https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Pinkbunny1228&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T13:44:21ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=User:Wuwang2002&diff=130387User:Wuwang20022020-08-04T01:55:06Z<p>Pinkbunny1228: /* Somewhat accurate user count */</p>
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<div>Hello! I am wuwang2002!<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About This Place</div></font>==<br />
<br />
What you saw above (the heading) was made by piphi.<br />
<br />
Uh, well...I have nothing else here to say.<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Links you should click NOW</div></font>==<br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">My own links from AoPS</div></font>===<br />
https://artofproblemsolving.com/community/c1132787_wuwang2002s_games_collection<br />
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Check that out: it is my own collection of my own fora.<br />
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https://artofproblemsolving.com/community/c1194307_the_land_of_empires<br />
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Check that out if you want.<br />
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Thanks!<br />
<br />
Sorry if this is not allowed @admins. I read that article, but I wasn't sure if this classified.<br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Outside links</div></font>===<br />
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<font color="red">[NOT PUT HERE BY WUWANG2002 BECAUSE OF POSSIBLE VIOLATIONS OF THE WIKI RULES]</font><br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">More info</div></font>==<br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Somewhat accurate user count</div></font>===<br />
Add 1 to this number whenever you visit for the first time: <br />
<center><font size="150 pts">9</center></font size><br />
<br />
Please put your name under this sentence when it is your first time visiting:<br />
*wuwang2002<br />
*Lcz<br />
*rocketsri<br />
*MIRB16<br />
*CreativeHedgehog<br />
*aops-g5-gethsemanea2<br />
*AndrewC10<br />
*LucasFan<br />
*pinkbunny1228<br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">My Talk</div></font>===<br />
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[[User talk:Wuwang2002|Click here for my talk]]<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">LaTeX 'n bbcode</div></font>==<br />
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C'mon! Ask me for any LaTeX and then I will give the code!<br />
Same with bbcode!<br />
Thanks!<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Games</div></font>==<br />
<br />
So far, my games are all in forums. But however, I hope to someday make some own games like Go! and Gomuku, which I don't have a translation for!<br />
<br />
EDIT: I am making a collaboration with [[User:Piphi|piphi]] to make more games.<br />
EDIT EDIT: I am now starting the template myself, too, along with piphi<br />
<br />
Go to [[Wuwang2002's Wiki Games]] to see it. It is very plain, however.<br />
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{{User:Wuwang2002/Template:19x19board|<!--<br />
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However, I can't seem to change it.<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Some Facts</div></font>==<br />
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1. wuwang2002 is the best!...or not<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Testing Stuff</div></font>==<br />
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This is free to anyone to edit. -[[User:Wuwang2002|Wuwang2002]] ([[User talk:Wuwang2002|talk]]) 21:26, 17 July 2020 (EDT)<br />
-[[User:Wuwang2002|Wuwang2002]] ([[User talk:Wuwang2002|talk]]) 21:26, 17 July 2020 (EDT)<br />
oh, I did it the other way!<br />
-[[User:Wuwang2002|Wuwang2002]] ([[User talk:Wuwang2002|talk]]) 21:26, 17 July 2020 (EDT)<br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Test Level 1</div></font>===<br />
====<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Test level 2</div></font>====<br />
=====<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Test level 3</div></font>=====<br />
======<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Last level</div></font>======<br />
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Go more!<br />
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Oh, you can't.<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Stories about bots</div></font>==<br />
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<div style="text-decoration:line-through;">To be made</div><br />
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Bots are basically the same as user, but, they do the same thing again and again. [https://artofproblemsolving.com/wiki/index.php/User:CreativeHedgehog CreativeHedgehog] 17:28, 31 July 2020 (EDT)<br />
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==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Math</div></font>==<br />
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Sorry this came so late, but...<br />
Put your opinions with your name under the latest post in the appropriate section. Feel free to add examples.<br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Arithmetic</div></font>===<br />
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====<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Opinions</div></font>====<br />
Bashy, annoying, drill-and-kill, and not that helpful or interesting. -wuwang2002<br />
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====<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Examples</div></font>====<br />
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<math>9+\dfrac{8}{7}\cdot5-\dfrac{19}{7}</math><br />
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===<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Algebra</div></font>===<br />
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====<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Opinions</div></font>====<br />
Much better than arithmetic, but sometimes bashy. -wuwang2002<br />
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====<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Examples</div></font>====<br />
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<math>\dfrac{9}{4}x+32=50</math><br />
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<math>x^2-8x+12=-3</math></div>Pinkbunny1228https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_17&diff=1284482010 AMC 12A Problems/Problem 172020-07-17T06:56:50Z<p>Pinkbunny1228: /* Solution 3 */</p>
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<div>{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #17]] and [[2010 AMC 10A Problems|2010 AMC 10A #19]]}}<br />
<br />
== Problem ==<br />
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math><br />
<br />
== Solution 1==<br />
It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on <math>\triangle ABC</math>, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>.<br />
<br />
If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore<br />
<br />
<math>\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)</math>.<br />
<br />
<br />
Based on the initial conditions,<br />
<br />
<cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath><br />
<br />
Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>.<br />
<br />
==Solution 2==<br />
Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get <math>AC=\sqrt{r^2+r+1}</math>. <br />
<br />
Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}</math>. The area of the hexagon is equal to <math>[ACE] + 3[ABC]</math>. We are given that <math>70\%</math> of this area is equal to <math>[ACE]</math>; solving for <math>AC</math> in terms of <math>r</math> gives <math>AC=\sqrt{7r}</math>.<br />
<br />
Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0</math> and by [[Vieta's Formulas]] , we get <math>\boxed{\textbf{E}}</math>.<br />
<br />
Note: Since <math>r</math> has to be positive we must first check that the discriminant is positive before applying Vieta's. And it indeed is.<br />
<br />
==Solution 3==<br />
Find the area of the triangle <math>ACE</math> as how it was done in solution 1. Find the sum of the areas of the congruent triangles <math>ABC, CDE, EFA</math> as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles <math>ABC, CDE, EFA</math> is <math>30\%</math> of the area of the hexagon. Hence <math>\frac{7}{3}</math> times the latter is equal to the triangle <math>ACE</math>. Hence <math>\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)</math>. We can simplify this to <math>7r=r^2+r+1\implies r^2-6r+1=0</math>. By Vieta's, we get the sum of all possible values of <math>r</math> is <math>-\frac{-6}{1}=6\text{ or } \boxed{\textbf{E}}</math>.<br />
-vsamc<br />
(Edited by pinkbunny1228)<br />
<br />
==Video Solution==<br />
https://youtu.be/rsURe5Xh-j0?t=961<br />
<br />
~IceMatrix<br />
<br />
== See also ==<br />
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Pinkbunny1228https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems&diff=1222302020 AMC 10B Problems2020-05-10T02:57:30Z<p>Pinkbunny1228: /* Problem 2 */ I added a solution link</p>
<hr />
<div>{{AMC10 Problems|year=2020|ab=B}}<br />
<br />
==Problem 1==<br />
<br />
What is the value of <cmath>1 - (-2) - 3 - (-4) - 5 - (-6)?</cmath><br />
<br />
<math>\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21</math><br />
<br />
[[2020 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
<br />
Carl has <math>5</math> cubes each having side length <math>1</math>, and Kate has <math>5</math> cubes each having side length <math>2</math>. What is the total volume of these <math>10</math> cubes?<br />
<br />
<math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45</math><br />
<br />
[[2020 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
The ratio of <math>w</math> to <math>x</math> is <math>4:3</math>, the ratio of <math>y</math> to <math>z</math> is <math>3:2</math>, and the ratio of <math>z</math> to <math>x</math> is <math>1:6</math>. What is the ratio of <math>w</math> to <math>y?</math><br />
<br />
<math>\textbf{(A)}\ 4:3 \qquad\textbf{(B)}\ 3:2 \qquad\textbf{(C)}\ 8:3 \qquad\textbf{(D)}\ 4:1 \qquad\textbf{(E)}\ 16:3</math><br />
<br />
[[2020 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
<br />
The acute angles of a right triangle are <math>a^{\circ}</math> and <math>b^{\circ}</math>, where <math>a>b</math> and both <math>a</math> and <math>b</math> are prime numbers. What is the least possible value of <math>b</math>?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11</math><br />
<br />
[[2020 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
<br />
How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)<br />
<br />
<math>\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050</math><br />
<br />
[[2020 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
<br />
Driving along a highway, Megan noticed that her odometer showed <math>15951</math> (miles). This number is a palindrome-it reads the same forward and backward. Then <math>2</math> hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this <math>2</math>-hour period?<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70</math><br />
<br />
[[2020 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
<br />
How many positive even multiples of <math>3</math> less than <math>2020</math> are perfect squares?<br />
<br />
<math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12</math><br />
<br />
[[2020 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
<br />
Points <math>P</math> and <math>Q</math> lie in a plane with <math>PQ=8</math>. How many locations for point <math>R</math> in this plane are there such that the triangle with vertices <math>P</math>, <math>Q</math>, and <math>R</math> is a right triangle with area <math>12</math> square units?<br />
<br />
<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12</math><br />
<br />
[[2020 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
<br />
How many ordered pairs of integers <math>(x,y)</math> satisfy the equation <cmath>x^{2020} + y^2 = 2y?</cmath><br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \text{infinitely many}</math><br />
<br />
[[2020 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
<br />
A three-quarter sector of a circle of radius <math>4</math> inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?<br />
<asy><br />
<br />
draw(Arc((0,0), 3, 0, 270));<br />
draw((0,-3)--(0,0)--(3,0));<br />
<br />
label("$4$", (2,0), S);<br />
<br />
<br />
</asy><br />
<math>\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7</math><br />
<br />
[[2020 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Ms.Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}</math><br />
<br />
[[2020 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
<br />
The decimal representation of <cmath>\frac{1}{20^{20}}</cmath><br />
consists of a string of zeros after the decimal point, followed by a <math>9</math> and then several more digits. How many zeros are in that initial string of zeros after the decimal point?<br />
<br />
<math>\textbf{(A)}\ 23 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 25 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 27</math><br />
<br />
[[2020 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
<br />
Andy the Ant lives on a coordinate plane and is currently at <math>(-20, 20)</math> facing east (that is, in the positive <math>x</math>-direction). Andy moves <math>1</math> unit and then turns <math>90^{\circ}</math> degrees left. From there, Andy moves <math>2</math> units (north) and then turns <math>90^{\circ}</math> degrees left. He then moves <math>3</math> units (west) and again turns <math>90^{\circ}</math> degrees left. Andy continues his progress, increasing his distance each time by <math>1</math> unit and always turning left. What is the location of the point at which Andy makes the <math>2020</math>th left turn?<br />
<br />
<math>\textbf{(A)}\ (-1030, -994)\qquad\textbf{(B)}\ (-1030, -990)\qquad\textbf{(C)}\ (-1026, -994)\qquad\textbf{(D)}\ (-1026, -990)\qquad\textbf{(E)}\ (-1022, -994)</math><br />
<br />
[[2020 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
<br />
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?<br />
<br />
<asy><br />
size(140);<br />
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));<br />
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);<br />
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);<br />
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);<br />
fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white);<br />
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);<br />
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);<br />
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));<br />
draw(arc((2,0),1,180,0)--(2,0)--cycle);<br />
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);<br />
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);<br />
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);<br />
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);<br />
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);<br />
label("$2$",(3.5,3sqrt(3)/2),NE);<br />
</asy><br />
<br />
<math> \textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \frac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \frac{3\sqrt{3}}{2} - \frac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \frac{9\sqrt{3}}{2} - \pi </math><br />
<br />
<br />
[[2020 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
<br />
Steve wrote the digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> in order repeatedly from left to right, forming a list of <math>10,000</math> digits, beginning <math>123451234512\ldots.</math> He then erased every third digit from his list (that is, the <math>3</math>rd, <math>6</math>th, <math>9</math>th, <math>\ldots</math> digits from the left), then erased every fourth digit from the resulting list (that is, the <math>4</math>th, <math>8</math>th, <math>12</math>th, <math>\ldots</math> digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions <math>2019, 2020, 2021</math>?<br />
<br />
<math>\textbf{(A)} \text{ 7} \qquad \textbf{(B)} \text{ 9} \qquad \textbf{(C)} \text{ 10} \qquad \textbf{(D)} \text{ 11} \qquad \textbf{(E)} \text{ 12}</math><br />
<br />
[[2020 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
<br />
Bela and Jenn play the following game on the closed interval <math>[0, n]</math> of the real number line, where <math>n</math> is a fixed integer greater than <math>4</math>. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval <math>[0, n]</math>. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?<br />
<br />
<math>\textbf{(A)} \text{ Bela will always win.} \qquad \textbf{(B)} \text{ Jenn will always win.} \qquad \textbf{(C)} \text{ Bela will win if and only if }n \text{ is odd.}</math><br />
<math>\textbf{(D)} \text{ Jenn will win if and only if }n \text{ is odd.} \qquad \textbf{(E)} \text { Jenn will win if and only if } n>8.</math><br />
<br />
[[2020 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
<br />
There are <math>10</math> people standing equally spaced around a circle. Each person knows exactly <math>3</math> of the other <math>9</math> people: the <math>2</math> people standing next to her or him, as well as the person directly across the circle. How many ways are there for the <math>10</math> people to split up into <math>5</math> pairs so that the members of each pair know each other?<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 />
[[2020 AMC 10B Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
<br />
An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?<br />
<br />
<math>\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12</math><br />
<br />
[[2020 AMC 10B Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
<br />
In a certain card game, a player is dealt a hand of <math>10</math> cards from a deck of <math>52</math> distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as <math>158A00A4AA0</math>. What is the digit <math>A</math>?<br />
<br />
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7</math><br />
<br />
[[2020 AMC 10B Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
<br />
Let <math>B</math> be a right rectangular prism (box) with edges lengths <math>1,</math> <math>3,</math> and <math>4</math>, together with its interior. For real <math>r\geq0</math>, let <math>S(r)</math> be the set of points in <math>3</math>-dimensional space that lie within a distance <math>r</math> of some point <math>B</math>. The volume of <math>S(r)</math> can be expressed as <math>ar^{3} + br^{2} + cr +d</math>, where <math>a,</math> <math>b,</math> <math>c,</math> and <math>d</math> are positive real numbers. What is <math>\frac{bc}{ad}?</math><br />
<br />
<math>\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38</math><br />
<br />
[[2020 AMC 10B Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
<br />
In square <math>ABCD</math>, points <math>E</math> and <math>H</math> lie on <math>\overline{AB}</math> and <math>\overline{DA}</math>, respectively, so that <math>AE=AH.</math> Points <math>F</math> and <math>G</math> lie on <math>\overline{BC}</math> and <math>\overline{CD}</math>, respectively, and points <math>I</math> and <math>J</math> lie on <math>\overline{EH}</math> so that <math>\overline{FI} \perp \overline{EH}</math> and <math>\overline{GJ} \perp \overline{EH}</math>. See the figure below. Triangle <math>AEH</math>, quadrilateral <math>BFIE</math>, quadrilateral <math>DHJG</math>, and pentagon <math>FCGJI</math> each has area <math>1.</math> What is <math>FI^2</math>?<br />
<asy><br />
real x=2sqrt(2);<br />
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br />
real z=2sqrt(8-4sqrt(2));<br />
pair A, B, C, D, E, F, G, H, I, J;<br />
A = (0,0);<br />
B = (4,0);<br />
C = (4,4);<br />
D = (0,4);<br />
E = (x,0);<br />
F = (4,y);<br />
G = (y,4);<br />
H = (0,x);<br />
I = F + z * dir(225);<br />
J = G + z * dir(225);<br />
<br />
draw(A--B--C--D--A);<br />
draw(H--E);<br />
draw(J--G^^F--I);<br />
draw(rightanglemark(G, J, I), linewidth(.5));<br />
draw(rightanglemark(F, I, E), linewidth(.5));<br />
<br />
dot("$A$", A, S);<br />
dot("$B$", B, S);<br />
dot("$C$", C, dir(90));<br />
dot("$D$", D, dir(90));<br />
dot("$E$", E, S);<br />
dot("$F$", F, dir(0));<br />
dot("$G$", G, N);<br />
dot("$H$", H, W);<br />
dot("$I$", I, SW);<br />
dot("$J$", J, SW);<br />
<br />
</asy><br />
<math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math><br />
<br />
[[2020 AMC 10B Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
<br />
What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>?<br />
<br />
<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math><br />
<br />
[[2020 AMC 10B Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
<br />
Square <math>ABCD</math> in the coordinate plane has vertices at the points <math>A(1,1), B(-1,1), C(-1,-1),</math> and <math>D(1,-1).</math> Consider the following four transformations:<br />
<br />
<br />
<math>L,</math> a rotation of <math>90^{\circ}</math> counterclockwise around the origin;<br />
<br />
<math>R,</math> a rotation of <math>90^{\circ}</math> clockwise around the origin;<br />
<br />
<math>H,</math> a reflection across the <math>x</math>-axis; and<br />
<br />
<math>V,</math> a reflection across the <math>y</math>-axis.<br />
<br />
<br />
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying <math>R</math> and then <math>V</math> would send the vertex <math>A</math> at <math>(1,1)</math> to <math>(-1,-1)</math> and would send the vertex <math>B</math> at <math>(-1,1)</math> to itself. How many sequences of <math>20</math> transformations chosen from <math>\{L, R, H, V\}</math> will send all of the labeled vertices back to their original positions? (For example, <math>R, R, V, H</math> is one sequence of <math>4</math> transformations that will send the vertices back to their original positions.)<br />
<br />
<math>\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}</math><br />
<br />
[[2020 AMC 10B Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
<br />
How many positive integers <math>n</math> satisfy<cmath>\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?<br />
<br />
</cmath>(Recall that <math>\lfloor x\rfloor</math> is the greatest integer not exceeding <math>x</math>.)<br />
<br />
<math>\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32</math><br />
<br />
[[2020 AMC 10B Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
<br />
Let <math>D(n)</math> denote the number of ways of writing the positive integer <math>n</math> as a product<cmath>n = f_1\cdot f_2\cdots f_k,</cmath><br />
<br />
where <math>k\ge1</math>, the <math>f_i</math> are integers strictly greater than <math>1</math>, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number <math>6</math> can be written as <math>6</math>, <math>2\cdot 3</math>, and <math>3\cdot2</math>, so <math>D(6) = 3</math>. What is <math>D(96)</math>?<br />
<br />
<math>\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184</math><br />
<br />
[[2020 AMC 10B Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC10 box|year=2020|ab=B|before=[[2020 AMC 10A Problems]]|after=[[2021 AMC 10A Problems]]}}<br />
{{MAA Notice}}</div>Pinkbunny1228https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=1110592016 AMC 10A Problems2019-11-10T03:13:05Z<p>Pinkbunny1228: /* Problem 11 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{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 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value <math>x</math> does <math>10^{x}\cdot 100^{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 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math><br />
<br />
<br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<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 \dfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\dfrac{x}{y}</math>. What is the value of <math>\text{rem}\left(\dfrac{3}{8}, -\dfrac{2}{5}\right)</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 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
<br />
[[2016 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
<br />
[[2016 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<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 10A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
<br />
[[2016 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<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 10A Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</math><br />
<br />
[[2016 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
Find the area of the shaded region.<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}{5} \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 10A Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
<br />
[[2016 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<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 10A Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
[[2016 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
<br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
[[2016 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<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 10A Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <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 10A Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
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<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math><br />
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[[2016 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
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<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
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[[2016 AMC 10A Problems/Problem 20|Solution]]<br />
==Problem 21==<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 />
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<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 10A Problems/Problem 21|Solution]]<br />
==Problem 22==<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 />
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<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
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[[2016 AMC 10A Problems/Problem 22|Solution]]<br />
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==Problem 23==<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 <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
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<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
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[[2016 AMC 10A Problems/Problem 23|Solution]]<br />
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==Problem 24==<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 the fourth side?<br />
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<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 />
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[[2016 AMC 10A Problems/Problem 24|Solution]]<br />
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==Problem 25==<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 />
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<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math><br />
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[[2016 AMC 10A Problems/Problem 25|Solution]]<br />
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==See also==<br />
{{AMC10 box|year=2016|ab=A|before=[[2015 AMC 10B Problems]]|after=[[2016 AMC 10B Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[2016 AMC 10A]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Pinkbunny1228https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_13&diff=1079222004 AMC 10B Problems/Problem 132019-07-25T20:01:26Z<p>Pinkbunny1228: Added another solution using congruence</p>
<hr />
<div>== Problem ==<br />
In the United States, coins have the following thicknesses: penny, <math>1.55</math> mm; nickel, <math>1.95</math> mm; dime, <math>1.35</math> mm; quarter, <math>1.75</math> mm. If a stack of these coins is exactly <math>14</math> mm high, how many coins are in the stack?<br />
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<math> \mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad \mathrm{(E) \ } 11 </math><br />
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== Solution 1 ==<br />
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All numbers in this solution will be in hundredths of a millimeter.<br />
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The thinnest coin is the dime, with thickness <math>135</math>. A stack of <math>n</math> dimes has height <math>135n</math>.<br />
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The other three coin types have thicknesses <math>135+20</math>, <math>135+40</math>, and <math>135+60</math>. By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set <math>\{135n, 135n+20, 135n+40, \dots, 195n\}</math>.<br />
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If we take an odd <math>n</math>, then all the possible heights will be odd, and thus none of them will be <math>1400</math>. Hence <math>n</math> is even.<br />
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If <math>n<8</math> the stack will be too low and if <math>n>10</math> it will be too high. Thus we are left with cases <math>n=8</math> and <math>n=10</math>.<br />
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If <math>n=10</math> the possible stack heights are <math>1350,1370,1390,\dots</math>, with the remaining ones exceeding <math>1400</math>.<br />
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Therefore there are <math>\boxed{\mathrm{(B)\ }8}</math> coins in the stack.<br />
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Using the above observation we can easily construct such a stack. A stack of <math>8</math> dimes would have height <math>8\cdot 135=1080</math>, thus we need to add <math>320</math>.<br />
This can be done for example by replacing five dimes by nickels (for <math>+60\cdot 5 = +300</math>), and one dime by a penny (for <math>+20</math>).<br />
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== Solution 2 ==<br />
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Let <math>p,n,d</math>, and <math>q</math> be the number of pennies, nickels, dimes, and quarters used in the stack. <br />
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From the conditions above, we get the following equation:<br />
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<cmath>155p+195n+135d+175q=1400.</cmath><br />
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Then we divide each side by five to get<br />
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<cmath>31p+39n+27d+35q=280.</cmath><br />
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Writing both sides in terms of mod 4, we have <math>-p-n-d-q \equiv 0 \pmod 4</math>. <br />
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This means that the sum <math>p+n+d+q</math> is divisible by 4. Therefore, the answer must be <math>\boxed{(B)\,\, 8}.</math><br />
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==Note==<br />
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We can easily add up <math>1.55\text{\ mm}</math> and <math>1.95\text{\ mm}</math> to get <math>3.50\text{\ mm}</math>. We multiply that by <math>4</math> to get <math>14\text{\ mm}</math>. Since this works and it requires 8 coins, the answer is clearly <math>\boxed{\mathrm{(B)\ }8}</math>.<br />
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Similarly, we can simply take <math>8</math> quarters to get <math>8\cdot 1.75=14</math>.<br />
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== See also ==<br />
<br />
{{AMC10 box|year=2004|ab=B|num-b=12|num-a=14}}<br />
{{MAA Notice}}</div>Pinkbunny1228