Answer: 0.0

Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$ , and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$ . Find the remainder when $m-n$ is divided by $1000$

Solution:
We can avoid computing $m$ and $n$ , instead we will compute $m-n$ directly.
Note that $4x+3y+2z=2009$ if and only if $4(x-1)+3(y-1)+2(z-1)=2000$ . Hence there is an almost 1-to-1 correspondence between the positive integer solutions of the two equations. The only exceptions are the solutions of the first equation in which at least one of the variables is equal to $1$ . The value $m-n$ is the number of such solutions.
If $x=1$ , we get the equation $3y+2z=2005$ . The variable $y$ must be odd, and it must be between $1$ and $667$ , inclusive. For each such $y$ there is exactly one valid $z$ . Hence in this case there are $334$ valid solutions.
If $y=1$ , we get the equation $4x+2z=2006$ , or equivalently $2x+z=1003$ . The variable $x$ must be between $1$ and $501$ , inclusive, and for each such $x$ there is exactly one valid $z$ . Hence in this case there are $501$ valid solutions.
If $z=1$ , we get the equation $4x+3y=2007$ . The variable $y$ must be odd, thus let $y=2u-1$ . We get $4x+6u=2010$ , or equivalently, $2x+3u=1005$ . Again, we see that $u$ must be odd, thus let $u=2v-1$ . We get $2x+6v=1008$ , which simplifies to $x+3v=504$ . Now, we see that $v$ must be between $1$ and $167$ , inclusive, and for each such $v$ we have exactly one valid $x$ . Hence in this case there are $167$ valid solutions.
Finally, we must note that there are two special solutions: one with $x=y=1$ , and one with $y=z=1$ . We counted each of them twice, hence we have to subtract two from the total.
Therefore $m-n = 334 + 501 + 167 - 2 = 1000$ , and the answer is $1000\bmod 1000 = \boxed{000}$


Answer: 19.0

Let $m$ be the smallest integer whose cube root is of the form $n+r$ , where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$ . Find $n$

Solution:
In order to keep $m$ as small as possible, we need to make $n$ as small as possible.
$m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3$ .  Since $r < \frac{1}{1000}$ and $m - n^3 = r(3n^2 + 3nr + r^2)$ is an integer, we must have that $3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000$ .  This means that the smallest possible $n$ should be less than 1000. In particular, $3nr + r^2$ should be less than 1, so $3n^2 > 999$ and $n > \sqrt{333}$ $18^2 = 324 < 333 < 361 = 19^2$ , so we must have $n \geq 19$ .  Since we want to minimize $n$ , we take $n = 19$ .  Then for any positive value of $r$ $3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000$ , so it is possible for $r$ to be less than $\frac{1}{1000}$ .  However, we still have to make sure a sufficiently small $r$ exists.
In light of the equation $m - n^3 = r(3n^2 + 3nr + r^2)$ , we need to choose $m - n^3$ as small as possible to ensure a small enough $r$ .  The smallest possible value for $m - n^3$ is 1, when $m = 19^3 + 1$ .  Then for this value of $m$ $r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}$ , and we're set.  The answer is $\boxed{019}$

Solution:
To minimize $m$ , we should minimize $n$ . We have that $(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}$ . For a given value of $n$ , if $(n + \frac{1}{1000})^3 - n^3 > 1$ , there exists an integer between $(n + \frac{1}{1000})^3$ and $n^3$ , and the cube root of this integer would be between $n$ and $n + \frac{1}{1000}$ as desired. We seek the smallest $n$ such that $(n + \frac{1}{1000})^3 - n^3 > 1$
\[(n + \frac{1}{1000})^3 - n^3 > 1\] \[\frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9} > 1\] \[3n^2 + \frac{3}{10^3} n + \frac{1}{10^6} > 10^3\]
Trying values of $n$ , we see that the smallest value of $n$ that works is $\boxed{019}$

Solution: Since $r$ is less than $1/1000$ , we have $\sqrt[3]{m} < n + \frac{1}{1000}$ . Notice that since we want $m$ minimized, $n$ should also be minimized. Also, $n^3$ should be as close as possible, but not exceeding $m$ . This means $m$ should be set to $n^3+1$ . Substituting and simplifying, we get \[\sqrt[3]{n^3+1} < n + \frac{1}{1000}\] \[n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}\] The last two terms in the right side can be ignored in the calculation because they are too small. This results in $1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}$ . The minimum positive integer $n$ that satisfies this is $\boxed{019}$ . ~ Hb10


Answer: 634.0

Let $m/n$ , in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$ .  Find $m + n$

Solution: $10^{99} = 2^{99}5^{99}$ , so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$ ; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$ . Our probability is $\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}$ , and $m + n = \boxed{634}$


Answer: 12.0

Let $m>1$ and $n>1$ be integers. Suppose that the product of the solutions for $x$ of the equation
\[8(\log_n x)(\log_m x)-7\log_n x-6 \log_m x-2013 = 0\]
is the smallest possible integer. What is $m+n$
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 272$

Solution:
Rearranging logs, the original equation becomes
\[\frac{8}{\log n \log m}(\log x)^2 - \left(\frac{7}{\log n}+\frac{6}{\log m}\right)\log x - 2013 = 0\]
By Vieta's Theorem, the sum of the possible values of $\log x$ is $\frac{\frac{7}{\log n}+\frac{6}{\log m}}{\frac{8}{\log n \log m}} = \frac{7\log m + 6 \log n}{8} = \log \sqrt[8]{m^7n^6}$ . But the sum of the possible values of $\log x$ is the logarithm of the product of the possible values of $x$ . Thus the product of the possible values of $x$ is equal to $\sqrt[8]{m^7n^6}$
It remains to minimize the integer value of $\sqrt[8]{m^7n^6}$ . Since $m, n>1$ , we can check that $m = 2^2$ and $n = 2^3$ work. Thus the answer is $4+8 = \boxed{12}$


Answer: 11.0

Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$ . There is a polynomial \[q(x) = c_3x^3+c_2x^2+c_1x+c_0\] such that $D(m) = q(m)$ for all odd integers $m\ge 5$ . What is $c_1?$
$\textbf{(A)}\ {-}6\qquad\textbf{(B)}\ {-}1\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 11$

Solution:
For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$
Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$
Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots,k+4(m-1)\}\] is the complete residue system modulo $m$ for all integers $k.$
Given any ordered quadruple $(a'_1, a'_2, a'_3, a'_4)$ in modulo $m,$ it follows that exactly one of these $m$ ordered quadruples has sum $0$ modulo $m:$ \[\begin{array}{c|c} & \\ [-2.5ex] \textbf{Ordered Quadruple} & \textbf{Sum Modulo }\boldsymbol{m} \\ [0.5ex] \hline & \\ [-2ex] (a'_1, a'_2, a'_3, a'_4) & S'+4(0) \\ [0.5ex] (a'_1+1, a'_2+1, a'_3+1, a'_4+1) & S'+4(1) \\ [0.5ex] (a'_1+2, a'_2+2, a'_3+2, a'_4+2) & S'+4(2) \\ [0.5ex] \cdots & \cdots \\ [0.5ex] (a'_1+m-1, a'_2+m-1, a'_3+m-1, a'_4+m-1) & S'+4(m-1) \\ [0.5ex] \end{array}\] We conclude that $q(m)=\frac1m\cdot[m(m-1)(m-2)(m-3)]=(m-1)(m-2)(m-3),$ so \[q(x)=(x-1)(x-2)(x-3)=c_3x^3+c_2x^2+c_1x+c_0.\] By Vieta's Formulas, we get $c_1=1\cdot2+1\cdot3+2\cdot3=\boxed{11}.$

Solution:
Define \[ b_i = \left\{ \begin{array}{ll} a_i & \mbox{ if } 1 \leq a_i \leq \frac{m-1}{2} \\ a_i - m & \mbox{ if } \frac{m-1}{2} + 1 \leq a_i \leq m - 1 \\ 0 & \mbox{ if } a_i = m \end{array} \right.. \]
Hence, $b_i$ is a one-to-one and onto function of $a_i$ , and the range of $b_i$ is $\left\{- \frac{m-1}{2}  , \cdots , \frac{m-1}{2} \right\}$
Therefore, to solve this problem, it is equivalent for us to count the number of tuples $\left( b_1 , b_2 , b_3 , b_4 \right)$ that are all distinct and satisfy $m | b_1 + b_2 + b_3 + b_4$
Denote by $d \left( m \right)$ the number of such tuples that are also subject to the constraint $b_1 < b_2 < b_3 < b_4$
Hence, $D \left( m \right) = 4! d \left( m \right) = 24 d \left( m \right)$
We do the following casework analysis to compute $d \left( m \right)$
$\textbf{Case 1}$ : There is one $0$ in $\left( b_1 , b_2 , b_3 , b_4 \right)$
Denote by $d_{1i} \left( m \right)$ the number of tuples with $b_i = 0$
By symmetry, $d_{11} \left( m \right)= d_{14} \left( m \right)$ and $d_{12} \left( m \right)= d_{13} \left( m \right)$
$\textbf{Case 2}$ : There is no $0$ in $\left( b_1 , b_2 , b_3 , b_4 \right)$
Denote by $d_{2i} \left( m \right)$ the number of tuples with $i$ positive entries.
By symmetry, $d_{20} \left( m \right) = d_{24} \left( m \right)$ and $d_{21} \left( m \right) = d_{23} \left( m \right)$
Therefore, \begin{align*} D \left( m \right) & = 24 d \left( m \right) \\ & = 24 \left( \sum_{i=1}^4 d_{1i} \left( m \right) + \sum_{i=0}^4 d_{2i} \left( m \right) \right) \\ & = 24 \left( 2 d_{11} \left( m \right) + 2 d_{12} \left( m \right) + 2 d_{24} \left( m \right) + 2 d_{23} \left( m \right) + d_{22} \left( m \right) \right) . \end{align*}
Now, we compute $D \left( m \right)$ for $m = 5 , 7 , 9 , 11$
$\underline{\textbf{SCENARIO}}$ $m = 5$
We have $b_i \in \left\{ - 2 , \cdots , 2 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
We cannot have $3$ distinct positive integers. So $d_{11} \left( 5 \right) = 0$
$\textbf{Case 1.2}$ $b_2 = 0$
Because there are $2$ positive integers, we must have $b_3 = 1$ $b_4 = 2$ .
Hence, $b_1 = - 3$ . However, this is out of the range of $b_i$ .
Thus, $d_{12} \left( 5 \right) = 0$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
We cannot have $4$ distinct positive integers. So $d_{24} \left( 5 \right) = 0$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
We cannot have $3$ distinct positive integers. So $d_{23} \left( 5 \right) = 0$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
The only solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left( - 2 , - 1 , 1 , 2 \right)$ . So $d_{22} \left( 5 \right) = 1$
Therefore, $D \left( 5 \right) = 24$
$\underline{\textbf{SCENARIO}}$ $m = 7$
We have $b_i \in \left\{ - 3 , \cdots , 3 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
We have no feasible solution. Thus, $d_{11} \left( 7 \right) = 0$
$\textbf{Case 1.2}$ $b_2 = 0$
The only solution is $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ .
Thus, $d_{12} \left( 7 \right) = 1$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
We cannot have $4$ distinct positive integers. So $d_{24} \left( 7 \right) = 0$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
To get $3$ distinct positive integers, we have $\left( b_2 , b_3 , b_4 \right) = \left( 1 , 2 , 3 \right)$ . This implies $b_1 = - 6$ . However, this is out of the range of $b_1$ . So $d_{23} \left( 6 \right) = 0$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
We have $d_{22} \left( 7 \right) = \binom{3}{2} = 3$
Therefore, $D \left( 7 \right) = 24 \cdot 5$
$\underline{\textbf{SCENARIO}}$ $m = 9$
We have $b_i \in \left\{ - 4 , \cdots , 4 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
The only solution is $\left( b_2 , b_3 , b_4 \right) = \left(  2, 3, 4 \right)$ .
Thus, $d_{11} \left( 9 \right) = 1$
$\textbf{Case 1.2}$ $b_2 = 0$
The feasible solutions are $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ $\left( - 4 , 1 , 3 \right)$ .
Thus, $d_{12} \left( 9 \right) = 2$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
There is no feasible solution. So $d_{24} \left( 9 \right) = 0$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
To get $3$ distinct positive integers, we have $b_2 + b_3 + b_4 \geq 1 + 2 + 3 = 6$ . This implies $b_1 = - 6$ . However, this is out of the range of $b_1$ . So $d_{23} \left( 9 \right) = 0$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
We have $d_{22} \left( 9 \right) = 8$
Therefore, $D \left( 9 \right) = 24 \cdot 14$
$\underline{\textbf{SCENARIO}}$ $m = 11$
We have $b_i \in \left\{ - 5 , \cdots , 5 \right\}$
$\textbf{Case 1}$
$\textbf{Case 1.1}$ $b_1 = 0$
The only solution is $\left( b_2 , b_3 , b_4 \right) = \left(  2, 4, 5 \right)$ .
Thus, $d_{11} \left( 11 \right) = 1$
$\textbf{Case 1.2}$ $b_2 = 0$
The feasible solutions are $\left( b_1 , b_3 , b_4 \right) = \left( - 3 , 1 , 2 \right)$ $\left( - 4 , 1 , 3 \right)$ $\left( - 5 , 1 , 4 \right)$ $\left( - 5 , 2 , 3 \right)$ .
Thus, $d_{12} \left( 11 \right) = 4$
$\textbf{Case 2}$
$\textbf{Case 2.1}$ $b_1 > 0$
The only feasible solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left(  1 , 2, 3, 5 \right)$ . So $d_{24} \left( 11 \right) = 1$
$\textbf{Case 2.2}$ $b_1 < 0 < b_2$
The only feasible solution is $\left( b_1 , b_2 , b_3 , b_4 \right) = \left(  -1 , 3, 4, 5 \right)$ .  So $d_{23} \left( 11 \right) = 1$
$\textbf{Case 2.3}$ $b_2 < 0 < b_3$
We have $d_{22} \left( 11 \right) = 16$
Therefore, $D \left( 11 \right) = 24 \cdot 30$
We know that $q \left( m \right) = D \left( m \right)$ for odd $m \geq 5$
Plugging $m = 5, 7, 9, 11$ into this equation, we get \begin{align*} c_3 5^3 + c_2 5^2 + c_1 5 + c_0 & = 24 \cdot 1 && (1.1) \\ c_3 7^3 + c_2 7^2 + c_1 7 + c_0 & = 24 \cdot 5 && (1.2) \\ c_3 9^3 + c_2 9^2 + c_1 9 + c_0 & = 24 \cdot 14 && (1.3) \\ c_3 11^3 + c_2 11^2 + c_1 11 + c_0 & = 24 \cdot 30 && (1.4) \end{align*}
Now, we solve this system of equations.
Taking $\frac{(1.2)-(1.1)}{2}$ $\frac{(1.3)-(1.2)}{2}$ $\frac{(1.4)-(1.2)}{2}$ , we get \begin{align*} c_3 109 + c_2 12 + c_1 & = 48 && (2.1) \\ c_3 193 + c_2 16 + c_1 & = 108 && (2.2) \\ c_3 301 + c_2 20 + c_1 & = 192 && (2.3) \end{align*}
Taking $\frac{(2.2)-(2.1)}{4}$ $\frac{(2.3)-(2.2)}{4}$ , we get \begin{align*} c_3 21 + c_2  & = 15 && (3.1) \\ c_3 27 + c_2 & = 21 && (3.2) \end{align*}
Taking $\frac{(3.2)-(3.1)}{6}$ , we get $c_3 = 1$
Plugging $c_3$ into Equation (3.1), we get $c_2 = - 6$
Plugging $c_2$ and $c_3$ into Equation (2.1), we get $c_1 = 11$
Therefore, the answer is $\boxed{11}$

Solution: As before, note that we have $m(m-1)(m-2)$ numbers we can choose as $a,b,c.$ From here, there is exactly one possible value of $1 \leq d \leq m$ that could make $S=a+b+c+d$ divisible by $m.$ However, there is a $\frac{3}{m}$ chance that this value of $d$ has already been chosen as $a,b$ or $c$ . Thus our polynomial is $m(m-1)(m-2)\left(1-\frac{3}{m}\right)=m(m-1)(m-2)\left(\frac{m-3}{m}\right)=(m-1)(m-2)(m-3)$ . By Vieta's, $c_1 = 2+3+6=\boxed{11}$


Answer: 8181.0

Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

Solution: $11|(q+r)$ implies that $11|(99q+q+r)$ and therefore $11|(100q+r)$ , so $11|n$ . Then, $n$ can range from $10010$ to $99990$ for a total of $\boxed{8181}$ numbers.


Answer: 8181.0

Solution:
When a $5$ -digit number is divided by $100$ , the first $3$ digits become the quotient, $q$ , and the last $2$ digits become the remainder, $r$
Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.
For each of the $9\cdot10\cdot10=900$ possible values of $q$ , there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$
Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$ , each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$
Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{8181}$

Solution: Notice that $q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}$ . This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are $\frac{99990-10010}{11}+1=8181$ possible values. The answer is $\boxed{8181}$

Solution: Let $abcde$ be the five digits of $n$ . Then $q = abc$ and $r = de$ . By the divisibility rules of $11$ $q = a - b + c \pmod{11}$ and $r = -d + e \pmod{11}$ , so $q + r = a - b + c - d + e = abcde = n \pmod{11}$ . Thus, $n$ must be divisble by $11$ . There are $\frac{99990 - 10010}{11} + 1 = 8181$ five-digit multiples of $11$ , so the answer is $\boxed{8181}$


Answer: 11.0

Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$ , and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$

Solution:
We can start by setting up an equation to convert $\underline{32d}$ base $n$ to base 10. To convert this to base 10, it would be $3{n}^2+2n+d.$ Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get $3{n}^2+2n = 263-d$
We can also set up equations to convert $\underline{324}$ base $n$ and $\underline{11d1}$ base 6 to base 10. The equation to covert $\underline{324}$ base $n$ to base 10 is $3{n}^2+2n+4.$ The equation to convert $\underline{11d1}$ base 6 to base 10 is ${6}^3+{6}^2+6d+1.$
Simplify ${6}^3+{6}^2+6d+1$ so it becomes $6d+253.$ Setting the above equations equal to each other, we have \[3{n}^2+2n+4 = 6d+253.\] Subtracting 4 from both sides gets $3{n}^2+2n = 6d+249.$
We can then use equations \[3{n}^2+2n = 263-d\] \[3{n}^2+2n = 6d+249\] to solve for $d$ . Set $263-d$ equal to $6d+249$ and solve to find that $d=2$
Plug $d=2$ back into the equation $3{n}^2+2n = 263-d$ . Subtract 261 from both sides to get your final equation of $3{n}^2+2n-261 = 0.$ We solve using the quadratic formula to find that the solutions are $9$ and $-29/3.$ Because the base must be positive, $n=9.$
Adding 2 to 9 gets $\boxed{11}$

Solution: $32d$ is greater than $263$ when both are interpreted in base 10, so $n$ is less than $10$ . Some trial and error gives $n=9$ $263$ in base 9 is $322$ , so the answer is $9+2=\boxed{11}$

Solution: We have \[3n^2 + 2n + d = 263\] \[3n^2 + 2n + 4 = 6^3 + 6^2 + 6d + 1\] Subtracting the 2nd from the 1st equation we get \begin{align*} d-4 &= 263 - (216 + 36 + 6d + 1) \\ &= 263 - 253 - 6d \\ &= 10 - 6d  \end{align*} Thus we have $d=2.$ Substituting into the first, we have $3n^2 + 2n + 2 = 263 \Rightarrow 3n^2 + 2n - 261 = 0.$ Factoring, we have $(n-9)(3n+29)=0.$ A digit cannot be negative, so we have $n=9.$ Thus, $d+n=2+9=\boxed{11}$

Solution: Find that $d=2$ using one of the methods above. Then we have that $3n^2 + 2n = 261$ . We know that $n$ is an integer, so we can solve the equation $n(2n+3)=261$ (this is guaranteed to have a solution if we did this correctly). The prime factorization of $261$ is $3^2 \cdot 29$ , so the corresponding factor pairs are $(1, 261)$ $(3,87)$ , and $(9,29)$ . If $n = 9$ , then the equation is true, so $n + d = 9 + 2 = \boxed{11}$


Answer: 8.0

Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution:
A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:
\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}\]
We first look at the case when $n!$ has $1$ zero and $(2n)!$ has $3$ zeros. If $n=5,6,7$ $(2n)!$ has only $2$ zeros. But for $n=8,9$ $(2n)!$ has $3$ zeros. Thus, $n=8$ and $n=9$ work.
Secondly, we look at the case when $n!$ has $2$ zeros and $(2n)!$ has $6$ zeros. If $n=10,11,12$ $(2n)!$ has only $4$ zeros. But for $n=13,14$ $(2n)!$ has $6$ zeros. Thus, the smallest four values of $n$ that work are $n=8,9,13,14$ , which sum to $44$ . The sum of the digits of $44$ is $\boxed{8}$

Solution:
By Legendre's Formula and the information given, we have that $3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor$
We have $n<100$ as there is no way that if $n>100$ $(2n)!$ would have $3$ times as many zeroes as $n!$
First, let's plug in the number $5$ .
We get that $3(1)=1$ , which is obviously not true. Hence, $n>5$
After several attempts, we realize that the RHS needs $1$ to $2$ more "extra" zeroes than the LHS. Hence, $n$ is greater than a multiple of $5$
We find that the least four possible $n$ are $8,9,13,14$
$8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{8}$

Solution:
Let $n=5m+k$ for some natural numbers $m$ $k$ such that $k\in\{0,1,2,3,4\}$ . Notice that $n<5^3=125$ . Thus \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor+\left\lfloor\frac{2n}{125}\right\rfloor\] For smaller $n$ , we temporarily let $\left\lfloor\frac{2n}{125}\right\rfloor=0$ \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{2(5m+k)}{5}\right\rfloor+\left\lfloor\frac{2(5m+k)}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{10m+2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[3m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=2m+\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] To minimize $n$ , we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{10m+2k}{25}\right\rfloor=0$ , then \[m=\left\lfloor\frac{2k}{5}\right\rfloor\] Since $k<5$ $m>0$ , the only integral value of $m$ is $1$ , from which we have $k=3,4\Longrightarrow n=8,9$
Now we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=0$ and $\left\lfloor\frac{10m+2k}{25}\right\rfloor=1$ , then \[m=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] Since $k<5$ $10m>15\Longrightarrow m\ge2$
If $m>2$ , then \[m>\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] which is a contradiction.
Thus $m=2\Longrightarrow\left\lfloor\frac{2k}{5}\right\rfloor=1\Longrightarrow n=13,14$
Finally, the sum of the four smallest possible $n=8+9+13+14=44$ and $4+4=8$ $\boxed{8}$

Solution:
We first note that the number of 0's in $n!$ is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
$\textbf{CASE ONE: } 5\leq 2n < 25.$
The only way we can fulfill the requirements is if $\lfloor{\dfrac{n}{5}}\rfloor = 1$ and $\lfloor{\dfrac{2n}{5}}\rfloor=3$ which means that $5\leq n <10$ and $15\leq 2n < 20$ . The only way this works is if $n = 8 \text{ or } 9.$
$\textbf{CASE TWO: } 25 \leq 2n$
Since we want the smallest values of $n$ , we first try it when $2n<30.$ Thus $(2n)!$ has 6 zeros, which implies that $n!$ must have 2. The only way to do this while maintaining our restrictions for $2n$ is if $n = 13 \text{ or } 14.$
So the sum of the four values is $8+9+13+14=44$ so the digit is sum is $\boxed{8}.$

Solution: We will use trial and error to determine the answer to this problem. If $n = 5$ , then $n!$ has $1$ zero, and $2n!$ will have $2$ zeros. But $3 \cdot 1 \neq 2$ so $n = 5$ does not work. Similarly $n = 6, 7$ do not work either. But $n = 8$ works because $8!$ has $1$ zero and $16!$ has $3$ zeros. Note that $n = 9$ also works because $9!$ has $1$ zero and $18!$ has $3$ zeros. After performing trial and error several times, we find that $n = 10, 11, 12$ do not work but $n = 13, 14$ do work. Therefore, the four smallest values of $n$ are $8, 9, 13, 14$ . Therefore adding them together gives $44$ and our answer is $\boxed{8}.$


Answer: 84.0

Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:
$\mathrm{(A)}\ 2\ \text{divides\ }n \qquad\mathrm{(B)}\ 3\ \text{divides\ }n \qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n  \qquad\mathrm{(D)}\ 7\ \text{divides\ }n \qquad\mathrm{(E)}\ n > 84$

Solution:
Since $\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}$ $0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2$
From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$ . The only answer choice that is not true is $\boxed{84}$

Solution: Since $\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}$ , it is very clear that $n=42$ makes the expression an integer. Because $n$ is a positive integer, $\frac{1}{n}$ must be less than or equal to $1$ . Thus, the only integer the expression can take is $1$ , which means the only value for $n$ is $42$ . Thus $\boxed{84}$


Answer: 84.0

Solution:
Since $\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}$ $0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2$
From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$ . The only answer choice that is not true is $\boxed{84}$

Solution: Since $\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}$ , it is very clear that $n=42$ makes the expression an integer. Because $n$ is a positive integer, $\frac{1}{n}$ must be less than or equal to $1$ . Thus, the only integer the expression can take is $1$ , which means the only value for $n$ is $42$ . Thus $\boxed{84}$


Answer: 12.0

Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$ $e$ , and $10d+e$ , where $d$ and $e$ are single digits. What is the sum of the digits of $n$
$\mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24$

Solution:
Since we want $n$ to be as large as possible, we would like $d$ in $10d+e$ to be as large as possible. So, $d=7,$ the greatest single-digit prime. Then, $e$ cannot be $5$ because $10(7)+5 = 75,$ which is not prime. So $e = 3$ . Therefore, $d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533$ .
So, the sum of the digits of $n$ is $1+5+3+3=12 \Rightarrow \boxed{12}$ ~ MathGenius_ (Edited by Sophia866)

Solution:
Since $d$ is a single digit prime number, the set of possible values of $d$ is $\{2,3,5,7\}$
Since $e$ is a single digit prime number and is the units digit of the prime number $10d+e$ , the set of possible values of $e$ is $\{3,7\}$
Using these values for $d$ and $e$ , the set of possible values of $10d+e$ is $\{23,27,33,37,53,57,73,77\}$
Out of this set, the prime values are $\{23,37,53,73\}$
Therefore the possible values of $n$ are:
$2\cdot3\cdot23=138$
$3\cdot7\cdot37=777$
$5\cdot3\cdot53=795$
$7\cdot3\cdot73=1533$
The largest possible value of $n$ is $1533$
So, the sum of the digits of $n$ is $1+5+3+3=12 \Rightarrow \boxed{12}$


Answer: 270.0

Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$

Solution:
As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$ Lifting the Exponent shows that \[3=v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$ . It also shows that \[7=v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$
Now, setting $n = 4c$ (necessitated by $149^n \equiv 2^n \pmod 5$ in order to set up LTE), we see \[v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})\] and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4c}-2^{4c})=v_5(149^4-16)+v_5(c)=1+v_5(c)$ meaning that we have that by LTE, $5^4 | c$ and $4 \cdot 5^4$ divides $n$
Since $3^2$ $7^5$ and $4\cdot 5^4$ all divide $n$ , the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$

Solution:
Note that for all $n$ $149^n - 2^n$ is divisible by $149-2 = 147$ by difference of $n$ th powers. That is $3\cdot7^2$ , so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$ . Similarly, we can reason that the smallest $n$ to make the expression divisible by $7^7$ is just $7^5$
Finally, for $5^5$ , take $\pmod {5}$ and $\pmod {25}$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$ , and other values are factors of $4$ . Testing all of them(just $1$ $2$ $4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$ , and this clearly is NOT divisible by $25$ .
Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \cdot 5^4$
Calculating the LCM of all these, one gets $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . Using the factor counting formula,
the answer is $3\cdot3\cdot5\cdot6$ $\boxed{270}$

Solution:
As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$ . For divisibility by $3^3$ , notice that $v_3(149^3 - 2^3) = 2$ as $149^3 - 2^3 =$ $(147)(149^2 + 2\cdot149 + 2^2)$ , and upon checking mods, $149^2 + 2\cdot149 + 2^2$ is divisible by $3$ but not $9$ . In addition, $149^9 - 2^9$ is divisible by $3^3$ because $149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\cdot2^3 + 2^6)$ , and the rightmost factor equates to $1 + 1 + 1 \pmod{3} \equiv 0 \pmod{3}$ . In fact, $n = 9 = 3^2$ is the least possible choice to ensure divisibility by $3^3$ because if $n = a \cdot 3^b$ , with $3 \nmid a$ and $b < 2$ , we write \[149^{a \cdot 3^b} - 2^{a \cdot 3^b} =  (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\cdot2^{3^b}+\cdots2^{3^b(a - 1)}).\] Then, the rightmost factor is equivalent to $\pm a \pmod{3} \not\equiv 0 \pmod{3}$ , and $v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3$
For divisibility by $7^7$ , we'll induct, claiming that $v_7(149^{7^k} - 2^{7^k}) = k + 2$ for whole numbers $k$ . The base case is clear. Then, \[v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k}).\] By the induction hypothesis, $v_7(149^{7^k} - 2^{7^k}) = k + 2$ . Then, notice that \[S(k) = 149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{7} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{49}.\] This tells us that $S(k)$ is divisible by $7$ , but not $49$ so that $v_7\left(S(k)\right) = 1$ , completing our induction. We can verify that $7^5$ is the least choice of $n$ to ensure divisibility by $7^7$ by arguing similarly to the $3^3$ case.
Finally, for $5^5$ , we take the powers of $149$ and $2$ in mod $5$ and mod $25$ . Writing out these mods, we have that $149^n \equiv 2^n \pmod{5}$ if and only if $4 | n$ , in which $149^n \equiv 2^n \equiv 1 \pmod{5}$ . So here we claim that $v_5(149^{4\cdot5^k} - 2^{4\cdot5^k}) = k + 1$ and perform yet another induction. The base case is true: $5 | 149^4 - 2^4$ , but $149^4 - 2^4 \equiv 1 - 16 \pmod{25}$ . Now then, assuming the induction statement to hold for some $k$ \[v_5(149^{4\cdot5^{k+1}} - 2^{4\cdot5^{k+1}}) = (k+1) + v_5(149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}).\] Note that $S'(k) =  149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}$ equates to $S''(k) = 1 + 2^{4\cdot5^k} + \cdots + 2^{16\cdot5^k}$ in both mod $5$ and mod $25$ . We notice that $S''(k) \equiv 0 \pmod{5}$ . Writing out the powers of $2$ mod $25$ , we have $S''(0) \equiv 5 \pmod{25}$ . Also $2^n \equiv 1 \pmod{25}$ when $n$ is a multiple of $20$ . Hence for $k > 0$ $S''(k) \equiv 5 \mod{25}$ . Thus, $v_5\left(S'(k)\right) = 1$ , completing our induction. Applying the same argument from the previous two cases, $4\cdot5^4$ is the least choice to ensure divisibility by $5^5$
Our answer is the number of divisors of $\text{lcm}(3^2, 7^5, 2^2\cdot5^4) = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$ . It is $(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}$


Answer: 18.0

Let $n$ be the least positive integer greater than $1000$ for which
\[\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.\]
What is the sum of the digits of $n$
$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

Solution:
We know that $\gcd(n+57,63)=21$ and $\gcd(n-57, 120)= 60$ by the Euclidean Algorithm. Hence, let $n+57=21\alpha$ and $n-57=60 \gamma$ , where $\gcd(\alpha,3)=1$ and $\gcd(\gamma,2)=1$ . Subtracting the two equations, $38=7\alpha-20\gamma$ . Letting $\gamma = 2s+1$ , we get $58=7\alpha-40s$ . Taking modulo $40$ , we have $\alpha \equiv{14} \pmod{40}$ . We are given that $n=21\alpha -57 >1000$ , so $\alpha \geq 51$ . Notice that if $\alpha =54$ then the condition $\gcd(\alpha,3)=1$ is violated. The next possible value of $\alpha = 94$ satisfies the given condition, giving us $n=1917$
Alternatively, we could have said $\alpha = 40k+14 \equiv{0} \pmod{3}$ for $k \equiv{1} \pmod{3}$ only, so $k \equiv{0,2} \pmod{3}$ , giving us our answer. Since the problem asks for the sum of the digits of $n$ , our answer is $1+9+1+7=\boxed{18}$

Solution: The conditions of the problem reduce to the following: $n+120 = 21k$ and $n+63 = 60\ell$ , where $\gcd(k,3) = 1$ and $\gcd(\ell,2) = 1$ . From these equations, we see that $21k - 60\ell = 57$ . Solving this Diophantine equation gives us that $k = 20a + 17$ and $\ell = 7a + 5$ . Since, $n>1000$ , we can do some bounding and get that $k > 53$ and $\ell > 17$ . Now we start bashing by plugging in numbers that satisfy these conditions: $a=4$ is the first number that works so we get $\ell = 33$ $k = 97$ . Therefore, we have $n=21(97)-120=60(33)-63=1917$ , and our answer is $1+9+1+7=\boxed{18}$

Solution: We first find that $n\equiv6\pmod{21}$ and $n\equiv57\pmod{60}$ , then we get $n=21x+6$ and $n=60y+57$ by definitions, where $x$ and $y$ are integers. It follows that $y$ must be odd, since the GCD will be $120$ instead of $60$ if $y$ is even. Also, the unit digit of $n$ must be $7$ , since the unit digit of $60y$ is always $0$ and the unit digit of $57$ is $7$ . Therefore, we find that $x$ must end in $1$ to satisfy $n$ having a unit digit of $7$ . Also, we find that $x$ must not be a multiple of $3$ or else the GCD will be $63$ . Therefore, we test values for $x$ and find that $x=91$ satisfies all these conditions. Therefore, $n=1917$ and $1+9+1+7 = \boxed{18}$

Solution:
We are given that $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60.$ By applying the Euclidean algorithm in reverse, we have \[\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21\] and \[\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.\]
We now know that $n+183$ must be divisible by $21$ and $60,$ so it is divisible by $\text{lcm}(21, 60) = 420.$ Therefore, $n+183 = 420k$ for some integer $k.$ We know that $3 \nmid k,$ or else the first condition won't hold ( $\gcd$ will be $63$ ) and $2 \nmid k,$ or else the second condition won't hold ( $\gcd$ will be $120$ ). Since $k = 1$ gives us too small of an answer, then $k=5,$ from which $n = 1917.$ So, the answer is $1+9+1+7 = \boxed{18}.$

Solution: $\gcd(n+63,120)=60$ tells us $n+63\equiv60\pmod {120}$ . The smallest $n+63$ that satisfies the previous condition and $n>1000$ is $1140$ , so we start from there. If $n+63=1140$ , then $n+120=1197$ . Because $\gcd(n+120,63)=21$ $n+120\equiv21\pmod {63}$ or $n+120\equiv42\pmod {63}$ . We see that $1197\equiv0\pmod {63}$ , which does not fulfill the requirement for $n+120$ , so we continue by keep on adding $120$ to $1197$ , in order to also fulfill the requirement for $n+63$ . Soon, we see that $n+120\pmod {63}$ decreases by $6$ every time we add $120$ , so we can quickly see that $n=1917$ because at that point $n+120\equiv21\pmod {63}$ . We add up all digits of $1917$ to get $\boxed{18}$

Solution: We are able to set up the following system of congruences: \begin{align*} n &\equiv 6 \pmod {21}, \\ n &\equiv 57 \pmod {60}. \end{align*} Therefore, by definition, we are able to set-up the following system of equations: \begin{align*} n &= 21a + 6, \\ n &= 60b + 57. \end{align*} Thus, we have $21a + 6 = 60b + 57,$ from which \[7a + 2 = 20b + 19.\] We know $7a \equiv 0 \pmod {7},$ and since $7a = 20b + 17,$ therefore $20b + 17 \equiv 0 \pmod{7}.$ Simplifying this congruence further, we have \[b\equiv 3 \pmod{7}.\] Thus, by definition, $b = 7x + 3.$ Substituting this back into our original equation, we get \[n = 60(7x + 3) + 57 = 420x + 237.\] By definition, we are able to set up the following congruence: \[n \equiv 237 \pmod{420}.\] Thus, $n = 1917$ , so our answer is simply $1+9+1+7=\boxed{18}$

Solution:
First, we find $n$ . We know that it is greater than $1000$ , so we first input $n = 1000$ . From the first equation, $\gcd(63, n + 120) = 21$ , we know that if $n$ is correct, after we add $120$ to it, it should be divisible by $21$ , but not $63$ \[\frac{n + 120}{21} = \frac{1120}{21} = 53\text{ R }7.\] This does not work. To get to the nearest number divisible by $21$ , we have to add $14$ to cancel out the remainder. (Note that we don't subtract $7$ to get to $53$ $n$ is already at its lowest possible value!)
Adding $14$ to $1000$ gives us $n = 1014$ . (Note: $n$ is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
Now using the second equation, $\gcd(n + 63, 120) = 60$ , we know that if $n$ is correct, then $n+63$ is divisible by $60$ but not $120$ \[\frac{n + 63}{60} = \frac{1077}{60} = 17\text{ R }57.\] Again, this does not work. This requires some guessing and checking. We can add $21$ over and over again until $n$ is valid. This changes $n$ while also maintaining that $\frac{n + 120}{21}$ has no remainders.
After adding $21$ once, we get $18 r 18$ . By pure luck, adding $21$ two more times gives us $19$ with no remainders.
We now have $1077 + 21 + 21 + 21 = 1140$ . However, this number is divisible by $120$ . To get the next possible number, we add the LCM of $21$ and $60$ (once again, to maintain divisibility), which is $420$ . Unfortunately, $1140 + 420 = 1560$ is still divisible by $120$ . Adding $420$ again gives us $1980$ , which is valid. However, remember that this is equal to $n + 63$ , so subtracting $63$ from $1980$ gives us $1917$ , which is $n$
The sum of its digits are $1 + 9 + 1 + 7 = \boxed{18}$

Solution:
By the Euclidean Algorithm, we have \begin{alignat*}{8} \gcd(63,n+120)&=\hspace{1mm}&&\gcd(63,\phantom{ }\underbrace{n+120-63k_1}_{(n+120) \ \mathrm{mod} \ 63}\phantom{ })&&=21, \\  \gcd(n+63,120)&=&&\gcd(\phantom{ }\underbrace{n+63-120k_2}_{(n+63) \ \mathrm{mod} \ 120}\phantom{ },120)&&=60.  \end{alignat*} Clearly, $n+120-63k_1$ must be either $21$ or $42,$ and $n+63-120k_2$ must be $60.$
More generally, let $t\in\{1,2\},$ so we get \begin{align*} n+120-63k_1&=21t, &\hspace{55.5mm}(1) \\ n+63-120k_2&=60. &\hspace{55.5mm}(2) \end{align*} Subtracting $(2)$ from $(1)$ and then simplifying give \begin{align*} 57-63k_1+120k_2&=21t-60 \\ 117-63k_1+120k_2&=21t \\ 39-21k_1+40k_2&=7t. \hspace{54mm}(\bigstar) \end{align*} Taking $(\bigstar)$ modulo $7$ produces \begin{align*} 4+5k_2&\equiv0\pmod{7} \\ k_2&\equiv2\pmod{7}. \end{align*} Recall that $n>1000.$ From $(2),$ it follows that \[1063-120k_2<n+63-120k_2=60,\] from which $k_2>8.$ Therefore, the possible values for $k_2$ are $9,16,23,\ldots.$
We need to check whether positive integers $k_1$ and $t$ (where $t\in\{1,2\}$ ) exist in $(1):$
Finally, the least such positive integer $n$ is $1917.$ The sum of its digits is $1+9+1+7=\boxed{18}.$

Solution:
Because we are finding value of $n$ for $n > 1000$ , let $n = 1000 + k$
Using the Euclidean Algorithm, \begin{align*} \gcd(63, n+120) &= \gcd(63, 1000 + k + 120) \\ &= \gcd(63, k + 1120 - 63 \cdot 18) \\ &= \gcd(63, k-14) \\ &= 21, \\ \gcd(n+63, 120) &= \gcd(k + 1000 + 63, 120) \\ &= \gcd(k + 1000 + 63 - 120 \cdot 9, 120) \\ &= \gcd(k-17, 120) \\ &= 60. \end{align*} So, we have \begin{align*} k &\equiv 14 \pmod{21}, \\ k &\equiv 17 \pmod{60}. \end{align*} Let $k = 21p + 14 = 60q + 17$ $7p= 20q + 1$ (by the Division Algorithm), $7p = 21q - (q - 1)$ $q - 1$ is a multiple of $7$
Let $q-1 = 7r$ $q = 7r+1$ $k = 60(7r+1) + 17 = 420r + 77$
$n = 1000 + k = 420r + 1077$ $n = 1077, 1497, 1917$
By substituting $1077$ $1497$ $1917$ into $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60$ $1077$ and $1497$ aren't valid answers, only $1917$ is. Therefore, the answer is $1+9+1+7=\boxed{18}$

Solution:
We know $63 = 3 \cdot 21$ , and $\gcd(63, n + 120) = 21 \Longrightarrow n + 120 = 21a$ , where $a$ is not a multiple of $3$
Also, $120 = 2 \cdot 60$ , and $gcd(n+63, 120) = 60 \Longrightarrow n + 63 = 60b$ , where $b$ is not a multiple of $2$
Let $n+63 = 60b = x$ $n = x - 63$ $n+120 = x+57 = 21a$
Now the problem becomes $\gcd(63, x+57) =21$ and $\gcd(x, 120)=60$
Meaning $x + 57$ has to be a multiple of $21$ but not $63$ , and $x$ is a multiple of $60$ but not $120$
Using trial and error, the least values are $x = 33\cdot60 = 1980$ and $n = x-63 = 1917$
Therefore, the answer is $1+9+1+7=\boxed{18}$

Solution:
We have \begin{align*} \gcd(63, n+120) &= 21 \Longrightarrow n + 120 \equiv 0 \pmod{21}, \text{ where } 9 \nmid n + 120, \\ \gcd(n+63, 120) &= 60 \Longrightarrow n + 63 \equiv 0 \pmod{60}, \text{ where } 8 \nmid n + 63. \end{align*} So, we conclude that $n \equiv 6 \pmod{21}$ and $n \equiv 57 \pmod{60}$ , respectively.
Because the $2$ moduli $21$ and $60$ are not relatively prime, namely $\gcd{(21, 60)} = 3$ $21 = 3 \cdot 7$ , and $60 = 3 \cdot 20$ , we convert the system of $2$ linear congruences with non-coprime moduli into a system of $3$ linear congruences with coprime moduli: \begin{align*} n &\equiv 0 \pmod{3}, \\ n &\equiv 6 \pmod{7}, \\ n &\equiv 17 \pmod{20}. \end{align*} By Chinese Remainder Theorem , the general solution of system of $3$ linear congruences is \[n = 420k + 1077.\] We construct the following table: \[\begin{array}{c|c|c|c} n & 1077 & 1497 & 1917\\ \hline n + 120 & 1197 & 1617 & 2037\\ \hline n + 63 & 1140 & 1560 & 1980\\ \end{array}\] Only $n = 1917$ satisfies $9 \nmid n + 120$ and $8 \nmid n + 63$ . Therefore, the answer is $1+9+1+7=\boxed{18}$

Solution:
From $\gcd(63, n + 120) = 21$ , we know that \begin{align} n + 120 &\equiv 0 \pmod{7} \Rightarrow n \equiv 6 \pmod{7} \tag{1} \\ n + 120 &\equiv 0 \pmod{3} \Rightarrow n \equiv 0 \pmod{3} \Rightarrow n \equiv 0, 3, 6 \pmod{9} \notag \end{align} Since \begin{align} n + 120 &\not\equiv 0 \pmod{9} \Rightarrow n \not\equiv 6 \pmod{9} \notag \end{align} We must have \begin{align} n &\equiv 0 \pmod{9} \tag{2} \end{align} Or \begin{align} n &\equiv 3 \pmod{9} \tag{3} \end{align} (Note that $n + 120 \equiv 0 \pmod{9}$ is impossible since then $9 | \gcd(63, n+120)$ .)
From $\gcd(n+63, 120) = 60$ , we have \begin{align} n + 63 &\equiv 0 \pmod{15} \Rightarrow n \equiv 12 \pmod {15} \tag{4} \\ n + 63 &\equiv 0 \pmod{4} \Rightarrow n \equiv 1 \pmod {4} \Rightarrow n \equiv 1, 5 \pmod{8} \notag \end{align} Since \begin{align} n + 63 &\not\equiv 0 \pmod{8} \Rightarrow n \not\equiv 1 \pmod{8} \notag \end{align} We must have \begin{align} n &\equiv 5 \pmod {8} \tag{5} \end{align} Solving equations $(1)$ $(4)$ , and $(5)$ , we get \begin{align} n &\equiv 237 \pmod{840} \tag{6} \end{align} Solving $(6)$ with $(2)$ , we have \begin{align} n &\equiv 1917 \pmod{2520} \tag{7} \end{align} Solving $(6)$ with $(3)$ , we have \begin{align} n &\equiv 237 \pmod{2520} \tag{8} \end{align} From $(7)$ , we obtain $1917$ , and from $(8)$ , we obtain $237 + 2520 = 2757$ as the smallest solutions above $1000$ . Thus, we get $1 + 9 + 1 + 7 = \boxed{18}$


Answer: 1.0

Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:
$\textbf{(A)}\ 4 \qquad  \textbf{(B) }\ 3 \qquad  \textbf{(C) }\ 2 \qquad  \textbf{(D) }\ 1 \qquad  \textbf{(E) }\ 0$

Solution:
First, we wish to factor $P$ into a more manageable form.
From the beginning of $P$ , we notice $x^4+6x^3$ , which gives us the idea to use $(x^2+3x)^2=x^4+6x^3+9x^2$
This gives us \[P=(x^2+3x)^2+2x^2+3x+31\] This is not useful, but it gives us a place to start from.
We can then try $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$ \[P=(x^2+3x+1)^2-3(x-10)\] This is much more useful, as it moves all non-linear terms inside of a squared expression.
We can then say $P=(x^2+3x+1)^2-3(x-10)=a^2$ , where $a^2$ is the square of an integer mentioned on the problem. Right from here, we can set $x=10$ , which cancels out the $3(x-10)$ , giving $(100+30+1)^2=a^2$ . This gives us one solution, $x=10$ $a=131$
We can then rearrange the expression, giving us $(x^2+3x+1)^2-a^2=3(x-10)$ . Factoring using difference of squares, we obtain \[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\]
We can then state that when $x$ is greater than $3$ and less than $-9$ $x^2+3x+1$ will be greater than $|3x-10|$ . This is obtained by setting $x^2+3x+1>|3x-10|$ and then solving the inequality. We can then conclude that $x^2+3x+1+|a|>|3x-10|$
Next, we claim that $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$ when $x \neq 10$ . We can prove this by first noting that since $x$ and $a$ are integers, $x^2+3x+1-|a|$ is an integer. Next, we shall assume that $x^2+3x+1-|a|=0$ . Solving this and plugging back into the original equation, we obtain $(2|a|)(0)=3(x-10)$ . Solving we obtain $x=10$ , which is a contraction to $x \neq 10$ . Therefore, $x^2+3x+1-|a| \neq 0$ and $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$
Finally, we can go back to the equation \[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\] We note that since $(x^2+3x+1+|a|)$ is larger than $3(x-10)$ , in order for there to be solutions, $(x^2+3x+1-|a|)$ must be in the range $(-1,1)$ . However, this contradicts what was proven earlier, so when $x \neq 10$ and $x < -9$ or $x > 3$ , there are no solutions for $x$
Now, all that remains to be checked are values of $x$ between $-9$ and $3$ . Using brute force and checking each value individually, we can assert that there are no such solutions for $x$ , leaving us with only $1$ solution, $x=10$ . Therefore, the answer is $\boxed{1}$


Answer: 196.0

Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution:
We want $x_1 +x_2+x_3+x_4 =98$ . This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$ , as for all integers $y_i$ $2y_i + 1$ will be odd. Substituting we get \[2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47\] Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\choose3}$ . Computing this and dividing by 100 gives us an answer of $\boxed{196}$ .
~smartguy888

Solution:
Define $x_i = 2y_i - 1$ . Then $2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98$ , so $\sum_{i = 1}^4 y_i = 51$
So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600$ , and $\frac n{100} = \boxed{196}$

Solution:
Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into $4$ boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have $94$ stones left. Because we want an odd number in each box, we pair the stones, creating $47$ sets of $2$ . Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).
Our problem can now be restated: how many ways are there to partition a line of $47$ stones? We can easily solve this by using $3$ sticks to separate the stones into $4$ groups, and this is the same as arranging a line of $3$ sticks and $47$ stones. \[\frac{50!}{47! \cdot 3!} = 19600\] \[\frac{50 * 49 * 48}{3 * 2} = 19600\] Our answer is therefore $\frac{19600}{100} = \boxed{196}$

Solution:
Let $x = a + b$ and $y = c + d$ . Then $x + y = 98$ , where $x, y$ are positive even integers ranging from $2-98$
We quickly see that the total number of acceptable ordered pairs $(a, b, c, d)$ is:
\begin{align*} &\mathrel{\phantom{=}} 1 \cdot 48 + 2 \cdot 47 + 3 \cdot 46 + ... + 48 \cdot 1\\ &= (24.5 - 23.5)(24.5) + (24.5 - 22.5)(24.5 + 22.5) + ... + (24.5 + 23.5)(24.5 - 23.5)\\ &= 48(24.5)^2 - 2(0.5^2 + 1.5^2 + ... + 23.5^2)\\ &= 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2}\\ &= 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2}\\ &= 28812 - \frac{\frac{47(47 + 1)(2(47) + 1)}{6} - \frac{4(23)(23 + 1)(2(23) + 1)}{6}}{2}\\ &= 19600 \end{align*}
Therefore, $\frac{n}{100} = \frac{19600}{100} = \boxed{196}$

Solution: We write the generating functions for each of the terms, and obtain $(x+x^3+x^5\cdots)^4$ as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\left(\frac{1}{1-x^2}\right)^4=\binom{3}{3} +\binom{4}{3}x^2+\binom{5}{3}x^4 \cdots$ and in general we want that the coefficient of $x^{2k}$ is $\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\binom{50}{3}=19600$ in which $\frac{n}{100}=\boxed{196}.$


Answer: 7.0

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

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


Answer: 4944.0

Let $n$ denote the smallest positive integer that is divisible by both $4$ and $9,$ and whose base- $10$ representation consists of only $4$ 's and $9$ 's, with at least one of each. What are the last four digits of $n?$
$\textbf{(A) } 4444 \qquad\textbf{(B) } 4494 \qquad\textbf{(C) } 4944 \qquad\textbf{(D) } 9444 \qquad\textbf{(E) } 9944$

Solution:
For a number to be divisible by $4,$ the last two digits have to be divisible by $4.$ That means the last two digits of this integer must be $4.$
For a number to be divisible by $9,$ the sum of all the digits must be divisible by $9.$ The only way to make this happen is with nine $4$ 's. However, we also need one $9.$
The smallest integer that meets all these conditions is $4444444944$ . The last four digits are $\boxed{4944}$


Answer: 589.0

Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$

Solution: We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\times (38+1)$ factors by its prime factorization . If we group all of these factors (excluding $n$ ) into pairs that multiply to $n^2$ , then one factor per pair is less than $n$ , and so there are $\frac{63\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$ . There are $32\times20-1 = 639$ factors of $n$ , which clearly are less than $n$ , but are still factors of $n$ . Therefore, using complementary counting, there are $1228-639=\boxed{589}$ factors of $n^2$ that do not divide $n$

Solution:
Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$ . Then $n^2$ has $\frac{(2k_1+1)(2k_2+1)-1}{2}$ factors less than $n$
This simplifies to $\frac{4k_1k_2+2k_1+2k_2}{2}=2k_1k_2+k_1+k_2$
The number of factors of $n$ less than $n$ is equal to $(k_1+1)(k_2+1)-1=k_1k_2+k_1+k_2$
Thus, our general formula for $n=p_1^{k_1}p_2^{k_2}$ is
Number of factors that satisfy the above $=(2k_1k_2+k_1+k_2)-(k_1k_2+k_1+k_2)=k_1k_2$
Incorporating this into our problem gives $19\times31=\boxed{589}$

Solution:
Consider divisors of $n^2: a,b$ such that $ab=n^2$ .
WLOG, let $b\ge{a}$ and $b=\frac{n}{a}$
Then, it is easy to see that $a$ will always be less than $b$ as we go down the divisor list of $n^2$ until we hit $n$
Therefore, the median divisor of $n^2$ is $n$
Then, there are $(63)(39)=2457$ divisors of $n^2$ . Exactly $\frac{2457-1}{2}=1228$ of these divisors are $<n$
There are $(32)(20)-1=639$ divisors of $n$ that are $<n$
Therefore, the answer is $1228-639=\boxed{589}$


Answer: 432.0

Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $\frac{n}{75}$

Solution: The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$ . For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}\cdots$ such that $e_1e_2 \cdots = 75$ . Since $75|n$ , two of the prime factors must be $3$ and $5$ . To minimize $n$ , we can introduce a third prime factor, $2$ . Also to minimize $n$ , we want $5$ , the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}$


Answer: 244.0

Let $p$ $q$ , and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67$ . It is given that there exist real numbers $A$ $B$ , and $C$ such that \[\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}\] for all $s\not\in\{p,q,r\}$ . What is $\tfrac1A+\tfrac1B+\tfrac1C$
$\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247$

Solution: Multiplying both sides by $(s-p)(s-q)(s-r)$ yields \[1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)\] As this is a polynomial identity, and it is true for infinitely many $s$ , it must be true for all $s$ (since a polynomial with infinitely many roots must in fact be the constant polynomial $0$ ). This means we can plug in $s = p$ to find that $\frac1A = (p-q)(p-r)$ .  Similarly, we can find $\frac1B = (q-p)(q-r)$ and $\frac1C = (r-p)(r-q)$ .  Summing them up, we get that \[\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - pq - qr - pr\] We can express $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + qr + pr)$ , and by Vieta's Formulas, we know that this expression is equal to $324$ . Vieta's also gives $pq + qr + pr = 80$ (which we also used to find $p^2+q^2+r^2$ ), so the answer is $324 -80 = \boxed{244}$

Solution:
Solution 1 uses a trick from Calculus that seemingly contradicts the restriction $s\not\in\{p,q,r\}$ . I am going to provide a solution with pure elementary algebra. \[A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)=1\] \[s^2(A+B+C)-s(Aq+Ar+Bp+Br+Cp+Cq)+(Aqr+Bpr+Cpq-1)=0\] \[\begin{cases} A+B+C=0 & (1)\\ Aq+Ar+Bp+Br+Cp+Cq=0 & (2)\\ Aqr+Bpr+Cpq=1 & (3) \end{cases}\] From $(1)$ we get $A=-(B+C)$ $B=-(A+C)$ $C=-(A+B)$ , substituting them in $(2)$ , we get $Ap + Bq + Cr=0$ $(4)$
$(4)- (1) \cdot r$ $A(p-r)+B(q-r)=0$ $(5)$
$(3) - (1) \cdot pq$ $Aq(r-p)+Bp(r-q)=1$ $(6)$
$(6) + (5) \cdot p$ $A(r-p)(q-p)=1$
$A = \frac{1}{(r-p)(q-p)}$ , by symmetry, $B = \frac{1}{(r-q)(p-q)}$ $C = \frac{1}{(q-r)(p-r)}$
The rest is similar to solution 1, we get $\boxed{244}$


Answer: 7.0

Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution:
More generally, let $a,b,c,d,p,$ and $q$ be positive integers such that $bc-ad=1$ and \[\frac ab < \frac pq < \frac cd.\] From $\frac ab < \frac pq,$ we have $bp-aq>0,$ or \[bp-aq\geq1. \hspace{15mm} (1)\] From $\frac pq < \frac cd,$ we have $cq-dp>0,$ or \[cq-dp\geq1. \hspace{15mm} (2)\] Since $bc-ad=1,$ note that:
To minimize $q,$ we set $q=b+d,$ from which $p=a+c.$ Together, we can prove that \[\frac{a+c-1}{b+d}\leq\frac ab<\frac{a+c}{b+d}<\frac cd\leq\frac{a+c+1}{b+d}. \hspace{15mm} (\bigstar)\] For this problem, we have $a=5,b=9,c=4,$ and $d=7,$ so $p=a+c=9$ and $q=b+d=16.$ The answer is $q-p=\boxed{7}.$

Solution:
Define variables $a,b,c,d,p,$ and $q$ as Solution 1 does. Moreover, this solution refers to inequalities $(1)$ and $(2)$ in Solution 1.
Note that \begin{align*} \frac{1}{bd}&=\frac{bc-ad}{bd} \\ &=\frac cd - \frac ab \\ &=\left(\frac cd - \frac pq\right)+\left(\frac pq - \frac ab\right) \\ &=\frac{cq-dp}{dq}+\frac{bp-aq}{bp} \\ &\geq\frac{1}{dq}+\frac{1}{bq}. \end{align*} Multiplying both sides of $\frac{1}{bd}\geq\frac{1}{dq}+\frac{1}{bq}$ by $bdq,$ we get $q\geq b+d.$
For this problem, we have $a=5,b=9,c=4,$ and $d=7,$ so $q\geq b+d=16.$ At $q=16,$ we have \[\frac{8}{16}<\frac59<\frac{9}{16}<\frac47<\frac{10}{16},\] from which $p=9.$ Therefore, the answer is $q-p=\boxed{7}.$

Solution:
Inverting the given inequality we get \[\frac{7}{4} < \frac{q}{p} < \frac{9}{5},\] which simplifies to \[35p < 20q < 36p.\] We can now substitute $q = p + k.$ Note we need to find $k:$ \[35p < 20p + 20k < 36p,\] which simplifies to \[15p < 20k < 16p.\] Clearly $p>k.$ We will now substitute $p = k + x$ to get \[15k + 15x < 20k < 16k + 16x.\] The inequality $15k + 15x < 20k$ simplifies to $3x < k.$
The inequality $20k < 16k + 16x$ simplifies to $k < 4x.$
Combining the two inequalities, we get \[3x < k < 4x.\] Since $x$ and $k$ are integers, the smallest values of $x$ and $k$ that satisfy the above equation are $2$ and $7$ respectively. Substituting these back in, we arrive with an answer of $\boxed{7}.$

Solution:
Because $q$ and $p$ are positive integers with $p<q,$ we can let $q=p+k$ where $k\in{\mathbb{Z}}.$ Now, the problem condition reduces to \[\frac{5}{9}<\frac{p}{p+k}<\frac{4}{7}.\]
Our first inequality is $\frac{5}{9}<\frac{p}{p+k},$ which gives us $5p+5k<9p,$ or $\frac{5}{4}k<p.$
Our second inequality is $\frac{p}{p+k}<\frac{4}{7},$ which gives us $7p<4p+4k,$ or $p<\frac{4}{3}k.$
Hence, we have $\frac{5}{4}k<p<\frac{4}{3}k,$ or $15k<12p<16k.$
It is clear that we are aiming to find the least positive integer value of $k$ such that there is at least one value of $p$ that satisfies the inequality.
Now, simple casework through the answer choices of the problem reveals that $q-p=p+k-p=k\ge{\boxed{7}.$

Solution: We subtract $\frac{1}{2}$ from both sides of the equation, so \[\frac{1}{18} < \frac{p}{q}-\frac{1}{2} < \frac{1}{14}.\] Then for $q$ to be as small as possible, $\frac{p}{q}-\frac{1}{2}$ has to be $\frac{1}{16},$ so $\frac{p}{q}$ is $\frac{9}{16}$ and $q-p$ is $\boxed{7}.$

Solution:
Cross-multiply the inequality to get \[35q < 63p < 36q.\] Then, we have $0 < 63p-35q < q,$ or \[0 < 7(9p-5q) < q.\] Since $p$ and $q$ are integers, $9p-5q$ is an integer. To minimize $q,$ start from $9p-5q=1,$ which gives $p=\frac{5q+1}{9}.$ This limits $q$ to be greater than $7,$ so test values of $q$ starting from $q=8.$ However, $q=8$ to $q=14$ do not give integer values of $p.$
Once $q>14,$ it is possible for $9p-5q$ to be equal to $2,$ so $p$ could also be equal to $\frac{5q+2}{9}.$ The next value, $q=15,$ is not a solution, but $q=16$ gives $p=\frac{5\cdot 16 + 1}{9} = 9.$ Thus, the smallest possible value of $q$ is $16,$ and the answer is $16-9= \boxed{7}.$

Solution:
Start with $\frac{5}{9}.$ Repeat the following process until you arrive at the answer: if the fraction is less than or equal to $\frac{5}{9},$ add $1$ to the numerator; otherwise, if it is greater than or equal to $\frac{4}{7},$ add one $1$ to the denominator. We have \[\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}.\]
Therefore, the answer is $16 - 9 = \boxed{7}.$

Solution:
Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.
The interval can also be written as $0.5556<x<0.5714.$ This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range.
The denominators to be considered are $9,10,11,12,\ldots.$ We check $\frac{6}{10}, \frac{6}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}.$ At this point we know that we've got our fraction and our answer is $16-9=\boxed{7}.$

Solution: Graph the regions $y > \frac{5}{9}x$ and $y < \frac{4}{7}x.$ Note that the lattice point $(16,9)$ is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is $\frac{9}{16}$ and the answer is $16-9= \boxed{7}.$

Solution: As the other solutions do, the mediant $\frac{9}{16}$ is between the two fractions, with a difference of $\boxed{7}\approx 0.6.$

Solution:
In ascending order, we can use answer choices, values for $q-p,$ as a method of figuring out our answer through the means of substitution. Let the assumed difference be $7.$ Then, $p=q-7.$ We thus have two inequalities: $\frac{5}{9} < \frac{q-7}{q}$ and $\frac{q-7}{q} < \frac{4}{7}.$
Solving for $q$ in these equalities, we get $\frac{63}{4} < q < \frac{49}{3}.$ So, $q$ is between $15.75$ and $16.\overline{3},$ making it $16$ as $q$ is a positive integer (again, at this point, this is still an assumption). This would set $p=16-7=9.$
Since $\frac{5}{9} < \frac{9}{16} < \frac{4}{7},$ the minimum difference is $\boxed{7}.$

Solution: Assume that the difference $\frac{p}{q} - \frac{5}{9}$ results in a fraction of the form $\frac{1}{9q}.$ Then, \[9p - 5q = 1.\] Also assume that the difference $\frac{4}{7} - \frac{p}{q}$ results in a fraction of the form $\frac{1}{7q}.$ Then, \[4q - 7p = 1.\] Solving the system of equations yields $q=16$ and $p=9.$ Therefore, the answer is $\boxed{7}.$


Answer: 110.0

Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$ . Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$

Solution:
If \(p=2\), then \(4\mid n^4+1\) for some integer \(n\). But \(\left(n^2\right)^2\equiv0\) or \(1\pmod4\), so it is impossible. Thus \(p\) is an odd prime.
For integer \(n\) such that \(p^2\mid n^4+1\), we have \(p\mid n^4+1\), hence \(p\nmid n^4-1\), but \(p\mid n^8-1\). By Fermat's Little Theorem , \(p\mid n^{p-1}-1\), so
\begin{equation*}
p\mid\gcd\left(n^{p-1}-1,n^8-1\right)=n^{\gcd(p-1,8)}-1.
\end{equation*}
Here, \(\gcd(p-1,8)\) mustn't be divide into \(4\) or otherwise \(p\mid n^{\gcd(p-1,8)}-1\mid n^4-1\), which contradicts. So \(\gcd(p-1,8)=8\), and so \(8\mid p-1\). The smallest such prime is clearly \(p=17=2\times8+1\).
So we have to find the smallest positive integer \(m\) such that \(17\mid m^4+1\). We first find the remainder of \(m\) divided by \(17\) by doing
\begin{array}{|c|cccccccccccccccc|}
\hline
\vphantom{\tfrac11}x\bmod{17}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\\hline
\vphantom{\dfrac11}\left(x^4\right)^2+1\bmod{17}&2&0&14&2&14&5&5&0&0&5&5&14&2&14&0&2\\\hline
\end{array}
So \(m\equiv\pm2\), \(\pm8\pmod{17}\). If \(m\equiv2\pmod{17}\), let \(m=17k+2\), by the binomial theorem,
\begin{align*}
0&\equiv(17k+2)^4+1\equiv\mathrm {4\choose 1}(17k)(2)^3+2^4+1=17(1+32k)\pmod{17^2}\\[3pt]
\implies0&\equiv1+32k\equiv1-2k\pmod{17}.
\end{align*}
So the smallest possible \(k=9\), and \(m=155\).
If \(m\equiv-2\pmod{17}\), let \(m=17k-2\), by the binomial theorem,
\begin{align*}
0&\equiv(17k-2)^4+1\equiv\mathrm {4\choose 1}(17k)(-2)^3+2^4+1=17(1-32k)\pmod{17^2}\\[3pt]
\implies0&\equiv1-32k\equiv1+2k\pmod{17}.
\end{align*}
So the smallest possible \(k=8\), and \(m=134\).
If \(m\equiv8\pmod{17}\), let \(m=17k+8\), by the binomial theorem,
\begin{align*}
0&\equiv(17k+8)^4+1\equiv\mathrm {4\choose 1}(17k)(8)^3+8^4+1=17(241+2048k)\pmod{17^2}\\[3pt]
\implies0&\equiv241+2048k\equiv3+8k\pmod{17}.
\end{align*}
So the smallest possible \(k=6\), and \(m=110\).
If \(m\equiv-8\pmod{17}\), let \(m=17k-8\), by the binomial theorem,
\begin{align*}
0&\equiv(17k-8)^4+1\equiv\mathrm {4\choose 1}(17k)(-8)^3+8^4+1=17(241-2048k)\pmod{17^2}\\[3pt]
\implies0&\equiv241+2048k\equiv3+9k\pmod{17}.
\end{align*}
So the smallest possible \(k=11\), and \(m=179\).
In conclusion, the smallest possible \(m\) is \(\boxed{110}\).

Solution: We work in the ring \(\mathbb Z/289\mathbb Z\) and use the formula \[\sqrt[4]{-1}=\pm\sqrt{\frac12}\pm\sqrt{-\frac12}.\] Since \(-\frac12=144\), the expression becomes \(\pm12\pm12i\), and it is easily calculated via Hensel that \(i=38\), thus giving an answer of \(\boxed{110}\).

Solution: Note that $n^4 + 1 \equiv 0 \pmod{p}$ means $\text{ord}_{p}(n) = 8 \mid p-1.$ The smallest prime that does this is $17$ and $2^4 + 1 = 17$ for example. Now let $g$ be a primitive root of $17^2.$ The satisfying $n$ are of the form, $g^{\frac{p(p-1)}{8}}, g^{3\frac{p(p-1)}{8}}, g^{5\frac{p(p-1)}{8}}, g^{7\frac{p(p-1)}{8}}.$ So if we find one such $n$ , then all $n$ are $n, n^3, n^5, n^7.$ Consider the $2$ from before. Note $17^2 \mid 2^{4 \cdot 17} + 1$ by LTE. Hence the possible $n$ are, $2^{17}, 2^{51}, 2^{85}, 2^{119}.$ Some modular arithmetic yields that $2^{51} \equiv \boxed{110}$ is the least value.


Answer: 8.0

Let $p(x) = x^3 + ax^2 + bx + c$ , where $a$ $b$ , and $c$ are complex numbers.  Suppose that
What is the number of nonreal zeros of $x^{12} + ax^8 + bx^4 + c$
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 6\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 12$

Solution: Consider the graph of $x^4$ . It is similar to a parabola, but with a wider "base". First examine $x^4-2009$ and $x^4-9002$ . Clearly they are just being translated down some large amount. This will result in the $x$ -axis being crossed twice, indicating $2$ real zeroes. From the Fundamental Theorem of Algebra we know that a polynomial must have exactly as many roots as its highest degree, so we are left with $4-2$ or $2$ nonreal roots for each of the graphs. For the graph of $x^4-(2009+9002\pi i)$ , it's not even possible to graph it on the Cartesian plane, so all $4$ roots will be nonreal. This is $2+2+4 = 8$ total nonreal roots $\Rightarrow \boxed{8}$


Answer: 37.0

Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of $5$ heads before one encounters a run of $2$ tails.  Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$

Solution:
Think of the problem as a sequence of 's and 's. No two 's can occur in a row, so the sequence is blocks of $1$ to $4$ 's separated by 's and ending in $5$ 's. Since the first letter could be or the sequence could start with a block of 's, the total probability is that $3/2$ of it has to start with an
The answer to the problem is then the sum of all numbers of the form $\frac 32 \left( \frac 1{2^a} \cdot \frac 12 \cdot \frac 1{2^b} \cdot \frac 12 \cdot \frac 1{2^c} \cdots \right) \cdot \left(\frac 12\right)^5$ , where $a,b,c \ldots$ are all numbers $1-4$ , since the blocks of 's can range from $1-4$ in length. The sum of all numbers of the form $(1/2)^a$ is $1/2+1/4+1/8+1/16=15/16$ , so if there are n blocks of 's before the final five 's, the answer can be rewritten as the sum of all numbers of the form $\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\right)=\frac 3{64}\left(\frac{15}{32}\right)^n$ , where $n$ ranges from $0$ to $\infty$ , since that's how many blocks of 's there can be before the final five. This is an infinite geometric series whose sum is $\frac{3/64}{1-(15/32)}=\frac{3}{34}$ , so the answer is $\boxed{037}$

Solution:
Let $p_H, p_T$ respectively denote the probabilities that a string beginning with 's and 's are successful. Thus,
A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with (as there cannot be $2$ 's in a row, or be the string HHHHH.
There is a $\frac{1}{16}$ probability we roll $4$ consecutive 's, and there is a $\frac{15}{16}$ probability we roll a . Thus,
The answer is $p_H + p_T = \frac{3}{2}p_H = \frac{3}{34}$ , and $m+n = \boxed{037}$

Solution:
For simplicity, let's compute the complement, namely the probability of getting to $2$ tails before $5$ heads.
Let $h_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive heads. Similarly, let $t_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive tails. Specifically, $h_{5} = 0$ and $t_{2} = 1$ . If we can solve for $h_{1}$ and $t_{1}$ , we are done; the answer is simply $1/2 * (h_{1} + t_{1})$ , since on our first roll, we have equal chances of getting a string with "1 consecutive head" or "1 consecutive tail."
Consider solving for $t_{1}$ . If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have:
Applying similar logic, we get the equations:
\begin{align*} h_{1} &= \frac{1}{2} h_{2} + \frac{1}{2} t_{1}\\ h_{2} &= \frac{1}{2} h_{3} + \frac{1}{2} t_{1}\\ h_{3} &= \frac{1}{2} h_{4} + \frac{1}{2} t_{1}\\ h_{4} &= \frac{1}{2} h_{5} + \frac{1}{2} t_{1} \end{align*}
Since $h_{5} = 0$ , we get $h_{4} = \frac{1}{2} t_{1}$ $\Rightarrow h_{3} = \frac{3}{4} t_{1}$ $\Rightarrow h_{2} = \frac{7}{8} t_{1}$ $\Rightarrow h_{1} = \frac{15}{16} t_{1}$ $\Rightarrow t_{1} = \frac{1}{2} t_{2} + \frac{1}{2} \cdot \frac{15}{16} t_{1} = \frac{1}{2} + \frac{15}{32} t_{1} \Rightarrow t_{1} = \frac{16}{17}$ $\Rightarrow h_{1} = \frac{15}{16} \cdot \frac{16}{17} = \frac{15}{17}$
So, the probability of reaching $2$ tails before $5$ heads is $\frac{1}{2} (h_{1} + t_{1}) = \frac{31}{34}$ ; we want the complement, $\frac{3}{34}$ , yielding an answer of $3 + 34 = \boxed{037}$

Solution: Consider what happens in the "endgame" or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are $\frac{1}{32},\frac{1}{64},\frac{1}{4},\frac{1}{8},\frac{1}{16},\frac{1}{32},\frac{1}{64}$ respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are $\frac{3}{64}$ . The sum of all these endgame outcomes are $\frac{34}{64}$ , hence the desired probability is $\frac{3}{34}$ , and in this case $m=3,n=34$ so we have $m+n=\boxed{037}$ -vsamc


Answer: 2.4

Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$ , as the number of terms increases without bound. Then $s$ equals:
$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$

Solution:
The infinite sum of a geometric series with first term $a$ and common ratio $r$ $-1<r<1$ ) is $\frac{a}{1-r}$ .
Now, in this geometric series, $a=4$ , and $r=-\frac23$ . Plugging these into the formula, we get $\frac4{1-(-\frac23)}$ , which simplifies to $\frac{12}5$ , or $\boxed{2.4}$


Answer: 10.0

Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution:
Let $p, q$ , and $r$ be the roots of the polynomial. Then,
$p^3 - 5p^2 + 8p - 13 = 0$
$q^3 - 5q^2 + 8q - 13 = 0$
$r^3 - 5r^2 + 8r - 13 = 0$
Adding these three equations, we get
$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$
$s_3 - 5s_2 + 8s_1 = 39$
$39$ can be written as $13s_0$ , giving
$s_3 = 5s_2 - 8s_1 + 13s_0$
We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$ $3$ $....$ , meaning it must be satisfied when $k = 2$ , giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$
Therefore, $a = 5, b = -8$ , and $c = 13$ by matching coefficients.
$5 - 8 + 13 = \boxed{10}$

Solution:
Let $p, q$ , and $r$ be the roots of the polynomial. By Vieta's Formulae, we have
$p+q+r = 5$
$pq+qr+rp = 8$
$pqr=13$
We know $s_k = p^k + q^k + r^k$ . Consider $(p+q+r)(s_k) =5s_k$
$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$
Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}$ , we see $13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}$
We have \[\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\ &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\ &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\ &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}\]
Rearrange to get $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$
So, $a+ b + c = 5 -8 + 13 = \boxed{10}$

Solution:
Let $r,s,t$ be the roots of $x^3-5x^2+8x-13$ . Then:
$r^3=5r^2-8r+13$ \\ $s^3=5s^2-8s+13$ \\ $t^3=5t^2-8t+13$
If we multiply both sides of the equation by $r^k$ , where $k$ is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find $r^4+s^4+t^4$ , but that is just to check. So then with the above information about $r^3,s^3,t^3$ , we see that:
$r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}$ $s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}$ $t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}$
$s_k=r^k+s^k+t^k$
Then: $s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}$
This means that $s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}$ , as expected. So we have $a=5, b=-8, c=13$ . So our answer is $5-8+13=\boxed{10}$


Answer: 21.0

Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$

Solution: Since $A = (u,v)$ , we can find the coordinates of the other points: $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\frac{1}{2}(2u)(u-v) = u^2 - uv$ . Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41$ . Since $u,v$ are positive, $u+3v>u$ , and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$ . Since $v < u$ the latter case is the answer, and $u+v = \boxed{021}$

Solution:
We find the coordinates like in the solution above: $A = (u,v)$ $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . Then we apply the Shoelace Theorem \[A =  \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451\] \[\frac{1}{2}(2u^2 + 6uv) = 451\] \[u(u + 3v) = 451\]
This means that $(u,v) = (11, 10)$ or $(1,150)$ , but since $v < u$ , then the answer is $\boxed{021}$


Answer: 582.0

Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ .  Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers .  Find $m+n$

Solution:
Define $\theta = 2\pi/1997$ . By De Moivre's Theorem the roots are given by
Now, let $v$ be the root corresponding to $m\theta=2m\pi/1997$ , and let $w$ be the root corresponding to $n\theta=2n\pi/ 1997$ . Then \begin{align*} |v+w|^2 &= \left(\cos(m\theta) + \cos(n\theta)\right)^2 + \left(\sin(m\theta) + \sin(n\theta)\right)^2 \\ &= 2 + 2\cos\left(m\theta\right)\cos\left(n\theta\right) + 2\sin\left(m\theta\right)\sin\left(n\theta\right) \end{align*} The cosine difference identity simplifies that to \[|v+w|^2 = 2+2\cos((m-n)\theta)\]
We need $|v+w|^2 \ge 2+\sqrt{3}$ , which simplifies to \[\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}\] Thus, \[|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166\]
Therefore, $m$ and $n$ cannot be more than $166$ away from each other.  This means that for a given value of $m$ , there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$ , and $166$ of them $< m$ .  Since $m$ and $n$ must be distinct, $n$ can have $1996$ possible values.  Therefore, the probability is $\frac{332}{1996}=\frac{83}{499}$ .  The answer is then $499+83=\boxed{582}$

Solution:
The solutions of the equation $z^{1997} = 1$ are the $1997$ th roots of unity and are equal to $\text{cis}(\theta_k)$ , where $\theta_k = \tfrac {2\pi k}{1997}$ for $k = 0,1,\ldots,1996.$ Thus, they are located at uniform intervals on the unit circle in the complex plane.
The quantity $|v+w|$ is unchanged upon rotation around the origin, so, WLOG, we can assume $v=1$ after rotating the axis till $v$ lies on the real axis. Let $w=\text{cis}(\theta_k)$ . Since $w\cdot \overline{w}=|w|^2=1$ and $w+\overline{w}=2\text{Re}(w) = 2\cos\theta_k$ , we have \[|v + w|^2  =  (1+w)(1+\overline{w}) = 2+2\cos\theta_k\] We want $|v + w|^2\ge 2 + \sqrt {3}.$ From what we just obtained, this is equivalent to \[\cos\theta_k\ge \frac {\sqrt {3}}2 \qquad \Leftrightarrow \qquad -\frac {\pi}6\le \theta_k \le  \frac {\pi}6\] which is satisfied by $k = 166,165,\ldots, - 165, - 166$ (we don't include 0 because that corresponds to $v$ ).  So out of the $1996$ possible $k$ $332$ work.  Thus, $m/n = 332/1996 = 83/499.$ So our answer is $83 + 499 = \boxed{582}.$

Solution:
We can solve a geometrical interpretation of this problem.
Without loss of generality, let $u = 1$ . We are now looking for a point exactly one unit away from $u$ such that the point is at least $\sqrt{2 + \sqrt{3}}$ units away from the origin. Note that the "boundary" condition is when the point will be exactly $\sqrt{2+\sqrt{3}}$ units away from the origin; these points will be the intersections of the circle centered at $(1,0)$ with radius $1$ and the circle centered at $(0,0)$ with radius $\sqrt{2+\sqrt{3}}$ . The equations of these circles are $(x-1)^2 = 1$ and $x^2 + y^2 = 2 + \sqrt{3}$ . Solving for $x$ yields $x = \frac{\sqrt{3}}{2}$ . Clearly, this means that the real part of $v$ is greater than $\frac{\sqrt{3}}{2}$ . Solving, we note that $332$ possible $v$ s exist, meaning that $\frac{m}{n} = \frac{332}{1996} = \frac{83}{499}$ . Therefore, the answer is $83 + 499 = \boxed{582}$

Solution:
Since $z^{1997}=1$ , the roots will have magnitude $1$ . Thus, the roots can be written as $\cos(\theta)+i\sin(\theta)$ and $\cos(\omega)+i\sin(\omega)$ for some angles $\theta$ and $\omega$ . We rewrite the requirement as $\sqrt{2+\sqrt3}\le|\cos(\theta)+\cos(\omega)+i\sin(\theta)+i\sin(\omega)|$ , which can now be easily manipulated to $2+\sqrt{3}\le(\cos(\theta)+\cos(\omega))^2+(\sin(\theta)+\sin(\omega))^2$
WLOG, let $\theta = 0$ . Thus, our inequality becomes $2+\sqrt{3}\le(1+\cos(\omega))^2+(\sin(\omega))^2$ $2+\sqrt{3}\le2+2\cos(\omega)$ , and finally $\cos(\omega)\ge\frac{\sqrt{3}}{2}$ . Obviously, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ , and thus it follows that, on either side of a given point, $\frac{1997}{12}\approx166$ points will work. The probability is $\frac{166\times2}{1996} = \frac{83}{499}$ , and thus our requested sum is $\boxed{582}$ ~SigmaPiE


Answer: 834.0

Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$

Solution:
We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*} for some integer $k.$
Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that \begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*} Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.
We apply casework to the values for $s+3k:$
Together, the answer is $264+306+264=\boxed{834}.$

Solution:
First we recognize that $w = \operatorname{cis}(30^{\circ})$ and $z = \operatorname{cis}(120^{\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$ , respectively. By De Moivre's theorem, $\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$ . When you multiply by $i$ , we can think of that as rotating the complex number $90^{\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle.
This means that \[30r + 90 \equiv 120s \pmod{360},\] which we can simplify to \[r+3 \equiv 4s \pmod{12}.\] Notice that this means that $r$ cycles by $12$ for every value of $s$ . This is because once $r$ hits $12$ , we get an angle of $360^{\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$ \[\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \ldots & (97,1) \\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \ldots & (97,4) \\  (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \ldots & (97,7) \\ [-1ex] \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \ldots & (97,100) \end{array}\] We have $25$ columns in total: $34$ values for the first column, $33$ for the second, $33$ for the third, and then $34$ for the fourth, $33$ for the fifth, $33$ for the sixth, etc. Therefore, this cycle repeats every $3$ columns and our total sum is $(34+33+33) \cdot 8 + 34 = 100 \cdot 8 + 34 = \boxed{834}$


Answer: 21.0

Let $w$ $x$ $y$ , and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$ , then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$

Solution:
The prime factorization of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$
\[2w+3x+5y+7z=4+3+0+14=\boxed{21}\]


Answer: 100.0

Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane)

Solution:
Let $w = \operatorname{cis}{(\alpha)}$ and $z = 10\operatorname{cis}{(\beta)}$ . Then, $\dfrac{w - z}{z} = \dfrac{\operatorname{cis}{(\alpha)} - 10\operatorname{cis}{(\beta)}}{10\operatorname{cis}{\beta}}$
Multiplying both the numerator and denominator of this fraction by $\operatorname{cis}{(-\beta)}$ gives us:
$\dfrac{w - z}{z} = \dfrac{1}{10}\operatorname{cis}{(\alpha - \beta)} - 1 = \dfrac{1}{10}\cos{(\alpha - \beta)} + \dfrac{1}{10}i\sin{(\alpha - \beta)} - 1$
We know that $\tan{\theta}$ is equal to the imaginary part of the above expression divided by the real part. Let $x = \alpha - \beta$ . Then, we have that:
$\tan{\theta} = \dfrac{\sin{x}}{\cos{x} - 10}.$
We need to find a maximum of this expression, so we take the derivative:
Note (not from author): To take the derivative, we need to use the Quotient Rule . In this case, \[\frac{d}{dx}\left(\frac{\sin x}{\cos x-10}\right)=\frac{\cos x(\cos x-10)-(-\sin x)\sin x}{(\cos x-10)^2}=\dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}\]
Thus, we see that the maximum occurs when $\cos{x} = \dfrac{1}{10}$ . Therefore, $\sin{x} = \pm\dfrac{\sqrt{99}}{10}$ , and $\tan{\theta} = \pm\dfrac{\sqrt{99}}{99}$ . Thus, the maximum value of $\tan^2{\theta}$ is $\dfrac{99}{99^2}$ , or $\dfrac{1}{99}$ , and our answer is $1 + 99 = \boxed{100}$

Solution:
Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$
using the Law of Cosines we get: $1^2=10^2+t^2-t*10*2\cos\theta$ rearranging: \[20t\cos\theta=t^2+99\] solving for $\cos\theta$ we get:
\[\frac{99}{20t}+\frac{t}{20}=\cos\theta\] if we want to maximize $\theta$ we need to minimize $\cos\theta$ , using AM-GM inequality we get that the minimum value for $\cos\theta= 2\left(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}}\right)=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}$ hence using the identity $\tan^2\theta=\sec^2\theta-1$ we get $\tan^2\theta=\frac{1}{99}$ and our answer is $1 + 99 = \boxed{100}$

Solution:
Note that $\frac{w-z}{z}=\frac{w}{z}-1$ , and that $\left|\frac{w}{z}\right|=\frac{1}{10}$ . Thus $\frac{w}{z}-1$ is a complex number on the circle with radius $\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\omega$ denote this circle.
Let $A$ and $C$ be the points that represent $\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In order to maximize $\tan^2(\theta)$ , we need to maximize $\angle{AOC}$ . This angle is maximized when $AO$ is tangent to $\omega$ . Using the Pythagorean Theorem, we get
\[AO^2=1^2-\left(\frac{1}{10}\right)^2=\frac{99}{100}\]
Thus
\[\tan^2(\theta)=\frac{AC^2}{AO^2}=\frac{1/100}{99/100}=\frac{1}{99}\]
And the answer is $1+99=\boxed{100}$


Answer: 169.0

Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$

Solution:
Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$
Let $w_3$ have center $(x,y)$ and radius $r$ .  Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$ , and if they are internally tangent, it is $|r_1 - r_2|$ .  So we have
\begin{align*} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Solving for $r$ in both equations and setting them equal, then simplifying, yields
\begin{align*} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{align*}
Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$
So the locus of points that can be the center of the circle with the desired properties is an ellipse
Since the center lies on the line $y = ax$ , we substitute for $y$ and expand: \[1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.\]
We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$ , so $(-96a)^2 - 4(3+4a^2)(276) = 0$
Solving yields $a^2 = \frac{69}{100}$ , so the answer is $\boxed{169}$

Solution:
As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$ . Let $F_1=(-5,12)$ and $F_2=(5,12)$ . If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$ , then $CF_1=16-r$ and $CF_2=4+r$ . Therefore, $CF_1+CF_2=20$ . In particular, the locus of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$
Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$ , and the line $y=ax$ is denoted by $\ell$ . Then we reflect the ellipse over $\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below.
By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$ $T$ , and $F_2'$ are collinear, and similarly, $F_2$ $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$ $F_1F_2=F_1'F_2'=10$ , and $F_1F_2'=F_1'F_2=20$ . Note that $\ell$ bisects $\angle F_1'OF_1$ . We can bisect this angle by bisecting $\angle F_1'OF_2$ and $F_2OF_1$ separately.
We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$ . The height of this from the base of $20$ is $\sqrt{69}$ . Therefore, the complex number $\sqrt{69}+10i$ represents the bisection of $\angle F_1'OF_2$
Similarly, using the 5-12-13 triangles, we easily see that $12+5i$ represents the bisection of the angle $F_2OF_1$ . Therefore, we can add these two angles together by multiplying the complex numbers, finding \[\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.\] Now the point $F_1$ is given by the complex number $-5+12i$ . Therefore, to find a point on line $\ell$ , we simply subtract $\frac{1}{2}\angle F_1'OF_1$ , which is the same as multiplying $-5+12i$ by the conjugate of $(\sqrt{69}+10i)(12+5i)$ . We find \[(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).\] In particular, note that the tangent of the argument of this complex number is $\sqrt{69}/10$ , which must be the slope of the tangent line. Hence $a^2=69/100$ , and the answer is $\boxed{169}$

Solution:
We use the same reflection as in Solution 2. As $OF_1'=OF_2=13$ , we know that $\triangle OF_1'F_2$ is isosceles. Hence $\angle F_2F_1'O=\angle F_1'F_2O$ . But by symmetry, we also know that $\angle OF_1T=\angle F_2F_1'O$ . Hence $\angle OF_1T=\angle F_1'F_2O$ . In particular, as $\angle OF_1T=\angle OF_2T$ , this implies that $O, F_1, F_2$ , and $T$ are concyclic.
Let $X$ be the intersection of $F_2F_1'$ with the $x$ -axis. As $F_1F_2$ is parallel to the $x$ -axis, we know that \[\angle TXO=180-\angle F_1F_2T.\tag{1}\] But \[180-\angle F_1F_2T=\angle F_2F_1T+\angle F_1TF_2.\tag{2}\] By the fact that $OF_1F_2T$ is cyclic, \[\angle F_2F_1T=\angle F_2OT\qquad\text{and}\qquad \angle F_1TF_2=\angle F_1OF_2.\tag{3}\] Therefore, combining (1), (2), and (3), we find that \[\angle TXO=\angle F_2OT+\angle F_1OF_2=\angle F_1OT.\tag{4}\]
By symmetry, we also know that \[\angle F_1TO=\angle OTF_1'.\tag{5}\] Therefore, (4) and (5) show by AA similarity that $\triangle F_1OT\sim \triangle OXT$ . Therefore, $\angle XOT=\angle OF_1T$
Now as $OF_1=OF_2'=13$ , we know that $\triangle OF_1F_2'$ is isosceles, and as $F_1F_2'=20$ , we can drop an altitude to $F_1F_2'$ to easily find that $\tan \angle OF_1T=\sqrt{69}/10$ . Therefore, $\tan\angle XOT$ , which is the desired slope, must also be $\sqrt{69}/10$ . As before, we conclude that the answer is $\boxed{169}$

Solution:
First, rewrite the equations for the circles as $(x+5)^2+(y-12)^2=16^2$ and $(x-5)^2+(y-12)^2=4^2$ .
Then, choose a point $(a,b)$ that is a distance of $x$ from both circles.  Use the distance formula between $(a,b)$ and each of $A$ and $C$ (in the diagram above).  The distances, as can be seen in the diagram above are $16-x$ and $4+x$ , respectively. \[(a-5)^2+(b-12)^2=(4+x)^2\] \[(a+5)^2+(b-12)^2=(16-x)^2\] Subtracting the first equation from the second gives \[20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2\] Substituting this into the first equation gives \[a^2-10a+25+b^2-24b+144=100-10a+\frac{a^2}4\] \[b^2-24b+69+\frac{3a^2}4=0\] Now, instead of converting this to the equation of an eclipse, solve for $b$ and then divide by $a$ \[b=\frac{24\pm\sqrt{300-3a^2}}{2}\] We take the smaller root to minimize $\frac b a$ \[\frac b a=\frac{24-\sqrt{300-3a^2}}{2a}=\frac{24-\sqrt3\cdot\sqrt{100-a^2}}{2a}=\frac{12}a-\frac{\sqrt3}{2a}\sqrt{100-a^2}\] Now, let $10\cos\theta=a$ .  This way, $\sqrt{100-a^2}=10\sin\theta$ .
Substitute this in. $\frac{b}{a}=\frac{12}{10\cos\theta}-\frac{\sqrt3\cdot\sin\theta}{2\cos\theta}=\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ Then, take the derivative of this and set it to 0 to find the minimum value. $\frac{6}{5}\sec\theta\tan\theta-\frac{\sqrt3}{2}\sec^2\theta=0\rightarrow\frac{6}{5}\sin\theta-frac{\sqrt3}{2}=0\rightarrow\sin\theta=\frac{5\sqrt3}{12}$ Then, use this value of $\sin\theta$ to find the minimum of $\frac65\sec\theta-\frac{\sqrt3}{2}\tan\theta$ to get $\frac{\sqrt{69}}{10}\rightarrow\left(\frac{\sqrt{69}}{10}\right)^2=\frac{69}{100}\rightarrow69+100=\boxed{169}$

Solution: First, obtain the equation of the ellipse as laid out in previous solutions. We now scale the coordinate plane in the $x$ direction by a factor of $\frac{\sqrt{3}}{2}$ centered at $x=0.$ This takes the ellipse to a circle centered at $(0,12)$ with radius $5\sqrt{3}$ and takes the line $y=ax$ to $y=\left( \frac{\sqrt{3}}{2} \right)^{-1} ax.$ The tangent point of our line to the circle with positive slope forms a right triangle with the origin and the center of the circle. Thus, the distance from this tangent point to the origin is $\sqrt{69}.$ By similar triangles, the slope of this line is then $\frac{\sqrt{69}}{5\sqrt{3}}.$ We multiply this by $\frac{\sqrt{3}}{2}$ to get $a=\frac{\sqrt{69}}{10},$ so our final answer is $\boxed{169.}$


Answer: 163.0

Let $w_1, w_2, \dots, w_n$ be complex numbers .  A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$ $w_2 = - 7 + 64i$ $w_3 = - 9 + 200i$ $w_4 = 1 + 27i$ , and $w_5 = - 14 + 43i$ , there is a unique mean line with $y$ -intercept 3.  Find the slope of this mean line.

Solution:
We know that
$\sum_{k=1}^5 w_k = 3 + 504i$
And because the sum of the 5 $z$ 's must cancel this out,
$\sum_{k=1}^5 z_k = 3 + 504i$
We write the numbers in the form $a + bi$ and we know that
$\sum_{k=1}^5 a_k = 3$ and $\sum_{k=1}^5 b_k = 504$
The line is of equation $y=mx+3$ . Substituting in the polar coordinates, we have $b_k = ma_k + 3$
Summing all 5 of the equations given for each $k$ , we get
$504 =  3m + 15$
Solving for $m$ , the slope, we get $\boxed{163}$

Solution: The mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\frac{3}{5}, \frac{504i}{5})$ . Since we now have two points, namely that one and $(0, 3i)$ , we can simply find the slope between them, which is $\boxed{163}$ by the good ol' slope formula.


Answer: 7.0

Let $x = .123456789101112....998999$ , where the digits are obtained by writing the integers $1$ through $999$ in order.
The $1983$ rd digit to the right of the decimal point is
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$

Solution: We consider the first $1983$ digits, letting the $1983$ rd digit be $z$ . We can break the string of digits into three segments: let $A$ denote $123456789$ (the $1$ -digit numbers), let $B$ denote $1011...9899$ (the $2$ -digit numbers), and let $C$ denote $100101...z$ (the $3$ -digit numbers). Clearly there are $9$ digits in $A$ ; in $B$ , there are $99-10+1 = 90$ numbers, so $90 \cdot 2 = 180$ digits. This leaves $1983 - 9 - 180 = 1794$ digits in $C$ . Notice that $1794 = 3 \cdot 598$ with no remainder, so $C$ consists of precisely the first $598$ $3$ -digit numbers. Since the first $3$ -digit number is $100$ , the $598$ th is $100 + 598 - 1 = 697$ , so as $z$ is the last digit, the answer is $\boxed{7}$