https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Oceanxia&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-08T17:12:32Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_15&diff=138237 2020 AMC 8 Problems/Problem 15 2020-11-23T19:19:03Z <p>Oceanxia: bdot</p> <hr /> <div>==Problem==<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 /> ==Solution 1==<br /> Since &lt;math&gt;20\% = \frac{1}{5}&lt;/math&gt;, multiplying the given condition by &lt;math&gt;5&lt;/math&gt; shows that &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;15 \cdot 5 = \boxed{\textbf{(C) }75}&lt;/math&gt; percent of &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Letting &lt;math&gt;x=100&lt;/math&gt; (without loss of generality), the condition becomes &lt;math&gt;0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75&lt;/math&gt;. Clearly, it follows that &lt;math&gt;y&lt;/math&gt; is &lt;math&gt;75\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }75}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We have &lt;math&gt;15\%=\frac{3}{20}&lt;/math&gt; and &lt;math&gt;20\%=\frac{1}{5}&lt;/math&gt;, so &lt;math&gt;\frac{3}{20}x=\frac{1}{5}y&lt;/math&gt;. Solving for &lt;math&gt;y&lt;/math&gt;, we multiply by &lt;math&gt;5&lt;/math&gt; to give &lt;math&gt;y = \frac{15}{20}x = \frac{3}{4}x&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) }75}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We are given &lt;math&gt;0.15x = 0.20y&lt;/math&gt;, so we may assume without loss of generality that &lt;math&gt;x=20&lt;/math&gt; and &lt;math&gt;y=15&lt;/math&gt;. This means &lt;math&gt;\frac{y}{x}=\frac{15}{20}=\frac{75}{100}&lt;/math&gt;, and thus answer is &lt;math&gt;\boxed{\textbf{(C) }75}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/mjS-PHTw-GE<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_15&diff=138236 2020 AMC 8 Problems/Problem 15 2020-11-23T19:18:23Z <p>Oceanxia: </p> <hr /> <div>bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot bdot yummy yummy bdot</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_25&diff=138235 2020 AMC 8 Problems/Problem 25 2020-11-23T19:10:34Z <p>Oceanxia: Replaced content with &quot;yum&quot;</p> <hr /> <div>yum</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=137918 2020 AMC 8 Problems/Problem 18 2020-11-20T03:16:25Z <p>Oceanxia: </p> <hr /> <div>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 /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<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 /> ==Solution 1==<br /> First, realize &lt;math&gt;ABCD&lt;/math&gt; is not a square. It can easily be seen that the diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so the radius is &lt;math&gt;\frac{34}{2}=17&lt;/math&gt;. Express the area of Rectangle &lt;math&gt;ABCD&lt;/math&gt; as &lt;math&gt;16h&lt;/math&gt;, where &lt;math&gt;h=AB&lt;/math&gt;. Notice that by the Pythagorean theorem &lt;math&gt;8^2+h^{2}=17^{2}\implies h=15&lt;/math&gt;. Then, the area of Rectangle &lt;math&gt;ABCD&lt;/math&gt; is equal to &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;. ~icematrix<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> We have &lt;math&gt;OC=17&lt;/math&gt;, as it is a radius, and &lt;math&gt;OD=8&lt;/math&gt; since it is half of &lt;math&gt;AD&lt;/math&gt;. This means that &lt;math&gt;CD=\sqrt{17^2-8^2}=15&lt;/math&gt;. So &lt;math&gt;16*15=\boxed{\textbf{(A)}240}&lt;/math&gt;<br /> <br /> ~yofro<br /> <br /> ==Solution 3 (coordinate bashing)==<br /> <br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D&lt;/math&gt; is at &lt;math&gt;(-8, 0)&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; is at &lt;math&gt;(8, 0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share x-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;, respectively, we can just find the y-coordinate of &lt;math&gt;B&lt;/math&gt; (which is just the width of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt;, or &lt;math&gt;16&lt;/math&gt;. Since the radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2}&lt;/math&gt;, or &lt;math&gt;17&lt;/math&gt;, the equation of the circle that our semicircle is a part of is &lt;math&gt;x^2+y^2=289&lt;/math&gt;. Since we know that the x-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;8&lt;/math&gt;, we can plug this into our equation to obtain that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt;, as the diagram suggests, we know that the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, our answer is &lt;math&gt;16\cdot 15&lt;/math&gt;, or &lt;math&gt;\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> NOTE: The synthetic solution is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier solution.<br /> <br /> - StarryNight7210<br /> <br /> ==Solution 4==<br /> <br /> First, realize that &lt;math&gt;ABCD&lt;/math&gt; is not a square. Let &lt;math&gt;O&lt;/math&gt; be the midpoint of &lt;math&gt;FE&lt;/math&gt;. Since &lt;math&gt;FE=9+9+16=34&lt;/math&gt;, we have &lt;math&gt;OF=OE=\frac{34}{2}=17=OB&lt;/math&gt; because they are all radii. Since &lt;math&gt;O&lt;/math&gt; is also the midpoint of &lt;math&gt;AD&lt;/math&gt;, we have &lt;math&gt;OA=\frac{16}2=8&lt;/math&gt;. By the Pythagorean Theorem on &lt;math&gt;\triangle BAO&lt;/math&gt;, we find that &lt;math&gt;AB=15&lt;/math&gt;. The answer is then &lt;math&gt;16\cdot 15=\textbf{(A) }240&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5 -SweetMango77==<br /> <br /> This is an example of a formula in the Introduction to Algebra book (a sidenote): with a semicircle: if the diameter is &lt;math&gt;1+n&lt;/math&gt; with the &lt;math&gt;1&lt;/math&gt; part at one side, and the &lt;math&gt;n&lt;/math&gt; part at the other side, then the height from the end of the &lt;math&gt;1&lt;/math&gt; side and the start of the &lt;math&gt;n&lt;/math&gt; side is &lt;math&gt;\sqrt{n}&lt;/math&gt;. <br /> <br /> Using this, we can scale the image down by &lt;math&gt;9&lt;/math&gt; to get what we note: The other side will be &lt;math&gt;\frac{16+9}{9}=\frac{25}{9}=\left(\frac{5}{3}\right)^2&lt;/math&gt;. Then, the height of that part will be &lt;math&gt;\frac{5}{3}&lt;/math&gt;. But, we have to scale it back up by &lt;math&gt;9&lt;/math&gt; to get a height of &lt;math&gt;15&lt;/math&gt;. Multiplying by &lt;math&gt;16&lt;/math&gt; gives our desired answer: &lt;math&gt;\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> A good diagram goes a long way. With a proper compass and ruler, this problem can be easily solved by a quick sketch. However, use this as a last resort.<br /> ==Solution 7== The other side will be &lt;math&gt;\frac{16+9}{9}=\frac{25} which we know it is a &lt;/math&gt;AB=15&lt;math&gt;. or so &lt;/math&gt;16\cdot 15=\textbf{(A) }240$. <br /> ~oceanxia<br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_20&diff=137917 2020 AMC 8 Problems/Problem 20 2020-11-20T03:14:12Z <p>Oceanxia: </p> <hr /> <div>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;\begin{tabular}{|c|c|} \hline Tree 1 &amp; \rule{0.2cm}{0.15mm} meters \\ Tree 2 &amp; 11 meters \\ Tree 3 &amp; \rule{0.2cm}{0.15mm} meters \\ Tree 4 &amp; \rule{0.2cm}{0.15mm} meters \\ Tree 5 &amp; \rule{0.2cm}{0.15mm} meters \\ \hline Average height &amp; \rule{0.2cm}{0.15mm}.2 meters \\ \hline \end{tabular}&lt;/cmath&gt;&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 /> ==Solution 1==<br /> It is not too hard to construct &lt;math&gt;22, 11, 22, 44, 22&lt;/math&gt; as the heights of the trees from left to right. The average is &lt;math&gt;\frac{121}{5}=24.2\rightarrow\boxed{\textbf{(B)}}&lt;/math&gt;. ~icematrix Note: this is the result of realizing that the 1st one must be greater then the 2nd, &lt;math&gt;\frac{121}{5}=24.2 or {\textbf{(B)}}&lt;/math&gt;. ~oceanxia <br /> <br /> <br /> <br /> ==Solution 2==<br /> For the heights of the trees to be integers, we must have the height of Tree 1, Tree 2, Tree 3 as &lt;math&gt;22, 11, 22&lt;/math&gt;. Now, there are two possible cases.<br /> <br /> <br /> <br /> Case 1: The length of tree 4 is &lt;math&gt;11&lt;/math&gt;-<br /> So, we have the first 4 tree lengths as &lt;math&gt;22, 11, 22, 11&lt;/math&gt;. For the length of Tree 5 to be an integer, we must have the length of Tree &lt;math&gt;5&lt;/math&gt; as &lt;math&gt;22&lt;/math&gt;. But, when we take the average of these &lt;math&gt;5&lt;/math&gt; integers, it results in &lt;math&gt;17.6&lt;/math&gt;, which doesn't satisfy our conditions.<br /> <br /> <br /> <br /> Case 2: The length of tree 4 is &lt;math&gt;44&lt;/math&gt;-<br /> So, we have the first &lt;math&gt;4&lt;/math&gt; tree lengths as &lt;math&gt;22, 11, 22, 44&lt;/math&gt;. Now, using quick modular arithmetic, we see that when the length of Tree 5 is &lt;math&gt;88&lt;/math&gt;, the average of the heights of the 5 trees is &lt;math&gt;\boxed{24.2}\rightarrow\boxed{\textbf{(B)}}&lt;/math&gt;. This is where our condition is satisfied.<br /> <br /> ~ATGY<br /> <br /> ==Solution 3==<br /> Let the sum of the heights of the trees be denoted by &lt;math&gt;S&lt;/math&gt;. The average height will then be &lt;math&gt;\frac{S}{5}&lt;/math&gt;. Since the average height has decimal part &lt;math&gt;.2&lt;/math&gt;, it follows that &lt;math&gt;S&lt;/math&gt; must have a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;5&lt;/math&gt;. Thus, &lt;math&gt;S&lt;/math&gt; must be &lt;math&gt;1&lt;/math&gt; more than a multiple of 5. Since &lt;math&gt;S&lt;/math&gt; is an integer, it follows that the heights of Tree 1 and Tree 3 are likely both 22. At this point, our table looks as follows:<br /> &lt;cmath&gt;\begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; 22 meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; 22 meters \\<br /> Tree 4 &amp; \rule{0.2cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.2cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.2cm}{0.15mm}.2 meters \\<br /> \hline<br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> If we guess that Tree 4 has a height of 11, we would need to make Tree 5 have a height of 22. In this case &lt;math&gt;S&lt;/math&gt; would equal &lt;math&gt;88&lt;/math&gt; which is not one more than a multiple of &lt;math&gt;5&lt;/math&gt;. So we instead guess that Tree 4 has a height of &lt;math&gt;44&lt;/math&gt;. Then we only need to try out heights of &lt;math&gt;22&lt;/math&gt; and &lt;math&gt;88&lt;/math&gt; for Tree 5. Using a height of &lt;math&gt;22&lt;/math&gt; for Tree 5 gives us &lt;math&gt;S=121&lt;/math&gt; which is &lt;math&gt;1&lt;/math&gt; more than a multiple of &lt;math&gt;5&lt;/math&gt;. Thus, the average height of the trees is &lt;math&gt;\frac{121}{5}=24.2&lt;/math&gt; meters &lt;math&gt;\implies\boxed{\textbf{(B) }24.2}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 4==<br /> <br /> The problem states that all tree heights are integers. Therefore, we can deduce that the first and third trees have a height of &lt;math&gt;22&lt;/math&gt; meters. Trees four and five must have heights of either &lt;math&gt;11,22&lt;/math&gt; or &lt;math&gt;44,88&lt;/math&gt; or &lt;math&gt;44,22&lt;/math&gt;. Checking which ones match the answer choices, we find that the trees four and five have heights of &lt;math&gt;44&lt;/math&gt; and &lt;math&gt;22&lt;/math&gt; meters, respectively. Thus, the answer is &lt;math&gt;\frac{22+11+22+44+22}5=\textbf{(B) }24.2&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_25&diff=137916 2020 AMC 8 Problems/Problem 25 2020-11-20T03:12:09Z <p>Oceanxia: </p> <hr /> <div>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$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> For each square &lt;math&gt;S_{k}&lt;/math&gt;, let the sidelength of this square be denoted by &lt;math&gt;s_{k}&lt;/math&gt;.<br /> <br /> As the diagram shows, &lt;math&gt;s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.&lt;/math&gt; We subtract the second equation from the first, getting &lt;math&gt;2s_{2}=1302&lt;/math&gt;, and thus &lt;math&gt;s_{2}=651&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\text{ }651}&lt;/math&gt; ~icematrix, edits by starrynight7210<br /> <br /> ==Solution 2==<br /> <br /> WLOG, assume that &lt;math&gt;S_1=S_3&lt;/math&gt; and &lt;math&gt;R_1=R_2&lt;/math&gt;. Let the sum of the lengths of &lt;math&gt;S_1&lt;/math&gt; and &lt;math&gt;S_2&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; and let the length of &lt;math&gt;S_2&lt;/math&gt; be &lt;math&gt;y&lt;/math&gt;. We have the system &lt;cmath&gt;x+y =3322&lt;/cmath&gt; &lt;cmath&gt;x-y=2020&lt;/cmath&gt;<br /> <br /> which we solve to find that &lt;math&gt;y=\textbf{(A) }651&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 3==<br /> Since each pair of boxes has a sum of &lt;math&gt;3322&lt;/math&gt; or &lt;math&gt;2020&lt;/math&gt; and a difference of &lt;math&gt;S_2&lt;/math&gt;, we see that the answer is &lt;math&gt;\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{(\text{A}) 651}.&lt;/math&gt;<br /> <br /> -A_MatheMagician. Note this is just a more quicker way to do it to get \boxed{(\text{A}) 651}.$<br /> https://artofproblemsolving.com/community/my-aops<br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_9&diff=137915 2020 AMC 8 Problems/Problem 9 2020-11-20T03:10:24Z <p>Oceanxia: </p> <hr /> <div>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 of the small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> /*<br /> Created by SirCalcsALot and sonone<br /> Code modfied from https://artofproblemsolving.com/community/c3114h2152994_the_old__aops_logo_with_asymptote<br /> */<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<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 /> 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 /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<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 /> // begin made by SirCalcsALot<br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<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 /> &lt;math&gt;\textbf{(A) }12 \qquad \textbf{(B) }16 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> its very easy to get trolled here if you dont read &quot;no icing on the bottom&quot;<br /> ==Solution 1==<br /> Notice that all the faces with the exception of the bottom faces have the two center edge pieces with 2 faces with icing on them. This is &lt;math&gt;8\cdot 2 = 16&lt;/math&gt;. Additionally, on the bottom face, the corners have 2 faces with icing, as the bottom face does not have icing. This is &lt;math&gt;4&lt;/math&gt; cubes. The total is &lt;math&gt;16+4 = 20, \textbf{(D) }20&lt;/math&gt;. Note we just need to get The total is &lt;math&gt;16+4 = 20, \textbf{(D) \textbf{(D) }20&lt;/math&gt; }20&lt;math&gt;<br /> <br /> ~Windgo and minor edits made by oceanxia<br /> <br /> ==Solution 2==<br /> This is just careful casework. Consider the following diagram: <br /> https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi82Lzk1NjIyMDEyYzEwMmU2MGRhM2U3OGMzYzA0MDNmOGFmZjdkMDk3LnBuZw==&amp;rn=YW1jIDggbm8gOS5wbmc<br /> <br /> The face on the opposite side of the front face (hidden) is an exact copy of the front face. So the answer is &lt;/math&gt;8+4+8=\textbf{(D)}20$.<br /> <br /> -franzliszt<br /> <br /> Franz Liszt is in 12th grade<br /> <br /> ==Video Solution==<br /> https://youtu.be/WyvmQUfxTfo<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_7&diff=137914 2020 AMC 8 Problems/Problem 7 2020-11-20T03:07:51Z <p>Oceanxia: </p> <hr /> <div>==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 /> ==Solution 1==<br /> First, observe that the second digit of the four digit number cannot be a &lt;math&gt;1&lt;/math&gt; or a &lt;math&gt;2&lt;/math&gt; because the digits must be distinct and increasing. The second digit also cannot be a &lt;math&gt;4&lt;/math&gt; because the number must be less than &lt;math&gt;2400&lt;/math&gt;. Thus, the second digit must be &lt;math&gt;3&lt;/math&gt;. If we place a &lt;math&gt;4&lt;/math&gt; in the third digit then there are 5 ways to select the last digit, namely the last digit could then be &lt;math&gt;5,6,7,8,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. If we place a &lt;math&gt;5&lt;/math&gt; in the third digit, there are 4 ways to select the last digit, namely the last digit could then be &lt;math&gt;6,7,8,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;. Similarly, if the third digit is &lt;math&gt;6&lt;/math&gt;, there are 3 ways to select the last digit, etc. Thus, it follows that the total number of valid numbers is &lt;math&gt;5+4+3+2+1=15\implies\boxed{\textbf{(C) }15}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 2==<br /> <br /> Notice that the number is of the form &lt;math&gt;23AB&lt;/math&gt; were &lt;math&gt;A&gt;B&gt;3&lt;/math&gt;. We have &lt;math&gt;A=4,B\in [5,9];A=5,B\in [6,9];A=6,B\in [7,9];A=7,B\in [8,9];A=8,B\in &lt;/math&gt;. Counting the numbers in the brackets, the answer is &lt;math&gt;5+4+3+2+1=\textbf{(C) }15&lt;/math&gt;.<br /> -oceanxia<br /> -franzliszt<br /> <br /> ==Video Solution==<br /> https://youtu.be/FjmBtgrGfCs<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> <br /> The thousands place (first digit) has to be a 2 (2020-2400). <br /> Since the thousands digit is 2, the next digit must be a 3 (not 4 or onwards because that will go over the range given). <br /> <br /> The next digit has to be from 4, 5, 6, 7, or 8. For each of the cases, you get a total of 15 possibilities, which gives you the answer C. <br /> <br /> <br /> ~itsmemasterS</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_13&diff=137913 2002 AIME II Problems/Problem 13 2020-11-20T03:06:47Z <p>Oceanxia: </p> <hr /> <div>== Problem ==<br /> In triangle &lt;math&gt;ABC,&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt; with &lt;math&gt;CD = 2&lt;/math&gt; and &lt;math&gt;DB = 5,&lt;/math&gt; point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{AC}&lt;/math&gt; with &lt;math&gt;CE = 1&lt;/math&gt; and &lt;math&gt;EA = 3,&lt;/math&gt; &lt;math&gt;AB = 8,&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; intersect at &lt;math&gt;P.&lt;/math&gt; Points &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;\overline{PQ}&lt;/math&gt; is parallel to &lt;math&gt;\overline{CA}&lt;/math&gt; and &lt;math&gt;\overline{PR}&lt;/math&gt; is parallel to &lt;math&gt;\overline{CB}.&lt;/math&gt; It is given that the ratio of the area of triangle &lt;math&gt;PQR&lt;/math&gt; to the area of triangle &lt;math&gt;ABC&lt;/math&gt; is &lt;math&gt;m/n,&lt;/math&gt; where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Let &lt;math&gt;X&lt;/math&gt; be the intersection of &lt;math&gt;\overline{CP}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;. <br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> pair A,B,C,D,E,X,P,Q,R; <br /> A=(0,0); <br /> B=(8,0); <br /> C=(1.9375,3.4994); <br /> D=(3.6696,2.4996); <br /> E=(1.4531,2.6246); <br /> X=(4.3636,0); <br /> P=(2.9639,2.0189); <br /> Q=(1.8462,0); <br /> R=(6.4615,0); <br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> dot(X);<br /> dot(P);<br /> dot(Q);<br /> dot(R);<br /> label(&quot;$A$&quot;,A,WSW);<br /> label(&quot;$B$&quot;,B,ESE);<br /> label(&quot;$C$&quot;,C,NNW);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,WNW);<br /> label(&quot;$X$&quot;,X,SSE);<br /> label(&quot;$P$&quot;,P,NNE);<br /> label(&quot;$Q$&quot;,Q,SSW);<br /> label(&quot;$R$&quot;,R,SE);<br /> draw(A--B--C--cycle);<br /> draw(P--Q--R--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> draw(C--X);<br /> &lt;/asy&gt;<br /> <br /> Since &lt;math&gt;\overline{PQ} \parallel \overline{CA}&lt;/math&gt; and &lt;math&gt;\overline{PR} \parallel \overline{CB}&lt;/math&gt;, &lt;math&gt;\angle CAB = \angle PQR&lt;/math&gt; and &lt;math&gt;\angle CBA = \angle PRQ&lt;/math&gt;. So &lt;math&gt;\Delta ABC \sim \Delta QRP&lt;/math&gt;, and thus, &lt;math&gt;\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\right)^2&lt;/math&gt;. <br /> <br /> Using [[mass points]]: <br /> <br /> [[WLOG]], let &lt;math&gt;W_C=15&lt;/math&gt;. <br /> <br /> Then: <br /> <br /> &lt;math&gt;W_A = \left(\frac{CE}{AE}\right)W_C = \frac{1}{3}\cdot15=5&lt;/math&gt;.<br /> <br /> &lt;math&gt;W_B = \left(\frac{CD}{BD}\right)W_C = \frac{2}{5}\cdot15=6&lt;/math&gt;. <br /> <br /> &lt;math&gt;W_X=W_A+W_B=5+6=11&lt;/math&gt;. <br /> <br /> &lt;math&gt;W_P=W_C+W_X=15+11=26&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{PX}{CX}=\frac{W_C}{W_P}=\frac{15}{26}&lt;/math&gt;. Therefore, &lt;math&gt;\frac{[\Delta PQR]}{[\Delta ABC]} = \left( \frac{15}{26} \right)^2 = \frac{225}{676}&lt;/math&gt;, and &lt;math&gt;m+n=\boxed{901}&lt;/math&gt;. Note we can just use mass points to get \left( \frac{15}{26} \right)^2= \frac{225}{676}&lt;math&gt; which is \boxed{901}&lt;/math&gt;.<br /> <br /> == Solution 2 == <br /> First draw &lt;math&gt;\overline{CP}&lt;/math&gt; and extend it so that it meets with &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;X&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> pair A,B,C,D,E,X,P,Q,R; <br /> A=(0,0); <br /> B=(8,0); <br /> C=(1.9375,3.4994); <br /> D=(3.6696,2.4996); <br /> E=(1.4531,2.6246); <br /> X=(4.3636,0); <br /> P=(2.9639,2.0189); <br /> Q=(1.8462,0); <br /> R=(6.4615,0); <br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> dot(X);<br /> dot(P);<br /> dot(Q);<br /> dot(R);<br /> label(&quot;$A$&quot;,A,WSW);<br /> label(&quot;$B$&quot;,B,ESE);<br /> label(&quot;$C$&quot;,C,NNW);<br /> label(&quot;$D$&quot;,D,NE);<br /> label(&quot;$E$&quot;,E,WNW);<br /> label(&quot;$X$&quot;,X,SSE);<br /> label(&quot;$P$&quot;,P,NNE);<br /> label(&quot;$Q$&quot;,Q,SSW);<br /> label(&quot;$R&quot;,R,SE);<br /> draw(A--B--C--cycle);<br /> draw(P--Q--R--cycle);<br /> draw(A--D);<br /> draw(B--E);<br /> draw(C--X);<br /> &lt;/asy&gt;<br /> <br /> We have that &lt;math&gt;[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}&lt;/math&gt;<br /> <br /> By Ceva's, &lt;cmath&gt;3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}&lt;/cmath&gt; That means that &lt;cmath&gt; \frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}&lt;/cmath&gt;<br /> <br /> Now we apply mass points. Assume WLOG that &lt;math&gt;W_{A}=1&lt;/math&gt;. That means that<br /> <br /> &lt;cmath&gt;W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}&lt;/cmath&gt;<br /> <br /> Notice now that &lt;math&gt;\triangle{PBQ}&lt;/math&gt; is similar to &lt;math&gt;\triangle{EBA}&lt;/math&gt;. Therefore,<br /> <br /> &lt;cmath&gt;\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}&lt;/cmath&gt;<br /> <br /> Also, &lt;math&gt;\triangle{PRA}&lt;/math&gt; is similar to &lt;math&gt;\triangle{DBA}&lt;/math&gt;. Therefore,<br /> <br /> &lt;cmath&gt;\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}&lt;/cmath&gt;<br /> <br /> Because &lt;math&gt;\triangle{PQR}&lt;/math&gt; is similar to &lt;math&gt;\triangle{CAB}&lt;/math&gt;, &lt;math&gt;\angle{C}=\angle{P}&lt;/math&gt;.<br /> <br /> As a result, &lt;math&gt;[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}&lt;/math&gt;.<br /> <br /> Therefore, &lt;cmath&gt;\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}&lt;/cmath&gt;<br /> <br /> *Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=12|num-a=14}}<br /> <br /> [[Category: Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_8&diff=137911 2002 AIME II Problems/Problem 8 2020-11-20T03:02:43Z <p>Oceanxia: </p> <hr /> <div>== Problem ==<br /> Find the least positive integer &lt;math&gt;k&lt;/math&gt; for which the equation &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=k&lt;/math&gt; has no integer solutions for &lt;math&gt;n&lt;/math&gt;. (The notation &lt;math&gt;\lfloor x\rfloor&lt;/math&gt; means the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> Note that if &lt;math&gt;\frac{2002}n - \frac{2002}{n+1}\leq 1&lt;/math&gt;, then either &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor&lt;/math&gt;,<br /> or &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1&lt;/math&gt;. Either way, we won't skip any natural numbers.<br /> <br /> The greatest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;\frac{2002}n - \frac{2002}{n+1} &gt; 1&lt;/math&gt; is &lt;math&gt;n=44&lt;/math&gt;. (The inequality simplifies to &lt;math&gt;n(n+1)&lt;2002&lt;/math&gt;, which is easy to solve by trial, as the solution is obviously &lt;math&gt;\simeq \sqrt{2002}&lt;/math&gt;.) Note Once we plug in a few values we can see easily that it is &lt;math&gt;\boxed{049}.&lt;/math&gt;<br /> <br /> <br /> We can now compute:<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{45}\right\rfloor=44 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{44}\right\rfloor=45 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{43}\right\rfloor=46 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{42}\right\rfloor=47 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{41}\right\rfloor=48 &lt;/cmath&gt;<br /> &lt;cmath&gt;\left\lfloor\frac{2002}{40}\right\rfloor=50 &lt;/cmath&gt;<br /> <br /> From the observation above (and the fact that &lt;math&gt;\left\lfloor\frac{2002}{2002}\right\rfloor=1&lt;/math&gt;) we know that all integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;44&lt;/math&gt; will be achieved for some values of &lt;math&gt;n&lt;/math&gt;. Similarly, for &lt;math&gt;n&lt;40&lt;/math&gt; we obviously have &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor &gt; 50&lt;/math&gt;.<br /> <br /> Hence the least positive integer &lt;math&gt;k&lt;/math&gt; for which the equation &lt;math&gt;\left\lfloor\frac{2002}{n}\right\rfloor=k&lt;/math&gt; has no integer solutions for &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\boxed{049}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Rewriting the given information and simplifying it a bit, we have<br /> &lt;cmath&gt; \begin{align*} <br /> k \le \frac{2002}{n} &lt; k+1 &amp;\implies \frac{1}{k} \ge \frac{n}{2002} &gt; \frac{1}{k+1}. \\ &amp;\implies \frac{2002}{k} \ge n &gt; \frac{2002}{k+1}. <br /> \end{align*} &lt;/cmath&gt;<br /> <br /> Now note that in order for there to be no integer solutions to &lt;math&gt;n,&lt;/math&gt; we must have &lt;math&gt;\left\lfloor \frac{2002}{k} \right\rfloor = \left\lfloor \frac{2002}{k+1} \right\rfloor.&lt;/math&gt; We seek the smallest such &lt;math&gt;k.&lt;/math&gt; A bit of experimentation yields that &lt;math&gt;k=49&lt;/math&gt; is the smallest solution, as for &lt;math&gt;k=49,&lt;/math&gt; it is true that &lt;math&gt;\left\lfloor \frac{2002}{49} \right\rfloor = \left\lfloor \frac{2002}{50} \right\rfloor = 40.&lt;/math&gt; Furthermore, &lt;math&gt;k=49&lt;/math&gt; is the smallest such case. (If unsure, we could check if the result holds for &lt;math&gt;k=48,&lt;/math&gt; and as it turns out, it doesn't.) Therefore, the answer is &lt;math&gt;\boxed{049}.&lt;/math&gt;<br /> <br /> <br /> ==Solution 3==<br /> In this solution we use inductive reasoning and a lot of trial and error. Depending on how accurately you can estimate, the solution will come quicker or slower.<br /> <br /> Using values of k as 1, 2, 3, 4, and 5, we can find the corresponding values of n relatively easily. For k = 1, n is in the range [2002-1002]; for k = 2, n is the the range [1001-668], etc: 3, [667,501]; 4, [500-401]; 5, [400-334]. For any positive integer k, n is in a range of &lt;math&gt;floor[2002/k]-ceiling[2002/(k+1)]&lt;/math&gt;.<br /> <br /> Now we try testing k = 1002 to get a better understanding of what our solution will look like. Obviously, there will be no solution for n, but we are more interested in how the range will compute to. Using the formula we got above, the range will be 1-2. Testing any integer k from 1002-2000 will result in the same range. Also, notice that each and every one of them have no solution for n. Testing 1001 gives a range of 2-2, and 2002 gives 1-1. They each have a solution for n, and their range is only one value. Therefore, we can assume with relative safety that the integer k we want is the lowest integer that follows this equation:<br /> <br /> floor[2002/k] + 1 = ceiling[2002/(k+1)]<br /> <br /> Now we can easily guess and check starting from k = 1. After a few tests it's not difficult to estimate a few jumps, and it took me only a few minutes to realize the answer was somewhere in the forties. A final test will show: We see that Naomi travels &lt;math&gt;6&lt;/math&gt; miles in &lt;math&gt;\dfrac{1}{6}&lt;/math&gt; of an hour. Thus, his speed is &lt;math&gt;36&lt;/math&gt; mph. Maya's speed is &lt;math&gt;12&lt;/math&gt; <br /> <br /> Alternatively, you could use the equation above and proceed with one of the other two solutions listed.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2002|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=User_talk:Oceanxia&diff=137906 User talk:Oceanxia 2020-11-20T01:46:38Z <p>Oceanxia: Created page with &quot;a noob&quot;</p> <hr /> <div>a noob</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_17&diff=137886 2020 AMC 8 Problems/Problem 17 2020-11-19T23:06:27Z <p>Oceanxia: </p> <hr /> <div>How many positive integer factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors?<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 /> ==Solution 1==<br /> We list out the factors of &lt;math&gt;2020&lt;/math&gt;: &lt;cmath&gt;1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.&lt;/cmath&gt; Of these, only &lt;math&gt;1, 2, 4, 5, 101&lt;/math&gt; (&lt;math&gt;5&lt;/math&gt; of them) do not have more than &lt;math&gt;3&lt;/math&gt; factors. Therefore the answer is &lt;math&gt;\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The prime factorization of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;2^2\cdot5\cdot101&lt;/math&gt;. The total number of factors of &lt;math&gt;2020&lt;/math&gt; is given by the product of one more than each of the prime powers which comes out to &lt;math&gt;3\cdot2\cdot2=12&lt;/math&gt;. Instead of finding how many factors of &lt;math&gt;2020&lt;/math&gt; have more than three factors, we will instead find how many have one, two, or three factors and subtract this number from &lt;math&gt;12&lt;/math&gt; to find the answer. The only number which has one factor is &lt;math&gt;1&lt;/math&gt;. For a number to have exactly two factors, it must be prime. From the prime factorization of &lt;math&gt;2020&lt;/math&gt;, we know that these can only be &lt;math&gt;2,5,&lt;/math&gt; and &lt;math&gt;101&lt;/math&gt;. For a number to have three factors, it must be a square of a prime. The only square of a prime that is a factor of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Our list of factors is &lt;math&gt;1,2,4,5,&lt;/math&gt; and &lt;math&gt;101&lt;/math&gt; which is a total five factors. Thus, the number of factors of &lt;math&gt;2020&lt;/math&gt; that have more than three factors is &lt;math&gt;12-5=7 \implies\boxed{\textbf{(B) }7}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> <br /> The prime factorization of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;2^2\cdot5\cdot101&lt;/math&gt; so it has &lt;math&gt;(2+1)(1+1)(1+1)=12&lt;/math&gt; factors. Then we can count that &lt;math&gt;1,2,4,5,101&lt;/math&gt; all have &lt;math&gt;3&lt;/math&gt; or fewer divisors so by complementary counting our answer is &lt;math&gt;12-5=\textbf{(B)}\ 7&lt;/math&gt;.<br /> <br /> -franzliszt<br /> ==Solution 4== <br /> The prime factorization of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;2^2\cdot5\cdot101&lt;/math&gt; so it has &lt;math&gt;(2+1)(1+1)(1+1)=12&lt;/math&gt; factors. Then we can count that &lt;math&gt;1,2,4,5,101&lt;/math&gt; which gets &lt;math&gt;12-5=\textbf{(B)}\ 7&lt;/math&gt;.<br /> ~oceanxia~<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_14&diff=137885 2020 AMC 8 Problems/Problem 14 2020-11-19T23:04:31Z <p>Oceanxia: </p> <hr /> <div>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 /> 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 /> Diagram by sircalcsalot<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 /> ==Solution 1==<br /> The average is given to be &lt;math&gt;4750&lt;/math&gt;. This is because the dotted line is halfway in between &lt;math&gt;4500&lt;/math&gt; and &lt;math&gt;5000&lt;/math&gt;. There are &lt;math&gt;20&lt;/math&gt; cities, so our answer is simply &lt;cmath&gt;4750\cdot20=95000==&gt;\boxed{\textbf{(D) }95,000}&lt;/cmath&gt;<br /> <br /> ==Solution 2==<br /> We know that the average (&lt;math&gt;a&lt;/math&gt;) of these group of numbers is the sum (&lt;math&gt;s&lt;/math&gt;) divided by &lt;math&gt;20&lt;/math&gt;, so we can make the equation &lt;math&gt;a = \frac{s}{20}&lt;/math&gt;. Since the average is &lt;math&gt;4750&lt;/math&gt;, we can solve for &lt;math&gt;s&lt;/math&gt; to get &lt;math&gt;\boxed{\textbf{(D) } 95,000}&lt;/math&gt;<br /> <br /> ~Pi_Pup<br /> <br /> ==Solution 3==<br /> <br /> After reading the question, we notice that the dashed line is the average population of each city. Also, that dashed line is slightly less than &lt;math&gt;5\,000&lt;/math&gt;. Since there are &lt;math&gt;20&lt;/math&gt; cities, the answer is slightly less than &lt;math&gt;20\cdot 5\,000\approx 100\,000&lt;/math&gt; which is closest to &lt;math&gt;\textbf{(D) }95{,}000&lt;/math&gt;.<br /> <br /> -franzliszt<br /> ==Solution 4== <br /> we get that it is &lt;math&gt;4750&lt;/math&gt; multiplied by &lt;math&gt;20&lt;/math&gt;,solve and get &lt;math&gt;\boxed{\textbf{(D) } 95,000}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/5y4uDwZEF0M<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_12&diff=137884 2020 AMC 8 Problems/Problem 12 2020-11-19T23:03:01Z <p>Oceanxia: </p> <hr /> <div>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 from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;. What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation? &lt;cmath&gt;5!\cdot 9!=12\cdot N!&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Notice that &lt;math&gt;5!&lt;/math&gt; = &lt;math&gt;2*3*4*5,&lt;/math&gt; and we can combine the numbers to create a larger factorial. To turn &lt;math&gt;9!&lt;/math&gt; into &lt;math&gt;10!,&lt;/math&gt; we need to multiply &lt;math&gt;9!&lt;/math&gt; by &lt;math&gt;2*5,&lt;/math&gt; which equals to &lt;math&gt;10!.&lt;/math&gt; <br /> <br /> Therefore, we have<br /> <br /> &lt;cmath&gt;10!*12=12*N!.&lt;/cmath&gt;<br /> We can cancel the &lt;math&gt;12&lt;/math&gt;'s, since we are multiplying them on both sides of the equation.<br /> <br /> We have<br /> <br /> &lt;cmath&gt;10!=N!.&lt;/cmath&gt;<br /> From here, it is obvious that &lt;math&gt;N=\boxed{10\textbf{(A)}}.&lt;/math&gt;<br /> <br /> -iiRishabii<br /> <br /> ==Solution 2==<br /> &lt;math&gt;5!\cdot 9!=12\cdot N!&lt;/math&gt;&lt;br&gt;&lt;math&gt;120\cdot 9!=12\cdot N!&lt;/math&gt;&lt;br&gt;&lt;math&gt;12\cdot 10\cdot 9!=12\cdot N!&lt;/math&gt;&lt;br&gt;&lt;math&gt;12 \cdot 10!=12\cdot N!&lt;/math&gt;&lt;br&gt;&lt;math&gt;N=10 \implies\boxed{\textbf{(A) }10}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3 (Non-rigorous)==<br /> We can see that the answers B through E have the factor 11, but there is no 11 in &lt;math&gt;5!\cdot9!&lt;/math&gt;. Therefore, the answer must be the only answer without a &lt;math&gt;11&lt;/math&gt; factor, &lt;math&gt;A&lt;/math&gt;.<br /> <br /> ~Windigo<br /> <br /> ==Solution 4==<br /> <br /> Notice that &lt;math&gt;5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!&lt;/math&gt;. We are also told that &lt;math&gt;12\cdot 10!=12*N!&lt;/math&gt; from where it is obvious that &lt;math&gt;N=\textbf{(A)}10&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5==<br /> We see that &lt;math&gt;5!\cdot9! = 5\cdot4\cdot3\cdot2\cdot1\cdot9! = 12\cdot{N!}&lt;/math&gt;. Notice that &lt;math&gt;12 = 3\cdot4&lt;/math&gt;, so:<br /> &lt;cmath&gt;5\cdot2\cdot1\cdot9! = N!&lt;/cmath&gt;<br /> We see that &lt;math&gt;5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!&lt;/math&gt;. So &lt;math&gt;N = \boxed{10} = \textbf{(A)}10&lt;/math&gt;.<br /> ==Solution 6==<br /> We note that &lt;math&gt;5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!&lt;/math&gt; we can actually get 120*9!= 12*N! which then you just get to your conclusion 10! which is equal to answer choice &lt;math&gt;N=\textbf{(A)}10&lt;/math&gt;.<br /> ==Video Solution==<br /> https://youtu.be/9k59v-Fr3aE<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_5&diff=137883 2020 AMC 8 Problems/Problem 5 2020-11-19T22:59:53Z <p>Oceanxia: </p> <hr /> <div>==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 /> <br /> &lt;math&gt;\frac{3}{4}\div 5 = \frac{3}{4}\cdot\frac{1}{5} =\frac{3}{20} = \frac{3}{20} \cdot 100 = \textbf{(C) }15&lt;/math&gt;<br /> <br /> ==Solution==<br /> To equally distribute to &lt;math&gt;5&lt;/math&gt; cups, we will simply divide &lt;math&gt;\dfrac{3}{4}&lt;/math&gt; by &lt;math&gt;5.&lt;/math&gt; Simplifying, we get: &lt;math&gt;\dfrac{3}{4} \cdot \dfrac{1}{5} = \dfrac{3}{20}.&lt;/math&gt; Converting that into a percent, we get an answer of &lt;math&gt;\boxed{\textbf{(C) }15}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Assume that the pitcher has a total capacity of &lt;math&gt;100&lt;/math&gt; ounces. Since the pitcher is filled three fourths with pineapple juice, it follows that it contains &lt;math&gt;75&lt;/math&gt; ounces of pineapple juice. The pineapple juice is then divided equally into 5 cups, which means that each cup will contain &lt;math&gt;\frac{75}{5}=15&lt;/math&gt; ounces of pineapple juice. Since the total capacity of the pitcher was &lt;math&gt;100&lt;/math&gt; ounces, it follows that each cup received &lt;math&gt;15\%&lt;/math&gt; of the total capacity of the pitcher &lt;math&gt;\implies\boxed{\textbf{(C) }15}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> Notice that each cup receives &lt;math&gt;\frac 34 \cdot \frac 15=\frac{3}{20}=\frac{15}{100}&lt;/math&gt;<br /> Notice that each cup receives &lt;math&gt;\frac 34 \cdot \frac 15=\frac{3}{20}=\frac{15}{100}&lt;/math&gt; of the entire pitcher which is &lt;math&gt;\textbf{(C) }15&lt;/math&gt; percent.<br /> <br /> -franzliszt<br /> <br /> ==Solution 4==<br /> In the problem, it states that the pitcher is &lt;math&gt;\frac{3}{4}&lt;/math&gt;, or &lt;math&gt;75\%&lt;/math&gt; full. So, we can just divide this by &lt;math&gt;5&lt;/math&gt; to get &lt;math&gt;\frac{75\%}{5}=15\%&lt;/math&gt;, which means that the answer is &lt;math&gt;\boxed{\textbf{(C) }15}&lt;/math&gt; <br /> ~aaja3427<br /> .<br /> ==Video Solution==<br /> https://youtu.be/ph_qAhXXKP4<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_4&diff=137882 2020 AMC 8 Problems/Problem 4 2020-11-19T22:59:08Z <p>Oceanxia: </p> <hr /> <div>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 /> 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 /> Diagram by sircalcsalot<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 /> ==Solution==<br /> We find the pattern &lt;math&gt;1, 6, 12, 18, \ldots&lt;/math&gt;. The sum of the first four numbers in this sequence is &lt;math&gt;\boxed{\textbf{(B) }37}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The first hexagon has &lt;math&gt;1&lt;/math&gt; dot. The second hexagon has &lt;math&gt;1+6&lt;/math&gt; dots. The third hexagon &lt;math&gt;1+6+12&lt;/math&gt; dots. Following this pattern, we predict that the fourth hexagon will have &lt;math&gt;1+6+12+18=37&lt;/math&gt; dots &lt;math&gt;\implies\boxed{\textbf{(B) }37}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> Each band adds &lt;math&gt;1, 6, 2(6), 3(6), 4(6) \cdots,&lt;/math&gt; so we have &lt;math&gt;1+6+2(6)+3(6)=1+6(6)=1+36=\boxed{\textbf{(B) }37}.&lt;/math&gt;<br /> <br /> [pog]<br /> <br /> ==Solution 4==<br /> <br /> Let the hexagon with &lt;math&gt;1&lt;/math&gt; dot be &lt;math&gt;h_0&lt;/math&gt;. Notice that the rest of the terms are generated by the recurrence relation &lt;math&gt;h_n=h_{n-1}+6n&lt;/math&gt; for &lt;math&gt;n&gt; 0&lt;/math&gt;. Using this, we find that &lt;math&gt;h_1=7,h_2=19,&lt;/math&gt; and &lt;math&gt;h_3=\textbf{(B) }37&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 5==<br /> <br /> Adding up the dots by rows in each hexagon, we see that the first hexagon has &lt;math&gt;1&lt;/math&gt; dot, the second has &lt;math&gt;2+3+2&lt;/math&gt; dots and the third has &lt;math&gt;3+4+5+4+3&lt;/math&gt; dots. Following the pattern, the fourth hexagon has &lt;math&gt;4+5+6+7+6+5+4=\boxed{\textbf{(B) }37}&lt;/math&gt;. Note yes this is what you do and if you don't undertstand just add them up to get 37.<br /> <br /> -vaporwave<br /> <br /> ==Video Solution==<br /> https://youtu.be/szWgrOPNw8c<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_3&diff=137881 2020 AMC 8 Problems/Problem 3 2020-11-19T22:57:54Z <p>Oceanxia: </p> <hr /> <div>==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 /> ==Solution==<br /> The answer is the product of the area of the field, the amount of strawberries per plant, and the amount of plants in one square feet. The answer is &lt;math&gt;6 \times 8 \times 10 \times 4 = 1920&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(D) }1920}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The area of the garden is &lt;math&gt;6&lt;/math&gt; ft &lt;math&gt;\times&lt;/math&gt; &lt;math&gt;8&lt;/math&gt; ft &lt;math&gt;= 48&lt;/math&gt; square feet. Since Carrie plants &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, it follows that she plants a total of &lt;math&gt;48 \times 4=192&lt;/math&gt; strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest &lt;math&gt;192 \times 10=1920&lt;/math&gt; strawberries &lt;math&gt;\implies\boxed{\textbf{(D) }1920}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> <br /> Note that &lt;math&gt;6\cdot 8 = 48&lt;/math&gt;, so Carrie has &lt;math&gt;4\cdot 48 = 192&lt;/math&gt; strawberry plants. Each plant produces &lt;math&gt;10&lt;/math&gt; strawberries, so the final answer is &lt;math&gt;192\cdot 10 = \textbf{(D)}\ 1920&lt;/math&gt;.'<br /> ==Solution 4== <br /> The answer is &lt;math&gt;6 \times 8 \times 10 \times 4 = 1920&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(D) }1920}&lt;/math&gt;.<br /> -oceanxia<br /> <br /> -franzliszt<br /> <br /> ==Video Solution==<br /> https://youtu.be/7S0wAZMy2ZQ<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_3&diff=137880 2020 AMC 8 Problems/Problem 3 2020-11-19T22:57:23Z <p>Oceanxia: </p> <hr /> <div>==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 /> ==Solution==<br /> The answer is the product of the area of the field, the amount of strawberries per plant, and the amount of plants in one square feet. The answer is &lt;math&gt;6 \times 8 \times 10 \times 4 = 1920&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(D) }1920}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> The area of the garden is &lt;math&gt;6&lt;/math&gt; ft &lt;math&gt;\times&lt;/math&gt; &lt;math&gt;8&lt;/math&gt; ft &lt;math&gt;= 48&lt;/math&gt; square feet. Since Carrie plants &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, it follows that she plants a total of &lt;math&gt;48 \times 4=192&lt;/math&gt; strawberry plants. Since each strawberry plant produces on average 10 strawberries, it follows that she can expect to harvest &lt;math&gt;192 \times 10=1920&lt;/math&gt; strawberries &lt;math&gt;\implies\boxed{\textbf{(D) }1920}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> <br /> Note that &lt;math&gt;6\cdot 8 = 48&lt;/math&gt;, so Carrie has &lt;math&gt;4\cdot 48 = 192&lt;/math&gt; strawberry plants. Each plant produces &lt;math&gt;10&lt;/math&gt; strawberries, so the final answer is &lt;math&gt;192\cdot 10 = \textbf{(D)}\ 1920&lt;/math&gt;.'<br /> ==Solution 4== <br /> The answer is &lt;math&gt;6 \times 8 \times 10 \times 4 = 1920&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(D) }1920}&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Video Solution==<br /> https://youtu.be/7S0wAZMy2ZQ<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_2&diff=137879 2020 AMC 8 Problems/Problem 2 2020-11-19T22:56:15Z <p>Oceanxia: </p> <hr /> <div>==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 /> ==Solution==<br /> First we average &lt;math&gt;15,20,25,40&lt;/math&gt; to get &lt;math&gt;25&lt;/math&gt;. Thus, &lt;math&gt;40 - 25 = \boxed{\textbf{(C) }\$15.}&lt;/math&gt;. ~~Spaced_Out Note if you get &lt;math&gt;100 once adding it up and then /4= &lt;/math&gt;15 or \boxed{\textbf{(C) }. <br /> &lt;math&gt;oceanxia<br /> ==Solution 2==<br /> The total earnings for the four friends is &lt;/math&gt;\$15+\$20+\$25+\$40=\$100&lt;math&gt;. Since they decided to split their earnings equally among themselves, it follows that each person will get &lt;/math&gt;\frac{\$100}{4}=\$25&lt;math&gt;. Since the friend who earned &lt;/math&gt;\$40&lt;math&gt; will need to leave with &lt;/math&gt;\$25&lt;math&gt;, he will have to give &lt;/math&gt;\$40-\$25=\$15&lt;math&gt; to the others &lt;/math&gt;\implies\boxed{\textbf{(C) }\$15}$.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Video Solution==<br /> https://youtu.be/-mSgttsOv2Y<br /> <br /> ~savannahsolver<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=1|num-a=3}}<br /> {{MAA Notice}}</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_1&diff=137878 2020 AMC 8 Problems/Problem 1 2020-11-19T22:52:58Z <p>Oceanxia: </p> <hr /> <div>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&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Luka will need &lt;math&gt;3\cdot 2=6&lt;/math&gt; cups of sugar and &lt;math&gt;6\cdot 4=24&lt;/math&gt; cups of water. The answer is &lt;math&gt;\boxed{\textbf{(E) } 24}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;W, S,&lt;/math&gt; and &lt;math&gt;L&lt;/math&gt; represent the number of cups of water, sugar, and lemon juice that Luka needs for his recipe, respectively. We are given that &lt;math&gt;W=4S&lt;/math&gt; and &lt;math&gt;S=2L&lt;/math&gt;. Since &lt;math&gt;L=3&lt;/math&gt;, it follows that &lt;math&gt;S=6&lt;/math&gt;, which in turn implies that &lt;math&gt;W=24 \implies\boxed{\textbf{(E) }24}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> We have that &lt;math&gt;\textsf{lemonade} : \textsf{water} : \textsf{lemon juice} = 8 : 2 : 1,&lt;/math&gt; so we have &lt;math&gt;3 \cdot 8 = \boxed{\textbf{(E) }24}.&lt;/math&gt;<br /> <br /> [pog]<br /> <br /> ==Solution 4==<br /> <br /> We are given that &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s=l&lt;/math&gt; which we combine to get &lt;math&gt;8w:2s:l&lt;/math&gt;. Letting all the variables equal &lt;math&gt;3&lt;/math&gt;, we find that the answer is &lt;math&gt;3\cdot 8=\textbf{(E)}\ 24&lt;/math&gt;.<br /> <br /> -franzliszt<br /> ==Solution 5==<br /> <br /> Put the numbers in ratios &lt;math&gt;4w:s&lt;/math&gt; and &lt;math&gt;2s:lj&lt;/math&gt; when w = water, s = sugar, and lj = lemon juice. then since we know there is &lt;math&gt;3&lt;/math&gt; cups of lemon juice, do the math. &lt;math&gt;3\cdot2\cdot4=6\cdot4=24&lt;/math&gt;<br /> <br /> ~ bsu1<br /> <br /> ==Video Solution==<br /> https://youtu.be/FPC792h-mGE<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|before=First problem|num-a=2}}<br /> {{MAA Notice}}<br /> we just get 4*3*2=24$ which is E<br /> -oceanxia</div> Oceanxia https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_10&diff=137877 2020 AMC 8 Problems/Problem 10 2020-11-19T22:51:52Z <p>Oceanxia: </p> <hr /> <div>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 /> ==Solution 1==<br /> By the [[Georgeooga-Harryooga Theorem]] there are &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{\textbf{(C) }12}&lt;/math&gt; way to arrange the marbles.<br /> <br /> <br /> Solution by [[User:RedFireTruck|RedFireTruck]]<br /> <br /> ==Solution 2==<br /> We can arrange our marbles like so &lt;math&gt;\square A\square B\square&lt;/math&gt;.<br /> <br /> To arrange the &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; we have &lt;math&gt;2!=2&lt;/math&gt; ways.<br /> <br /> To place the &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the blanks we have &lt;math&gt;_3P_2=6&lt;/math&gt; ways.<br /> <br /> By fundamental counting principle our final answer is &lt;math&gt;2\cdot6=\boxed{\textbf{(C) }12}&lt;/math&gt;<br /> <br /> <br /> Solution by [[User:Redfiretruck|RedFireTruck]]<br /> <br /> ==Solution 3==<br /> Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by &lt;math&gt;A,B,S,&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt;, respectively. If we ignore the constraint that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; cannot be next to each other, we get a total of &lt;math&gt;4!=24&lt;/math&gt; ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; can be next to each other. If we place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. &lt;math&gt;ST\square\square, \square ST\square, \square\square ST&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;TS\square\square, \square TS\square, \square\square TS&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; directly next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; next to each other. Subtracting this from the total number of arrangements gives us &lt;math&gt;24-12=12&lt;/math&gt; total arrangements &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> <br /> We can also solve this problem directly by looking at the number of ways that we can place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not directly next to each other. Observe that there are three ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; (in that order) into the four slots so they are not next to each other (i.e. &lt;math&gt;S\square T\square, \square S\square T, S\square\square T&lt;/math&gt;). However, we could also have placed &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; in the opposite order (i.e. &lt;math&gt;T\square S\square, \square T\square S, T\square\square S&lt;/math&gt;). Thus there are 6 ways of placing &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; so that they are not next to each other. Next, notice that for each of these placements, we have two open slots for placing &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. Specifically, we can place &lt;math&gt;A&lt;/math&gt; in the first open slot and &lt;math&gt;B&lt;/math&gt; in the second open slot or switch their order and place &lt;math&gt;B&lt;/math&gt; in the first open slot and &lt;math&gt;A&lt;/math&gt; in the second open slot. This gives us a total of &lt;math&gt;6\times 2=12&lt;/math&gt; ways to place &lt;math&gt;S&lt;/math&gt; and &lt;math&gt;T&lt;/math&gt; such that they are not next to each other &lt;math&gt;\implies\boxed{\textbf{(C) }12}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 4==<br /> Let's try complementary counting. There &lt;math&gt;4!&lt;/math&gt; ways to arrange the 4 marbles. However, there are &lt;math&gt;2\cdot3!&lt;/math&gt; arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, &lt;cmath&gt;4!-2\cdot3!=\boxed{12 \textbf{(C)}}&lt;/cmath&gt;<br /> <br /> ==Solution 5==<br /> <br /> We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a &quot;super marble.&quot; There are &lt;math&gt;2!&lt;/math&gt; ways to arrange Steelie and Tiger within this &quot;super marble.&quot; Then there are &lt;math&gt;3!&lt;/math&gt; ways to arrange the &quot;super marble&quot; and Zara's two other marbles in a row. Since there are &lt;math&gt;4!&lt;/math&gt; ways to arrange the marbles without any restrictions, the answer is given by &lt;math&gt;4!-2!\cdot 3!=\textbf{(C) }12&lt;/math&gt;<br /> <br /> -franzliszt<br /> <br /> ==Solution 6==<br /> <br /> We will use the following<br /> <br /> &lt;math&gt;\textbf{Georgeooga-Harryooga Theorem:}&lt;/math&gt; The [[Georgeooga-Harryooga Theorem]] states that if you have &lt;math&gt;a&lt;/math&gt; distinguishable objects and &lt;math&gt;b&lt;/math&gt; of them cannot be together, then there are &lt;math&gt;\frac{(a-b)!(a-b+1)!}{b!}&lt;/math&gt; ways to arrange the objects.<br /> <br /> &lt;math&gt;\textit{Proof. (Created by AoPS user redfiretruck)}&lt;/math&gt;<br /> <br /> Let our group of &lt;math&gt;a&lt;/math&gt; objects be represented like so &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, ..., &lt;math&gt;a-1&lt;/math&gt;, &lt;math&gt;a&lt;/math&gt;. Let the last &lt;math&gt;b&lt;/math&gt; objects be the ones we can't have together.<br /> <br /> Then we can organize our objects like so &lt;math&gt;\square1\square2\square3\square...\square a-b-1\square a-b\square&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;(a-b)!&lt;/math&gt; ways to arrange the objects in that list.<br /> <br /> Now we have &lt;math&gt;a-b+1&lt;/math&gt; blanks and &lt;math&gt;b&lt;/math&gt; other objects so we have &lt;math&gt;_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}&lt;/math&gt; ways to arrange the objects we can't put together.<br /> <br /> By fundamental counting principal our answer is &lt;math&gt;\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}&lt;/math&gt;.<br /> <br /> <br /> Proof by [[User:RedFireTruck|redfiretruck]]<br /> <br /> <br /> Back to the problem. By the [[Georgeooga-Harryooga Theorem]], our answer is &lt;math&gt;\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\textbf{(C) }12&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Video Solution==<br /> https://youtu.be/pB46JzBNM6g<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=9|num-a=11}}<br /> {{MAA Notice}}<br /> Solution 6 we can just get &lt;math&gt;\frac{(4!)/2=\textbf{(C) }12&lt;/math&gt;.</div> Oceanxia