https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Elbertpark&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-14T05:53:33Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Slalom_conjuncture&diff=143319 Slalom conjuncture 2021-01-26T18:26:51Z <p>Elbertpark: </p> <hr /> <div>&lt;h1&gt;The Slalom Conjuncture&lt;/h1&gt;<br /> &lt;h2&gt;As discovered by Elbertpark&lt;/h2&gt;<br /> &lt;h3&gt;Written by Elbertpark&lt;/h3&gt;<br /> &lt;h4&gt;Idea made by Elbertpark...&lt;/h4&gt;<br /> &lt;h5&gt;and so on&lt;/h5&gt;<br /> <br /> &lt;h1&gt;What IS the Slalom Conjuncture?&lt;/h1&gt;<br /> &lt;p&gt;The Slalom Conjuncture was discovered during a math assignment. It states that if there is an odd square &lt;math&gt;n^2&lt;/math&gt;, then this square has a maximum of &lt;math&gt;n^2 - 2n&lt;/math&gt; factors starting from 3.&lt;/p&gt;<br /> <br /> Listed is a table of squares and factors up to 11.<br /> &lt;table style=&quot;width:17%&quot;&gt;<br /> &lt;tr&gt;<br /> &lt;th&gt;Number&lt;/th&gt;<br /> &lt;th&gt;&lt;math&gt;n^2&lt;/math&gt;&lt;/th&gt;<br /> &lt;th&gt;# of factors&lt;/th&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;td&gt;25&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;7&lt;/td&gt;<br /> &lt;td&gt;49&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;11&lt;/td&gt;<br /> &lt;td&gt;121&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;6561&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;4001&lt;/td&gt;<br /> &lt;td&gt;16008001&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;/table&gt;<br /> Notice that most of the squares, even 4001, have only 3 factors.<br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Unfortunately, only Doggo and Gmaas have the logical, solid proof to this conjuncture. That is why this is a conjuncture.<br /> &lt;h2&gt;Broken proof&lt;/h2&gt;<br /> For now we can agree that because soon the squares will be growing exponentially, this conjuncture cannot be wrong... yet.<br /> <br /> &lt;code&gt; This article is a stub. Help us by expanding it.&lt;code&gt;</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=Slalom_conjuncture&diff=142941 Slalom conjuncture 2021-01-21T23:02:41Z <p>Elbertpark: </p> <hr /> <div>&lt;h1&gt;The Slalom Conjuncture&lt;/h1&gt;<br /> &lt;h2&gt;As discovered by Elbertpark&lt;/h2&gt;<br /> &lt;h3&gt;Written by Elbertpark&lt;/h3&gt;<br /> &lt;h4&gt;Idea made by Elbertpark...&lt;/h4&gt;<br /> &lt;h5&gt;and so on&lt;/h5&gt;<br /> <br /> &lt;h1&gt;What IS the Slalom Conjuncture?&lt;/h1&gt;<br /> &lt;p&gt;The Slalom Conjuncture was discovered during a math assignment. It states that if there is an odd square &lt;math&gt;n^2&lt;/math&gt;, then this square has a maximum of &lt;math&gt;n^2 - 2n&lt;/math&gt; factors starting from 3.&lt;/p&gt;<br /> <br /> Listed is a table of squares and factors up to 11.<br /> &lt;table style=&quot;width:17%&quot;&gt;<br /> &lt;tr&gt;<br /> &lt;th&gt;Number&lt;/th&gt;<br /> &lt;th&gt;&lt;math&gt;n^2&lt;/math&gt;&lt;/th&gt;<br /> &lt;th&gt;# of factors&lt;/th&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;td&gt;25&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;7&lt;/td&gt;<br /> &lt;td&gt;49&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;11&lt;/td&gt;<br /> &lt;td&gt;121&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;6561&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;4001&lt;/td&gt;<br /> &lt;td&gt;16008001&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;/table&gt;<br /> Note that most of the squares, even 4001, have only 3 factors.<br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Unfortunately, only Doggo and Gmaas have the logical, solid proof to this conjuncture. That is why this is a conjuncture.<br /> &lt;h2&gt;Broken proof&lt;/h2&gt;<br /> For now we can agree that because soon the squares will be growing exponentially, this conjuncture cannot be wrong... yet.<br /> <br /> &lt;code&gt; This article is a stub. Help us by expanding it.&lt;code&gt;</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=Slalom_conjuncture&diff=142938 Slalom conjuncture 2021-01-21T22:39:21Z <p>Elbertpark: </p> <hr /> <div>&lt;h1&gt;The Slalom Conjuncture&lt;/h1&gt;<br /> &lt;h2&gt;As discovered by Elbertpark&lt;/h2&gt;<br /> &lt;h3&gt;Written by Elbertpark&lt;/h3&gt;<br /> &lt;h4&gt;Idea made by Elbertpark...&lt;/h4&gt;<br /> &lt;h5&gt;and so on&lt;/h5&gt;<br /> <br /> &lt;h1&gt;What IS the Slalom Conjuncture?&lt;/h1&gt;<br /> &lt;p&gt;The Slalom Conjuncture was discovered during a math assignment. It states that if there is an odd square &lt;math&gt;n^2&lt;/math&gt;, then this square has a maximum of &lt;math&gt;n^2 - 2n&lt;/math&gt; factors starting from 3.&lt;/p&gt;<br /> <br /> Listed is a table of squares and factors up to 11.<br /> &lt;table style=&quot;width:17%&quot;&gt;<br /> &lt;tr&gt;<br /> &lt;th&gt;Number&lt;/th&gt;<br /> &lt;th&gt;&lt;math&gt;n^2&lt;/math&gt;&lt;/th&gt;<br /> &lt;th&gt;# of factors&lt;/th&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;td&gt;25&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;7&lt;/td&gt;<br /> &lt;td&gt;49&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;11&lt;/td&gt;<br /> &lt;td&gt;121&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;6561&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;4001&lt;/td&gt;<br /> &lt;td&gt;16008001&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;/table&gt;<br /> Note that most of the squares, even 4001, have only 3 factors.<br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Unfortunately, only Doggo and Gmaas have the logical, solid proof to this conjuncture. That is why this is a conjuncture.<br /> &lt;h2&gt;Broken proof&lt;/h2&gt;<br /> For now we can agree that because soon the squares will be growing exponentially, this conjuncture cannot be wrong... yet.</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=Slalom_conjuncture&diff=142937 Slalom conjuncture 2021-01-21T22:37:06Z <p>Elbertpark: Created page with &quot;&lt;h1&gt;The Slalom Conjuncture&lt;/h1&gt; &lt;h2&gt;As discovered by Elbertpark&lt;/h2&gt; &lt;h3&gt;Written by Elbertpark&lt;/h3&gt; &lt;h4&gt;Idea made by Elbertpark...&lt;/h4&gt; &lt;h5&gt;and so on&lt;/h5&gt; &lt;h1&gt;What IS the Sla...&quot;</p> <hr /> <div>&lt;h1&gt;The Slalom Conjuncture&lt;/h1&gt;<br /> &lt;h2&gt;As discovered by Elbertpark&lt;/h2&gt;<br /> &lt;h3&gt;Written by Elbertpark&lt;/h3&gt;<br /> &lt;h4&gt;Idea made by Elbertpark...&lt;/h4&gt;<br /> &lt;h5&gt;and so on&lt;/h5&gt;<br /> <br /> &lt;h1&gt;What IS the Slalom Conjuncture?&lt;/h1&gt;<br /> &lt;p&gt;The Slalom Conjuncture was discovered during a math assignment. It states that if there is an odd square &lt;math&gt;n^2&lt;/math&gt;, then this square has a maximum of &lt;math&gt;n^2 - 2n&lt;/math&gt; factors.&lt;/p&gt;<br /> <br /> Listed is a table of squares and factors up to 11.<br /> &lt;table style=&quot;width:17%&quot;&gt;<br /> &lt;tr&gt;<br /> &lt;th&gt;Number&lt;/th&gt;<br /> &lt;th&gt;&lt;math&gt;n^2&lt;/math&gt;&lt;/th&gt;<br /> &lt;th&gt;# of factors&lt;/th&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;td&gt;1&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;td&gt;25&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;7&lt;/td&gt;<br /> &lt;td&gt;49&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;5&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;11&lt;/td&gt;<br /> &lt;td&gt;121&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;td&gt;...&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;81&lt;/td&gt;<br /> &lt;td&gt;6561&lt;/td&gt;<br /> &lt;td&gt;9&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;tr&gt;<br /> &lt;td&gt;4001&lt;/td&gt;<br /> &lt;td&gt;16008001&lt;/td&gt;<br /> &lt;td&gt;3&lt;/td&gt;<br /> &lt;/tr&gt;<br /> &lt;/table&gt;<br /> Note that most of the squares, even 4001, have only 3 factors.<br /> &lt;h1&gt;Proof&lt;/h1&gt;<br /> Unfortunately, only Doggo and Gmaas have the logical, solid proof to this conjuncture. That is why this is a conjuncture.<br /> &lt;h2&gt;Broken proof&lt;/h2&gt;<br /> For now we can agree that because soon the squares will be growing exponentially, this conjuncture cannot be wrong... yet.</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=User:Rusczyk&diff=142936 User:Rusczyk 2021-01-21T22:07:34Z <p>Elbertpark: /* User Count */</p> <hr /> <div>Rusczyk's Page:<br /> &lt;br&gt;<br /> __NOTOC__&lt;div style=&quot;border:2px solid black; -webkit-border-radius: 10px; background:#4EC284&quot;&gt;<br /> ==&lt;font color=&quot;white&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;white&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;&lt;font color=&quot;white&quot;&gt;<br /> &lt;center&gt;&lt;font size=&quot;101px&quot;&gt;80&lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> <br /> I can edit<br /> <br /> ==&lt;font color=&quot;white&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;white&quot;&gt;Rusczyk is currently borderline AIME.&lt;br&gt;<br /> <br /> Rusczyk just turned 13 years old.&lt;br&gt;<br /> <br /> Rusczyk scored 44/46 when mocking the 2018 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State held on AoPS.&lt;br&gt;<br /> <br /> Rusczyk is a pro at maths and physics<br /> <br /> Rusczyk has come world and country #1 in various international tournaments and competitions starting from 2017<br /> <br /> I am better than Rusczyk at math<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#4EC284;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;white&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;&lt;font color=&quot;white&quot;&gt; A User Count of &lt;math&gt;\color{white}{\infty}&lt;/math&gt;<br /> <br /> Make AIME 2021 (Currently borderline)<br /> <br /> Pass AP Calculus AB, BC and AP Physics exam<br /> <br /> Get in the Alcumus HoF in the next 6 months<br /> <br /> Convince OlympusHero that he is better than Rusczyk (at Math)<br /> <br /> Get &lt;math&gt;\color{white}{2 \times}&lt;/math&gt; medals this year as compared to what they did last year. That is &lt;math&gt;\color{white}{2 \times 14 = \boxed{28}}&lt;/math&gt; which is nearly impossible.<br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=Doggo&diff=142929 Doggo 2021-01-21T18:57:03Z <p>Elbertpark: Created page with &quot;Doggo is Gmaas' rival. He is &lt;math&gt;legendary^{9999999999999999999999999999999999999}&lt;/math&gt;. Doggo is the only one to have defeated Gmaas, in a 0-second game of chess. Doggo...&quot;</p> <hr /> <div>Doggo is Gmaas' rival. He is &lt;math&gt;legendary^{9999999999999999999999999999999999999}&lt;/math&gt;. Doggo is the only one to have defeated Gmaas, in a 0-second game of chess.<br /> <br /> Doggo spawned from an especially tasty cookie (that Gmaas accidentally made) in a mature status.<br /> <br /> FACTS:<br /> 1. Doggo is the god of chess. He has never lost</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=136749 2017 AMC 8 Problems/Problem 5 2020-11-06T17:43:15Z <p>Elbertpark: /* Solution 3 */</p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> <br /> Directly calculating:<br /> <br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> It is well known that the sum of all numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt;. Therefore, the denominator is equal to &lt;math&gt;\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6&lt;/math&gt;. Now we can cancel the factors of &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; from both the numerator and denominator, only leaving &lt;math&gt;8 \cdot 7 \cdot 5 \cdot 4 \cdot 1&lt;/math&gt;. This evaluates to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> First, we evaluate &lt;math&gt;1 + 2 + 3 + 4 + 5 + 6 + 7 + 8&lt;/math&gt;, to get 36. We notice that 36 is 6 squared, so we can factor the denominator like &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}&lt;/math&gt; then cancel the 6s out, to get &lt;math&gt;\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}&lt;/math&gt;. Now that we have escaped fraction form, multiplying &lt;math&gt;4 \cdot 5 \cdot 7 \cdot 8 \cdot&lt;/math&gt;. Multiplying these, we get &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/cY4NYSAD0vQ<br /> <br /> ~[[User:icematrix2|icematrix2]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136696 2017 AMC 8 Problems/Problem 9 2020-11-06T06:02:00Z <p>Elbertpark: </p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles, just like as in the previous solution. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136695 2017 AMC 8 Problems/Problem 9 2020-11-06T06:01:29Z <p>Elbertpark: </p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136694 2017 AMC 8 Problems/Problem 9 2020-11-06T06:00:59Z <p>Elbertpark: </p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;1 - \(frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136693 2017 AMC 8 Problems/Problem 9 2020-11-06T06:00:29Z <p>Elbertpark: Fixed LaTeX bug, part 3</p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;1 - \(frac{1}{3} + \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136692 2017 AMC 8 Problems/Problem 9 2020-11-06T05:59:36Z <p>Elbertpark: Fixed LaTeX bug part 2</p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;\frac{12}{12} - \(frac{1}{3} + \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136691 2017 AMC 8 Problems/Problem 9 2020-11-06T05:58:57Z <p>Elbertpark: /* Solution 2 */</p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;\frac{12}{12} (\(frac{1}{3} + \frac{1}{4}) = \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136690 2017 AMC 8 Problems/Problem 9 2020-11-06T05:58:36Z <p>Elbertpark: Fixed LaTeX bug</p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;\(frac{12}{12} \(frac{1}{3} + \frac{1}{4}) = \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_9&diff=136689 2017 AMC 8 Problems/Problem 9 2020-11-06T05:57:00Z <p>Elbertpark: Added Solution 2</p> <hr /> <div>==Problem 9==<br /> All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?<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 /> ==Video Solution==<br /> https://youtu.be/rQUwNC0gqdg?t=770<br /> <br /> ==Solution 1==<br /> <br /> The &lt;math&gt;6&lt;/math&gt; green marbles and yellow marbles form &lt;math&gt;1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}&lt;/math&gt; of the total marbles. Now suppose the total number of marbles is &lt;math&gt;x&lt;/math&gt;. We know the number of yellow marbles is &lt;math&gt;\frac{5}{12}x - 6&lt;/math&gt; and a positive integer. Therefore, &lt;math&gt;12&lt;/math&gt; must divide &lt;math&gt;x&lt;/math&gt;. Trying the smallest multiples of &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt;, we see that when &lt;math&gt;x = 12&lt;/math&gt;, we get there are &lt;math&gt;-1&lt;/math&gt; yellow marbles, which is impossible. However when &lt;math&gt;x = 24&lt;/math&gt;, there are &lt;math&gt;\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles, which must be the smallest possible.<br /> <br /> ==Solution 2==<br /> The 6 green and yellow marbles make up &lt;math&gt;\frac{1}{3} + \frac{1}{4} = \frac{7}{12} → \frac{5}{12}&lt;/math&gt; of the total marbles. Now we know that there are &lt;math&gt;\frac{5}{12} - 6&lt;/math&gt; yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are &lt;math&gt;\frac{10}{24} - 6&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) }4}&lt;/math&gt; yellow marbles. <br /> <br /> -[[User:elbertpark|elbertpark]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=8|num-a=10}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=136686 2017 AMC 8 Problems/Problem 5 2020-11-06T05:25:43Z <p>Elbertpark: /* Solution 3 */</p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> <br /> Directly calculating:<br /> <br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> It is well known that the sum of all numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt;. Therefore, the denominator is equal to &lt;math&gt;\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6&lt;/math&gt;. Now we can cancel the factors of &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; from both the numerator and denominator, only leaving &lt;math&gt;8 \cdot 7 \cdot 5 \cdot 4 \cdot 1&lt;/math&gt;. This evaluates to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> First, we evaluate &lt;math&gt;1 + 2 + 3 + 4 + 5 + 6 + 7 + 8&lt;/math&gt;, to get 36. We notice that 36 is 6 squared, so we can factor the denominator like &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}&lt;/math&gt; then cancel the 6s out, to get &lt;math&gt;\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}&lt;/math&gt;. Now that we have escaped fraction form, multiplying &lt;math&gt;4 \cdot 5 \cdot 7 \cdot 8 \cdot&lt;/math&gt;. Multiplying these, we get &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ~[[User:elbertpark|elbertpark]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/cY4NYSAD0vQ<br /> <br /> ~[[User:icematrix2|icematrix2]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=136684 2017 AMC 8 Problems/Problem 5 2020-11-06T05:24:07Z <p>Elbertpark: </p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> <br /> Directly calculating:<br /> <br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> It is well known that the sum of all numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt;. Therefore, the denominator is equal to &lt;math&gt;\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6&lt;/math&gt;. Now we can cancel the factors of &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; from both the numerator and denominator, only leaving &lt;math&gt;8 \cdot 7 \cdot 5 \cdot 4 \cdot 1&lt;/math&gt;. This evaluates to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> First, we evaluate &lt;math&gt;1 + 2 + 3 + 4 + 5 + 6 + 7 + 8&lt;/math&gt;, to get 36. We notice that 36 is 6 squared, so we can go like &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}&lt;/math&gt; then cancel the 6s out, to get &lt;math&gt;\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}&lt;/math&gt;. Now that we have escaped fraction form, multiplying &lt;math&gt;4 \cdot 5 \cdot 7 \cdot 8 \cdot&lt;/math&gt;. Multiplying these, we get &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ~[[User:elbertpark|elbertpark]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/cY4NYSAD0vQ<br /> <br /> ~[[User:icematrix2|icematrix2]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=User:Elbertpark&diff=136683 User:Elbertpark 2020-11-06T05:23:20Z <p>Elbertpark: Created page with &quot;Welcome to my page! I am studying for the AMC 8 and sometimes contribute to its wiki. Thanks for visiting!&quot;</p> <hr /> <div>Welcome to my page! I am studying for the AMC 8 and sometimes contribute to its wiki. Thanks for visiting!</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=136682 2017 AMC 8 Problems/Problem 5 2020-11-06T05:22:04Z <p>Elbertpark: </p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> <br /> Directly calculating:<br /> <br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> It is well known that the sum of all numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt;. Therefore, the denominator is equal to &lt;math&gt;\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6&lt;/math&gt;. Now we can cancel the factors of &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; from both the numerator and denominator, only leaving &lt;math&gt;8 \cdot 7 \cdot 5 \cdot 4 \cdot 1&lt;/math&gt;. This evaluates to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> First, we evaluate 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8, to get 36. We notice that 36 is 6 squared, so we can go like &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}&lt;/math&gt; then cancel the 6s out, to get &lt;math&gt;\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}&lt;/math&gt;. Now that we have escaped fraction form, multiplying &lt;math&gt;4 \cdot 5 \cdot 7 \cdot 8 \cdot&lt;/math&gt;. Multiplying these, we get &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ~[[User:elbertpark|elbertpark]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/cY4NYSAD0vQ<br /> <br /> ~[[User:icematrix2|icematrix2]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=136680 2017 AMC 8 Problems/Problem 5 2020-11-06T05:20:41Z <p>Elbertpark: Added a Solution 3</p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> <br /> Directly calculating:<br /> <br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> It is well known that the sum of all numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt;. Therefore, the denominator is equal to &lt;math&gt;\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6&lt;/math&gt;. Now we can cancel the factors of &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; from both the numerator and denominator, only leaving &lt;math&gt;8 \cdot 7 \cdot 5 \cdot 4 \cdot 1&lt;/math&gt;. This evaluates to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> First, we evaluate 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8, to get 36. We notice that 36 is 6 squared, so we can go like &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}&lt;/math&gt; then cancel the 6s out, to get &lt;math&gt;\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}&lt;/math&gt;. Now that we have escaped fraction form, multiplying &lt;math&gt;4 \cdot 5 \cdot 7 \cdot 8 \cdot&lt;/math&gt;. Multiplying these, we get &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/cY4NYSAD0vQ<br /> <br /> ~[[User:icematrix2|icematrix2]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_5&diff=136679 2017 AMC 8 Problems/Problem 5 2020-11-06T05:19:05Z <p>Elbertpark: </p> <hr /> <div>==Problem 5==<br /> What is the value of the expression &lt;math&gt;\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360&lt;/math&gt; <br /> <br /> ==Solution 1==<br /> <br /> Directly calculating:<br /> <br /> We evaluate both the top and bottom: &lt;math&gt;\frac{40320}{36}&lt;/math&gt;. This simplifies to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> It is well known that the sum of all numbers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n(n+1)}{2}&lt;/math&gt;. Therefore, the denominator is equal to &lt;math&gt;\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6&lt;/math&gt;. Now we can cancel the factors of &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;6&lt;/math&gt; from both the numerator and denominator, only leaving &lt;math&gt;8 \cdot 7 \cdot 5 \cdot 4 \cdot 1&lt;/math&gt;. This evaluates to &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> First, we evaluate 1+2+3+4+5+6+7+8, to get 36. We notice that 36 is 6 squared, so we can go like &lt;math&gt;\frac{\cdot 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}&lt;/math&gt; then cancel the 6s out, to get &lt;math&gt;\frac{\cdot 4 \cdot 5 \cdot 7 \cdot 8}{1}&lt;/math&gt;. Now that we have escaped fraction form, multiplying \cdot 4 \cdot 5 \cdot 7 \cdot 8. Multiplying these, we get &lt;math&gt;\boxed{\textbf{(B)}\ 1120}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/cY4NYSAD0vQ<br /> <br /> ~[[User:icematrix2|icematrix2]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=4|num-a=6}}<br /> <br /> {{MAA Notice}}</div> Elbertpark https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_7&diff=136352 2019 AMC 8 Problems/Problem 7 2020-11-01T19:04:05Z <p>Elbertpark: /* Solution 3 */</p> <hr /> <div>==Problem 7==<br /> Shauna takes five tests, each worth a maximum of &lt;math&gt;100&lt;/math&gt; points. Her scores on the first three tests are &lt;math&gt;76&lt;/math&gt;, &lt;math&gt;94&lt;/math&gt;, and &lt;math&gt;87&lt;/math&gt;. In order to average &lt;math&gt;81&lt;/math&gt; for all five tests, what is the lowest score she could earn on one of the other two tests?<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; for the scores on the last two tests. &lt;cmath&gt;\frac{76+94+87+x+y}{5} = 81,&lt;/cmath&gt; &lt;cmath&gt;\frac{257+x+y}{5} = 81.&lt;/cmath&gt; We can now cross multiply to get rid of the denominator. &lt;cmath&gt;257+x+y = 405,&lt;/cmath&gt; &lt;cmath&gt;x+y = 148.&lt;/cmath&gt; Now that we have this equation, we will assign &lt;math&gt;y&lt;/math&gt; as the lowest score of the two other tests, and so: &lt;cmath&gt;x = 100,&lt;/cmath&gt; &lt;cmath&gt;y=48.&lt;/cmath&gt; Now we know that the lowest score on the two other tests is &lt;math&gt;\boxed{48}&lt;/math&gt;.<br /> <br /> ~ aopsav<br /> <br /> ==Solution 2==<br /> Right now, she scored &lt;math&gt;76, 94,&lt;/math&gt; and &lt;math&gt;87&lt;/math&gt; points, with a total of &lt;math&gt;257&lt;/math&gt; points. She wants her average to be &lt;math&gt;81&lt;/math&gt; for her &lt;math&gt;5&lt;/math&gt; tests so she needs to score &lt;math&gt;405&lt;/math&gt; points in total. She needs to score a total of &lt;math&gt;(405-257) <br /> 148&lt;/math&gt; points in her &lt;math&gt;2&lt;/math&gt; tests. So the minimum score she can get is when one of her &lt;math&gt;2&lt;/math&gt; scores is &lt;math&gt;100&lt;/math&gt;. So the least possible score she can get is &lt;math&gt;\boxed{\textbf{(A)}\ 48}&lt;/math&gt;. &lt;math&gt;3.1415926&lt;/math&gt;<br /> <br /> <br /> Note: You can verify that &lt;math&gt;\boxed{48}&lt;/math&gt; is the right answer because it is the lowest answer out of the 5. Since it is possible to get 48, we are guaranteed that that is the right answer.<br /> <br /> ==Solution 3==<br /> We can compare each of the scores with the average of &lt;math&gt;81&lt;/math&gt;:<br /> &lt;math&gt;76&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;-5&lt;/math&gt;,<br /> &lt;math&gt;94&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+13&lt;/math&gt;,<br /> &lt;math&gt;87&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+6&lt;/math&gt;,<br /> &lt;math&gt;100&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;+19&lt;/math&gt;;<br /> <br /> So the last one has to be &lt;math&gt;-33&lt;/math&gt; (since all the differences have to sum to &lt;math&gt;0&lt;/math&gt;), which corresponds to &lt;math&gt;81-33 = \boxed{48}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=6|num-a=8}}<br /> <br /> {{MAA Notice}}</div> Elbertpark