https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ultrajj1&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-10T20:17:47Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_5&diff=111325 2016 AMC 8 Problems/Problem 5 2019-11-12T22:19:17Z <p>Ultrajj1: /* Solution */</p> <hr /> <div>The number &lt;math&gt;N&lt;/math&gt; is a two-digit number.<br /> <br /> • When &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;9&lt;/math&gt;, the remainder is &lt;math&gt;1&lt;/math&gt;.<br /> <br /> • When &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;10&lt;/math&gt;, the remainder is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;11&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> when x equals y what is 2345432x?<br /> <br /> IF U DONT KNOW DIS U ARE FREGING IDIOT</div> Ultrajj1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_4&diff=111324 2016 AMC 8 Problems/Problem 4 2019-11-12T22:18:05Z <p>Ultrajj1: /* Solution */</p> <hr /> <div>When Cheenu was a boy he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;3&lt;/math&gt; hours and &lt;math&gt;30&lt;/math&gt; minutes. As an old man he can now walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;4&lt;/math&gt; hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30&lt;/math&gt;</div> Ultrajj1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_25&diff=111060 2017 AMC 8 Problems/Problem 25 2019-11-10T03:23:20Z <p>Ultrajj1: /* Solution */</p> <hr /> <div>==Problem 25==<br /> <br /> In the figure shown, &lt;math&gt;\overline{US}&lt;/math&gt; and &lt;math&gt;\overline{UT}&lt;/math&gt; are line segments each of length 2, and &lt;math&gt;m\angle TUS = 60^\circ&lt;/math&gt;. Arcs &lt;math&gt;\overarc{TR}&lt;/math&gt; and &lt;math&gt;\overarc{SR}&lt;/math&gt; are each one-sixth of a circle with radius 2. What is the area of the region shown? <br /> <br /> &lt;asy&gt;draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label(&quot;$U$&quot;, (2,3.464), N); label(&quot;$S$&quot;, (1,1.732), W); label(&quot;$T$&quot;, (3,1.732), E); label(&quot;$R$&quot;, (2,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> Ultrajj1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_23&diff=111057 2017 AMC 8 Problems/Problem 23 2019-11-10T03:00:34Z <p>Ultrajj1: </p> <hr /> <div>==Problem 23==<br /> <br /> Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Ultrajj1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_23&diff=111056 2017 AMC 8 Problems/Problem 23 2019-11-10T02:58:33Z <p>Ultrajj1: /* Solution */</p> <hr /> <div>==Problem 23==<br /> <br /> Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> It is well known that &lt;math&gt;\text{Distance}=\text{Speed} \cdot \text{Time}&lt;/math&gt;. In the question, we want distance. From the question, we have that the time is &lt;math&gt;60&lt;/math&gt; minutes or &lt;math&gt;1&lt;/math&gt; hour. By the equation derived from &lt;math&gt;\text{Distance}=\text{Speed} \cdot \text{Time}&lt;/math&gt;, we have &lt;math&gt;\text{Speed}=\frac{\text{Distance}}{\text{Time}}&lt;/math&gt;, so the speed is &lt;math&gt;1&lt;/math&gt; mile per &lt;math&gt;x&lt;/math&gt; minutes. Because we want the distance, we multiply the time and speed together yielding &lt;math&gt;60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}&lt;/math&gt;. The minutes cancel out, so now we have &lt;math&gt;\dfrac{60}{x}&lt;/math&gt; as our distance for the first day. The distance for the following days are:<br /> &lt;cmath&gt;\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}&lt;/cmath&gt;<br /> We know that &lt;math&gt;x,x+5,x+10,x+15&lt;/math&gt; are all factors of &lt;math&gt;60&lt;/math&gt;, therefore, &lt;math&gt;x=5&lt;/math&gt; because the factors have to be in an arithmetic sequence with the common difference being &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;x=5&lt;/math&gt; is the only solution.<br /> &lt;cmath&gt;\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}&lt;/cmath&gt; &lt;math&gt;\LaTeX&lt;/math&gt; YEET<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Ultrajj1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_23&diff=111055 2017 AMC 8 Problems/Problem 23 2019-11-10T02:58:21Z <p>Ultrajj1: /* Solution */</p> <hr /> <div>==Problem 23==<br /> <br /> Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> It is well known that &lt;math&gt;\text{Distance}=\text{Speed} \cdot \text{Time}&lt;/math&gt;. In the question, we want distance. From the question, we have that the time is &lt;math&gt;60&lt;/math&gt; minutes or &lt;math&gt;1&lt;/math&gt; hour. By the equation derived from &lt;math&gt;\text{Distance}=\text{Speed} \cdot \text{Time}&lt;/math&gt;, we have &lt;math&gt;\text{Speed}=\frac{\text{Distance}}{\text{Time}}&lt;/math&gt;, so the speed is &lt;math&gt;1&lt;/math&gt; mile per &lt;math&gt;x&lt;/math&gt; minutes. Because we want the distance, we multiply the time and speed together yielding &lt;math&gt;60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}&lt;/math&gt;. The minutes cancel out, so now we have &lt;math&gt;\dfrac{60}{x}&lt;/math&gt; as our distance for the first day. The distance for the following days are:<br /> &lt;cmath&gt;\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}&lt;/cmath&gt;<br /> We know that &lt;math&gt;x,x+5,x+10,x+15&lt;/math&gt; are all factors of &lt;math&gt;60&lt;/math&gt;, therefore, &lt;math&gt;x=5&lt;/math&gt; because the factors have to be in an arithmetic sequence with the common difference being &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;x=5&lt;/math&gt; is the only solution.<br /> &lt;cmath&gt;\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}&lt;/cmath&gt; &lt;math&gt;LaTeX&lt;/math&gt; YEET<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Ultrajj1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_23&diff=111054 2017 AMC 8 Problems/Problem 23 2019-11-10T02:57:57Z <p>Ultrajj1: /* Solution */</p> <hr /> <div>==Problem 23==<br /> <br /> Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?<br /> <br /> &lt;math&gt;\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> It is well known that &lt;math&gt;\text{Distance}=\text{Speed} \cdot \text{Time}&lt;/math&gt;. In the question, we want distance. From the question, we have that the time is &lt;math&gt;60&lt;/math&gt; minutes or &lt;math&gt;1&lt;/math&gt; hour. By the equation derived from &lt;math&gt;\text{Distance}=\text{Speed} \cdot \text{Time}&lt;/math&gt;, we have &lt;math&gt;\text{Speed}=\frac{\text{Distance}}{\text{Time}}&lt;/math&gt;, so the speed is &lt;math&gt;1&lt;/math&gt; mile per &lt;math&gt;x&lt;/math&gt; minutes. Because we want the distance, we multiply the time and speed together yielding &lt;math&gt;60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}&lt;/math&gt;. The minutes cancel out, so now we have &lt;math&gt;\dfrac{60}{x}&lt;/math&gt; as our distance for the first day. The distance for the following days are:<br /> &lt;cmath&gt;\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}&lt;/cmath&gt;<br /> We know that &lt;math&gt;x,x+5,x+10,x+15&lt;/math&gt; are all factors of &lt;math&gt;60&lt;/math&gt;, therefore, &lt;math&gt;x=5&lt;/math&gt; because the factors have to be in an arithmetic sequence with the common difference being &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;x=5&lt;/math&gt; is the only solution.<br /> &lt;cmath&gt;\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}&lt;/cmath&gt; &lt;math&gt;\Latex&lt;/math&gt; YEET<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=22|num-a=24}}<br /> <br /> {{MAA Notice}}</div> Ultrajj1