Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution: Another way to solve this problem solve for x. First, we can divide both sides by 7 to get:
\[x^5 = \frac{11y^{13}}{7}\]
Next, we take the fifth root on both sides, which gets us:
\[x = \sqrt[5]{\frac{11y^{13}}{7}}\]
\[x = y^2 \cdot \sqrt[5]{\frac{11y^{3}}{7}}\]
Since we know x is a positive integer that we are trying to minimize, we can let y equal the smallest number that will make x an integer. In this case, we let $y = 11^3 \cdot 7^2$ (Make sure you see why this makes x the smallest integer possible!), which when plugged in, results in:
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot (11^3 \cdot 7^2)^3}\]
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot 11^9 \cdot 7^6}\]
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{11^{10} \cdot 7^5}\]
\[x = (11^3 \cdot 7^2)^2 \cdot 11^2 \cdot 7^1\]
\[x = 7^5 \cdot 11^ 8\]
This gets us $a^cb^d = 7^5\cdot 11^8$ , so $a + b + c + d = 7 + 5 + 11 + 8 = \boxed{31}$ ~lucaswujc, $\LaTeX$ help from Technodoggo

Answer: B 31
Problem: Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution: According to the problem, we have that $x^5$ and $y^{13}$ must be a multiple of both $7$ and $11$ . Thus, in their prime factorisation, there must be $7$ and $11$ . Thus, we have $a=7$ and $b=11$ . Now, let $x=7^c11^d\implies x^5=7^{5c}11^{5d}\implies7x^5=7^{5c+1}11^{5d}$ .
We can now divide both sides by 11 in our original equation $7x^5=11y^{13}$ to get $y^{13}=7^{5c+1}11^{5d-1}$ . As we are only considering integers, we have \[5c+1\equiv0~(\text{mod}~13)\implies5c\equiv12~(\text{mod}~13)\] and \[5d-1\equiv0~(\text{mod}~13)\implies5d\equiv1~(\text{mod}~13).\]
We can apply brute force to solve for $c$ and $d$ as the numbers aren't big. For the first congruence, we find that $25$ is the smallest number that satisfies, thus $c=5$ . For the second congruence, we find that $40$ is the smallest number that satisfies, thus $d=8$ . Summarising, we get $a+b+c+d=7+11+5+8=\boxed{31}$

Answer: D 31
Problem: Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b$
2003 12B AMC-20.png
$\mathrm{(A)}\ -4 \qquad\mathrm{(B)}\ -2 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 4$

Solution: The roots of this equation are $-1, 1, \text{ and } x$ , letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\frac{b}{a}$ and $-1\cdot 1\cdot x=-x=-\frac{d}{a}$ . Therefore, $x=\frac{d}{a}$ . Setting the two equations for $x$ equal to each other, $\frac{d}{a}=-\frac{b}{a}$ . We know that the y-intercept of the polynomial is $d$ , so $d=2$ . Plugging in for $d$ $\frac{2}{a}=-\frac{b}{a}$
Therefore, $b=-2 \Rightarrow \boxed{2}$

Answer: B 2
Problem: An object moves $8$ cm in a straight line from $A$ to $B$ , turns at an angle $\alpha$ , measured in radians and chosen at random from the interval $(0,\pi)$ , and moves $5$ cm in a straight line to $C$ . What is the probability that $AC < 7$
$\mathrm{(A)}\ \frac{1}{6} \qquad\mathrm{(B)}\ \frac{1}{5} \qquad\mathrm{(C)}\ \frac{1}{4} \qquad\mathrm{(D)}\ \frac{1}{3} \qquad\mathrm{(E)}\ \frac{1}{2}$

Solution: $WLOG$ , let the object turn clockwise.
Let $B = (0, 0)$ $A = (0, -8)$
Note that the possible points of $C$ create a semi-circle of radius $5$ and center $B$ . The area where $AC < 7$ is enclosed by a circle of radius $7$ and center $A$ . The probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ}$
The function of $\odot B$ is $x^2 + y^2 = 25$ , the function of $\odot A$ is $x^2 + (y+8)^2 = 49$
$O$ is the point that satisfies the system of equations: $\begin{cases} x^2 + y^2 = 25 \\ x^2 + (y+8)^2 = 49 \end{cases}$
$x^2 + (y+8)^2 - x^2 - y^2 = 49 - 25$ $64 + 16y =24$ $y = - \frac52$ $x = \frac{5 \sqrt{3}}{2}$ $O = (\frac{5 \sqrt{3}}{2}, - \frac52)$
Note that $\triangle BDO$ is a $30-60-90$ triangle, as $BO = 5$ $BD = \frac{5 \sqrt{3}}{2}$ $DO = \frac52$ . As a result $\angle CBO = 30 ^\circ$ $\angle ABO = 60 ^\circ$
Therefore the probability that $AC < 7$ is $\frac{\angle ABO}{180 ^\circ} = \frac{60 ^\circ}{180 ^\circ} = \boxed{13}$

Answer: D 13
Problem: Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$ . Let $N$ be a point on $\overline{AB}$ , and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$ , respectively. Which of the following is closest to the minimum possible value of $PQ$
$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75  \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$

Solution: Let the intersection of $\overline{AC}$ and $\overline{BD}$ be $E$ . Since $ABCD$ is a rhombus, we have $\overline{AC} \perp \overline{BD}$ and $AE = CE = \dfrac{AC}{2} = 8$ . Since $\overline{NQ} \perp \overline{BD}$ , we have $\overline{NQ} \parallel \overline{AC}$ , so $\triangle{BNQ} \sim \triangle{BAE} \sim \triangle{NAP}$ . Therefore, \[\dfrac{NP}{AP} = \dfrac{NP}{8 - NQ} = \dfrac{BE}{AE} = \dfrac{15}{8} \Rightarrow NP = 15 - \dfrac{15}{8} NQ.\] By Pythagorean Theorem, \[PQ^2 = NQ^2 + NP^2 = NQ^2 + \left(15 - \dfrac{15}{8} NQ \right)^2 = \dfrac{289}{64} NQ^2 - \dfrac{225}{4} NQ + 225.\] The minimum value of $PQ^2$ would give the minimum value of $PQ$ , so we take the derivative (or use vertex form) to find that the minimum occurs when $NQ = \dfrac{225 \cdot 8}{289}$ which gives $PQ^2 = \dfrac{225 \cdot 64}{289}$ . Hence, the minimum value of $PQ$ is $\sqrt{\dfrac{225 \cdot 64}{289}} = \dfrac{120}{17}$ , which is closest to $7 \Rightarrow \boxed{7}$

Answer: C 7
Problem: Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$ . Let $N$ be a point on $\overline{AB}$ , and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$ , respectively. Which of the following is closest to the minimum possible value of $PQ$
$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75  \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$

Solution: Let the intersection $\overline{AC}$ and $\overline{BD}$ be $E.$ Let $\overline{PE}=y \implies \overline{AP}=8-y$ and $\overline{EQ}=x \implies \overline {QB}=15-x.$ Since $\triangle NQB \sim \triangle APN,$ we have $\frac{8-y}{x}=\frac{y}{15-x} \implies 120=15y+8x.$ We want to minimize $\overline{PQ}=\sqrt{x^2+y^2}.$ By Cauchy, $17^2(x^2+y^2)=(x^2+y^2)(8^2+15^2) \ge (8x+15y)^2=120^2 \implies \sqrt{x^2+y^2} \ge \frac{120}{17} \approx 7.$ So, choice $\boxed{7}$ is closest to the minimum.

Answer: C 7
Problem: The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$

Solution: The function $f(x) = \sin x$ has roots in the form of $\pi n$ for all integers $n$ . Therefore, we want $\frac{1}{x} = \pi n$ on $\frac{1}{10000} \le x \le \frac{1}{1000}$ , so $1000 \le \frac 1x = \pi n \le 10000$ . There are $\frac{10000-1000}{\pi} \approx \boxed{2900}$ solutions for $n$ on this interval.

Answer: A 2900
Problem: The number of $x$ -intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to
$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$

Solution: We know that $x$ belongs to the interval $(0.0001,0.001)$ for $\sin(1/x)$ . We see that when we plug in $x$ into $\sin(1/x)$ , the argument $(1/x)$ is always from the range $(1000, 10000)$ . Therefore, the problem simply asks for all the zeros of $\sin(x)$ with $x$ values between $(1000, 10000)$ . We know that the $x$ values of any sine graph is $\pi(n-1)$ so, we see that values of $n$ are any integer value from $320$ to $3184$ and therefore gives us an answer of approximately $2865$ which is answer $\boxed{2900}$

Answer: A 2900
Problem: Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations
has exactly one solution. What is the minimum value of $c$
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$

Solution: Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$
When $x<a$ , the slope is $-3$
When $a<x<b$ , the slope is $-1$
When $b<x<c$ , the slope is $1$
When $c<x$ , the slope is $3$
Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$ , so $(b,c-a)$ is a point on $f(x)$ . In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$ . Thus, graphing this will produce a figure that looks like a cup: [asy] import graph; size(100); draw((0,6)--(3,0)); xaxis(0,8.5); yaxis(0,10); real f(real x) { return -3(x-2)+5; } real f2(real x) { return -1(x-2)+5; } real f3(real x) { return 1(x-4)+3; } real f4(real x) { return 3(x-7)+6; } draw(graph(f,0,2)); draw(graph(f2,2,4)); draw(graph(f3,4,7)); draw(graph(f4,7,8.5)); draw((2,-0.25)--(2,0.25)); label("a",(2,0),N); draw((4,-0.25)--(4,0.25)); label("b",(4,0),N); draw((7,-0.25)--(7,0.25)); label("c",(7,0),N); dot((2,5)); label("P",(1.9,5.2),E); [/asy] From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$ . Since the line where $a<x<b$ has slope $-1$ , the positive difference in $y$ -coordinates from $x=a$ to $x=b$ must be $b-a$ . Together with the fact that $(b,c-a)$ is on $f(x)$ , we see that $P=(a,c-a+b-a)$ . Since this point is on $x=a$ , the only intersection point with $2x+y=2003$ , we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$ . As $c>b$ , the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$ . This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$ , and thus the answer is $\boxed{1002}$

Answer: C 1002
Problem: Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations
has exactly one solution. What is the minimum value of $c$
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$

Solution: We just showed that whenever $c<1002$ , the system has at least two different solutions: one with $x\leq a$ and one with $x>a$
We also showed that for $c=1002$ there are some $a,b$ for which the system has exactly one solution.
Hence the optimal value of $c$ is $\boxed{1002}$

Answer: C 1002
Problem: Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$ . What is $a+b$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }\sqrt 5 \qquad \text{(D) }\sqrt 6 \qquad \text{(E) }3$

Solution: Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$
Hence it is either a solution to $x - \frac 1x = 1$ , or to $\frac 1x - x = 1$ . Then it must be a solution either to $x^2 - x - 1 = 0$ , or to $x^2 + x - 1 = 0$
There are in total four such values of $x$ , namely $\frac{\pm 1 \pm \sqrt 5}2$
Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$ . We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{5}$

Answer: C 5
Problem: For all positive integers $n$ , let $f(n)=\log_{2002} n^2$ . Let $N=f(11)+f(13)+f(14)$ . Which of the following relations is true?
$\text{(A) }N<1 \qquad \text{(B) }N=1 \qquad \text{(C) }1<N<2 \qquad \text{(D) }N=2 \qquad \text{(E) }N>2$

Solution: First, note that $2002 = 11 \cdot 13 \cdot 14$
Using the fact that for any base we have $\log a + \log b = \log ab$ , we get that $N = \log_{2002} (11^2 \cdot 13^2 \cdot 14^2) = \log_{2002} 2002^2 = \boxed{2}$

Answer: D 2
Problem: Several sets of prime numbers, such as $\{7,83,421,659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
$\text{(A) }193 \qquad \text{(B) }207 \qquad \text{(C) }225 \qquad \text{(D) }252 \qquad \text{(E) }447$

Solution: Neither of the digits $4$ $6$ , and $8$ can be a units digit of a prime. Therefore the sum of the set is at least $40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207$
We can indeed create a set of primes with this sum, for example the following sets work: $\{ 41, 67, 89, 2, 3, 5 \}$ or $\{ 43, 61, 89, 2, 5, 7 \}$
Thus the answer is $207\implies \boxed{207}$

Answer: B 207
Problem: Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$

Solution: First examine the formula $(x-10)^2+y^2=36$ , for the circle $C_1$ . Its center, $D_1$ , is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$ , at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: [asy] unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9);  draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0));  pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0));  label("$C_1$",(10,0) + 6*dir(-45), SE ); label("$C_2$",(-15,0) + 9*dir(225), SW ); label("$D_1$",(10,0), SE ); label("$D_2$",(-15,0), SW ); label("$Q$", p2[0], NE ); label("$P$", p1[1], SW ); label("$O$", (0,0), SW ); [/asy] Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_2QO$ and $D_1PO$ similar by AA, with a scale factor of $6:9$ , or $2:3$ . Next, we must subdivide the line $D_2D_1$ in a 2:3 ratio to get the length of the segments $D_2O$ and $D_1O$ . The total length is $10 - (-15)$ , or $25$ , so applying the ratio, $D_2O$ 15 and $D_1O$ 10 . These are the hypotenuses of the triangles. We already know the length of $D_2Q$ and $D_1P$ and (they're radii). So in order to find $PQ$ , we must find the length of the longer legs of the two triangles and add them.
$15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$
$\sqrt{144} = 12$
$10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$
$\sqrt{64} = 8$
Finally, the length of PQ is $12+8=\boxed{20}$ , or (C)

Answer:  20
Problem: Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$

Solution: Using the above diagram, imagine that segment $\overline{QD_2}$ is shifted to the right to match up with $\overline{PD_1}$ . Then shift $\overline{QP}$ downwards to make a right triangle. We know $\overline{D_2D_1} = 25$ from the given information and the newly created leg has length $\overline{QD_2} + \overline{PD_1} = 9 + 6 = 15$ . Hence by Pythagorean theorem $15^2 + {\overline{QP}}^2 = 25^2$
$\overline{QP} = \boxed{20}$ , or C.

Answer:  20
Problem: The graph of the function $f$ is shown below. How many solutions does the equation $f(f(x))=6$ have?
[asy] size(200); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4;  pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};  draw(P1--P2--P3--P4--P5);  dot("(-7, -4)",P1); dot("(-2, 6)",P2,LeftSide); dot("(1, 6)",P4); dot("(5, -6)",P5);  xaxis("$x$",-7.5,7,Ticks(xticks),EndArrow(6)); yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6)); [/asy]
$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }6 \qquad \text{(E) }7$

Solution: First of all, note that the equation $f(t)=6$ has two solutions: $t=-2$ and $t=1$
Given an $x$ , let $f(x)=t$ . Obviously, to have $f(f(x))=6$ , we need to have $f(t)=6$ , and we already know when that happens. In other words, the solutions to $f(f(x))=6$ are precisely the solutions to ( $f(x)=-2$ or $f(x)=1$ ).
Without actually computing the exact values, it is obvious from the graph that the equation $f(x)=-2$ has two and $f(x)=1$ has four different solutions, giving us a total of $2+4=\boxed{6}$ solutions.

Answer: D 6
Problem: Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$

Solution: The repeating decimal $0.\overline{ab}$ is equal to \[\frac{10a+b}{100} + \frac{10a+b}{10000} + \cdots = (10a+b)\cdot\left(\frac 1{10^2} + \frac 1{10^4} + \cdots \right) =  (10a+b) \cdot \frac 1{99} = \frac{10a+b}{99}\]
When expressed in the lowest terms, the denominator of this fraction will always be a divisor of the number $99 = 3\cdot 3\cdot 11$ . This gives us the possibilities $\{1,3,9,11,33,99\}$ . As $a$ and $b$ are not both nine and not both zero, the denominator $1$ can not be achieved, leaving us with $\boxed{5}$ possible denominators.

Answer: C 5
Problem: Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$

Solution: Another way to convert the decimal into a fraction (simplifying, I guess?). We have \[100(0.\overline{ab}) = ab.\overline{ab}\] \[99(0.\overline{ab}) = 100(0.\overline{ab}) - 0.\overline{ab} = ab.\overline{ab} - 0.\overline{ab} = ab\] \[0.\overline{ab} = \frac{ab}{99}\] where $a, b$ are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denominator. $\boxed{5}$

Answer: C 5
Problem: Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$

Solution: Since $\frac{1}{99}=0.\overline{01}$ , we know that $0.\overline{ab} = \frac{ab}{99}$ . From here, we wish to find the number of factors of $99$ , which is $6$ . However, notice that $1$ is not a possible denominator, so our answer is $6-1=\boxed{5}$ \[\] ~AopsUser101

Answer:  5
Problem: Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?
$\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9$

Solution: Since $0.\overline{ab} = \frac{ab}{99}$ , the denominator must be a factor of $99 = 3^2 \cdot 11$ . The factors of $99$ are $1,$ $3,$ $9,$ $11,$ $33,$ and $99$ . Since $a$ and $b$ are not both nine, the denominator cannot be $1$ . By choosing $a$ and $b$ appropriately, we can make fractions with each of the other denominators.
Thus, the answer is $\boxed{5}$

Answer:  5
Problem: Consider the sequence of numbers: $4,7,1,8,9,7,6,\dots$ For $n>2$ , the $n$ -th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. The smallest value of $n$ for which $S_n>10,000$ is:
$\text{(A) }1992 \qquad \text{(B) }1999 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution: The sequence is infinite. As there are only $100$ pairs of digits, sooner or later a pair of consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic.
(Additionally, as every two consecutive digits uniquely determine the previous one as well, the first pair of digits that will occur twice must be the first pair $4,7$ .)
Hence it is a good idea to find the period. Writing down more terms of the sequence, we get:
\[4,7,1,8,9,7,6,3,9,2,1,3,4,7,\dots\]
and we found the period. The length of the period is $12$ , and its sum is $4+7+\cdots+1+3 = 60$ . Hence for each $k$ we have $S_{12k} = 60k$
We have $\lfloor 10000/60 \rfloor = 166$ and $166\cdot 12 = 1992$ , therefore $S_{1992} = 60\cdot 166 = 9960$ .
The rest can now be computed by hand, we get $S_{1998} = 9960+4+7+1+8+9+7= 9996$ , and $S_{1999}=9996 + 6 = 10002$ , thus the answer is $\boxed{1999}$

Answer: B 1999
Problem: Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution: Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$ . Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$ , while the magnitude of $a-bi$ is $s$ . We get that $s^{2002}=s$ , hence either $s=0$ or $s=1$
For $s=0$ we get a single solution $(a,b)=(0,0)$
Let's now assume that $s=1$ . Multiply both sides by $a+bi$ . The left hand side becomes $(a+bi)^{2003}$ , the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$ . Hence the solutions for this case are precisely all the $2003$ rd complex roots of unity, and there are $2003$ of those.
The total number of solutions is therefore $1+2003 = \boxed{2004}$

Answer:  2004
Problem: Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution: As in the other solution, split the problem into when $s=0$ and when $s=1$ . When $s=1$ and $a+bi=\cos\theta+i\sin\theta$
$(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)$ $=a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)$
so we must have $2002\theta=-\theta+2\pi k$ and hence $\theta=\frac{2\pi k}{2003}$ . Since $\theta$ is restricted to $[0,2\pi)$ $k$ can range from $0$ to $2002$ inclusive, which is $2002-0+1=2003$ values. Thus the total is $1+2003 = \boxed{2004}$

Answer: E 2004
Problem: Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution: Let $r$ be the magnitude of $a+bi$ . Notice that $r$ must be either $0$ or $1$ for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the $2003$ rd power, and if the magnitude of $a+bi$ is larger than $1$ , it will increase and if it is smaller than $1$ it will decrease. However, the magnitude on the RHS is still $r$ , so this is not possible. Again, only $r=0$ and $r=1$ satisfy the equation.
Now if $r=0$ , then $(a,b)$ must be $(0,0)$
However if $r=1$ , we then have:
$\cos(2002 \theta) = \cos(-\theta)$ . This has solution of $\theta = 0$ . This would represent the number $1+0i$ , with conjugate $1-0i$ . This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by $2002$ through De Moivre's Theorem and also we do $-\theta$ because it is a reflection, angles therefore is negative.
We then write:
$\cos(2002 \theta) = \cos(360-\theta)$ which has solution of $\theta = \frac{360}{2003}$
We can also write:
$\cos(2002 \theta) = \cos(720-\theta)$ which has solution $\theta = \frac{720}{2003}$
We notice that it is simply headed upwards and the answer is of the form $\frac{720}{2003} n$ , where n is some integer from $0$ to infinity, inclusive.
Well wait, it repeats itself $n=2003$ , that is $360$ which is also $0$ ! Hence we only have $n=0$ to $2002$ as original solutions, or $2003$ solutions.
$1+2003 = \boxed{2004}$

Answer:  2004