 \equiv -19 \pmod {99}$ yields $n=89$ as the smallest integer solution.
$2n+1 \equiv 19 \pmod {99}$ yields $n=9$ as the smallest integer solution. However, $n$ must be greater than $10$
The smallest positive integer solution greater than $10$ is $n=\boxed{045}$

Solution:
$a_n=a_{n-1} + n \pmod{99}$ .  Using the steps of the previous solution we get up to $n^2+n \equiv 90 \pmod{99}$ . This gives away the fact that $(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}$ so either $n$ or $n+1$ must be a multiple of 9.
Case 1 ( $9 \mid n$ ): Say $n=9x$ and after simplification $x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}$
Case 2: ( $9 \mid n+1$ ): Say $n=9a-1$ and after simplification $(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}$
As a result $a$ must be a divisor of $10$ and after doing some testing in both cases the smallest value that works is $x=5 \implies \boxed{045}$

Solution:
We just notice that $100 \equiv 1 \pmod{99}$ , so we are just trying to find $10 + 11 + 12 + \cdots + n$ modulo $99$ , or $\dfrac{n(n+1)}{2} - 45$ modulo $99$ . Also, the sum to $44$ is divisible by $99$ , and is the first one that is. Thus, if we sum to $45$ the $45$ is cut off and thus is just a sum to $44$
Without checking whether there are other sums congruent to $45 \pmod{99}$ , we can just write the answer to be $\boxed{045}$

Solution:
Let $b_n = 2a_{n+10}$ . We can find a formula for $b_n$
$b_n = (20+n)(n+1)$
Notice that both can't have a factor of 3. Thus we can limit our search range of n to $n \equiv 7,8 \pmod{9}$ . Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, $b_n = 2a_{n+10}$ , so, the 35th term in b corresponds to the 45th term in a. Thus our answer is $\boxed{045}$

Solution:
The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. $10, 1011, 101112, \dots$ . Now the fact that the sequence starts at $10$ can be completely discarded for this solution. Just consider $a(10)$ then same as $a(1)$ , and we can add nine to the answer at the end.
The second step is to split $99$ as $9$ and $11$ and solve for divisibility rules individually. Let's start with $11$ because it gives us the most information to continue.
In any number generated, if the numbers don't go beyond $20$ , then the highest number we can get is $10111213141516171819$ , with every odd digit being $1$ . This is a little risky because we are assuming that it doesn't exceed $20$ . If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by $11$ . The expression being $\frac{((n-1) + 0)\cdot n)}{2}-n$
The concept here is to add $0$ to the $n-1^{th}$ term altogether, then subtract the number of ones in it, which is $n$ . Simplify to $\frac{n(n-3)}{2}$ congruent to $0 \pmod{11}$ . Now notice the divide by two can be discarded because one of $n$ or $(n-3)$ will be even. So if $n$ or $n-3$ is to be divisible by $11$ , we can make a simple list.
\[n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, \dots\]
Now we test each $n$ for divisibility by $9$ .
This is done by making a list that ultimately calculates the sum of every digit in the large number. $n(1)$ to $n(10)$ has the first digit $1$ $n(11)$ to $n(20)$ has the first digit $2$ , and so on.
The necessary thing to realize is that the sum of all digits $0-9$ is divisible by $9$ , so we only have to solve for the sum of the first digits, and then the short list of second digits.
For example, let's test $n=25$
So we know that $25$ include both $1-10$ and $11-20$ , so that's $10 + 20$ right away. $21-25$ contains $5$ numbers that have the first digit $3$ , so $+15$ . Then we add $0-4$ together, which is $10$ $10+20+15+10=55$ , which is not divisible by $9$ , so it is not the answer.
Do this for just a minute you get that $36$ sums to $99$ , a multiple of nine! So $n(36)$ is the answer, right? Don't forget we have to add $9$ because we translated $n(10)$ to $n(1)$ at the very beginning!
Finally, after a short bash, we get $\boxed{045}$

Solution: Taking the recurrence mod $99$ , we have \[a_n=a_{n-1}+n\] for $a_{10}=10$ . Then, we have \[a_n=10+11+12+\cdots+n \implies a_n=\frac{n(n+1)}2-45 \equiv 0 \pmod{99} \implies n(n+1)-90 \equiv 0 \pmod{99} \implies n(n+1)+9 \equiv 0 \pmod{99} \implies n \equiv 0 \pmod{9}.\] Then, we simply can test these values of $n$ to find that $n=\boxed{045}$ produces a value that is also divisible by $11$


Answer: 1.0

Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{10}{11} \qquad\textbf{(E)}\ 1$

Solution: Plugging in $c$ , we get \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}.\] Since $\sin(x+2\pi)=\sin(x),$ $\sin(2\pi-x)=\sin(-x),$ and $\sin(-x)=-\sin(x),$ we get \[\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{-10\pi}{11} \cdot \sin \frac{-4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}}=\boxed{1}.\]

Solution:
Eisenstein used such a quotient in his proof of quadratic reciprocity . Let $c=\frac{2\pi}{p}$ where $p$ is an odd prime number and $q$ is any integer.
Then $\dfrac{\sin(qc)\sin(2qc)\cdots\sin(\frac{p-1}{2}qc)}{\sin(c)\sin(2c)\cdots\sin(\frac{p-1}{2}c)}$ is the Legendre symbol $\left(\frac{q}{p}\right)$ . Legendre symbol is calculated using quadratic reciprocity which is $\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$ . The Legendre symbol $\left(\frac{3}{11}\right)=(-1)\left(\frac{11}{3}\right)=(-1)\left(\frac{-1}{3}\right)=(-1)(-1)=\boxed{1}$


Answer: 4.5

Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$ . Points $z_1$ $z_2$ $\frac{1}{z_1}$ , and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$

Solution:
Because $c$ is real, $z_2 = \bar z_1$ .
We have \begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*} where the first equality follows from Vieta's formula.
Thus, $|z_1| = \sqrt{10}$
We have \begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*} where the first equality follows from Vieta's formula.
Thus, ${\rm Re}(z_1) = \frac{c}{2}$
We have \begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*} where the second equality follows from Vieta's formula.
We have \begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*} where the second equality follows from Vieta's formula.
Therefore, \begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*} where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if $c^2 = 40 - c^2$ .
Thus, $|c| = 2 \sqrt{5} \approx \boxed{4.5}$

Solution:
Like the solutions above we can know that $|z_1| = |z_2| = \sqrt{10}$ and $z_1=\overline{z_2}$
Let $z_1=\sqrt{10}e^{i\theta}$ where $0<\theta<\pi$ , then $z_2=\sqrt{10}e^{-i\theta}$ $\frac{1}{z_1}=\frac{1}{\sqrt{10}}e^{-i\theta}$ $\frac{1}{z_2}= \frac{1}{\sqrt{10}}e^{i\theta}$
On the basis of symmetry, the area $A$ of $\mathcal{Q}$ is the difference between two isoceles triangles,so
$2A=10\sin2\theta-\frac{1}{10}\sin2\theta\leq10-\frac{1}{10}$ . The inequality holds when $2\theta=\frac{\pi}{2}$ , or $\theta=\frac{\pi}{4}$
Thus, $c= 2 {\rm Re} \ z_1 =2 \sqrt{10} \cos\frac{\pi}{4}=\sqrt{20} \approx \boxed{4.5}$

Solution:
Like in Solution 3, we find that $Q = \frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}= \frac{99}{100}ab$ , thus, $Q$ is maximized when $ab$ is maximized. $ab = a \cdot \sqrt{10 - a^2} = \sqrt{10a^2 - a^4}$ , let $f(a) =  \sqrt{10a^2 - a^4}$
By the Chain Rule and the Power Rule, $f'(a) = \frac12 \cdot (10a^2 - a^4)^{-\frac12} \cdot (10(2a)-4a^3)) = \frac{20a-4a^3}{ 2\sqrt{10a^2 - a^4} } = \frac{10a-2a^3}{ \sqrt{10a^2 - a^4} }$
$f'(a) = 0$ $10a-2a^3 = 0$ $a \neq 0$ $a^2 = 5$ $a = \sqrt{5}$
$\because f'(a) = 0$ when $a = \sqrt{5}$ $f'(a)$ is positive when $a < \sqrt{5}$ , and $f'(a)$ is negative when $a > \sqrt{5}$
$\therefore f(a)$ has a local maximum when $a = \sqrt{5}$
Notice that $ab = \frac{ (a+b)^2 - (a^2+b^2) }{2} = \frac{c^2 - 10}{2}$ $\frac{c^2 - 10}{2} = 5$ $c = \sqrt{2 \cdot 5 + 10} = \sqrt{20} \approx \boxed{4.5}$


Answer: 4.5

Solution:
Since $c$ , which is the sum of roots $z_1$ and $z_2$ , is real, $z_1=\overline{z_2}$
Let $z_1=a+bi$ . Then $z_2=a-bi$ . Note that the product of the roots is $10$ by Vieta's, so $z_1z_2=(a+bi)(a-bi)=a^2+b^2=10$
Thus, $\frac{1}{z_1}=\frac{1}{a-bi}\cdot\frac{a+bi}{a+bi}=\frac{a+bi}{a^2+b^2}=\frac{a+bi}{10}$ . With the same process, $\frac{1}{z_2}=\frac{a-bi}{10}$
So, our four points are $a+bi,\frac{a+bi}{10},a-bi,$ and $\frac{a-bi}{10}$ . WLOG let $a+bi$ be in the first quadrant and graph these four points on the complex plane. Notice how quadrilateral Q is a trapezoid with the real axis as its axis of symmetry. It has a short base with endpoints $\frac{a+bi}{10}$ and $\frac{a-bi}{10}$ , so its length is $\left|\frac{a+bi}{10} - \frac{a-bi}{10}\right| =\frac{b}{5}$ . Likewise, its long base has endpoints $a+bi$ and $a-bi$ , so its length is $|(a+bi)-(a-bi)|=2b$
The height, which is the distance between the two lines, is the difference between the real values of the two bases $\implies h= a-\frac{a}{10}=\frac{9a}{10}$
Plugging these into the area formula for a trapezoid, we are trying to maximize $\frac12\cdot\left(2b+\frac{b}{5}\right)\cdot\frac{9a}{10}=\frac{99ab}{100}$ . Thus, the only thing we need to maximize is $ab$
With the restriction that $a^2+b^2=(a-b)^2+2ab=10$ $ab$ is maximized when $a=b=\sqrt{5}$
Remember, $c$ is the sum of the roots, so $c=z_1+z_2=2a=2\sqrt5=\sqrt{20}\approx\boxed{4.5}$


Answer: 0.0

Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$ . What is the value of $(d-1)(e-1)$
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution:
Using factoring:
$2x^{2}+3x-5=0$
$(2x+5)(x-1)=0$
$x = -\frac{5}{2}$ or $x=1$
So $d$ and $e$ are $-\frac{5}{2}$ and $1$
Therefore the answer is $\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{0}$

Solution:
We can use the sum and product of a quadratic (a.k.a Vieta):
$(d-1)(e-1)=de-(d+e)+1 \implies\text{product}-\text{sum}+1 \implies \dfrac{c}{a}-\left(-\dfrac{b}{a}\right)+1 \implies \dfrac{b+c}{a}+1= \dfrac{5}{-5}+1=\boxed{0}$

Solution:
By inspection, we quickly note that $x=1$ is a solution to the equation, therefore the answer is
$(d-1)(e-1)=(1-1)(e-1)=\boxed{0}$

Solution: The form $(d-1)(e-1)$ resembles the factored form for the quadratic, namely $(x-d)(x-e)$ based on the information given. Note putting in 1 for $x$ in the that quadratic immediately yields the desired expression. Thus, $2(1)^2+3(1)-5 = \boxed{0}$


Answer: 9.0

Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

Solution:
We consider the prime factorization of $n:$ \[n=\prod_{i=1}^{k}p_i^{e_i}.\] By the Multiplication Principle, we have \[d(n)=\prod_{i=1}^{k}(e_i+1).\] Now, we rewrite $f(n)$ as \[f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_i+1}{p_i^{{e_i}/3}}.\] As $f(n)>0$ for all positive integers $n,$ note that $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3$ for all positive integers $a$ and $b.$ So, $f(n)$ is maximized if and only if \[f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}\] is maximized.
For each independent factor $\frac{(e_i+1)^3}{p_i^{e_i}}$ with a fixed prime $p_i,$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\ldots\right)=\left(2,3,5,7,\ldots\right),$ we look for the nonnegative integer $e_i$ such that $\frac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum: \[\begin{array}{c|c|c|c|c}  & & & & \\ [-2.25ex] \boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{\dfrac{(e_i+1)^3}{p_i^{e_i}}} & \textbf{Max?} \\ [2.5ex] \hline\hline  & & & & \\ [-2ex] 1 & 2 & 0 & 1 & \\      & & 1 & 4 & \\     & & 2 & 27/4 &\\     & & 3 & 8 & \checkmark\\     & & 4 & 125/16 & \\ [0.5ex] \hline   & & & & \\ [-2ex] 2 & 3 & 0 & 1 &\\     & & 1 & 8/3 & \\     & & 2 & 3 &  \checkmark\\     & & 3 & 64/27 &  \\ [0.5ex] \hline   & & & & \\ [-2ex] 3 & 5 & 0 & 1 &  \\     & & 1 & 8/5 &  \checkmark\\     & & 2 & 27/25 & \\ [0.5ex] \hline   & & & & \\ [-2ex] 4 & 7 & 0 & 1 &  \\     & & 1 & 8/7 &  \checkmark\\     & & 2 & 27/49 & \\ [0.5ex] \hline   & & & & \\ [-2ex] \geq5 & \geq11 & 0 & 1 & \checkmark \\     & & \geq1 & \leq8/11 &   \\ [0.5ex] \end{array}\] Finally, the positive integer we seek is $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{9}.$

Solution:
The question statement asks for the value of $N$ that maximizes $f(N)$ . Let $N$ start out at $1$ ; we will find what factors to multiply $N$ by, in order for $N$ to maximize the function.
First, we will find what power of $2$ to multiply $N$ by. If we multiply $N$ by $2^{a}$ , the numerator of $f$ $d(N)$ , will multiply by a factor of $a+1$ ; this is because the number $2^{a}$ has $a+1$ divisors. The denominator, $\sqrt[3]{N}$ , will simply multiply by $\sqrt[3]{2^{a}}$ . Therefore, the entire function multiplies by a factor of $\frac{a+1}{\sqrt[3]{2^{a}}}$ . We want to find the integer value of $a$ that maximizes this value. By inspection, this is $3$ . Therefore, we multiply $N$ by $8$ ; right now, $N$ is $8$
Next, we will find what power of $3$ to multiply $N$ by. Similar to the previous step, we wish to find the integer value of $a$ that maximizes $\frac{a+1}{\sqrt[3]{3^{a}}}$ . This value, also by inspection, is $2$
We can repeat this step on the rest of the primes to get \[N = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7\] but from $11$ on, $a=0$ will maximize the value of the function, so the prime is not a factor in $N$ .
We evaluate $N$ to be $2520$ , so the answer is $\boxed{9}$

Solution:
Using the answer choices to our advantage, we can show that $N$ must be divisible by 9 without explicitly computing $N$ , by exploiting the following fact:
Claim : If $n$ is not divisible by 3, then $f(9n) > f(3n) > f(n)$
Proof : Since $d(\cdot)$ is a multiplicative function , we have $d(3n) = d(3)d(n) = 2d(n)$ and $d(9n) = 3d(n)$ . Then \begin{align*} f(3n) &= \frac{2d(n)}{\sqrt[3]{3n}} \approx 1.38 f(n)\\ f(9n) &= \frac{3d(n)}{\sqrt[3]{9n}} \approx 1.44 f(n) \end{align*} Note that the values $\frac{2}{\sqrt[3]{3}}$ and $\frac{3}{\sqrt[3]{9}}$ do not have to be explicitly computed; we only need the fact that $\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1$ which is easy to show by hand.
The above claim automatically implies $N$ is a multiple of 9: if $N$ was not divisible by 9, then $f(9N) > f(N)$ which is a contradiction, and if $N$ was divisible by 3 and not 9, then $f(3N) > f(N) > f\left(\frac{N}{3}\right)$ , also a contradiction. Then the sum of digits of $N$ must be a multiple of 9, so only choice $\boxed{9}$ works.

Solution: The problem mentions the sum of digits - recall that if a number is divisible by 9, then so is the sum of its digits. Guess that the answer must therefore be $\boxed{9}$


Answer: 19.0

Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]

Solution: Let $y = \frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\frac{1}{3}, \frac{2}{9}$ . These sum up to $\boxed{19}$

Solution: This is quite trivially solved, as $3x = \dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has solutions $-\frac{1}{3}$ and $\frac{2}{9}$ . Adding these yields a solution of $\boxed{19}$

Solution: Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{19}$

Solution: Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-\frac{1}{9}$ or $\boxed{19}$

Solution:
Since we have $f(x/3)$ $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7:
$81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-\frac{b}{a}$ . Thus, the sum of the roots of this equation is $-\frac{9}{81} = \boxed{19}$


Answer: 19.0

Solution: Let $y = \frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\frac{1}{3}, \frac{2}{9}$ . These sum up to $\boxed{19}$

Solution: This is quite trivially solved, as $3x = \dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has solutions $-\frac{1}{3}$ and $\frac{2}{9}$ . Adding these yields a solution of $\boxed{19}$

Solution: Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{19}$

Solution: Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-\frac{1}{9}$ or $\boxed{19}$

Solution:
Since we have $f(x/3)$ $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7:
$81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-\frac{b}{a}$ . Thus, the sum of the roots of this equation is $-\frac{9}{81} = \boxed{19}$