AMC12 2011 B

Add up the numbers in each fraction to get $\frac{12}{9} - \frac{9}{12}$, which equals $\frac{4}{3} - \frac{3}{4}$. Doing the subtraction yields $\boxed{\frac{7}{12}\ \textbf{(C)}}$

Take the average of her current test scores, which is \[\frac{90+80+70+60+85}{5} = \frac{385}{5} = 77\] This means that she wants her test average after the sixth test to be $80.$ Let $x$ be the score that Josanna receives on her sixth test. Thus, our equation is \[\frac{90+80+70+60+85+x}{6} = 80\] \[385+x = 480\] \[x = \boxed{95 \textbf{(E)}}\]

The total amount of money that was spent during the trip was \[A + B\] So each person should pay \[\frac{A+B}{2}\] if they were to share the costs equally. Because LeRoy has already paid $A$ dollars of his part, he still has to pay \[\frac{A+B}{2} - A =\] \[= \boxed{\frac{B-A}{2} \textbf{(C)}}\]

Taking the prime factorization of $161$ reveals that it is equal to $23*7.$ Therefore, the only ways to represent $161$ as a product of two positive integers is $161*1$ and $23*7.$ Because neither $161$ nor $1$ is a two-digit number, we know that $a$ and $b$ are $23$ and $7.$ Because $23$ is a two-digit number, we know that a, with its two digits reversed, gives $23.$ Therefore, $a = 32$ and $b = 7.$ Multiplying our two correct values of $a$ and $b$ yields \[a*b = 32*7 =\] \[= \boxed{224 \textbf{(E)}}\]

$N$ must be divisible by every positive integer less than $7$, or $1, 2, 3, 4, 5,$ and $6$. Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$, so each number divisible by these is a multiple of $60$. The smallest multiple of $60$ is $60$, so the second smallest multiple of $60$ is $2\times60=120$. Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3\ \textbf{(A)}}$

In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°). In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d. Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°. Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer: 1/2 (216°-144°) = 1/2 (72°) $=\boxed{36 \textbf{(C)}}.$

If $x$ and $y$ have a mean of $60$, then $\frac{x+y}{2}=60$ and $x+y=120$. To maximize $\frac{x}{y}$, we need to maximize $x$ and minimize $y$. Since they are both two-digit positive integers, the maximum of $x$ is $99$ which gives $y=21$. $y$ cannot be decreased because doing so would increase $x$, so this gives the maximum value of $\frac{x}{y}$, which is $\frac{99}{21}=\boxed{\frac{33}{7}\ \textbf{(B)}}$

To find Keiko's speed, all we need to find is the difference between the distance around the inside edge of the track and the distance around the outside edge of the track, and divide it by the difference in the time it takes her for each distance. We are given the difference in time, so all we need to find is the difference between the distances. The track is divided into lengths and curves. The lengths of the track will exhibit no difference in distance between the inside and outside edges, so we only need to concern ourselves with the curves. The curves of the track are semicircles, but since there are two of them, we can consider both of the at the same time by treating them as a single circle. We need to find the difference in the circumferences of the inside and outside edges of the circle. The formula for the circumference of a circle is $C = 2 * \pi * r$ where $r$ is the radius of the circle. Let's define the circumference of the inside circle as $C_1$ and the circumference of the outside circle as $C_2$. If the radius of the inside circle ($r_1$) is $n$, then given the thickness of the track is 6 meters, the radius of the outside circle ($r_2$) is $n + 6$. Using this, the difference in the circumferences is: $C_2 - C_1 = 2 * \pi * (r_2 - r_1) = 2 * \pi * (n + 6 - n) = 2 * \pi * 6 = 12\pi$ $12\pi$ is the difference between the inside and outside lengths of the track. Divided by the time differential, we get: $12\pi \div 36 = \boxed {\textbf{(A)}\ \frac{\pi}{3}}$

For the product to be greater than zero, we must have either both numbers negative or both positive. Both numbers are negative with a $\frac{2}{3}*\frac{2}{3}=\frac{4}{9}$ chance. Both numbers are positive with a $\frac{1}{3}*\frac{1}{3}=\frac{1}{9}$ chance. Therefore, the total probability is $\frac{4}{9}+\frac{1}{9}=\frac{5}{9}$ and we are done. $\boxed{D}$

Since $AB \parallel CD$, $\angle AMD = \angle CDM$, so $\angle AMD = \angle CMD = \angle CDM$, so $\bigtriangleup CMD$ is isosceles, and hence $CM=CD=6$. Therefore, $\triangle BMC$ is a 30-60-90 triangle with $\angle BMC = 30^\circ$. Therefore $\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}$

Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle). Because either $1$, $5$, or $7$ is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from $(0,0)$ to $(1,0)$ in an even number of moves. Therefore, the frog cannot reach $(1,0)$ in two moves. However, a path is possible in 3 moves: from $(0,0)$ to $(3,4)$ to $(6,0)$ to $(1,0)$. Thus, the answer is $= \boxed{3 \textbf{}}$.

Let's assume that the side length of the octagon is $x$. The area of the center square is just $x^2$. The triangles are all $45-45-90$ triangles, with a side length ratio of $1:1:\sqrt{2}$. The area of each of the $4$ identical triangles is $\left(\dfrac{x}{\sqrt{2}}\right)^2\times\dfrac{1}{2}=\dfrac{x^2}{4}$, so the total area of all of the triangles is also $x^2$. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is $x$ and the other side length is $\dfrac{x}{\sqrt{2}}=\dfrac{x\sqrt{2}}{2}$, so the area of all of the rectangles is $2x^2\sqrt{2}$. The ratio of the area of the square to the area of the octagon is $\dfrac{x^2}{2x^2+2x^2\sqrt{2}}$. Cancelling $x^2$ from the fraction, the ratio becomes $\dfrac{1}{2\sqrt2+2}$. Multiplying the numerator and the denominator each by $2\sqrt{2}-2$ will cancel out the radical, so the fraction is now $\dfrac{1}{2\sqrt2+2}\times\dfrac{2\sqrt{2}-2}{2\sqrt{2}-2}=\dfrac{2\sqrt{2}-2}{4}=\boxed{\mathrm{(A)}\ \dfrac{\sqrt{2}-1}{2}}$

Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. This means $a+b+c=9$. $a,b,c$ must be in the set ${1,3,4,5,6}$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. The ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$. \begin{align*} z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ 4z + a + (a + b) + 9 &= 44\\ \text{if} \hspace{1cm} a &= 3 \\ a + b &= 4\\ 4z &= 44 - 9 - 3 - 4\\ z &= 7\\ w &= 16\\ \end{align*} \begin{align*} \text{if} \hspace{1cm} a &= 5\\ a + b &= 6\\ 4z &= 44 - 9 - 5 - 6\\ z &= 6\\ w &= 15\\ \end{align*} The sum of the two w's is $15+16=31$ $\boxed{B}$

Name the directrix of the parabola $l$. Define $d(X,k)$ to be the distance between a point $X$ and a line $k$. Now we remember the geometric definition of a parabola: given any line $l$ (called the directrix) and any point $F$ (called the focus), the parabola corresponding to the given directrix and focus is the locus of the points that are equidistant from $F$ and $l$. Therefore $FV=d(V,l)$. Let this distance be $d$. Now note that $d(F,l)=2d$, so $d(A,l)=d(B,l)=2d$. Therefore $AF=BF=2d$. We now use the Pythagorean Theorem on triangle $AFV$; $AV=\sqrt{AF^2+FV^2}=d\sqrt{5}$. Similarly, $BV=d\sqrt{5}$. We now use the Law of Cosines: \[AB^2=AV^2+VB^2-2AV\cdot VB\cos{\angle AVB}\Rightarrow 16d^2=10d^2-10d^2\cos{\angle AVB}\] \[\Rightarrow \cos{\angle AVB}=-\frac{3}{5}\] This shows that the answer is $\boxed{\textbf{(D)}}$.

Repeating difference of squares: $2^{24}-1=(2^{12}+1)(2^{6}+1)(2^{3}+1)(2^{3}-1)$ $2^{24}-1=(2^{12}+1)\cdot65\cdot9\cdot7$ $2^{24}-1 = (2^{12} +1) * 5 * 13 * 3^2 * 7$ The sum of cubes formula gives us: $2^{12}+1=(2^4+1)(2^8-2^4+1)$ $2^{12}+1 = 17\cdot241$ A quick check shows $241$ is prime. Thus, the only factors to be concerned about are $3^2\cdot5\cdot7\cdot13\cdot17$, since multiplying by $241$ will make any factor too large. Multiplying $17$ by $3$ or $5$ will give a two-digit factor; $17$ itself will also work. The next smallest factor, $7$, gives a three-digit number. Thus, there are $3$ factors that are multiples of $17$. Multiplying $13$ by $3$, $5$, or $7$ will also give a two-digit factor, as well as $13$ itself. Higher numbers will not work, giving $4$ additional factors. Multiply $7$ by $3$, $5$, or $3^2$ for a two-digit factor. There are no more factors to check, as all factors which include $13$ are already counted. Thus, there are an additional $3$ factors. Multiply $5$ by $3$ or $3^2$ for a two-digit factor. All higher factors have been counted already, so there are $2$ more factors. Thus, the total number of factors is $3+4+3+2=\boxed{\textbf{(D) }12}$ The 12 two-digit factors are 13, 15, 17, 21, 35, 39, 45, 51, 63, 65, 85, and 91.

Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define similarly $\ell_{BD},S_{BD}$ and $\ell_{BA},S_{BA}$. Then $R$ is equal to $ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}$. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles: Since $\triangle BCD$ and $\triangle BAD$ are equilateral, $\ell_{BC}$ contains $D$, $\ell_{BD}$ contains $A$ and $C$, and $\ell_{BA}$ contains $D$. Then $\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH$ with $BE = 1$ and $EF = \frac{1}{\sqrt{3}}$ so $[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$. Multiply this by 4 and it turns out that the pentagon has area $\boxed{(C)\frac{2\sqrt{3}}{3}}$.

$g(x)=\log_{10}\left(\frac{x}{10}\right)=\log_{10}\left({x}\right) - 1$ $h_{1}(x)=g(f(x))\text{ = }g(10^{10x})=\log_{10}\left({10^{10x}}\right){ - 1 = 10x - 1}$ Proof by induction that $h_{n}(x)\text{ = }10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$: For $n=1$, $h_{1}(x)=10x - 1$ Assume $h_{n}(x)=10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})$ is true for n: \begin{align*} h_{n+1}(x)&= h_{1}(h_{n}(x))\\ &=10 h_{n}(x) - 1\\ &=10 (10^n x - (1 + 10 + 10^2 + ... + 10^{n-1})) - 1\\ &= 10^{n+1} x - (10 + 10^2 + ... + 10^{n}) - 1\\ &= 10^{n+1} x - (1 + 10 + 10^2 + ... + 10^{(n+1)-1}) \end{align*} Therefore, if it is true for n, then it is true for n+1; since it is also true for n = 1, it is true for all positive integers n. $h_{2011}(1) = 10^{2011}\times 1{ - }(1 + 10 + 10^2 + ... + 10^{2010})$, which is the 2011-digit number 8888...8889 The sum of the digits is 8 times 2010 plus 9, or $\boxed{16089\textbf{(B)}}$

We can use the Pythagorean Theorem to split one of the triangular faces into two 30-60-90 triangles with side lengths $\frac{1}{2}, 1$ and $\frac{\sqrt{3}}{2}$. Next, take a cross-section of the pyramid, forming a triangle with the top of the pyramid and the midpoints of two opposite sides of the square base. This triangle is isosceles with a base of 1 and two sides of length $\frac{\sqrt{3}}{2}$. The height of this triangle will equal the height of the pyramid. To find this height, split the triangle into two right triangles, with sides $\frac{1}{2}, \frac{\sqrt2}{2}$ and $\frac{\sqrt{3}}{2}$. The cube, touching all four triangular faces, will form a similar pyramid that sits on top of the cube. If the cube has side length $x$, the small pyramid has height $\frac{x\sqrt{2}}{2}$ (because the pyramids are similar). Thus, the height of the cube plus the height of the smaller pyramid equals the height of the larger pyramid. $x +\frac{x\sqrt{2}}{2} = \frac{\sqrt2}{2}$. $x\left(1+\frac{\sqrt{2}}{2} \right) =\frac{\sqrt{2}}{2}$ $x\left(2+\sqrt{2}\right) = \sqrt{2}$ $x = \frac{\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2\sqrt{2}-2}{4-2} = \sqrt{2}-1 =$side length of cube. $\left(\sqrt{2}-1\right)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(-1) + 3(\sqrt{2})(-1)^2 + (-1)^3 = 2\sqrt{2} - 6 +3\sqrt{2} - 1 =\textbf{(A)} 5\sqrt{2} - 7$

It is very easy to see that the $+2$ in the graph does not impact whether it passes through the lattice. We need to make sure that $m$ cannot be in the form of $\frac{a}{b}$ for $1\le b\le 100$. Otherwise, the graph $y= mx$ passes through the lattice point at $x = b$. We only need to worry about $\frac{a}{b}$ very close to $\frac{1}{2}$, $\frac{n+1}{2n+1}$, $\frac{n+1}{2n}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\le b\le 100$, the smallest is $\frac{50}{99}$, so answer is $\boxed{\frac{50}{99} \textbf{(B)}}$ (In other words we are trying to find the smallest $m>\frac{1}{2}$ such that $b\le 100$.)

Let us also consider the circumcircle of $\triangle ADF$. Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of $\triangle ABC$ which is $P$, Also, since $m\angle ADP = m\angle AFP = 90^\circ$. $ADPF$ is cyclic, similarly, $BDPE$ and $CEPF$ are also cyclic. With this, we know that the circumcircles of $\triangle ADF$, $\triangle BDE$ and $\triangle CEF$ all intersect at $P$, so $P$ is $X$. The question now becomes calculating the sum of the distance from each vertex to the circumcenter. We can calculate the distances with coordinate geometry. (Note that $XA = XB = XC$ because $X$ is the circumcenter.) Let $A = (5,12)$, $B = (0,0)$, $C = (14, 0)$, $X= (x_0, y_0)$ Then $X$ is on the line $x = 7$ and also the line with slope $-\frac{5}{12}$ that passes through $(2.5, 6)$ (realize this is due to the fact that $XD$ is the perpendicular bisector of $AB$). $y_0 = 6-\frac{45}{24} = \frac{33}{8}$ So $X = (7, \frac{33}{8})$ and $XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\boxed{\frac{195}{8}}$ Remark: the intersection of the three circles is called a Miquel point.

Answer: (D) $\frac{x + y}{2} = 10 a+b$ for some $1\le a\le 9$, $0\le b\le 9$. $\sqrt{xy} = 10 b+a$ Squaring the first and second equations, $\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2$ $xy = 100b^2 + 20ab + a^2$ Subtracting the previous two equations, $\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$ $|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}$ Note that for x-y to be an integer, $(a^2 - b^2)$ has to be $11n$ for some perfect square $n$. Since $a$ is at most $9$, $n = 1$ or $4$ If $n = 1$, $|x-y| = 66$, if $n = 4$, $|x-y| = 132$. In AMC, we are done. Otherwise, we need to show that $n=4$, or $a^2 -b^2 = 44$ is impossible. We can arrive at $|x-y| = 6\sqrt{11(a^2 - b^2)}$ using the method above. Because we know that $|x-y|$ is an integer, it must be a multiple of 6 and 11. Hence the answer is $66.$

Answer: (D) Let $AB = c$, $BC = a$, and $AC = b$ Then $AD = AF$, $BE = BD$ and $CF = CE$ Then $a = BE + CF$, $b = AD + CF$, $c = AD + BE$ Hence: Note that $a + 1 = b$ and $a - 1 = c$ for $n = 1$, I claim that it is true for all $n$, assume for induction that it is true for some $n$, then Furthermore, the average for the sides is decreased by a factor of 2 each time. So $T_n$ is a triangle with side length $\frac{2012}{2^{n- 1}} - 1$, $\frac{2012}{2^{n-1}}$, $\frac{2012}{2^{n-1}} + 1$ and the perimeter of such $T_n$ is $\frac{(3)(2012)}{2^{n-1}}$ Now we need to find when $T_n$ fails the triangle inequality. So we need to find the last $n$ such that $\frac{2012}{2^{n-1}} > 2$ For $n = 10$, perimeter is $\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}$

We declare a point $(x, y)$ to make up for the extra steps that the bug has to move. If the point $(x, y)$ satisfies the property that $|x - 3| + |y + 2| + |x + 3| + |y - 2| \le 20$, then it is in the desirable range because $|x - 3| + |y + 2|$ is the length of the shortest path from $(x,y)$ to $(3, -2)$ and $|x + 3| + |y - 2|$ is the length of the shortest path from $(x,y)$ to $(-3, 2)$. If $-3\le x \le 3$, then $-7\le y \le 7$ satisfy the property. there are $15 \times 7 = 105$ lattice points here. else let $3< x \le 8$ (and for $-8 \le x < -3$ because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from $(-3, 2)$ to $(x, y)$ added to the shortest distance from $(x, y)$ to $(3, -2)$ is $|x - 3| + |y + 2| + |x + 3| + |y - 2|$. Since the minimum value for the difference between the y-coordinates is at $y = 0$, we get $2x + 4 = 20$ or $-2x + 4 = 20$. Thus, the upper and lower bounds for $x$ are $8$ and $-8$, respectively. Now we test each value for x satisfying $3< x \le 8$ and double the result because of symmetry. For $x = 4$, the possibles values of y are such that $|2y| \le 12$ for a total of $13$ lattice points, for $x = 5$, the possibles values of y are such that $|2y| \le 10$ for a total of $11$ lattice points, for $x = 6$, the possibles values of y are such that $|2y| \le 8$ for a total of $9$ lattice points, for $x = 7$, the possibles values of y are such that $|2y| \le 6$ for a total of $7$ lattice points, for $x = 8$, the possibles values of y are such that $|2y| \le 4$ for a total of $5$ lattice points, Hence, there are a total of $105 + 2 ( 13 + 11 + 9 + 7 + 5) = \boxed{(C) 195}$ lattice points. One may also obtain the result by using Pick's Theorem(how?). $i = a - \frac{b}{2} - 1$ (Suggestion)

Answer: (B) First of all, we need to find all $z$ such that $P(z) = 0$ $P(z) = \left(z^4 - 1\right)\left(z^4 + \left(4\sqrt{3} + 7\right)\right)$ So $z^4 = 1 \implies z = e^{i\frac{n\pi}{2}}$ or $z^4 = - \left(4\sqrt{3} + 7\right)$ $z^2 = \pm i \sqrt{4\sqrt{3} + 7} = e^{i\frac{(2n+1)\pi}{2}} \left(\sqrt{3} + 2\right)$ $z = e^{i\frac{(2n+1)\pi}{4}} \sqrt{\sqrt{3} + 2} = e^{i\frac{(2n+1)\pi}{4}} \left(\frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)$ Now we have a solution at $\frac{n\pi}{4}$ if we look at them in polar coordinate, further more, the 8-gon is symmetric (it is an equilateral octagon) . So we only need to find the side length of one and multiply by $8$. So answer $= 8 \times$ distance from $1$ to $\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)\left(1 + i\right)$ Side length $= \sqrt{\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^2} = \sqrt{2\left(\frac{3}{4} + \frac{1}{4}\right)} = \sqrt{2}$ Hence, answer is $8\sqrt{2}$.

Answer: $(D) \frac{34}{67}$ First of all, you have to realize that if $\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$ then $\left[\frac{n - k}{k}\right] + \left[\frac{100 - (n - k)}{k}\right] = \left[\frac{100}{k}\right]$ So, we can consider what happen in $1\le n \le k$ and it will repeat. Also since range of $n$ is $1$ to $99!$, it is always a multiple of $k$. So we can just consider $P(k)$ for $1\le n \le k$. Let $\text{fpart}(x)$ be the fractional part function This is an AMC exam, so use the given choices wisely. With the given choices, and the previous explanation, we only need to consider $k = 99$, $87$, $67$, $13$. $1\le n \le k$ For $k > \frac{200}{3}$, $\left[\frac{100}{k}\right] = 1$. 3 of the $k$ that we should consider land in here. For $n < \frac{k}{2}$, $\left[\frac{n}{k}\right] = 0$, then we need $\left[\frac{100 - n}{k}\right] = 1$ else for $\frac{k}{2}< n < k$, $\left[\frac{n}{k}\right] = 1$, then we need $\left[\frac{100 - n}{k}\right] = 0$ For $n < \frac{k}{2}$, $\left[\frac{100 - n}{k}\right] = \left[\frac{100}{k} - \frac{n}{k}\right]= 1$ So, for the condition to be true, $100 - n > \frac{k}{2}$ . ( $k > \frac{200}{3}$, no worry for the rounding to be $> 1$) $100 > k > \frac{k}{2} + n$, so this is always true. For $\frac{k}{2}< n < k$, $\left[\frac{100 - n}{k}\right] = 0$, so we want $100 - n < \frac{k}{2}$, or $100 < \frac{k}{2} + n$ $100 <\frac{k}{2} + n < \frac{3k}{2}$ For k = 67, $67 > n > 100 - \frac{67}{2} = 66.5$ For k = 69, $69 > n > 100 - \frac{69}{2} = 67.5$ etc. We can clearly see that for this case, $k = 67$ has the minimum $P(k)$, which is $\frac{34}{67}$. Also, $\frac{7}{13} > \frac{34}{67}$ . So for AMC purpose, answer is $\boxed{\textbf{(D) }\frac{34}{67}}$. Notice that for these integers $99,87,67$: $0\rightarrow 49,50,51\rightarrow 98$ $100\rightarrow 51,50,49\rightarrow 2$ $P=\frac{98}{99}$ $0\rightarrow 43,44\rightarrow 56,57\rightarrow 86$ $87\rightarrow 57,56\rightarrow 44,43\rightarrow 14$ $P=\frac{74}{87}$ $0\rightarrow 33,34\rightarrow 66$ $100\rightarrow 67,66\rightarrow 34$ $P=\frac{34}{67}$ That the probability is $\frac{2k-100}{k}=2-\frac{100}{k}$. Even for $k=13$, $P(13)=\frac{9}{13}=\frac{100}{13}-7$. And $P(11)=\frac{10}{11}=10-\frac{100}{11}$. Perhaps the probability for a given $k$ is $\left\lceil{\frac{100}{k}}\right\rceil-\frac{100}{k}$ if $\left[\frac{100}{k}\right]=\left\lfloor{\frac{100}{k}}\right\rfloor$ and $\frac{100}{k}-\left\lfloor{\frac{100}{k}}\right\rfloor$ if $\left[\frac{100}{k}\right]=\left\lceil{\frac{100}{k}}\right\rceil$. So $P>\frac{1}{2}$ and $P_\text{min}=\frac{k_\text{min}+1}{2k_\text{min}}=\frac{101}{201}$. Because $201=3\cdot 67\mid 99!$ ! Now, let's say we are not given any answer, we need to consider $k < \frac{200}{3}$. I claim that $P(k) \ge \frac{1}{2} + \frac{1}{2k}$ If $\left[\frac{100}{k}\right]$ got round down, then $1 \le n \le \frac{k}{2}$ all satisfy the condition along with $n = k$ because if $\text{fpart}\left(\frac{100}{k}\right) < \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) < \frac{1}{2}$, so must $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$ and for $n = k$, it is the same as $n = 0$. , which makes $P(k) \ge \frac{1}{2} + \frac{1}{2k}$. If $\left[\frac{100}{k}\right]$ got round up, then $\frac{k}{2} \le n \le k$ all satisfy the condition along with $n = 1$ because if $\text{fpart}\left(\frac{100}{k}\right) > \frac{1}{2}$ and $\text{fpart} \left(\frac{n}{k}\right) > \frac{1}{2}$ Case 1) $\text{fpart} \left(\frac{100 - n}{k}\right) < \frac{1}{2}$ Case 2)