https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Moony+eyed&feedformat=atom AoPS Wiki - User contributions [en] 2021-07-30T15:14:41Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_19&diff=137568 2020 AMC 8 Problems/Problem 19 2020-11-18T15:56:07Z <p>Moony eyed: /* Solution 1 */</p> <hr /> <div>==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> To be divisible by &lt;math&gt;15&lt;/math&gt;, a number must first be divisible by &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. By divisibility rules, the last digit must be either &lt;math&gt;5&lt;/math&gt; or &lt;math&gt;0&lt;/math&gt;, and the sum of the digits must be divisible by &lt;math&gt;3&lt;/math&gt;. If the last digit is &lt;math&gt;0&lt;/math&gt;, the first digit would be &lt;math&gt;0&lt;/math&gt; (because the digits alternate). So, the last digit must be &lt;math&gt;5&lt;/math&gt;, and we have &lt;cmath&gt;5+x+5+x+5 \equiv 0 \pmod{3}&lt;/cmath&gt; &lt;cmath&gt;2x \equiv 0 \pmod{3}&lt;/cmath&gt; We know the inverse exists because 2 is relatively prime to 3, and thus we can conclude that &lt;math&gt;x&lt;/math&gt; (or the second and fourth digits) is always a multiple of &lt;math&gt;3&lt;/math&gt;. We have 4 options: &lt;math&gt;0, 3, 6, 9&lt;/math&gt;, and our answer is &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;\boxed{\textbf{(B) } 4}&lt;/math&gt; ~samrocksnature<br /> <br /> ==Solution 2==<br /> A number that is divisible by &lt;math&gt;15&lt;/math&gt; must be divisible by &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. To be divisible by &lt;math&gt;3&lt;/math&gt;, the sum of its digits must be divisible by &lt;math&gt;3&lt;/math&gt; and to be divisible by &lt;math&gt;5&lt;/math&gt;, it must end in a &lt;math&gt;5&lt;/math&gt; or a &lt;math&gt;0&lt;/math&gt;. Observe that a five-digit flippy number must start and end with the same digit. Since a five-digit number cannot start with &lt;math&gt;0&lt;/math&gt;, it also cannot end in &lt;math&gt;0&lt;/math&gt;. This means that the numbers that we are looking for must end in &lt;math&gt;5&lt;/math&gt;. This also means that they must start with &lt;math&gt;5&lt;/math&gt; and alternate with &lt;math&gt;5&lt;/math&gt; (i.e. the number must be of the form &lt;math&gt;5\square5\square5&lt;/math&gt;). The two digits between the &lt;math&gt;5s&lt;/math&gt; must be the same. Let's call that digit &lt;math&gt;x&lt;/math&gt;. We know that the sum of the digits must be a multiple of &lt;math&gt;3&lt;/math&gt;. Since the sum of the three &lt;math&gt;5s&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt; which is already a multiple of &lt;math&gt;3&lt;/math&gt;, for the entire five-digit number to be a multiple of &lt;math&gt;3&lt;/math&gt;, it must also be the case that &lt;math&gt;x+x=2x&lt;/math&gt; is also a multiple of &lt;math&gt;3&lt;/math&gt;. Thus, the problem reduces to finding the number of digits from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt; for which &lt;math&gt;2x&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. This leads to &lt;math&gt;x=0,3,6,&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; and we have four valid answers (i.e. &lt;math&gt;50505,53535,56565,&lt;/math&gt; and &lt;math&gt;59595&lt;/math&gt;) &lt;math&gt;\implies\boxed{\textbf{(B) } 4}&lt;/math&gt;.&lt;br&gt;<br /> ~ junaidmansuri<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Moony eyed https://artofproblemsolving.com/wiki/index.php?title=User:Edud_looc&diff=131737 User:Edud looc 2020-08-13T17:43:23Z <p>Moony eyed: </p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;102px&quot;&gt;10<br /> &lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;eduD_looC is 11 years old.&lt;br&gt;<br /> <br /> eduD_looC is a pro at maths and chess<br /> <br /> eduD_looC is much better than OlympusHero at maths<br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;A User Count of 400<br /> <br /> Get at least 8 on AIME 2021 (Got 5 in AOIME)<br /> <br /> Get distinguished on AMC 10 2021<br /> <br /> &lt;/div&gt;<br /> &lt;/div&gt;</div> Moony eyed https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=129303 2018 AMC 8 Problems/Problem 22 2020-07-26T02:04:31Z <p>Moony eyed: I just added another solution =)</p> <hr /> <div>==Problem 22==<br /> Point &lt;math&gt;E&lt;/math&gt; is the midpoint of side &lt;math&gt;\overline{CD}&lt;/math&gt; in square &lt;math&gt;ABCD,&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; meets diagonal &lt;math&gt;\overline{AC}&lt;/math&gt; at &lt;math&gt;F.&lt;/math&gt; The area of quadrilateral &lt;math&gt;AFED&lt;/math&gt; is &lt;math&gt;45.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> size(5cm);<br /> draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);<br /> draw((0,6)--(6,0)); draw((3,0)--(6,6));<br /> label(&quot;$A$&quot;,(0,6),NW);<br /> label(&quot;$B$&quot;,(6,6),NE);<br /> label(&quot;$C$&quot;,(6,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$E$&quot;,(3,0),S);<br /> label(&quot;$F$&quot;,(4,2),E);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let the area of &lt;math&gt;\triangle CEF&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt;. Thus, the area of triangle &lt;math&gt;\triangle ACD&lt;/math&gt; is &lt;math&gt;45+x&lt;/math&gt; and the area of the square is &lt;math&gt;2(45+x) = 90+2x&lt;/math&gt;.<br /> <br /> By AA similarity, &lt;math&gt;\triangle CEF \sim \triangle ABF&lt;/math&gt; with a 1:2 ratio, so the area of triangle &lt;math&gt;\triangle ABF&lt;/math&gt; is &lt;math&gt;4x&lt;/math&gt;. Now consider trapezoid &lt;math&gt;ABED&lt;/math&gt;. Its area is &lt;math&gt;45+4x&lt;/math&gt;, which is three-fourths the area of the square. We set up an equation in &lt;math&gt;x&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 45+4x = \frac{3}{4}\left(90+2x\right) &lt;/cmath&gt;<br /> Solving, we get &lt;math&gt;x = 9&lt;/math&gt;. The area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We can use analytic geometry for this problem.<br /> <br /> Let us start by giving &lt;math&gt;D&lt;/math&gt; the coordinate &lt;math&gt;(0,0)&lt;/math&gt;, &lt;math&gt;A&lt;/math&gt; the coordinate &lt;math&gt;(0,1)&lt;/math&gt;, and so forth. &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{EB}&lt;/math&gt; can be represented by the equations &lt;math&gt;y=-x+1&lt;/math&gt; and &lt;math&gt;y=2x-1&lt;/math&gt;, respectively. Solving for their intersection gives point &lt;math&gt;F&lt;/math&gt; coordinates &lt;math&gt;\left(\frac{2}{3},\frac{1}{3}\right)&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\triangle&lt;/math&gt;&lt;math&gt;EFC&lt;/math&gt;’s area is simply &lt;math&gt;\frac{\frac{1}{2}\cdot\frac{1}{3}}{2}&lt;/math&gt; or &lt;math&gt;\frac{1}{12}&lt;/math&gt;. This means that pentagon &lt;math&gt;ABCEF&lt;/math&gt;’s area is &lt;math&gt;\frac{1}{2}+\frac{1}{12}=\frac{7}{12}&lt;/math&gt; of the entire square, and it follows that quadrilateral &lt;math&gt;AFED&lt;/math&gt;’s area is &lt;math&gt;\frac{5}{12}&lt;/math&gt; of the square. <br /> <br /> The area of the square is then &lt;math&gt;\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}&lt;/math&gt;.<br /> ==Solution 3==<br /> Note that triangle &lt;math&gt;ABC&lt;/math&gt; has half the area of the square and triangle &lt;math&gt;FEC&lt;/math&gt; has &lt;math&gt;\dfrac1{12}&lt;/math&gt;th. Thus the area of the quadrilateral is &lt;math&gt;1-1/2-1/12=5/12&lt;/math&gt;th the area of the square. The area of the square is then &lt;math&gt;45\cdot\dfrac{12}{5}=\boxed{\textbf{(B.)}108}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Extend &lt;math&gt;\overline{AD}&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; to meet at &lt;math&gt;X&lt;/math&gt;. Drop an altitude from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{CE}&lt;/math&gt; and call it &lt;math&gt;h&lt;/math&gt;. Also, call &lt;math&gt;\overline{CE}&lt;/math&gt; &lt;math&gt;x&lt;/math&gt;. As stated before, we have &lt;math&gt;\triangle ABF \sim \triangle CEF&lt;/math&gt;, so the ratio of their heights is in a &lt;math&gt;1:2&lt;/math&gt; ratio, making the altitude from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;\overline{AB}&lt;/math&gt; &lt;math&gt;2h&lt;/math&gt;. Note that this means that the side of the square is &lt;math&gt;3h&lt;/math&gt;. In addition, &lt;math&gt;\triangle XDE \sim \triangle XAB&lt;/math&gt; by AA Similarity in a &lt;math&gt;1:2&lt;/math&gt; ratio. This means that the side length of the square is &lt;math&gt;2x&lt;/math&gt;, making &lt;math&gt;3h=2x&lt;/math&gt;.<br /> <br /> Now, note that &lt;math&gt;[ADEF]=[XAB]-[XDE]-[ABF]&lt;/math&gt;. We have &lt;math&gt;[\triangle XAB]=(4x)(2x)/2=4x^2,&lt;/math&gt; &lt;math&gt;[\triangle XDE]=(x)(2x)/2=x^2,&lt;/math&gt; and &lt;math&gt;[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.&lt;/math&gt; Subtracting makes &lt;math&gt;[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.&lt;/math&gt; We are given that &lt;math&gt;[ADEF]=45,&lt;/math&gt; so &lt;math&gt;5x^2/3=45 \Rightarrow x^2=27.&lt;/math&gt; Therefore, &lt;math&gt;x= 3 \sqrt{3},&lt;/math&gt; so our answer is &lt;math&gt;(2x)^2=4x^2=4(27)=\boxed{\textbf{(B) }108}.&lt;/math&gt; - moony_eyed<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=c4_-h7DsZFg - Happytwin<br /> <br /> =See Also=<br /> {{AMC8 box|year=2018|num-b=21|num-a=23}}<br /> Set s to be the bottom left triangle.<br /> {{MAA Notice}}</div> Moony eyed