https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Beastej&feedformat=atom AoPS Wiki - User contributions [en] 2022-07-07T11:37:08Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_19&diff=164278 2001 AMC 10 Problems/Problem 19 2021-10-30T00:09:52Z <p>Beastej: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> == Solution==<br /> <br /> Let's use [[stars and bars]].<br /> Let the donuts be represented by &lt;math&gt; O &lt;/math&gt;s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us &lt;math&gt; 4 &lt;/math&gt; in all. The four donuts we want can be represented as &lt;math&gt; OOOO &lt;/math&gt;. Notice that we can add two &quot;dividers&quot; to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, &lt;math&gt; O|OO|O &lt;/math&gt; represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in &lt;math&gt; \binom{6}{2}=15 &lt;/math&gt; ways. Our answer is hence &lt;math&gt; \boxed{\textbf{(D)}\ 15} &lt;/math&gt;. Notice that this can be generalized to get the stars and bars (balls and urns) identity.<br /> <br /> ==Solution 2==<br /> <br /> Simple casework works here as well:<br /> Set up the following ratios:<br /> &lt;cmath&gt;4:0:0&lt;/cmath&gt;<br /> &lt;cmath&gt;3:1:0&lt;/cmath&gt;<br /> &lt;cmath&gt;2:2:0&lt;/cmath&gt;<br /> &lt;cmath&gt;2:1:1&lt;/cmath&gt;<br /> <br /> In three of these cases we see that there are two of the same ratios (so like two boxes would have &lt;math&gt;0&lt;/math&gt;), and so if we swapped those two donuts, we would have the same case. Thus we get &lt;math&gt;\frac{4!}{3!2!}&lt;/math&gt; for those &lt;math&gt;3&lt;/math&gt; (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal &lt;math&gt;\binom{4}{3}=6.&lt;/math&gt; Thus, our answer is &lt;math&gt;3 \cdot 3+6 = \boxed{15}&lt;/math&gt;.<br /> <br /> Solution by IronicNinja<br /> <br /> Edit by virjoy2001 (Reason LaTeX mistake)<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2001|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Beastej https://artofproblemsolving.com/wiki/index.php?title=User:Flamekhoemberish&diff=160828 User:Flamekhoemberish 2021-08-24T01:14:47Z <p>Beastej: /* User Count */</p> <hr /> <div>==User Count==<br /> If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;7&lt;/font&gt;&lt;/center&gt;<br /> <br /> ==Introduction==<br /> Hi Everyone,<br /> <br /> I am Flame Kho. I am currently working in England and quite fluent in both English and Mandarin.<br /> <br /> I have a broad range of interests: mathematics, Cover Chess, classic Chinese songs, travels, etc. I also enjoy creating certificates for chess winners. Visit this page for my interests and artworks. :)<br /> <br /> Below is my upcoming avatar, which I will change two weeks from now.<br /> <br /> [[File:FlameKhoAvatar.jpeg|center|300px]]<br /> ~FlameKhoEmberish<br /> <br /> ==Mathematics==<br /> I am an amateur problem-solver. I am so proud of my contribution to [[2021_AMC_10A_Problems/Problem_25|2021 AMC 10A Problem 25]] (Solution 3).<br /> <br /> ==Cover Chess==<br /> Cover Chess is a variant of Xiangqi (Chinese Chess), where all pieces are covered initially. The game has a considerable amount of luck--even amateur players have a chance of beating masters.<br /> <br /> I managed many Cover Chess tournaments and participated in certificate-making. See my artworks below.<br /> <br /> ==Guess Songs==<br /> I am a fan of Chinese classic music. I recognize a lot of songs. If you want to challenge me on guessing songs, I will always be here! :)<br /> <br /> ==Artworks==<br /> In this section, I will show you my artworks across different areas. I am the original author for all of the artworks below.<br /> <br /> Hope you enjoy them! :)<br /> <br /> ===Certificates===<br /> ====Cover Chess====<br /> &lt;center&gt;[[File:Cover Chess Certificate 1.jpeg|450px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Cover Chess Certificate 2.jpeg|450px]]&lt;/center&gt;&lt;p&gt;<br /> &lt;center&gt;[[File:Cover Chess Certificate 3.jpeg|450px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Cover Chess Certificate 7.jpeg|450px]]&lt;/center&gt;&lt;p&gt;<br /> &lt;center&gt;[[File:Cover Chess Certificate 8.jpeg|center|450px]]&lt;/center&gt;&lt;p&gt;<br /> &lt;center&gt;[[File:Cover Chess Certificate 4.jpeg|300px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Cover Chess Certificate 5.jpeg|300px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Cover Chess Certificate 6.jpeg|300px]]&lt;/center&gt;<br /> <br /> ====Guess Songs====<br /> [[File:GuessSongs.jpeg|center|300px]]<br /> <br /> ===Posters===<br /> ====Valentine Speed Chess Tournament====<br /> &lt;center&gt;[[File:Season 2.jpeg|300px]][[File:Season 3.jpeg|300px]][[File:Season 4.jpeg|300px]]&lt;/center&gt;<br /> &lt;center&gt;[[File:Season 5.jpeg|300px]][[File:Season 6.jpeg|300px]][[File:Season 7.jpeg|300px]]&lt;/center&gt;<br /> &lt;center&gt;[[File:Season 8.jpeg|300px]][[File:Season 9.jpeg|300px]][[File:Season 10.jpeg|300px]]&lt;/center&gt;<br /> <br /> ====Master Challenges====<br /> [[File:Master 1.jpeg|center|600px]]&lt;p&gt;<br /> [[File:Master 2.png|center|600px]]&lt;p&gt;<br /> [[File:Master 3.jpeg|center|600px]]<br /> <br /> ====Other Chess Tournaments====<br /> &lt;center&gt;[[File:Chess Poster 1.jpeg|400px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Chess Poster 2.jpeg|400px]]&lt;/center&gt;&lt;p&gt;<br /> &lt;center&gt;[[File:Chess Poster 3.jpeg|400px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Chess Poster 4.jpeg|400px]]&lt;/center&gt;<br /> <br /> ===Photographs===<br /> [[File:Church.jpeg|center]]&lt;p&gt;<br /> [[File:Outdoors.jpeg|center]]<br /> <br /> ===Drawings===<br /> [[File:Drawing.jpeg|center|300px]]<br /> <br /> ==Cartoons==<br /> [[File:Cartoon.jpeg|center]]<br /> <br /> ==Stickers==<br /> Finally, I am a huge fan of stickers. Below are my favorite ones. As you can tell, I love cats in particular.<br /> &lt;center&gt;[[File:Coming.png|250px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Hahahahaha.png|250px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Bad Words.jpg|250px]]&lt;/center&gt;&lt;p&gt;<br /> &lt;center&gt;[[File:Yes.gif|250px]]&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;[[File:Fighting.gif|250px]]&lt;/center&gt;<br /> <br /> ==Acknowledgements==<br /> I especially thank [[User:MRENTHUSIASM|MRENTHUSIASM]] for creating and editing my user page.</div> Beastej https://artofproblemsolving.com/wiki/index.php?title=User:MRENTHUSIASM&diff=160827 User:MRENTHUSIASM 2021-08-24T01:12:51Z <p>Beastej: /* User Count */</p> <hr /> <div>==User Count==<br /> If this is your first time visiting this page, then edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;45&lt;/font&gt;&lt;/center&gt;<br /> <br /> ==Introduction==<br /> Hi Everyone,<br /> <br /> I am a hedgehog who likes Message Board halping, funny pictures, emoji wars, and AoPS Wiki contributions. Keep these fun things coming!<br /> <br /> This hedgehog represents my attitude towards mathematics--victorious and enthusiastic!<br /> [[File:Victorious-Hedgehog.png|center]]<br /> And here is my personified self-portrait:<br /> [[File:Enthusiastic.gif|center]]<br /> ~MRENTHUSIASM<br /> <br /> ==AoPS Bio==<br /> Jerry graduated from Northeastern University in 2018 as a mathematics and computer science combined major. He has ten years of experience working with students. During his last two years in high school, he served as the math club president and prepared lively lectures on contest math. As a result, he inspired many kids to get interested in math and participate in math competitions (such as AMC and ARML). For the last eight years, he has worked with Math League to write math contests for students from grades 4 through 12 in the USA. Starting in 2019, he has served as an instructor and a Message Board Halper for many AoPS online classes and truly enjoyed that experience! In his mind, nothing is more rewarding than educating tomorrow’s bright minds. In his spare time, he loves contributing to the AoPS Wiki and playing chess and all sorts of board games, especially Scotland Yard and Samurai.<br /> <br /> ==AoPS Wiki Contributions==<br /> Below are my contributions to the AMC/AIME Problems' Solutions in the AoPS Wiki. I understand that in this communal Wiki, we all have to collaborate and make compromises. &lt;i&gt;&lt;b&gt;In any of my solutions, if you find flaws and/or want major revisions, then please contact me via my [[User_talk:MRENTHUSIASM|user talk page]] or the private messaging system before you take action. I am sure that we can work things out.&lt;/b&gt;&lt;/i&gt;<br /> <br /> Thank you for your cooperation, and hope you enjoy reading my solutions. :)<br /> <br /> ~MRENTHUSIASM<br /> <br /> ===AMC 8===<br /> * [[2007_AMC_8_Problems/Problem_8|2007 AMC 8 Problem 8]] (Solutions 1, 2)<br /> * [[2007_AMC_8_Problems/Problem_10|2007 AMC 8 Problem 10]] (Solution)<br /> * [[2017_AMC_8_Problems/Problem_24|2017 AMC 8 Problem 24]] (Solution 1)<br /> * [[2020_AMC_8_Problems/Problem_1|2020 AMC 8 Problem 1]] (Solution 2)<br /> * [[2020_AMC_8_Problems/Problem_3|2020 AMC 8 Problem 3]] (Solution 3)<br /> <br /> ===AMC 10===<br /> * [[2004_AMC_10A_Problems/Problem_24|2004 AMC 10A Problem 24]] (Solutions 1, 2)<br /> * [[2007_AMC_10A_Problems/Problem_20|2007 AMC 10A Problem 20]] (Solutions 1, 2, 3, 4, 5, 6, 7)<br /> * [[2009_AMC_10B_Problems/Problem_1|2009 AMC 10B Problem 1]] (Solutions 1, 2, 3)<br /> * [[2018_AMC_10A_Problems/Problem_1|2018 AMC 10A Problem 1]] (Solution)<br /> * [[2018_AMC_10A_Problems/Problem_5|2018 AMC 10A Problem 5]] (Solution 3)<br /> * [[2018_AMC_10A_Problems/Problem_7|2018 AMC 10A Problem 7]] (Solution 1)<br /> * [[2018_AMC_10A_Problems/Problem_21|2018 AMC 10A Problem 21]] (Remark)<br /> * [[2018_AMC_10A_Problems/Problem_23|2018 AMC 10A Problem 23]] (Solution 1)<br /> * [[2020_AMC_10A_Problems/Problem_3|2020 AMC 10A Problem 3]] (Solution 3)<br /> * [[2020_AMC_10A_Problems/Problem_5|2020 AMC 10A Problem 5]] (Solution 3)<br /> * [[2020_AMC_10A_Problems/Problem_11|2020 AMC 10A Problem 11]] (Solution 5)<br /> * [[2020_AMC_10A_Problems/Problem_17|2020 AMC 10A Problem 17]] (Solution 4)<br /> * [[2020_AMC_10A_Problems/Problem_24|2020 AMC 10A Problem 24]] (Solution 11)<br /> * [[2020_AMC_10A_Problems/Problem_25|2020 AMC 10A Problem 25]] (Solution 2)<br /> * [[2020_AMC_10B_Problems/Problem_3|2020 AMC 10B Problem 3]] (Solution 4)<br /> * [[2020_AMC_10B_Problems/Problem_4|2020 AMC 10B Problem 4]] (Solution 3)<br /> * [[2020_AMC_10B_Problems/Problem_8|2020 AMC 10B Problem 8]] (Solution 4)<br /> * [[2020_AMC_10B_Problems/Problem_13|2020 AMC 10B Problem 13]] (Solution 2)<br /> * [[2020_AMC_10B_Problems/Problem_15|2020 AMC 10B Problem 15]] (Solution 3)<br /> * [[2020_AMC_10B_Problems/Problem_23|2020 AMC 10B Problem 23]] (Solution 5)<br /> * [[2021_AMC_10A_Problems/Problem_1|2021 AMC 10A Problem 1]] (Solution 2)<br /> * [[2021_AMC_10A_Problems/Problem_2|2021 AMC 10A Problem 2]] (Solutions 2, 3, 4)<br /> * [[2021_AMC_10A_Problems/Problem_3|2021 AMC 10A Problem 3]] (Solution 3)<br /> * [[2021_AMC_10A_Problems/Problem_4|2021 AMC 10A Problem 4]] (Solutions 1, 2)<br /> * [[2021_AMC_10A_Problems/Problem_5|2021 AMC 10A Problem 5]] (Solutions 1, 2, 3)<br /> * [[2021_AMC_10A_Problems/Problem_6|2021 AMC 10A Problem 6]] (Solutions 1, 2)<br /> * [[2021_AMC_10A_Problems/Problem_7|2021 AMC 10A Problem 7]] (Solutions 2, 4)<br /> * [[2021_AMC_10A_Problems/Problem_8|2021 AMC 10A Problem 8]] (Solutions 1, 2, 3)<br /> * [[2021_AMC_10A_Problems/Problem_9|2021 AMC 10A Problem 9]] (Solution 3)<br /> * [[2021_AMC_10A_Problems/Problem_10|2021 AMC 10A Problem 10]] (Solution 7)<br /> * [[2021_AMC_10A_Problems/Problem_11|2021 AMC 10A Problem 11]] (Solutions 1, 3)<br /> * [[2021_AMC_10A_Problems/Problem_12|2021 AMC 10A Problem 12]] (Solution 1)<br /> * [[2021_AMC_10A_Problems/Problem_13|2021 AMC 10A Problem 13]] (Solutions 1, 2, 3)<br /> * [[2021_AMC_10A_Problems/Problem_17|2021 AMC 10A Problem 17]] (Diagram, Solution 2)<br /> * [[2021_AMC_10A_Problems/Problem_18|2021 AMC 10A Problem 18]] (Solutions 1, 4)<br /> * [[2021_AMC_10A_Problems/Problem_20|2021 AMC 10A Problem 20]] (Solutions 2, 3)<br /> * [[2021_AMC_10A_Problems/Problem_21|2021 AMC 10A Problem 21]] (Diagram, Solution, Video Solution)<br /> * [[2021_AMC_10A_Problems/Problem_22|2021 AMC 10A Problem 22]] (Solution 2)<br /> * [[2021_AMC_10A_Problems/Problem_23|2021 AMC 10A Problem 23]] (Solution 3)<br /> * [[2021_AMC_10A_Problems/Problem_24|2021 AMC 10A Problem 24]] (Diagram, Solutions 1, 2, 3)<br /> * [[2021_AMC_10A_Problems/Problem_25|2021 AMC 10A Problem 25]] (Solutions 2, 3)<br /> * [[2021_AMC_10B_Problems/Problem_6|2021 AMC 10B Problem 6]] (Solution 3)<br /> * [[2021_AMC_10B_Problems/Problem_7|2021 AMC 10B Problem 7]] (Solution 2)<br /> * [[2021_AMC_10B_Problems/Problem_8|2021 AMC 10B Problem 8]] (Solution 3)<br /> * [[2021_AMC_10B_Problems/Problem_16|2021 AMC 10B Problem 16]] (Solution 4)<br /> * [[2021_AMC_10B_Problems/Problem_17|2021 AMC 10B Problem 17]] (Solution 3)<br /> <br /> ===AMC 12===<br /> * [[2004_AMC_12A_Problems/Problem_16|2004 AMC 12A Problem 16]] (Solution)<br /> * [[2004_AMC_12A_Problems/Problem_17|2004 AMC 12A Problem 17]]: same as [[2004_AMC_10A_Problems/Problem_24|2004 AMC 10A Problem 24]] (Solutions 1, 2)<br /> * [[2009_AMC_12B_Problems/Problem_1|2009 AMC 12B Problem 1]]: same as [[2009_AMC_10B_Problems/Problem_1|2009 AMC 10B Problem 1]] (Solutions 1, 2, 3)<br /> * [[2014_AMC_12A_Problems/Problem_18|2014 AMC 12A Problem 18]] (Solution 1)<br /> * [[2018_AMC_12A_Problems/Problem_2|2018 AMC 12A Problem 2]] (Solutions 1, 2)<br /> * [[2018_AMC_12A_Problems/Problem 4|2018 AMC 12A Problem 4]]: same as [[2018_AMC_10A_Problems/Problem_5|2018 AMC 10A Problem 5]] (Solution 3)<br /> * [[2018_AMC_12A_Problems/Problem_7|2018 AMC 12A Problem 7]]: same as [[2018_AMC_10A_Problems/Problem_7|2018 AMC 10A Problem 7]] (Solution 1)<br /> * [[2018_AMC_12A_Problems/Problem_14|2018 AMC 12A Problem 14]] (Solutions 1, 2, 3, 4, 5)<br /> * [[2018_AMC_12A_Problems/Problem_16|2018 AMC 12A Problem 16]]: same as [[2018_AMC_10A_Problems/Problem_21|2018 AMC 10A Problem 21]] (Remark)<br /> * [[2018_AMC_12A_Problems/Problem_17|2018 AMC 12A Problem 17]]: same as [[2018_AMC_10A_Problems/Problem_23|2018 AMC 10A Problem 23]] (Solution 1)<br /> * [[2018_AMC_12A_Problems/Problem_19|2018 AMC 12A Problem 19]] (Solution 1)<br /> * [[2018_AMC_12A_Problems/Problem_21|2018 AMC 12A Problem 21]] (Solution 1)<br /> * [[2018_AMC_12A_Problems/Problem_23|2018 AMC 12A Problem 23]] (Diagram, Solution 2)<br /> * [[2018_AMC_12A_Problems/Problem_24|2018 AMC 12A Problem 24]] (Solutions 1, 2, 3)<br /> * [[2020_AMC_12A_Problems/Problem_1|2020 AMC 12A Problem 1]] (Solution 3)<br /> * [[2020_AMC_12A_Problems/Problem_6|2020 AMC 12A Problem 6]] (Solution 2)<br /> * [[2020_AMC_12A_Problems/Problem_8|2020 AMC 12A Problem 8]]: same as [[2020_AMC_10A_Problems/Problem_11|2020 AMC 10A Problem 11]] (Solution 5)<br /> * [[2020_AMC_12A_Problems/Problem_9|2020 AMC 12A Problem 9]] (Remark)<br /> * [[2020_AMC_12A_Problems/Problem_10|2020 AMC 12A Problem 10]] (Solutions 1, 2, 5)<br /> * [[2020_AMC_12A_Problems/Problem_15|2020 AMC 12A Problem 15]] (Remark)<br /> * [[2020_AMC_12A_Problems/Problem_23|2020 AMC 12A Problem 23]]: same as [[2020_AMC_10A_Problems/Problem_25|2020 AMC 10A Problem 25]] (Solution 2)<br /> * [[2020_AMC_12A_Problems/Problem_24|2020 AMC 12A Problem 24]] (Remark)<br /> * [[2020_AMC_12A_Problems/Problem_25|2020 AMC 12A Problem 25]] (Solution 1, Remark)<br /> * [[2020_AMC_12B_Problems/Problem_3|2020 AMC 12B Problem 3]]: same as [[2020_AMC_10B_Problems/Problem_3|2020 AMC 10B Problem 3]] (Solution 4)<br /> * [[2020_AMC_12B_Problems/Problem_4|2020 AMC 12B Problem 4]]: same as [[2020_AMC_10B_Problems/Problem_4|2020 AMC 10B Problem 4]] (Solution 3)<br /> * [[2020_AMC_12B_Problems/Problem_5|2020 AMC 12B Problem 5]] (Solution 1)<br /> * [[2020_AMC_12B_Problems/Problem_6|2020 AMC 12B Problem 6]] (Solution 2)<br /> * [[2020_AMC_12B_Problems/Problem_10|2020 AMC 12B Problem 10]] (Diagram, Solution 7)<br /> * [[2020_AMC_12B_Problems/Problem_12|2020 AMC 12B Problem 12]] (Diagram)<br /> * [[2020_AMC_12B_Problems/Problem_13|2020 AMC 12B Problem 13]] (Solutions 1, 4)<br /> * [[2020_AMC_12B_Problems/Problem_19|2020 AMC 12B Problem 19]]: same as [[2020_AMC_10B_Problems/Problem_23|2020 AMC 10B Problem 23]] (Solution 5)<br /> * [[2021_AMC_12A_Problems/Problem_2|2021 AMC 12A Problem 2]] (Solutions 2, 3)<br /> * [[2021_AMC_12A_Problems/Problem_3|2021 AMC 12A Problem 3]]: same as [[2021_AMC_10A_Problems/Problem_3|2021 AMC 10A Problem 3]] (Solution 3)<br /> * [[2021_AMC_12A_Problems/Problem_4|2021 AMC 12A Problem 4]]: same as [[2021_AMC_10A_Problems/Problem_7|2021 AMC 10A Problem 7]] (Solutions 2, 4)<br /> * [[2021_AMC_12A_Problems/Problem_5|2021 AMC 12A Problem 5]]: same as [[2021_AMC_10A_Problems/Problem_8|2021 AMC 10A Problem 8]] (Solutions 1, 2, 3)<br /> * [[2021_AMC_12A_Problems/Problem_6|2021 AMC 12A Problem 6]] (Solutions 2, 3)<br /> * [[2021_AMC_12A_Problems/Problem_7|2021 AMC 12A Problem 7]]: same as [[2021_AMC_10A_Problems/Problem_9|2021 AMC 10A Problem 9]] (Solution 3)<br /> * [[2021_AMC_12A_Problems/Problem_9|2021 AMC 12A Problem 9]]: same as [[2021_AMC_10A_Problems/Problem_10|2021 AMC 10A Problem 10]] (Solution 7)<br /> * [[2021_AMC_12A_Problems/Problem_10|2021 AMC 12A Problem 10]]: same as [[2021_AMC_10A_Problems/Problem_12|2021 AMC 10A Problem 12]] (Solution 1)<br /> * [[2021_AMC_12A_Problems/Problem_11|2021 AMC 12A Problem 11]] (Solutions 2, 3, 4)<br /> * [[2021_AMC_12A_Problems/Problem_13|2021 AMC 12A Problem 13]] (Solutions 2, 3)<br /> * [[2021_AMC_12A_Problems/Problem_14|2021 AMC 12A Problem 14]] (Solutions 2, 3, 4)<br /> * [[2021_AMC_12A_Problems/Problem_15|2021 AMC 12A Problem 15]] (Solutions 3, 4)<br /> * [[2021_AMC_12A_Problems/Problem_17|2021 AMC 12A Problem 17]]: same as [[2021_AMC_10A_Problems/Problem_17|2021 AMC 10A Problem 17]] (Diagram, Solution 2)<br /> * [[2021_AMC_12A_Problems/Problem_18|2021 AMC 12A Problem 18]]: same as [[2021_AMC_10A_Problems/Problem_18|2021 AMC 10A Problem 18]] (Solutions 1, 4)<br /> * [[2021_AMC_12A_Problems/Problem_19|2021 AMC 12A Problem 19]] (Solutions 2, 3, Remark)<br /> * [[2021_AMC_12A_Problems/Problem_21|2021 AMC 12A Problem 21]] (Solution 2, Remark)<br /> * [[2021_AMC_12A_Problems/Problem_22|2021 AMC 12A Problem 22]] (Solution 4)<br /> * [[2021_AMC_12A_Problems/Problem_23|2021 AMC 12A Problem 23]]: same as [[2021_AMC_10A_Problems/Problem_23|2021 AMC 10A Problem 23]] (Solution 3)<br /> * [[2021_AMC_12A_Problems/Problem_24|2021 AMC 12A Problem 24]] (Solution 1)<br /> * [[2021_AMC_12A_Problems/Problem_25|2021 AMC 12A Problem 25]] (Solution 1)<br /> * [[2021_AMC_12B_Problems/Problem_4|2021 AMC 12B Problem 4]]: same as [[2021_AMC_10B_Problems/Problem_6|2021 AMC 10B Problem 6]] (Solution 3)<br /> * [[2021_AMC_12B_Problems/Problem_11|2021 AMC 12B Problem 11]] (Solution 4)<br /> * [[2021_AMC_12B_Problems/Problem_14|2021 AMC 12B Problem 14]] (Solution 3)<br /> * [[2021_AMC_12B_Problems/Problem_20|2021 AMC 12B Problem 20]] (Solution 4)<br /> * [[2021_AMC_12B_Problems/Problem_21|2021 AMC 12B Problem 21]] (Solution 3)<br /> * [[2021_AMC_12B_Problems/Problem_23|2021 AMC 12B Problem 23]] (Solution 4)<br /> <br /> ===AIME===<br /> * [[1984_AIME_Problems/Problem_4|1984 AIME Problem 4]] (Solutions 1, 2)<br /> * [[1984_AIME_Problems/Problem_10|1984 AIME Problem 10]] (Solution 3)<br /> * [[1985_AIME_Problems/Problem_12|1985 AIME Problem 12]] (Solutions 1, 2, 3)<br /> * [[1986_AIME_Problems/Problem_2|1986 AIME Problem 2]] (Solutions 1, 2)<br /> * [[1987_AIME_Problems/Problem_14|1987 AIME Problem 14]] (Solutions 2, 3)<br /> * [[1988_AIME_Problems/Problem_8|1988 AIME Problem 8]] (Solution 1)<br /> * [[1988_AIME_Problems/Problem_13|1988 AIME Problem 13]] (Solution 6)<br /> * [[1989_AIME_Problems/Problem_1|1989 AIME Problem 1]] (Solutions 4, 5)<br /> * [[1989_AIME_Problems/Problem_8|1989 AIME Problem 8]] (Solutions 1, 2, 3)<br /> * [[1989_AIME_Problems/Problem_9|1989 AIME Problem 9]] (Solutions 1, 2, 4)<br /> * [[1990_AIME_Problems/Problem_8|1990 AIME Problem 8]] (Solution, Remark)<br /> * [[1992_AIME_Problems/Problem_6|1992 AIME Problem 6]] (Solution 1)<br /> * [[1993_AIME_Problems/Problem_7|1993 AIME Problem 7]] (Solution 4)<br /> * [[2019_AIME_II_Problems/Problem_7|2019 AIME II Problem 7]] (Diagram)<br /> * [[2020_AIME_I_Problems/Problem_5|2020 AIME I Problem 5]] (Solution 9)<br /> * [[2021_AIME_I_Problems/Problem_1|2021 AIME I Problem 1]] (Solution 1)<br /> * [[2021_AIME_I_Problems/Problem_2|2021 AIME I Problem 2]] (Solution 2)<br /> * [[2021_AIME_I_Problems/Problem_3|2021 AIME I Problem 3]] (Solution 3)<br /> * [[2021_AIME_I_Problems/Problem_7|2021 AIME I Problem 7]] (Remark)<br /> * [[2021_AIME_I_Problems/Problem_8|2021 AIME I Problem 8]] (Solution 3, Remark)<br /> * [[2021_AIME_I_Problems/Problem_9|2021 AIME I Problem 9]] (Diagram, Solution 6)<br /> * [[2021_AIME_I_Problems/Problem_10|2021 AIME I Problem 10]] (Solution 2)<br /> * [[2021_AIME_I_Problems/Problem_11|2021 AIME I Problem 11]] (Solution 4)<br /> * [[2021_AIME_I_Problems/Problem_12|2021 AIME I Problem 12]] (Solution)<br /> * [[2021_AIME_I_Problems/Problem_14|2021 AIME I Problem 14]] (Solutions 2, 3)<br /> * [[2021_AIME_I_Problems/Problem_15|2021 AIME I Problem 15]] (Diagram, Solution 1)<br /> * [[2021_AIME_II_Problems/Problem_1|2021 AIME II Problem 1]] (Solution 3, Remark)<br /> * [[2021_AIME_II_Problems/Problem_2|2021 AIME II Problem 2]] (Solution 1)<br /> * [[2021_AIME_II_Problems/Problem_3|2021 AIME II Problem 3]] (Solution 2)<br /> * [[2021_AIME_II_Problems/Problem_4|2021 AIME II Problem 4]] (Solution 1)<br /> * [[2021_AIME_II_Problems/Problem_5|2021 AIME II Problem 5]] (Solutions 2, 5)<br /> * [[2021_AIME_II_Problems/Problem_6|2021 AIME II Problem 6]] (Solution 3)<br /> * [[2021_AIME_II_Problems/Problem_7|2021 AIME II Problem 7]] (Solution 4)<br /> * [[2021_AIME_II_Problems/Problem_8|2021 AIME II Problem 8]] (Solution 1)<br /> * [[2021_AIME_II_Problems/Problem_9|2021 AIME II Problem 9]] (Solution 2)<br /> * [[2021_AIME_II_Problems/Problem_10|2021 AIME II Problem 10]] (Diagram, Solution 3)<br /> * [[2021_AIME_II_Problems/Problem_11|2021 AIME II Problem 11]] (Solution 2)<br /> * [[2021_AIME_II_Problems/Problem_12|2021 AIME II Problem 12]] (Diagram, Solution 2)<br /> * [[2021_AIME_II_Problems/Problem_13|2021 AIME II Problem 13]] (Solutions 1, 3)<br /> * [[2021_AIME_II_Problems/Problem_14|2021 AIME II Problem 14]] (Diagram, Solution 1)<br /> * [[2021_AIME_II_Problems/Problem_15|2021 AIME II Problem 15]] (Solution 2)<br /> <br /> ==External Links==<br /> * [[User_talk:MRENTHUSIASM|User Talk Page]] &lt;br&gt; Welcome to my user talk page here. Feel free to leave any message you like.<br /> * [[User:Flamekhoemberish|Flame Kho's User Page]] &lt;br&gt; Recently I am helping my friend Flame Kho to create his user page. Enjoy his AMC Solutions and artworks! &lt;math&gt;\smiley{}&lt;/math&gt;</div> Beastej https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=143254 2020 AMC 8 Problems 2021-01-25T21:28:12Z <p>Beastej: /* Problem 14 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/em 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive? <br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;. (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1&lt;/math&gt;.) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by B?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;<br /> , the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Beastej https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=143253 2020 AMC 8 Problems 2021-01-25T21:27:44Z <p>Beastej: /* Problem 14 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/em 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive? <br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;. (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1&lt;/math&gt;.) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities??<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by B?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;<br /> , the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Beastej https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=135404 Simon's Favorite Factoring Trick 2020-10-19T20:27:05Z <p>Beastej: /* The General Statement */</p> <hr /> <div><br /> ==The General Statement==<br /> Simon's Favorite Factoring Trick (SFFT) is often used in a diophantine equation where factoring is needed. The most common form it appears is when there is a constant on one side of the equation and a product of variables with each of those variables in a linear term on the other side. A simple example would be: &lt;cmath&gt;xy+66x-88y=23333&lt;/cmath&gt;where &lt;math&gt;23333&lt;/math&gt; is the constant term, &lt;math&gt;xy&lt;/math&gt; is the product of the variables, &lt;math&gt;66x&lt;/math&gt; and &lt;math&gt;-88y&lt;/math&gt; are the variables in linear terms.<br /> <br /> <br /> Let's put it in general terms. We have an equation &lt;math&gt;xy+jx+ky=a&lt;/math&gt;, where &lt;math&gt;j&lt;/math&gt;, &lt;math&gt;k&lt;/math&gt;, and &lt;math&gt;a&lt;/math&gt; are integral constants. According to Simon's Favorite Factoring Trick, this equation can be transformed into: &lt;cmath&gt;(x+k)(y+j)=a+jk&lt;/cmath&gt; <br /> Using the previous example, &lt;math&gt;xy+66x-88y=23333&lt;/math&gt; is the same as: &lt;cmath&gt;(x-88)(y+66)=(23333)+(-88)(66)&lt;/cmath&gt;<br /> <br /> <br /> If this is confusing or you would like to know the thought process behind SFFT, see this eight-minute video by Richard Rusczyk from AoPS: https://www.youtube.com/watch?v=0nN3H7w2LnI. For the thought process, start from https://youtu.be/0nN3H7w2LnI?t=366<br /> <br /> == Applications ==<br /> This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are variables and &lt;math&gt;j,k&lt;/math&gt; are known constants. Also, it is typically necessary to add the &lt;math&gt;jk&lt;/math&gt; term to both sides to perform the factorization.<br /> <br /> == Fun Practice Problems ==<br /> ===Introductory===<br /> *Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ([[2000 AMC 12/Problem 6|Source]])<br /> <br /> ===Intermediate===<br /> *&lt;math&gt;m, n&lt;/math&gt; are integers such that &lt;math&gt;m^2 + 3m^2n^2 = 30n^2 + 517&lt;/math&gt;. Find &lt;math&gt;3m^2n^2&lt;/math&gt;.<br /> <br /> ([[1987 AIME Problems/Problem 5|Source]])<br /> <br /> ===Olympiad===<br /> <br /> *The integer &lt;math&gt;N&lt;/math&gt; is positive. There are exactly 2005 ordered pairs &lt;math&gt;(x, y)&lt;/math&gt; of positive integers satisfying:<br /> <br /> &lt;cmath&gt;\frac 1x +\frac 1y = \frac 1N&lt;/cmath&gt;<br /> <br /> Prove that &lt;math&gt;N&lt;/math&gt; is a perfect square. <br /> <br /> Source: (British Mathematical Olympiad Round 3, 2005)<br /> <br /> == See More==<br /> * [[Algebra]]<br /> * [[Factoring]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Theorems]]</div> Beastej https://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=133782 User:Piphi 2020-09-18T01:41:21Z <p>Beastej: /* User Count */</p> <hr /> <div>{{User:Piphi/Template:Header}}<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#dddddd&quot;&gt;<br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;100px&quot;&gt;325&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;Piphi is legendary and made the USA IMO team in 2019.&lt;br&gt;<br /> <br /> Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.&lt;br&gt;<br /> <br /> Piphi started the signature trend at around May 2020.&lt;br&gt;<br /> <br /> Piphi is an extremely OP person - LJCoder619. &lt;br&gt;<br /> <br /> Piphi is OP --[[User:Aray10|Aray10]] ([[User talk:Aray10|talk]]) 23:22, 17 June 2020 (EDT) &lt;br&gt;<br /> <br /> Piphi is 100 percent OP compared to the rest of us --[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] ([[User talk:Aops-g5-gethsemanea2|talk]]) 09:24, 3 September 2020 (EDT) &lt;br&gt;<br /> <br /> Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].&lt;br&gt;<br /> <br /> Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].&lt;br&gt;<br /> <br /> Piphi has been called op by many AoPSers, including the legendary [[User:Radio2|Radio2]] himself [https://artofproblemsolving.com/community/c19451h1826745p15526800 here]. (note: Radio2 calls many users op.)&lt;br&gt;<br /> <br /> Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.&lt;br&gt;<br /> <br /> Piphi has also added a lot of the info that is in the [[Reaper Archives]].&lt;br&gt;<br /> <br /> Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].&lt;br&gt;<br /> <br /> Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br /> <br /> Piphi has a post that was made an announcement on a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br /> <br /> Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;<br /> You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br /> <br /> A User Count of 330<br /> {{User:Piphi/Template:Progress_Bar|96.97|width=100%}}<br /> <br /> 200 subpages of [[User:Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|61.5|width=100%}}<br /> <br /> 200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br /> {{User:Piphi/Template:Progress_Bar|42|width=100%}}<br /> <br /> Make 10,000 edits<br /> {{User:Piphi/Template:Progress_Bar|18.8|width=100%}}&lt;/div&gt;<br /> &lt;/div&gt;</div> Beastej