https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mishai&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T09:32:17ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_10&diff=824822009 AIME II Problems/Problem 102017-01-22T01:10:07Z<p>Mishai: </p>
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<div>Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. <br />
<br />
<br />
== Solution 1==<br />
<br />
Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>.<br />
<br />
==Solution 2==<br />
<br />
Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>.<br />
<br />
By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>:<br />
<br />
<cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\<br />
13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\<br />
AD^2&=130-\dfrac{130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\<br />
AD^2&=\dfrac{130(23^2-169)}{23^2}\\<br />
AD^2&=\dfrac{130(360)}{23^2}\\<br />
AD&=\dfrac{60\sqrt{13}}{23}\\<br />
\end{align*}</cmath><br />
<br />
and our final answer is <math>60+13+23=\boxed{096}</math>.<br />
<br />
==Solution 3==<br />
<br />
Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems&diff=755462016 AMC 12A Problems2016-02-05T16:25:46Z<p>Mishai: /* Problem 20 */</p>
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<div>{{AMC12 Problems|year=2016|ab=A}}<br />
<br />
==Problem 1==<br />
<br />
What is the value of <math>\frac{11!-10!}{9!}</math>?<br />
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<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
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[[2016 AMC 12A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
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For what value of <math>x</math> does <math>10^x\cdot100^{2x}=1000^5</math>?<br />
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
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[[2016 AMC 12A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
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<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
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[[2016 AMC 12A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
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<math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 100</math><br />
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[[2016 AMC 12A Problems/Problem 4|Solution]]<br />
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==Problem 5==<br />
<br />
Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, <math>2016=13+2003</math>). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?<br />
<br />
<math> \textbf{(A)}\ \text{an odd integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br />
\qquad\textbf{(B)}\ \text{an odd integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}\\<br />
\qquad\textbf{(C)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two numbers that are not prime}\\<br />
\qquad\textbf{(D)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br />
\qquad\textbf{(E)}\ \text{an even integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}</math><br />
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[[2016 AMC 12A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math> ?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
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[[2016 AMC 12A Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ?<br />
<br />
<math> \textbf{(A)}\ \text{two parallel lines}\\<br />
\qquad\textbf{(B)}\ \text{two intersecting lines}\\<br />
\qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\<br />
\qquad\textbf{(D)}\ \text{three lines that do not all pass through a common point}\\<br />
\qquad\textbf{(E)}\ \text{a line and a parabola}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
What is the area of the shaded region of the given <math>8\times 5</math> rectangle?<br />
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<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
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label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
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label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
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<math>\textbf{(A)}\ 4\dfrac{3}{4}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 5\dfrac{1}{4}\qquad\textbf{(D)}\ 6\dfrac{1}{2}\qquad\textbf{(E)}\ 8</math><br />
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[[2016 AMC 12A Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is <math>\tfrac{a-\sqrt{2}}{b}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math> ?<br />
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<asy><br />
real x=.369;<br />
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br />
filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray);<br />
filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray);<br />
filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray);<br />
filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray);<br />
filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray);<br />
</asy><br />
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<math>\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11</math><br />
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[[2016 AMC 12A Problems/Problem 9|Solution]]<br />
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==Problem 10==<br />
<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
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[[2016 AMC 12A Problems/Problem 10|Solution]]<br />
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==Problem 11==<br />
<br />
Each of the <math>100</math> students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are <math>42</math> students who cannot sing, <math>65</math> students who cannot dance, and <math>29</math> students who cannot act. How many students have two of these talents?<br />
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<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64</math><br />
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[[2016 AMC 12A Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?<br />
<br />
<asy><br />
pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C);<br />
draw(A--B--C--A--D^^B--E);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,NE);<br />
label("$E$",E,NW);<br />
label("$F$",F,1.5*N);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math><br />
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[[2016 AMC 12A Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A)}\ 12\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 20</math><br />
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[[2016 AMC 12A Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Each vertex of a cube is to be labeled with an integer from <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24</math><br />
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[[2016 AMC 12A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
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[[2016 AMC 12A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
The graphs of <math>y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,</math> and <math>y=\log_x \dfrac{1}{3}</math> are plotted on the same set of axes. How many points in the plane with positive <math>x</math>-coordinates lie on two or more of the graphs? <br />
<br />
<math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6</math><br />
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[[2016 AMC 12A Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Let <math>ABCD</math> be a square. Let <math>E, F, G</math> and <math>H</math> be the centers, respectively, of equilateral triangles with bases <math>\overline{AB}, \overline{BC}, \overline{CD},</math> and <math>\overline{DA},</math> each exterior to the square. What is the ratio of the area of square <math>EFGH</math> to the area of square <math>ABCD</math>? <br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}</math><br />
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[[2016 AMC 12A Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
For some positive integer <math>n,</math> the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3.</math> How many positive integer divisors does the number <math>81n^4</math> have? <br />
<br />
<math>\textbf{(A)}\ 110\qquad\textbf{(B)}\ 191\qquad\textbf{(C)}\ 261\qquad\textbf{(D)}\ 325\qquad\textbf{(E)}\ 425</math><br />
<br />
[[2016 AMC 12A Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
Jerry starts at <math>0</math> on the real number line. He tosses a fair coin <math>8</math> times. When he gets heads, he moves <math>1</math> unit in the positive direction; when he gets tails, he moves <math>1</math> unit in the negative direction. The probability that he reaches <math>4</math> at some time during this process <math>\frac{a}{b},</math> where <math>a</math> and <math>b</math> are relatively prime positive integers. What is <math>a + b?</math> (For example, he succeeds if his sequence of tosses is <math>HTHHHHHH.</math>)<br />
<br />
<math>\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313</math><br />
<br />
[[2016 AMC 12A Problems/Problem 19|Solution]]<br />
<br />
<br />
==Problem 20==<br />
<br />
A binary operation <math>\diamondsuit </math> has the properties that <math>a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\cdot c</math> and that <math>a\ \diamondsuit\ a = 1</math> for all nonzero real numbers <math>a, b</math> and <math>c.</math> (Here the dot <math>\cdot</math> represents the usual multiplication operation.) The solution to the equation <math>2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100</math> can be written as <math>\frac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p + q?</math> <br />
<br />
<math>\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601</math><br />
<br />
[[2016 AMC 12A Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}.</math> Three of the sides of this quadrilateral have length <math>200.</math> What is the length of its fourth side? <br />
<br />
<math>\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500</math><br />
<br />
[[2016 AMC 12A Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math><br />
<br />
[[2016 AMC 12A Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Three numbers in the interval <math>\left[0,1\right]</math> are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6}\qquad\textbf{(B)}\ \dfrac{1}{3}\qquad\textbf{(C)}\ \dfrac{1}{2}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{5}{6}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
There is a smallest positive real number <math>a</math> such that there exists a positive real number <math>b</math> such that all the roots of the polynomial <math>x^3-ax^2+bx-a</math> are real. In fact, for this value of <math>a</math> the value of <math>b</math> is unique. What is the value of <math>b?</math><br />
<br />
<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math><br />
<br />
[[2016 AMC 12A Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
Let <math>k</math> be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with <math>k+1</math> digits. Every time Bernardo writes a number, Silvia erases the last <math>k</math> digits of it. Bernardo then writes the next perfect square, Silvia erases the last <math>k</math> digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let <math>f(k)</math> be the smallest positive integer not written on the board. For example, if <math>k = 1</math>, then the numbers that Bernardo writes are <math>16, 25, 36, 49, 64</math>, and the numbers showing on the board after Silvia erases are <math>1, 2, 3, 4,</math> and <math>6</math>, and thus <math>f(1) = 5</math>. What is the sum of the digits of <math>f(2) + f(4)+ f(6) + ... + f(2016)</math>?<br />
<br />
<math>\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064</math><br />
<br />
[[2016 AMC 12A Problems/Problem 25|Solution]]<br />
<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems&diff=755422016 AMC 12A Problems2016-02-05T16:23:47Z<p>Mishai: /* Problem 20 */</p>
<hr />
<div>{{AMC12 Problems|year=2016|ab=A}}<br />
<br />
==Problem 1==<br />
<br />
What is the value of <math>\frac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 12A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
For what value of <math>x</math> does <math>10^x\cdot100^{2x}=1000^5</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
[[2016 AMC 12A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<math>\textbf{(A)}\ 50\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 75\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 100</math><br />
<br />
[[2016 AMC 12A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, <math>2016=13+2003</math>). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?<br />
<br />
<math> \textbf{(A)}\ \text{an odd integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br />
\qquad\textbf{(B)}\ \text{an odd integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}\\<br />
\qquad\textbf{(C)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two numbers that are not prime}\\<br />
\qquad\textbf{(D)}\ \text{an even integer greater than } 2 \text{ that can be written as the sum of two prime numbers}\\<br />
\qquad\textbf{(E)}\ \text{an even integer greater than } 2 \text{ that cannot be written as the sum of two prime numbers}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math> ?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
<br />
[[2016 AMC 12A Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
Which of these describes the graph of <math>x^2(x+y+1)=y^2(x+y+1)</math> ?<br />
<br />
<math> \textbf{(A)}\ \text{two parallel lines}\\<br />
\qquad\textbf{(B)}\ \text{two intersecting lines}\\<br />
\qquad\textbf{(C)}\ \text{three lines that all pass through a common point}\\<br />
\qquad\textbf{(D)}\ \text{three lines that do not all pass through a common point}\\<br />
\qquad\textbf{(E)}\ \text{a line and a parabola}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
What is the area of the shaded region of the given <math>8\times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{4}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 5\dfrac{1}{4}\qquad\textbf{(D)}\ 6\dfrac{1}{2}\qquad\textbf{(E)}\ 8</math><br />
<br />
[[2016 AMC 12A Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is <math>\tfrac{a-\sqrt{2}}{b}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math> ?<br />
<br />
<asy><br />
real x=.369;<br />
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br />
filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray);<br />
filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray);<br />
filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray);<br />
filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray);<br />
filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11</math><br />
<br />
[[2016 AMC 12A Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
[[2016 AMC 12A Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
<br />
Each of the <math>100</math> students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are <math>42</math> students who cannot sing, <math>65</math> students who cannot dance, and <math>29</math> students who cannot act. How many students have two of these talents?<br />
<br />
<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64</math><br />
<br />
[[2016 AMC 12A Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
In <math>\triangle ABC</math>, <math>AB = 6</math>, <math>BC = 7</math>, and <math>CA = 8</math>. Point <math>D</math> lies on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects <math>\angle BAC</math>. Point <math>E</math> lies on <math>\overline{AC}</math>, and <math>\overline{BE}</math> bisects <math>\angle ABC</math>. The bisectors intersect at <math>F</math>. What is the ratio <math>AF</math> : <math>FD</math>?<br />
<br />
<asy><br />
pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C);<br />
draw(A--B--C--A--D^^B--E);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,N);<br />
label("$D$",D,NE);<br />
label("$E$",E,NW);<br />
label("$F$",F,1.5*N);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 3:2\qquad\textbf{(B)}\ 5:3\qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 7:3\qquad\textbf{(E)}\ 5:2</math><br />
<br />
[[2016 AMC 12A Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A)}\ 12\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 20</math><br />
<br />
[[2016 AMC 12A Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
Each vertex of a cube is to be labeled with an integer from <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 24</math><br />
<br />
[[2016 AMC 12A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
The graphs of <math>y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,</math> and <math>y=\log_x \dfrac{1}{3}</math> are plotted on the same set of axes. How many points in the plane with positive <math>x</math>-coordinates lie on two or more of the graphs? <br />
<br />
<math>\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6</math><br />
<br />
[[2016 AMC 12A Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
Let <math>ABCD</math> be a square. Let <math>E, F, G</math> and <math>H</math> be the centers, respectively, of equilateral triangles with bases <math>\overline{AB}, \overline{BC}, \overline{CD},</math> and <math>\overline{DA},</math> each exterior to the square. What is the ratio of the area of square <math>EFGH</math> to the area of square <math>ABCD</math>? <br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
For some positive integer <math>n,</math> the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3.</math> How many positive integer divisors does the number <math>81n^4</math> have? <br />
<br />
<math>\textbf{(A)}\ 110\qquad\textbf{(B)}\ 191\qquad\textbf{(C)}\ 261\qquad\textbf{(D)}\ 325\qquad\textbf{(E)}\ 425</math><br />
<br />
[[2016 AMC 12A Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
Jerry starts at <math>0</math> on the real number line. He tosses a fair coin <math>8</math> times. When he gets heads, he moves <math>1</math> unit in the positive direction; when he gets tails, he moves <math>1</math> unit in the negative direction. The probability that he reaches <math>4</math> at some time during this process <math>\frac{a}{b},</math> where <math>a</math> and <math>b</math> are relatively prime positive integers. What is <math>a + b?</math> (For example, he succeeds if his sequence of tosses is <math>HTHHHHHH.</math>)<br />
<br />
<math>\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313</math><br />
<br />
[[2016 AMC 12A Problems/Problem 19|Solution]]<br />
<br />
<br />
==Problem 20==<br />
<br />
A binary operation <math>\diamondsuit </math> has the properties that <math>a\ \diamondsuit\ (b\ \diamondsuit\ c) = (a\ \diamondsuit\ b)\cdot c</math> and that <math>a\ \diamondsuit\ a = 1</math> for all nonzero real numbers <math>a, b</math> and <math>c.</math> (Here the dot <math>\ \cdot</math> represents the usual multiplication operation.) The solution to the equation <math>2016\ \diamondsuit\ (6\ \diamondsuit\ x) = 100</math> can be written as <math>\frac{p}{q},</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p + q?</math> <br />
<br />
<math>\textbf{(A)}\ 109\qquad\textbf{(B)}\ 201\qquad\textbf{(C)}\ 301\qquad\textbf{(D)}\ 3049\qquad\textbf{(E)}\ 33,601</math><br />
<br />
[[2016 AMC 12A Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}.</math> Three of the sides of this quadrilateral have length <math>200.</math> What is the length of its fourth side? <br />
<br />
<math>\textbf{(A)}\ 200\qquad\textbf{(B)}\ 200\sqrt{2} \qquad\textbf{(C)}\ 200\sqrt{3} \qquad\textbf{(D)}\ 300\sqrt{2} \qquad\textbf{(E)}\ 500</math><br />
<br />
[[2016 AMC 12A Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math><br />
<br />
[[2016 AMC 12A Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Three numbers in the interval <math>\left[0,1\right]</math> are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br />
<br />
<math>\textbf{(A)}\ \dfrac{1}{6}\qquad\textbf{(B)}\ \dfrac{1}{3}\qquad\textbf{(C)}\ \dfrac{1}{2}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{5}{6}</math><br />
<br />
[[2016 AMC 12A Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
There is a smallest positive real number <math>a</math> such that there exists a positive real number <math>b</math> such that all the roots of the polynomial <math>x^3-ax^2+bx-a</math> are real. In fact, for this value of <math>a</math> the value of <math>b</math> is unique. What is the value of <math>b?</math><br />
<br />
<math>\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12</math><br />
<br />
[[2016 AMC 12A Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
Let <math>k</math> be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with <math>k+1</math> digits. Every time Bernardo writes a number, Silvia erases the last <math>k</math> digits of it. Bernardo then writes the next perfect square, Silvia erases the last <math>k</math> digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let <math>f(k)</math> be the smallest positive integer not written on the board. For example, if <math>k = 1</math>, then the numbers that Bernardo writes are <math>16, 25, 36, 49, 64</math>, and the numbers showing on the board after Silvia erases are <math>1, 2, 3, 4,</math> and <math>6</math>, and thus <math>f(1) = 5</math>. What is the sum of the digits of <math>f(2) + f(4)+ f(6) + ... + f(2016)</math>?<br />
<br />
<math>\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064</math><br />
<br />
[[2016 AMC 12A Problems/Problem 25|Solution]]<br />
<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_12&diff=599882014 AMC 10B Problems/Problem 122014-02-20T18:20:42Z<p>Mishai: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The largest divisor of <math>2,014,000,000</math> is itself. What is the fifth-largest divisor?<br />
<br />
<math> \textbf {(A) } 125, 875, 000 \qquad \textbf {(B) } 201, 400, 000 \qquad \textbf {(C) } 251, 750, 000 \qquad \textbf {(D) } 402, 800, 000 \qquad \textbf {(E) } 503, 500, 000 </math><br />
<br />
==Solution==<br />
<br />
Note that <math>2,014,000,000</math> is divisible by <math>2,\ 4,\ 5,\ 8</math>. Then, the fifth largest factor would come from divisibility by 8, or <math>251,750,000</math>, or <math>\textbf{(C)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_12&diff=599872014 AMC 10B Problems/Problem 122014-02-20T18:20:13Z<p>Mishai: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The largest divisor of <math>2,014,000,000</math> is itself. What is the fifth-largest divisor?<br />
<br />
<math> \textbf {(A) } 125, 875, 000 \qquad \textbf {(B) } 201, 400, 000 \qquad \textbf {(C) } 251, 750, 000 \qquad \textbf {(D) } 402, 800, 000 \qquad \textbf {(E) } 503, 500, 000 </math><br />
<br />
==Solution==<br />
<br />
Note that <math>2\ ,014\ ,000\ ,000</math> is divisible by <math>2,\ 4,\ 5,\ 8</math>. Then, the fifth largest factor would come from divisibility by 8, or <math>251\ ,750\ ,000</math>, or <math>\textbf{(C)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_12&diff=599862014 AMC 10B Problems/Problem 122014-02-20T18:19:24Z<p>Mishai: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The largest divisor of <math>2,014,000,000</math> is itself. What is the fifth-largest divisor?<br />
<br />
<math> \textbf {(A) } 125, 875, 000 \qquad \textbf {(B) } 201, 400, 000 \qquad \textbf {(C) } 251, 750, 000 \qquad \textbf {(D) } 402, 800, 000 \qquad \textbf {(E) } 503, 500, 000 </math><br />
<br />
==Solution==<br />
<br />
Note that <math>2014000000</math> is divisible by <math>2,\ 4,\ 5,\ 8</math>. Then, the fifth largest factor would come from divisibility by 8, or <math>251750000</math>, or <math>\textbf{(C)}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_12&diff=599852014 AMC 10B Problems/Problem 122014-02-20T18:17:26Z<p>Mishai: /* Problem */</p>
<hr />
<div>==Problem==<br />
<br />
The largest divisor of <math>2,014,000,000</math> is itself. What is the fifth-largest divisor?<br />
<br />
<math> \textbf {(A) } 125, 875, 000 \qquad \textbf {(B) } 201, 400, 000 \qquad \textbf {(C) } 251, 750, 000 \qquad \textbf {(D) } 402, 800, 000 \qquad \textbf {(E) } 503, 500, 000 </math><br />
<br />
==Solution==<br />
<br />
==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_2&diff=590732014 AMC 10A Problems/Problem 22014-02-07T02:53:49Z<p>Mishai: /* Problem */</p>
<hr />
<div>==Problem ==<br />
<br />
Roy's cat eats <math>\frac{1}{3}</math> of a can of cat food every morning and <math>\frac{1}{4}</math> of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing <math>6</math> cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?<br />
<br />
<math> \textbf{(A)}\ \text{Tuesday}\qquad\textbf{(B)}\ \text{Wednesday}\qquad\textbf{(C)}\ \text{Thursday}\qquad\textbf{(D)}}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}</math><br />
<br />
== Solution ==<br />
Each day, the cat eats <math>\dfrac13+\dfrac14=\dfrac7{12}</math> of a can of cat food. Therefore, the cat food will last for <math>\dfrac{6}{\dfrac7{12}}=\dfrac{72}7>10</math> days.<br />
Because the number of days is greater than 10, the cat will finish eating in 11 days, which is <math>4 \pmod 7</math>, so the cat will finish on <math>\textbf{(C)}</math> Thursday.</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_8&diff=590622014 AMC 10A Problems/Problem 82014-02-07T02:50:43Z<p>Mishai: </p>
<hr />
<div>==Problem==<br />
<br />
Which of the following number is a perfect square?<br />
<br />
<math> \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 </math><br />
<br />
== Solution ==<br />
<br />
Note that<br />
<math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2*(n+1)}{2}=(n!)^2*\dfrac{n+1}{2}</math>. Therefore, the product will only be a perfect square if the second term is a perfect square.<br />
The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2*9</math>.</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_1&diff=590492014 AMC 10A Problems/Problem 12014-02-07T02:46:01Z<p>Mishai: Created page with "== Solution == <math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45\\ \\ 10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math>"</p>
<hr />
<div>== Solution ==<br />
<math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45\\ \\<br />
10*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math></div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Answer_Key&diff=590342014 AMC 10A Answer Key2014-02-07T02:30:02Z<p>Mishai: Answers</p>
<hr />
<div>1. C<br />
<br />
2. C<br />
<br />
3. E<br />
<br />
4. B<br />
<br />
5. C<br />
<br />
6. A<br />
<br />
7. B<br />
<br />
8. D<br />
<br />
9. C<br />
<br />
10. B<br />
<br />
11. C<br />
<br />
12. C<br />
<br />
13. C<br />
<br />
14. D<br />
<br />
15. C<br />
<br />
16. E<br />
<br />
17. D<br />
<br />
18. B<br />
<br />
19. A<br />
<br />
20. D<br />
<br />
21. E<br />
<br />
22. E<br />
<br />
23. C<br />
<br />
24. A<br />
<br />
25. B</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_3&diff=587002003 AIME II Problems/Problem 32014-01-13T17:29:14Z<p>Mishai: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Define a <math>good~word</math> as a sequence of letters that consists only of the letters <math>A</math>, <math>B</math>, and <math>C</math> - some of these letters may not appear in the sequence - and in which <math>A</math> is never immediately followed by <math>B</math>, <math>B</math> is never immediately followed by <math>C</math>, and <math>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there?<br />
<br />
== Solution == <br />
<br />
There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is <math>3*2^6=192</math><br />
<br />
Therefore, there are <math>\boxed{192}</math> seven-letter good words.<br />
<br />
{{MAA Notice}}<br />
<br />
== See also ==<br />
{{AIME box|year=2003|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_7&diff=586982003 AIME I Problems/Problem 72014-01-13T07:20:34Z<p>Mishai: /* Solution */</p>
<hr />
<div>== Problem ==<br />
[[Point]] <math> B </math> is on <math> \overline{AC} </math> with <math> AB = 9 </math> and <math> BC = 21. </math> Point <math> D </math> is not on <math> \overline{AC} </math> so that <math> AD = CD, </math> and <math> AD </math> and <math> BD </math> are [[integer]]s. Let <math> s </math> be the sum of all possible [[perimeter]]s of <math> \triangle ACD</math>. Find <math> s. </math><br />
<br />
== Solution ==<br />
<center><asy><br />
size(220);<br />
pointpen = black; pathpen = black + linewidth(0.7);<br />
pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8);<br />
D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle);<br />
D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7));<br />
MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2);<br />
</asy></center> <!-- Asymptote replacement for Image:2003_I_AIME-7.png by azjps --><br />
<br />
Denote the height of <math>\triangle ACD</math> as <math>h</math>, <math>x = AD = CD</math>, and <math>y = BD</math>. Using the [[Pythagorean theorem]], we find that <math>h^2 = y^2 - 6^2</math> and <math>h^2 = x^2 - 15^2</math>. Thus, <math>y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189</math>. The LHS is [[difference of squares]], so <math>(x + y)(x - y) = 189</math>. As both <math>x,\ y</math> are integers, <math>x+y,\ x-y</math> must be integral divisors of <math>189</math>. <br />
<br />
The pairs of divisors of <math>189</math> are <math>(1,189)\ (3,63)\ (7,27)\ (9,21)</math>. This yields the four potential sets for <math>(x,y)</math> as <math>(95,94)\ (33,30)\ (17,10)\ (15,6)</math>. The last is not a possibility since it simply [[degenerate]]s into a [[line]]. The sum of the three possible perimeters of <math><br />
\triangle ACD</math> is equal to <math>3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
Using [[Stewart's Theorem]], letting the side length be c, and the cevian be d, then we have<br />
<math>30d^2+21*9*30=9c^2+21c^2=30c^2</math>. Dividing both sides by thirty leaves<br />
<math>d^2+189=c^2</math>.<br />
The solution follows as above.<br />
<br />
== See also ==<br />
*[[Stewart's Theorem]]<br />
{{AIME box|year=2003|n=I|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems&diff=583222008 AMC 12B Problems2013-12-14T19:22:53Z<p>Mishai: /* Problem 7 */</p>
<hr />
<div>==Problem 1==<br />
A basketball player made <math>5</math> baskets during a game. Each basket was worth either <math>2</math> or <math>3</math> points. How many different numbers could represent the total points scored by the player?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math><br />
<br />
([[2008 AMC 12B Problems/Problem 1|Solution]])<br />
==Problem 2==<br />
A <math>4\times 4</math> block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?<br />
<br />
<math>\begin{tabular}[t]{|c|c|c|c|}<br />
\multicolumn{4}{c}{}\\\hline<br />
1&2&3&4\\\hline<br />
8&9&10&11\\\hline<br />
15&16&17&18\\\hline<br />
22&23&24&25\\\hline<br />
\end{tabular}</math><br />
<br />
<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math><br />
<br />
([[2008 AMC 12B Problems/Problem 2|Solution]])<br />
<br />
==Problem 3==<br />
A semipro baseball league has teams with <math>21</math> players each. League rules state that a player must be paid at least <math>15,000</math> dollars, and that the total of all players' salaries for each team cannot exceed <math>700,000</math> dollars. What is the maximum possible salary, in dollars, for a single player?<br />
<br />
<math>\textbf{(A)}\ 270,000 \qquad \textbf{(B)}\ 385,000 \qquad \textbf{(C)}\ 400,000 \qquad \textbf{(D)}\ 430,000 \qquad \textbf{(E)}\ 700,000</math><br />
<br />
([[2008 AMC 12B Problems/Problem 3|Solution]])<br />
<br />
==Problem 4==<br />
On circle <math>O</math>, points <math>C</math> and <math>D</math> are on the same side of diameter <math>\overline{AB}</math>, <math>\angle AOC = 30^\circ</math>, and <math>\angle DOB = 45^\circ</math>. What is the ratio of the area of the smaller sector <math>COD</math> to the area of the circle?<br />
<br />
<asy><br />
unitsize(6mm);<br />
defaultpen(linewidth(0.7)+fontsize(8pt));<br />
<br />
pair C = 3*dir (30);<br />
pair D = 3*dir (135);<br />
pair A = 3*dir (0);<br />
pair B = 3*dir(180);<br />
pair O = (0,0);<br />
draw (Circle ((0, 0), 3));<br />
label ("\(C\)", C, NE);<br />
label ("\(D\)", D, NW);<br />
label ("\(B\)", B, W);<br />
label ("\(A\)", A, E);<br />
label ("\(O\)", O, S);<br />
label ("\(45^\circ\)", (-0.3,0.1), WNW);<br />
label ("\(30^\circ\)", (0.5,0.1), ENE);<br />
draw (A--B);<br />
draw (O--D);<br />
draw (O--C);<br />
</asy><br />
<br />
<math>\textbf{(A)}\ \frac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 4|Solution]])<br />
==Problem 5==<br />
A class collects <math>50</math> dollars to buy flowers for a classmate who is in the hospital. Roses cost <math>3</math> dollars each, and carnations cost <math>2</math> dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly <math>50</math> dollars?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17</math><br />
<br />
([[2008 AMC 12B Problems/Problem 5|Solution]])<br />
==Problem 6==<br />
Postman Pete has a pedometer to count his steps. The pedometer records up to <math>99999</math> steps, then flips over to <math>00000</math> on the next step. Pete plans to determine his mileage for a year. On January <math>1</math> Pete sets the pedometer to <math>00000</math>. During the year, the pedometer flips from <math>99999</math> to <math>00000</math> forty-four times. On December <math>31</math> the pedometer reads <math>50000</math>. Pete takes <math>1800</math> steps per mile. Which of the following is closest to the number of miles Pete walked during the year?<br />
<br />
<math>\textbf{(A)}\ 2500 \qquad \textbf{(B)}\ 3000 \qquad \textbf{(C)}\ 3500 \qquad \textbf{(D)}\ 4000 \qquad \textbf{(E)}\ 4500</math><br />
<br />
([[2008 AMC 12B Problems/Problem 6|Solution]])<br />
==Problem 7==<br />
For real numbers <math>a</math> and <math>b</math>, define <math>a</math>&#036;<math>b = (a - b)^2</math>. What is <math>(x - y)^2</math>&#036;<math>(y - x)^2</math>?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy</math><br />
<br />
([[2008 AMC 12B Problems/Problem 7|Solution]])<br />
<br />
==Problem 8==<br />
Points <math>B</math> and <math>C</math> lie on <math>\overline{AD}</math>. The length of <math>\overline{AB}</math> is <math>4</math> times the length of <math>\overline{BD}</math>, and the length of <math>\overline{AC}</math> is <math>9</math> times the length of <math>\overline{CD}</math>. The length of <math>\overline{BC}</math> is what fraction of the length of <math>\overline{AD}</math>?<br />
<br />
<math>\textbf{(A)}\ \frac {1}{36} \qquad \textbf{(B)}\ \frac {1}{13} \qquad \textbf{(C)}\ \frac {1}{10} \qquad \textbf{(D)}\ \frac {5}{36} \qquad \textbf{(E)}\ \frac {1}{5}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 8|Solution]])<br />
==Problem 9==<br />
Points <math>A</math> and <math>B</math> are on a circle of radius <math>5</math> and <math>AB = 6</math>. Point <math>C</math> is the midpoint of the minor arc <math>AB</math>. What is the length of the line segment <math>AC</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4</math><br />
<br />
([[2008 AMC 12B Problems/Problem 9|Solution]])<br />
==Problem 10==<br />
Bricklayer Brenda would take <math>9</math> hours to build a chimney alone, and bricklayer Brandon would take <math>10</math> hours to build it alone. When they work together they talk a lot, and their combined output is decreased by <math>10</math> bricks per hour. Working together, they build the chimney in <math>5</math> hours. How many bricks are in the chimney?<br />
<br />
<math>\textbf{(A)}\ 500 \qquad \textbf{(B)}\ 900 \qquad \textbf{(C)}\ 950 \qquad \textbf{(D)}\ 1000 \qquad \textbf{(E)}\ 1900</math><br />
<br />
([[2008 AMC 12B Problems/Problem 10|Solution]])<br />
==Problem 11==<br />
A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top <math>\frac{1}{8}</math> of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?<br />
<br />
<math>\textbf{(A)}\ 4000 \qquad \textbf{(B)}\ 2000(4-\sqrt{2}) \qquad \textbf{(C)}\ 6000 \qquad \textbf{(D)}\ 6400 \qquad \textbf{(E)}\ 7000</math><br />
<br />
([[2008 AMC 12B Problems/Problem 11|Solution]])<br />
==Problem 12==<br />
For each positive integer <math>n</math>, the mean of the first <math>n</math> terms of a sequence is <math>n</math>. What is the <math>2008</math>th term of the sequence?<br />
<br />
<math>\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064</math><br />
<br />
([[2008 AMC 12B Problems/Problem 12|Solution]])<br />
==Problem 13==<br />
Vertex <math>E</math> of equilateral triangle <math>\triangle ABE</math> is in the interior of unit square <math>ABCD</math>. Let <math>R</math> be the region consisting of all points inside <math>ABCD</math> and outside <math>\triangle ABE</math> whose distance from <math>\overline{AD}</math> is between <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math>. What is the area of <math>R</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{12-5\sqrt{3}}{72} \qquad \textbf{(B)}\ \frac{12-5\sqrt{3}}{36} \qquad \textbf{(C)}\ \frac{\sqrt{3}}{18} \qquad \textbf{(D)}\ \frac{3-\sqrt{3}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{12}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 13|Solution]])<br />
==Problem 14==<br />
A circle has a radius of <math>\log_{10}{(a^2)}</math> and a circumference of <math>\log_{10}{(b^4)}</math>. What is <math>\log_{a}{b}</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{4\pi} \qquad \textbf{(B)}\ \frac{1}{\pi} \qquad \textbf{(C)}\ \pi \qquad \textbf{(D)}\ 2\pi \qquad \textbf{(E)}\ 10^{2\pi}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 14|Solution]])<br />
==Problem 15==<br />
On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let <math>R</math> be the region formed by the union of the square and all the triangles, and <math>S</math> be the smallest convex polygon that contains <math>R</math>. What is the area of the region that is inside <math>S</math> but outside <math>R</math>?<br />
<br />
<math>\textbf{(A)} \; \frac {1}{4} \qquad \textbf{(B)} \; \frac {\sqrt {2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt {3} \qquad \textbf{(E)} \; 2 \sqrt {3}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 15|Solution]])<br />
==Problem 16==<br />
A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width <math>1</math> foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?<br />
<br />
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
([[2008 AMC 12B Problems/Problem 16|Solution]])<br />
==Problem 17==<br />
Let <math>A</math>, <math>B</math> and <math>C</math> be three distinct points on the graph of <math>y=x^2</math> such that line <math>AB</math> is parallel to the <math>x</math>-axis and <math>\triangle ABC</math> is a right triangle with area <math>2008</math>. What is the sum of the digits of the <math>y</math>-coordinate of <math>C</math>?<br />
<br />
<math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20</math><br />
<br />
([[2008 AMC 12B Problems/Problem 17|Solution]])<br />
==Problem 18==<br />
A pyramid has a square base <math>ABCD</math> and vertex <math>E</math>. The area of square <math>ABCD</math> is <math>196</math>, and the areas of <math>\triangle ABE</math> and <math>\triangle CDE</math> are <math>105</math> and <math>91</math>, respectively. What is the volume of the pyramid?<br />
<br />
<math>\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784</math><br />
<br />
([[2008 AMC 12B Problems/Problem 18|Solution]])<br />
==Problem 19==<br />
A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math>?<br />
<br />
<math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math><br />
<br />
([[2008 AMC 12B Problems/Problem 19|Solution]])<br />
<br />
==Problem 20==<br />
Michael walks at the rate of <math>5</math> feet per second on a long straight path. Trash pails are located every <math>200</math> feet along the path. A garbage truck travels at <math>10</math> feet per second in the same direction as Michael and stops for <math>30</math> seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?<br />
<br />
<math>\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8</math><br />
<br />
([[2008 AMC 12B Problems/Problem 20|Solution]])<br />
==Problem 21==<br />
Two circles of radius 1 are to be constructed as follows. The center of circle <math>A</math> is chosen uniformly and at random from the line segment joining <math>(0,0)</math> and <math>(2,0)</math>. The center of circle <math>B</math> is chosen uniformly and at random, and independently of the first choice, from the line segment joining <math>(0,1)</math> to <math>(2,1)</math>. What is the probability that circles <math>A</math> and <math>B</math> intersect?<br />
<br />
<math>\textbf{(A)} \; \frac {2 + \sqrt {2}}{4} \qquad \textbf{(B)} \; \frac {3\sqrt {3} + 2}{8} \qquad \textbf{(C)} \; \frac {2 \sqrt {2} - 1}{2} \qquad \textbf{(D)} \; \frac {2 + \sqrt {3}}{4} \qquad \textbf{(E)} \; \frac {4 \sqrt {3} - 3}{4}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 21|Solution]])<br />
==Problem 22==<br />
A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?<br />
<br />
<math>\textbf{(A)} \; \frac {11}{20} \qquad \textbf{(B)} \; \frac {4}{7} \qquad \textbf{(C)} \; \frac {81}{140} \qquad \textbf{(D)} \; \frac {3}{5} \qquad \textbf{(E)} \; \frac {17}{28}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 22|Solution]])<br />
==Problem 23==<br />
The sum of the base-<math>10</math> logarithms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>?<br />
<br />
<math>\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15</math><br />
<br />
([[2008 AMC 12B Problems/Problem 23|Solution]])<br />
==Problem 24==<br />
Let <math>A_0 = (0,0)</math>. Distinct points <math>A_1,A_2,\ldots</math> lie on the <math>x</math>-axis, and distinct points <math>B_1,B_2,\ldots</math> lie on the graph of <math>y = \sqrt {x}</math>. For every positive integer <math>n</math>, <math>A_{n - 1}B_nA_n</math> is an equilateral triangle. What is the least <math>n</math> for which the length <math>A_0A_n\ge100</math>?<br />
<br />
<math>\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21</math><br />
<br />
([[2008 AMC 12B Problems/Problem 24|Solution]])<br />
==Problem 25==<br />
Let <math>ABCD</math> be a trapezoid with <math>AB||CD</math>, <math>AB = 11</math>, <math>BC = 5</math>, <math>CD = 19</math>, and <math>DA = 7</math>. Bisectors of <math>\angle A</math> and <math>\angle D</math> meet at <math>P</math>, and bisectors of <math>\angle B</math> and <math>\angle C</math> meet at <math>Q</math>. What is the area of hexagon <math>ABQCDP</math>?<br />
<br />
<math>\textbf{(A)}\ 28\sqrt {3}\qquad \textbf{(B)}\ 30\sqrt {3}\qquad \textbf{(C)}\ 32\sqrt {3}\qquad \textbf{(D)}\ 35\sqrt {3}\qquad \textbf{(E)}\ 36\sqrt {3}</math><br />
<br />
([[2008 AMC 12B Problems/Problem 25|Solution]])<br />
<br />
==See also==<br />
* [[AMC 12]]<br />
* [[AMC 12 Problems and Solutions]]<br />
* [[2008 AMC 10B]]<br />
* [http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=219 2008 AMC B Math Jam Transcript]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=Derivative/Definition&diff=44126Derivative/Definition2012-01-03T07:06:15Z<p>Mishai: /* Problems */</p>
<hr />
<div>The '''[[derivative]]''' of a [[function]] is defined as the instantaneous rate of change of the function at a certain [[point]]. For a [[line]], this is just the [[slope]]. For more complex [[curves]], we can find the rate of change between two points on the curve easily since we can draw a line through them.<br />
<br />
<center>[[Image:derivative1.PNG]]</center><br />
<br />
In the image above, the average rate of change between the two points is the slope of the line that goes through them: <math>\frac{f(x+h)-f(x)}h</math>.<br />
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We can move the second point closer to the first one to find a more accurate value of the derivative. Thus, taking the [[limit]] as <math>h</math> goes to 0 will give us the derivative of the function at <math>x</math>:<br />
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<center>[[Image:derivative2.PNG]]</center><br />
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<center><math> f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}h. </math></center> <br />
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If this limit exists, it is the derivative of <math>f</math> at <math>x</math>. If it does not exist, we say that <math>f</math> is not differentiable at <math>x</math>. This limit is called '''Fermat's difference quotient'''.<br />
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== Examples ==<br />
We can apply the Fermat's difference quotient to a polynomial of the form <math>f(x)=ax^n+bx^{n-1}+cx^{n-2}+ \cdots + z=0</math> in order to find its derivative. If we imagine the secant line intersecting a curve at the points <math>A</math> and <math>B</math>. Then we can change this to the tangent by setting <math>B</math> on top of <math>A</math>. Let us call the horizontal or vertical distance as <math>h</math>.<br />
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<math>\lim_{h\to0} \frac{f(x+h)-f(x)}{h}</math><br />
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<math>\implies \lim_{h\to0} \frac{a(x+h)^n+b(x+h)^{n-1}+c(x+h)^{n-2}+ \cdots z-(ax^n+bx^{n-1}+cx^{n-2}+ \cdots z)}{h}</math><br />
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After canceling like terms we should have all terms contain an <math>h</math>. We can then cancel out the <math>h</math> and set <math>h=0</math>. Our end result is the first-derivative.<br />
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The first derivative is denoted as <math>f'(x)</math>.<br />
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This would be some tedious work so instead there is a much nicer way to find the derivative.<br />
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Let <math>f(x)=3x^n</math>. Let <math>g(x)=x^t+x^{n-1}+5x^{3}</math><br />
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1. Find <math>f'(x)</math>.<br />
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Any function like this is:<br />
<math>f'(x)=3 \cdot n \cdot x^{n-1}=3n \cdot x^{n-1}</math><br />
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2. Find <math>g'(x)</math>.<br />
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Breaking apart on what we used above. <br />
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<math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+ 5 \cdot 3 \cdot x^2</math><br />
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<math>g'(x)=t \cdot x^{t-1}+(n-1) \cdot x^{n-2}+15x^2</math><br />
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Let <math>f(x)=-147</math>. Find <math>f'(x)</math>.<br />
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If the function <math>f(x)</math> is a constant then its derivative will always be <math>0</math>.<br />
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Notation: <math>f'(x)</math> denotes the first derivative for <math>f(x)</math>. The symbol for the second derivative is just <math>f''(x)</math>. For the third derivative it is just <math>f'''(x)</math>. Derivatives are also written as <math>\frac{d}{dx} f(x)</math>. Or if for the nth derivative they are written as <math>\frac{d^n}{dx^n} f(x)</math>.<br />
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Maximum and Minimum: We can use the first derivative to determine the maximum and the minimum points of a graph.<br />
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If <math>f'(x)=6x^2-24</math>. Then the maximum and the minimum occur when:<br />
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<math>6x^2-24=</math>, <math>x=2</math> or <math>x=-2</math>. We can plug each back in to the original <math>f(x)</math> if it was given, and the one with the higher y-coordinate is the maximum, while the smaller y-coordinate gives the minimum.<br />
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Below are problems for Part I. In Part II(see link below) we will begin to actually "start" the calculus with this.<br />
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=== Problems ===<br />
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<math>\boxed{\text{Problem 1}}</math>: Find the first derivative of <math>f(x)</math>, where <math>f(x)=2x^2-15x+7</math>. <br />
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<math>\boxed{\text{Solution 1}}</math>:<br />
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<math>f'(x)=2 \cdot 2 \cdot x^1-15 \cdot 1 \cdot x^0+0</math><br />
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<math>f'(x)=4x-15</math>.<br />
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<math>\boxed{\text{Problem 2}}</math>: Find the equation of the line tangent to the function <math>f(x)=3x^3-5x^2+12</math> at <math>(-1,14)</math>.<br />
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<math>\boxed{\text{Solution 2}}</math>:<br />
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We will take the first derivative to determine the slope of the tangent line.<br />
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<math>f'(x)=9x^2-10x</math>. If this is the slope of the tangent point then we can just plug <math>-1</math> into the <math>x</math> coordinate to find the actual slope.<br />
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<math>f'(x)=9+10=19</math>. The slope of the line is <math>19</math>.<br />
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Let the equation be:<br />
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<math>y=19x+b</math>.<br />
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Plugging <math>(-1,14)</math> in gives:<br />
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<math>14=-19+b</math><br />
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<math>\implies b=33</math>.<br />
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<math>\therefore</math> The equation of the line is <math>y=19x+33</math>.<br />
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<math>\boxed{\text{Problem 3}}</math>: Find the nth derivative of <math>f(x)=x^n</math><br />
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<math>\boxed{\text{Solution 3}}</math>:<br />
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<math>\frac{d}{dx} f(x)=nx^{n-1}</math><br />
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<math>\frac{d^2}{dx^2} f(x)=n(n-1) x^{n-2}</math><br />
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<math>\vdots</math><br />
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<math>\frac{d^{n}}{dx^{n}} f(x)=n(n-1)(n-2) \cdots 1</math><br />
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<math>\frac{d^{n}}{dx^{n}} f(x)=n!</math><br />
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<math>\therefore</math> The nth derivative of <math>f(x)</math> is <math>n!</math>.<br />
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== See also ==<br />
* [[Calculus]]<br />
* [[Derivative]]<br />
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[[Category:Calculus]]</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=California_mathematics_competitions&diff=44125California mathematics competitions2012-01-03T06:35:30Z<p>Mishai: /* California middle school mathematics competitions */</p>
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<div>==California middle school mathematics competitions==<br />
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[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=298 middle school math forums] where students can discuss contest problems and mathematics.<br />
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* [[California MathCounts]] is part of the national [[MathCounts]] competition.<br />
* [[Bay Area Mathematical Olympiad (BAMO)]] [http://mathcircle.berkeley.edu/BMC6/pages/BAMO/bamo.html website]<br />
* [[California Mathematics League]]<br />
* [[Math Field Day for San Diego Middle Schools]] [http://best.me.berkeley.edu/%7Eaagogino/IU/mdrules/mathsenior.html website]<br />
* [[Polya Competition]] [http://www.castilleja.org/faculty/josh_zucker/polya/ website]<br />
* [[Redwood Empire Mathematics Tournament hosted by Humboldt State]]- middle and high school [http://www.humboldt.edu/%7Eremt/ website]<br />
* The [[San Diego Math League]] and [[San Diego Math Olympiad]] hosted by the [[San Diego Math Circle]] [http://www.artofproblemsolving.com/SDMC/AoPS_S_Schedule.php website]<br />
* [[San Jose JACL-Tokutomi Mathematics Contest]]- for grades 7-12. [http://www.sanjosejacl.org/math.html website]<br />
* [[Mathleague]]- elementary, middle, high school. [http://www.mathleague.org website]<br />
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==California high school mathematics competitions==<br />
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[[AoPS]] hosts [http://www.artofproblemsolving.com/Forum/index.php?f=214 high school math forums] where students can discuss contest problems and mathematics.<br />
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* [[Bay Area Mathematical Olympiad (BAMO)]] [http://mathcircle.berkeley.edu/BMC6/pages/BAMO/bamo.html website]<br />
* [[Bay Area Math Meet]] [http://www.usfca.edu/math/bamm/ website]<br />
* [[California Mathematics League]]<br />
* [[College of Creative Studies Math Competition]] [http://www.ccs.ucsb.edu:81/index.php?Itemid=36%20 website]<br />
* [[LACC Math Contest]] [http://www.lacitycollege.edu/academic/departments/mathdept/contest.html website]<br />
* [[Lawrence Livermore National Laboratories Annual High School Math Challenge]] [http://www.lacitycollege.edu/academic/departments/math/mathdept/contest/index.html website]<br />
* [[Math Day at the Beach]] hosted by CSULB [http://www.csulb.edu/depts/math/mathday/index.htm website]<br />
* [[Mesa Day Math Contest at UC Berkeley]]<br />
* [[Polya Competition]] [http://www.castilleja.org/faculty/josh_zucker/polya/ website]<br />
* [[Pomona College Mathematical Talent Search]] [http://www.math.pomona.edu/talentsearch.html website]<br />
* [[Redwood Empire Mathematics Tournament hosted by Humboldt State]]- middle and high school [http://www.humboldt.edu/%7Eremt/ website]<br />
* The [[San Diego Math League]] and [[San Diego Math Olympiad]] hosted by the [[San Diego Math Circle]] [http://www.artofproblemsolving.com/SDMC/AoPS_S_Schedule.php website]<br />
* [[San Jose JACL-Tokutomi Mathematics Contest]]- for grades 7-12. [http://www.sanjosejacl.org/math.html website]<br />
* [[Santa Clara University High School Mathematics Contest]] [http://math.scu.edu/activities/hscontest.html website]<br />
* [[Stanford Mathematics Tournament]] [http://www.stanford.edu/group/sumo/smt/ website]<br />
* [[UCSD/GSDMC High School Honors Mathematics Contest]] [http://www.math.ucsd.edu/resources/honors_math_contest/ website]<br />
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== National math contests ==<br />
* [[List of United States elementary school mathematics competitions]] for national contests.<br />
* [[List of United States middle school mathematics competitions]] for national contests.<br />
* [[List of United States high school mathematics competitions]] for national contests.<br />
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== See also ==<br />
* [[List of mathematics competitions]]<br />
* [[Mathematics competitions resources]]<br />
* [[Math circles]] -- There are many in California.</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=User:Mishai&diff=40302User:Mishai2011-07-08T07:42:15Z<p>Mishai: </p>
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<div>Oyddoyddoy!!!!<br />
Wild Bananas!<br />
Pokemon!<br />
Asymptote!<br />
Latex!</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=User_talk:Mishai&diff=38367User talk:Mishai2011-05-08T01:52:43Z<p>Mishai: Created page with 'Lalalala'</p>
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<div>Lalalala</div>Mishaihttps://artofproblemsolving.com/wiki/index.php?title=User:Mishai&diff=36360User:Mishai2011-01-19T06:32:35Z<p>Mishai: Created page with 'Oyddoyddoy!!!!'</p>
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<div>Oyddoyddoy!!!!</div>Mishai