Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$

Solution: First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$
The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$ , and the probability both are blue is $\frac{6}{10}\cdot\frac{N}{16+N}$ , so \[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}\] Solving this equation, \[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\] Multiplying both sides by $16+N$ , we get
\begin{align*} 20\cdot16+30\cdot N&=29(16+N)\\ 320+30N&=464+29N\\ N&=\boxed{144}

Answer:  144
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: Let the numerator and denominator $x,y$ with $\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\gcd(x,y) = 1$ then $\gcd(x,y) =\gcd(x,1000-x)= \gcd(x,1000-x-(-1)x)=\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\frac{x}{y}<1$ and note by Euclidean algorithm $\gcd(1000,1000-x)=1$ , so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $\boxed{200}$ desired fractions.

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of using $\frac{n}{m}$ as the fraction to use the Euclidean Algorithm on, we can rewrite this as $\frac{500-x}{500+x}$ or \[\gcd(500+x,500-x)=\gcd((500+x)+(500-x),500-x)=\gcd(1000,500-x).\] Thus, we want $\gcd(1000,500-x)=1$ . You can either proceed as Solution $1$ , or consider that no even numbers work, limiting us to $250$ choices of numbers and restricting $x$ to be odd. If $x$ is odd, $500-x$ is odd, so the only possible common factors $1000$ and $500-x$ can share are multiples of $5$ . Thus, we want to avoid these. There are $50$ odd multiples of $5$ less than $500$ , so the answer is $250-50=\boxed{200}$

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: Say $r=\frac{d}{1000-d}$ ; then $1\leq d\leq499$ . If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$ . Since $1000=2^3\cdot5^3$ , that modulus can only be $2$ or $5$ . This implies that if $d\mid2$ or $d\mid5$ , the fraction is reducible. There are $249$ cases where $d\mid2$ $99$ where $d\mid5$ , and $49$ where $d\mid(2\cdot5=10)$ , so by PIE, the number of fails is $299$ , so our answer is $\boxed{200}$

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of $5$ because the denominator would also be a multiple of $5$ . Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have $499$ fractions to start with, and $250$ with odd numerators. Subtract $50$ to account for the multiples of $5$ , and we get $\boxed{200}$

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: We know that the set of these rational numbers is from $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.
We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ .
Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreducible, and $\dfrac{1000}{m}$ is irreducible if $m$ is not divisible by $2$ or $5$ . Thus, the answer to the question is the number of integers between $501$ and $999$ inclusive that are not divisible by $2$ or $5$
We note there are $499$ numbers between $501$ and $999$ , and
Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$ , so our answer is $\boxed{200}$

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: We notice that there are a total of $400$ fractions that are in simplest form where the numerator and denominator add up to $1000$ . Because the numerator and denominator have to be relatively prime, there are $\varphi(1000)=400$ fractions. Half of these are greater than $1$ , so the answer is $400\div2=\boxed{200}$

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: Our fraction can be written in the form $\frac{1000 - a}{a} = \frac{1000}{a} - 1.$ Thus the fraction is reducible when $a$ divides $1000.$ We also want $500 < a < 1000.$ By PIE , the total values of $a$ that make the fraction reducible is, \[249 + 99 - 49 = 299.\] By complementary counting , the answer we want is $499 - 299 = \boxed{200}.$

Answer:  200
Problem: Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.

Solution: Suppose our fraction is $\frac{a}{b}$ . The given condition means $a+b=1000$ . Now, if $a$ and $b$ share a common factor greater than $1$ , then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even.
There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$ , which means the answer is $250-50=\boxed{200}$

Answer:  200
Problem: Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\dfrac{1}{3}$ mi/min.
Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2.
Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of $r_1 - \dfrac{1}{3}$ . Similarly, the second train has to cover a distance equal to its own length, at a rate of $r_2 + \dfrac{1}{3}$ . Since the times are equal and $d = rt$ , we have that $\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}$ . Solving for $r_1$ in terms of $r_2$ , we get that $r_1 = r_2 + \dfrac{2}{3}$
Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by $\dfrac{1}{3}$ , and decrease train 2's speed by $\dfrac{1}{3}$ . Thus, we have that $\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}$
Multiplying this out and simplifying, we get that $r_1 = 10r_2 - \dfrac{11}{3}$ . Since we now have 2 expressions for $r_1$ in terms of $r_2$ , we can set them equal to each other:
$r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}$ . Solving for $r_2$ , we get that $r_2 = \dfrac{13}{27}$ . Since we know that it took train 2 1 minute to pass Jon, we know that $1 = \dfrac{d}{r_2 + \dfrac{1}{3}}$ . Plugging in $\dfrac{13}{27}$ for $r_2$ and solving for $d$ , we get that $d = \dfrac{22}{27}$ , and our answer is $27 + 22 = \boxed{049}$

Answer:  049
Problem: Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Using a similar approach to Solution 1, let the speed of the east bound train be $a$ and the speed of the west bound train be $b$
So $a-20=b+20$ and $a+20=10(b-20)$
From the first equation, $a=b+40$ . Substituting into the second equation, \[b+60=10b-200\] \[260=9b\] \[b=\frac{260}{9}\text{ mph}\] This means that \[a=\frac{260}{9}+40=\frac{620}{9}\text{ mph}\] Checking, we get that the common difference in Jon's speed and trains' speeds is $\frac{440}{9}$ and the differences for Steve is $\frac{800}{9}$ and $\frac{80}{9}$
This question assumes the trains' lengths in MILES: \[\frac{440}{9}\cdot \frac{1}{60}=\frac{440}{540}=\frac{22}{27}\text{ miles}\] Adding up, we get $22+27=\boxed{049}$

Answer:  049
Problem: Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the length of the trains be $L$ , let the rate of the westward train be $W_{R}$ , ;et the rate of the eastward train be $E_{R}$ , and let the time it takes for the eastward train to pass Steve be $E_{T}$
We have that
$L=(\frac{1}{60})(W_{R}+20)$
$L=(\frac{1}{60})(E_{R}-20)$
Adding both of the equations together, we get that
$2L=\frac{W_{R}}{60}+\frac{E_{R}}{60}\implies 120L=W_{R}+E_{R}$
Now, from the second part of the problem, we acquire that
$L=(E_{T})(E_{R}+20)$
$L=(10E_{T})(W_{R}-20)$
Dividing the second equation by the first, we get that... $1=\frac{10(W_{R}-20)}{E_{R}+20}\implies E_{R}+20=10W_{R}-200\implies E_{R}+220=10W_{R}\implies E_{R}=10W_{R}-220$
Now, substituting into the $120L=W_{R}+E_{R}$
$120L=W_{R}+(10W_{R}-220)\implies 120L= 11W_{R}-220\implies W_{R}=\frac{120L+220}{11}$
Finally, plugging this back into our very first equation..
$L=(\frac{1}{60})((\frac{120L+220}{11})+20)\implies 660L=120L+440\implies 540L=440\implies L=\frac{22}{27}$
Hence, the answer is $22+27=\boxed{049}$

Answer:  049
Problem: Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called "communal" if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)

Solution: By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers $2$ disjoint areas of the $12$ -gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent to one another. We now count the number of ways to select a row of adjacent vertices. We notice that for any subset size between $1$ and $11$ , there are $12$ possible subsets like this (this is true because we can pick any of the $12$ vertices as a "starting" vertex, include some number of vertices counterclockwise from that vertex, and generate all possible configurations). However, we also have to include the set of all $12$ vertices, as well as the empty set. Thus, the total number is $12\cdot11 + 2 = \boxed{134}$

Answer:  134
Problem: The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution: Begin by setting $x$ to 0, then set both equations to $h^2=\frac{2013-j}{3}$ and $h^2=\frac{2014-k}{2}$ , respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\ge32$
We see that $h^2=\frac{2014-k}{2}$ , so we now need to find a positive integer $h$ which has positive integer x-intercepts for both equations.
Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $\boxed{036}$

Answer:  036
Problem: The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution: Let $x=0$ and $y=2013$ for the first equation, resulting in $j=2013-3h^2$ .  Substituting back in to the original equation, we get $y=3(x-h)^2+2013-3h^2$
Now we set $y$ equal to zero, since there are two distinct positive integer roots.  Rearranging, we get $2013=3h^2-3(x-h)^2$ , which simplifies to $671=h^2-(x-h)^2$ .  Applying difference of squares, we get $671=(2h-x)(x)$
Now, we know that $x$ and $h$ are both integers, so we can use the fact that $671=61\times11$ , and set $2h-x=11$ and $x=61$ (note that letting $x=11$ gets the same result).  Therefore, $h=\boxed{036}$

Answer:  036
Problem: The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution: Similar to the first two solutions, we deduce that $\text{(-)}j$ and $\text{(-)}k$ are of the form $3a^2$ and $2b^2$ , respectively, because the roots are integers and so is the $y$ -intercept of both equations. So the $x$ -intercepts should be integers also.
The first parabola gives \[3h^2+j=3\left(h^2-a^2\right)=2013\] \[h^2-a^2=671\] And the second parabola gives \[2h^2+k=2\left(h^2-b^2\right)=2014\] \[h^2-b^2=1007\]
We know that $671=11\cdot 61$ and that $1007=19\cdot 53$ . It is just a fitting coincidence that the average of $11$ and $61$ is the same as the average of $19$ and $53$ . That is $\boxed{036}$

Answer:  036
Problem: The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution: First, we expand both equations to get $y=3x^2-6hx+3h^2+j$ and $y=2x^2-4hx+2h^2+k$ . The $y$ -intercept for the first equation can be expressed as $3h^2+j$ . From this, the x-intercepts for the first equation can be written as
\[x=h \pm \sqrt{(-6h)^2-4*3(3h^2+j)}=h \pm \sqrt{36h^2-12(2013)}=h \pm \sqrt{36h^2-24156}\]
Since the $x$ -intercepts must be integers, $\sqrt{36h^2-24156}$ must also be an integer. From solution 1, we know $h$ must be greater than or equal to 32. We can substitute increasing integer values for $h$ starting from 32; we find that $h=36$
We can test this result using the second equation, whose $x$ -intercepts are \[x=h \pm \sqrt{(-4h)^2-4*2(2h^2+k)}=h \pm \sqrt{16h^2-8(2014)}=h \pm \sqrt{16h^2-16112}\] Substituting 36 in for $h$ , we get $h=36 \pm 68$ , which satisfies the requirement that all x-intercepts must be (positive) integers.
Thus, $h=\boxed{036}$

Answer:  036
Problem: The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution: We have the equation $y=3(x-h)^2 + j.$
We know: $(x,y):(0,2013)$ , so $h^2=2013/3 - j/3$ after plugging in the values and isolating $h^2$ . Therefore, $h^2=671-j/3$
Lets call the x-intercepts $x_1$ $x_2$ . Since both $x_1$ and $x_2$ are positive there is a relationship between $x_1$ $x_2$ and $h$ . Namely, $x_1+x_2=2h$ . The is because: $x_1-h=-(x_2-h)$
Similarly, we know: $(x,y):(x_1,0)$ , so $j=-3(x_1-h)^2$ . Combining the two equations gives us \[h^2=671+(x_1-h)^2\] \[h^2=671+x_1^2-2x_1h+h^2\] \[h=(671+x_1^2)/2x_1.\]
Now since we have this relationship, $2h=x_1+x_2$ , we can just multiply the last equation by 2(so that we get $2h$ on the left side) which gives us \[2h=671/x_1+x_1^2/x^1\] \[2h=671/x_1+x_1\] \[x_1+x_2=671/x_1+x_1\] \[x_2=671/x_1\] \[x_1x_2=671.\] Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$ . Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$ , so $\boxed{36}$

Answer:  36
Problem: The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution: First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$ . We can apply Vieta's. Let the roots of the first equation be $a, b$ and the roots of the second equation be $c, d$ . Thus, we have that $a\cdot b=1007$ $a+b=2h$ and $c\cdot d=671$ $c+d=2h$ . Simple evaluations finds that $\boxed{36}$

Answer:  36
Problem: 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 $1 + 99 = \boxed{100}$

Answer:  100
Problem: 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: 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 $BOA$ [asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot("$A$",A,dir(45)); dot("$B$",B,dir(45)); dot("$O$",O,dir(135)); dot("$ \theta$",O,(7,1.2));  label("$1$", ( A--B )); label("$10$",(O--A)); label("$t$",(O--B)); [/asy]
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 $1 + 99 = \boxed{100}$

Answer:  100
Problem: 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: 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:  100
Problem: The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$

Solution: We have that $N^2 - N = N(N - 1)\equiv 0\mod{10000}$
Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or $5$ . Thus, one of them is divisible by $2^4 = 16$ , and the other is divisible by $5^4 = 625$ . Noting that $625 \equiv 1\mod{16}$ , we see that $625$ would work for $N$ , except the thousands digit is $0$ . The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$ . In order for this to happen, \[N-1 \equiv -1 \pmod {16}.\] Since $625 \equiv 1 \pmod{16}$ , we know that $15 \cdot 625 = 9375 \equiv 15 \equiv -1 \mod{16}$ . Thus, $N-1 = 9375$ , so $N = 9376$ , and our answer is $\boxed{937}$

Answer:  937