https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ggs&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-20T14:19:27Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_14&diff=131862 2019 AMC 8 Problems/Problem 14 2020-08-15T18:50:43Z <p>Ggs: /* Solution 1 */</p> <hr /> <div>==Problem 14==<br /> Isabella has &lt;math&gt;6&lt;/math&gt; coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every &lt;math&gt;10&lt;/math&gt; days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the &lt;math&gt;6&lt;/math&gt; dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Monday}\qquad\textbf{(B) }\text{Tuesday}\qquad\textbf{(C) }\text{Wednesday}\qquad\textbf{(D) }\text{Thursday}\qquad\textbf{(E) }\text{Friday}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;\text{Day }1&lt;/math&gt; to &lt;math&gt;\text{Day }2&lt;/math&gt; denote a day where one coupon is redeemed and the day when the second coupon is redeemed. <br /> <br /> If she starts on a &lt;math&gt;\text{Monday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Thursday}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(A)}\ \text{Monday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Tuesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(B)}\ \text{Tuesday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Wednesday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Saturday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Saturday}&lt;/math&gt; to &lt;math&gt;\text{Tuesday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Tuesday}&lt;/math&gt; to &lt;math&gt;\text{Friday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Friday}&lt;/math&gt; to &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> And on &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her last coupon. <br /> <br /> <br /> No sunday occured thus &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt; is correct. <br /> <br /> <br /> Checking for the other options, <br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Thursday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;\textbf{(D)}\ \text{Thursday}&lt;/math&gt; is incorrect.<br /> <br /> <br /> If she starts on a &lt;math&gt;\text{Friday}&lt;/math&gt; she redeems her next coupon on &lt;math&gt;\text{Monday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Monday}&lt;/math&gt; to &lt;math&gt;\text{Thursday}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Thursday}&lt;/math&gt; to &lt;math&gt;\text{Sunday}&lt;/math&gt;.<br /> <br /> <br /> Checking for the other options gave us negative results, thus the answer is &lt;math&gt;\boxed{\textbf{(C)}\ \text{Wednesday}}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 2==<br /> Let <br /> <br /> &lt;math&gt;Sunday \equiv 0 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Monday \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Tuesday \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Wednesday \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Thursday \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Friday \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;Saturday \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;10 \equiv 3 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;20 \equiv 6 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;30 \equiv 2 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;40 \equiv 5 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;50 \equiv 1 \pmod{7}&lt;/math&gt;<br /> <br /> &lt;math&gt;60 \equiv 4 \pmod{7}&lt;/math&gt;<br /> <br /> <br /> Which clearly indicates if you start form a &lt;math&gt;x \equiv 3 \pmod{7}&lt;/math&gt; you will not get a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Any other starting value may lead to a &lt;math&gt;y \equiv 0 \pmod{7}&lt;/math&gt;.<br /> <br /> Which means our answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> ~phoenixfire<br /> <br /> == Solution 3 ==<br /> Like Solution 2, let the days of the week be numbers&lt;math&gt;\pmod 7&lt;/math&gt;. &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are coprime, so continuously adding &lt;math&gt;3&lt;/math&gt; to a number&lt;math&gt;\pmod 7&lt;/math&gt; will cycle through all numbers from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt;. If a string of 6 numbers in this cycle does not contain &lt;math&gt;0&lt;/math&gt;, then if you minus 3 from the first number of this cycle, it will always be &lt;math&gt;0&lt;/math&gt;. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. ~~SmileKat32<br /> <br /> == Solution 4 ==<br /> Since Sunday is the only day that has not been counted yet. We can just add the 3 days as it will become &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;. <br /> ~~ gorefeebuddie<br /> Note: This only works when 7 and 3 are relatively prime.<br /> <br /> == Solution 5 ==<br /> Let Sunday be Day 0, Monday be Day 1, Tuesday be Day 2, and so forth. We see that Sundays fall on Day &lt;math&gt;n&lt;/math&gt;, where n is a multiple of seven. If Isabella starts using her coupons on Monday (Day 1), she will fall on a Day that is a multiple of seven, a Sunday (her third coupon will be &quot;used&quot; on Day 21). Similarly, if she starts using her coupons on Tuesday (Day 2), Isabella will fall on a Day that is a multiple of seven (Day 42). Repeating this process, if she starts on Wednesday (Day 3), Isabella will first fall on a Day that is a multiple of seven, Day 63 (13, 23, 33, 43, 53 are not multiples of seven), but on her eleventh coupon, of which she only has ten. So, the answer is &lt;math&gt;\boxed{\textbf{(C)}\ Wednesday}&lt;/math&gt;.<br /> <br /> == Solution 6 ==<br /> Associated video - https://www.youtube.com/watch?v=LktgMtgb_8E<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=13|num-a=15}}<br /> <br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_22&diff=111281 2018 AMC 8 Problems/Problem 22 2019-11-12T15:38:51Z <p>Ggs: /* Solution 1 */</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 /> <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> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106190 1999 AMC 8 Problems/Problem 19 2019-06-08T23:37:37Z <p>Ggs: /* Problem &quot;Doesn't Make Sense&quot; */</p> <hr /> <div>==Problem==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans. Since fractional pans are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; pans.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; tablespoons of butter per pan, meaning &lt;math&gt;3 \cdot 15 = 45&lt;/math&gt; tablespoons of butter are required for &lt;math&gt;15&lt;/math&gt; pans.<br /> <br /> Each stick of butter has &lt;math&gt;8&lt;/math&gt; tablespoons, so we need &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of butter. However, we must round up again because partial sticks of butter are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106180 1999 AMC 8 Problems/Problem 19 2019-06-08T16:24:38Z <p>Ggs: /* Solution */</p> <hr /> <div>==Problem &quot;Doesn't Make Sense&quot;==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans. Since fractional pans are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; pans.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; tablespoons of butter per pan, meaning &lt;math&gt;3 \cdot 15 = 45&lt;/math&gt; tablespoons of butter are required for &lt;math&gt;15&lt;/math&gt; pans.<br /> <br /> Each stick of butter has &lt;math&gt;8&lt;/math&gt; tablespoons, so we need &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of butter. However, we must round up again because partial sticks of butter are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106179 1999 AMC 8 Problems/Problem 19 2019-06-08T16:23:52Z <p>Ggs: /* Solution */</p> <hr /> <div>==Problem &quot;Doesn't Make Sense&quot;==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans. Since fractional pans are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; pans.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; tablespoons of butter per pan, meaning &lt;math&gt;3 \cdot 15 = 45&lt;/math&gt; tables of butter are required for &lt;math&gt;15&lt;/math&gt; pans.<br /> <br /> Each stick of butter has &lt;math&gt;8&lt;/math&gt; tablespoons, so we need &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of butter. However, we must round up again because partial sticks of butter are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106178 1999 AMC 8 Problems/Problem 19 2019-06-08T16:23:16Z <p>Ggs: /* Solution */</p> <hr /> <div>==Problem &quot;Doesn't Make Sense&quot;==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans. Since fractional pans are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; pans.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; tablespoons of butter per pan, meaning &lt;math&gt;3 \cdot 15 = 45&lt;/math&gt; tables of butter are required for &lt;math&gt;15&lt;/math&gt; pans.<br /> <br /> Each stick of butter has &lt;math&gt;8&lt;/math&gt; tablespoons, so we need to replace the butter with &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of soy sauce. However, we must round up again because partial sticks of butter are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106177 1999 AMC 8 Problems/Problem 19 2019-06-08T16:22:00Z <p>Ggs: /* Solution */</p> <hr /> <div>==Problem &quot;Doesn't Make Sense&quot;==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans(For whapping, of course). Since fractional pans are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; BROWNies.<br /> <br /> There are &lt;math&gt;300&lt;/math&gt; tons of butter per cookie, meaning &lt;math&gt;3 \cdot 15 = 45&lt;/math&gt; tables of butter are required for &lt;math&gt;15&lt;/math&gt; pans.<br /> <br /> Each stick of butter has &lt;math&gt;8&lt;/math&gt; tablespoons, so we need to replace the butter with &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of soy sauce. However, we must run away again because partial sticks of butter are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106176 1999 AMC 8 Problems/Problem 19 2019-06-08T16:15:10Z <p>Ggs: /* Problem &quot;Doesn't Make Sense&quot; */</p> <hr /> <div>==Problem &quot;Doesn't Make Sense&quot;==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans(For whapping, of course). Since people are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; BROWNies.<br /> <br /> There are &lt;math&gt;300&lt;/math&gt; tons of butter per cookie, meaning &lt;math&gt;3 \cdot 15 = 4500676&lt;/math&gt; tables of butter are required for &lt;math&gt;1&lt;/math&gt; cookie crumb.<br /> <br /> Each stick of butter has no real butter, so we need to replace the butter with &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of soy sauce. However, we must run away again because we are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=1999_AMC_8_Problems/Problem_19&diff=106175 1999 AMC 8 Problems/Problem 19 2019-06-08T16:12:58Z <p>Ggs: Undo revision 104582 by Galaxyfighter09 (talk)</p> <hr /> <div>==Problem &quot;Doesn't Make Sense&quot;==<br /> <br /> At Central Middle School the 108 students who take the AMC10 meet in the evening to talk about food and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie's Best Bar Cookies this year. Their recipe, which makes a pan of 15 cookies, lists this items: &lt;math&gt;1\frac{1}{2}&lt;/math&gt; cups flour, &lt;math&gt;2&lt;/math&gt; eggs, &lt;math&gt;3&lt;/math&gt; tablespoons butter, &lt;math&gt;\frac{3}{4}&lt;/math&gt; cups sugar, and &lt;math&gt;1&lt;/math&gt; package of chocolate drops. They will not make full recipes, not partial recipes.<br /> <br /> The drummer gets sick. The concert is cancelled. Walter and Gretel must make enough pans of cookies to supply 216 cookies. There are 8 tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)<br /> <br /> &lt;math&gt;\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> For &lt;math&gt;216&lt;/math&gt; cookies, you need to make &lt;math&gt;\frac{216}{15} = 14.4&lt;/math&gt; pans(For whapping, of course). Since people are forbidden, round up to make &lt;math&gt;\lceil \frac{216}{15} \rceil = 15&lt;/math&gt; BROWNies.<br /> <br /> There are &lt;math&gt;300&lt;/math&gt; tons of butter per cookie, meaning &lt;math&gt;3 \cdot 15 = 4500676&lt;/math&gt; tables of butter are required for &lt;math&gt;1&lt;/math&gt; cookie crumb.<br /> <br /> Each stick of butter has no real butter, so we need to replace the butter with &lt;math&gt;\frac{45}{8} = 5.625&lt;/math&gt; sticks of soy sauce. However, we must run away again because we are forbidden! Thus, we need &lt;math&gt;\lceil \frac{45}{8} \rceil = 6&lt;/math&gt; sticks of butter, and the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=1999|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_17&diff=106112 2011 AMC 10A Problems/Problem 17 2019-06-04T16:58:52Z <p>Ggs: /* Solution 2 */</p> <hr /> <div>==Problem 17==<br /> In the eight-term sequence &lt;math&gt;A,B,C,D,E,F,G,H&lt;/math&gt;, the value of &lt;math&gt;C&lt;/math&gt; is 5 and the sum of any three consecutive terms is 30. What is &lt;math&gt;A+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We consider the sum &lt;math&gt;A+B+C+D+E+F+G+H&lt;/math&gt; and use the fact that &lt;math&gt;C=5&lt;/math&gt;, and hence &lt;math&gt;A+B=25&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\<br /> &amp;A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the two values we get for the sum, we get the answer &lt;math&gt;A+H+60=85&lt;/math&gt; &lt;math&gt;\Longrightarrow&lt;/math&gt; &lt;math&gt;A+H=\boxed{25 \ \mathbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Given that the sum of 3 consecutive terms is 30, we have<br /> &lt;math&gt;(A+B+C)+(C+D+E)+(F+G+H)=90&lt;/math&gt; and &lt;math&gt;(B+C+D)+(E+F+G)=60&lt;/math&gt;<br /> <br /> It follows that &lt;math&gt;A+B+C+D+E+F+G+H=85&lt;/math&gt; because &lt;math&gt;C=5&lt;/math&gt;.<br /> <br /> Subtracting, we have that &lt;math&gt;A+H=25\rightarrow \boxed{\textbf{C}}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> We see that &lt;math&gt;A+B+C=30&lt;/math&gt;, and by substituting the given &lt;math&gt;C=5&lt;/math&gt;, we find that &lt;math&gt;A+B=25&lt;/math&gt;. Similarly, &lt;math&gt;B+D=25&lt;/math&gt; and &lt;math&gt;D+E=25&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;(A+B)-(B+D)=A-D=0\\<br /> &amp;A=D\\<br /> &amp;(B+D)-(D+E)=B-E=0\\<br /> &amp;B=E\\<br /> &amp;A, B, 5, A, B, 5, G, H<br /> \end {align*}&lt;/cmath&gt;<br /> <br /> Similarly, &lt;math&gt;G=A&lt;/math&gt; and &lt;math&gt;H=B&lt;/math&gt;, giving us &lt;math&gt;A, B, 5, A, B, 5, A, B&lt;/math&gt;. Since &lt;math&gt;H=B&lt;/math&gt;, &lt;math&gt;A+H=A+B=\boxed{25 \ \mathbf{(C)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_17&diff=106111 2011 AMC 10A Problems/Problem 17 2019-06-04T16:58:42Z <p>Ggs: /* Solution 4 */</p> <hr /> <div>==Problem 17==<br /> In the eight-term sequence &lt;math&gt;A,B,C,D,E,F,G,H&lt;/math&gt;, the value of &lt;math&gt;C&lt;/math&gt; is 5 and the sum of any three consecutive terms is 30. What is &lt;math&gt;A+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We consider the sum &lt;math&gt;A+B+C+D+E+F+G+H&lt;/math&gt; and use the fact that &lt;math&gt;C=5&lt;/math&gt;, and hence &lt;math&gt;A+B=25&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\<br /> &amp;A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the two values we get for the sum, we get the answer &lt;math&gt;A+H+60=85&lt;/math&gt; &lt;math&gt;\Longrightarrow&lt;/math&gt; &lt;math&gt;A+H=\boxed{25 \ \mathbf{(C)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Given that the sum of 3 consecutive terms is 30, we have<br /> &lt;math&gt;(A+B+C)+(C+D+E)+(F+G+H)=90&lt;/math&gt; and &lt;math&gt;(B+C+D)+(E+F+G)=60&lt;/math&gt;<br /> <br /> It follows that &lt;math&gt;A+B+C+D+E+F+G+H=85&lt;/math&gt; because &lt;math&gt;C=5&lt;/math&gt;.<br /> <br /> Subtracting, we have that &lt;math&gt;A+H=25\rightarrow \boxed{\textbf{C}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> We see that &lt;math&gt;A+B+C=30&lt;/math&gt;, and by substituting the given &lt;math&gt;C=5&lt;/math&gt;, we find that &lt;math&gt;A+B=25&lt;/math&gt;. Similarly, &lt;math&gt;B+D=25&lt;/math&gt; and &lt;math&gt;D+E=25&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;(A+B)-(B+D)=A-D=0\\<br /> &amp;A=D\\<br /> &amp;(B+D)-(D+E)=B-E=0\\<br /> &amp;B=E\\<br /> &amp;A, B, 5, A, B, 5, G, H<br /> \end {align*}&lt;/cmath&gt;<br /> <br /> Similarly, &lt;math&gt;G=A&lt;/math&gt; and &lt;math&gt;H=B&lt;/math&gt;, giving us &lt;math&gt;A, B, 5, A, B, 5, A, B&lt;/math&gt;. Since &lt;math&gt;H=B&lt;/math&gt;, &lt;math&gt;A+H=A+B=\boxed{25 \ \mathbf{(C)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_17&diff=106110 2011 AMC 10A Problems/Problem 17 2019-06-04T16:58:16Z <p>Ggs: /* Solution 3 */</p> <hr /> <div>==Problem 17==<br /> In the eight-term sequence &lt;math&gt;A,B,C,D,E,F,G,H&lt;/math&gt;, the value of &lt;math&gt;C&lt;/math&gt; is 5 and the sum of any three consecutive terms is 30. What is &lt;math&gt;A+H&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A)}\,17 \qquad\text{(B)}\,18 \qquad\text{(C)}\,25 \qquad\text{(D)}\,26 \qquad\text{(E)}\,43&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We consider the sum &lt;math&gt;A+B+C+D+E+F+G+H&lt;/math&gt; and use the fact that &lt;math&gt;C=5&lt;/math&gt;, and hence &lt;math&gt;A+B=25&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;A+B+C+D+E+F+G+H=A+(B+C+D)+(E+F+G)+H=A+30+30+H=A+H+60\\<br /> &amp;A+B+C+D+E+F+G+H=(A+B)+(C+D+E)+(F+G+H)=25+30+30=85<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> Equating the two values we get for the sum, we get the answer &lt;math&gt;A+H+60=85&lt;/math&gt; &lt;math&gt;\Longrightarrow&lt;/math&gt; &lt;math&gt;A+H=\boxed{25 \ \mathbf{(C)}}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> Given that the sum of 3 consecutive terms is 30, we have<br /> &lt;math&gt;(A+B+C)+(C+D+E)+(F+G+H)=90&lt;/math&gt; and &lt;math&gt;(B+C+D)+(E+F+G)=60&lt;/math&gt;<br /> <br /> It follows that &lt;math&gt;A+B+C+D+E+F+G+H=85&lt;/math&gt; because &lt;math&gt;C=5&lt;/math&gt;.<br /> <br /> Subtracting, we have that &lt;math&gt;A+H=25\rightarrow \boxed{\textbf{C}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> We see that &lt;math&gt;A+B+C=30&lt;/math&gt;, and by substituting the given &lt;math&gt;C=5&lt;/math&gt;, we find that &lt;math&gt;A+B=25&lt;/math&gt;. Similarly, &lt;math&gt;B+D=25&lt;/math&gt; and &lt;math&gt;D+E=25&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> &amp;(A+B)-(B+D)=A-D=0\\<br /> &amp;A=D\\<br /> &amp;(B+D)-(D+E)=B-E=0\\<br /> &amp;B=E\\<br /> &amp;A, B, 5, A, B, 5, G, H<br /> \end {align*}&lt;/cmath&gt;<br /> <br /> Similarly, &lt;math&gt;G=A&lt;/math&gt; and &lt;math&gt;H=B&lt;/math&gt;, giving us &lt;math&gt;A, B, 5, A, B, 5, A, B&lt;/math&gt;. Since &lt;math&gt;H=B&lt;/math&gt;, &lt;math&gt;A+H=A+B=\boxed{25 \ \mathbf{(C)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2011|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_22&diff=106008 2016 AMC 10B Problems/Problem 22 2019-05-30T01:28:12Z <p>Ggs: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won &lt;math&gt;10&lt;/math&gt; games and lost &lt;math&gt;10&lt;/math&gt; games; there were no ties. How many sets of three teams &lt;math&gt;\{A, B, C\}&lt;/math&gt; were there in which &lt;math&gt;A&lt;/math&gt; beat &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; beat &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; beat &lt;math&gt;A?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 385 \qquad<br /> \textbf{(B)}\ 665 \qquad<br /> \textbf{(C)}\ 945 \qquad<br /> \textbf{(D)}\ 1140 \qquad<br /> \textbf{(E)}\ 1330&lt;/math&gt;<br /> <br /> ==Solution==<br /> There are &lt;math&gt;21&lt;/math&gt; teams. Any of the &lt;math&gt;\tbinom{21}3=1330&lt;/math&gt; sets of three teams must either be a fork (in which one team beat both the others) or a cycle:<br /> <br /> &lt;asy&gt;size(7cm);label(&quot;X&quot;,(5,5));label(&quot;Z&quot;,(10,0));label(&quot;Y&quot;,(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow);<br /> label(&quot;X&quot;,(20,5));label(&quot;Z&quot;,(25,0));label(&quot;Y&quot;,(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow);<br /> &lt;/asy&gt;<br /> But we know that every team beat exactly &lt;math&gt;10&lt;/math&gt; other teams, so for each possible &lt;math&gt;X&lt;/math&gt; at the head of a fork, there are always exactly &lt;math&gt;\tbinom{10}2&lt;/math&gt; choices for &lt;math&gt;Y&lt;/math&gt; and &lt;math&gt;Z&lt;/math&gt;. Therefore there are &lt;math&gt;21\cdot\tbinom{10}2=945&lt;/math&gt; forks, and all the rest must be cycles.<br /> <br /> Thus the answer is &lt;math&gt;1330-945=385&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Since there are &lt;math&gt;21&lt;/math&gt; teams and for each set of three teams there is a cycle, there are a total of &lt;math&gt;\tbinom{21}3=1330&lt;/math&gt; cycles of three teams. Because about &lt;math&gt;1/4&lt;/math&gt; of the cycles &lt;math&gt;\{A, B, C\}&lt;/math&gt; satisfy the conditions of the problems, our answer is close to &lt;math&gt;1/4*1330=332.5&lt;/math&gt;. Looking at the answer choices, we find that &lt;math&gt;332.5&lt;/math&gt; is closer to &lt;math&gt;385&lt;/math&gt; than any other answer choices, so our answer is &lt;math&gt;385&lt;/math&gt; which is &lt;math&gt;\boxed{\textbf{(A)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_24&diff=106001 2015 AMC 8 Problems/Problem 24 2019-05-29T16:50:59Z <p>Ggs: /* Solution 2 */</p> <hr /> <div>A baseball league consists of two four-team divisions. Each team plays every other team in its division &lt;math&gt;N&lt;/math&gt; games. Each team plays every team in the other division &lt;math&gt;M&lt;/math&gt; games with &lt;math&gt;N&gt;2M&lt;/math&gt; and &lt;math&gt;M&gt;4&lt;/math&gt;. Each team plays a 76 game schedule. How many games does a team play within its own division?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 36 \qquad<br /> \textbf{(B) } 48 \qquad<br /> \textbf{(C) } 54 \qquad<br /> \textbf{(D) } 60 \qquad<br /> \textbf{(E) } 72<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> On one team they play &lt;math&gt;\binom{3}{2}N&lt;/math&gt; games in their division and &lt;math&gt;4(M)&lt;/math&gt; games in the other. This gives &lt;math&gt;3N+4M=76&lt;/math&gt; <br /> <br /> Since &lt;math&gt;M&gt;4&lt;/math&gt; we start by trying &lt;math&gt;M=5&lt;/math&gt;. This doesn't work because &lt;math&gt;56&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Next, &lt;math&gt;M=6&lt;/math&gt; does not work because &lt;math&gt;52&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;<br /> <br /> We try &lt;math&gt;M=7&lt;/math&gt; this does work giving &lt;math&gt;N=16,~M=7&lt;/math&gt; and thus &lt;math&gt;3\times 16=\boxed{\textbf{(B)}~48}&lt;/math&gt; games in their division.<br /> <br /> &lt;math&gt;M=10&lt;/math&gt; seems to work,until we realize this gives &lt;math&gt;N=12&lt;/math&gt;, but &lt;math&gt;N&gt;2M&lt;/math&gt; so this will not work.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;76=3N+4M &gt; 10M&lt;/math&gt;, giving &lt;math&gt;M \le 7&lt;/math&gt;.<br /> Since &lt;math&gt;M&gt;4&lt;/math&gt;, we have &lt;math&gt;M=5,6,7&lt;/math&gt;.<br /> Since &lt;math&gt;4M&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\pmod{3}&lt;/math&gt;, we must have &lt;math&gt;M&lt;/math&gt; equal to &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\pmod{3}&lt;/math&gt;, so &lt;math&gt;M=7&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;3N=48&lt;/math&gt;, as desired. The answer is &lt;math&gt;\boxed{\textbf{(B)}~48}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_24&diff=106000 2015 AMC 8 Problems/Problem 24 2019-05-29T16:50:49Z <p>Ggs: /* Solution 1 */</p> <hr /> <div>A baseball league consists of two four-team divisions. Each team plays every other team in its division &lt;math&gt;N&lt;/math&gt; games. Each team plays every team in the other division &lt;math&gt;M&lt;/math&gt; games with &lt;math&gt;N&gt;2M&lt;/math&gt; and &lt;math&gt;M&gt;4&lt;/math&gt;. Each team plays a 76 game schedule. How many games does a team play within its own division?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 36 \qquad<br /> \textbf{(B) } 48 \qquad<br /> \textbf{(C) } 54 \qquad<br /> \textbf{(D) } 60 \qquad<br /> \textbf{(E) } 72<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> On one team they play &lt;math&gt;\binom{3}{2}N&lt;/math&gt; games in their division and &lt;math&gt;4(M)&lt;/math&gt; games in the other. This gives &lt;math&gt;3N+4M=76&lt;/math&gt; <br /> <br /> Since &lt;math&gt;M&gt;4&lt;/math&gt; we start by trying &lt;math&gt;M=5&lt;/math&gt;. This doesn't work because &lt;math&gt;56&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Next, &lt;math&gt;M=6&lt;/math&gt; does not work because &lt;math&gt;52&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;<br /> <br /> We try &lt;math&gt;M=7&lt;/math&gt; this does work giving &lt;math&gt;N=16,~M=7&lt;/math&gt; and thus &lt;math&gt;3\times 16=\boxed{\textbf{(B)}~48}&lt;/math&gt; games in their division.<br /> <br /> &lt;math&gt;M=10&lt;/math&gt; seems to work,until we realize this gives &lt;math&gt;N=12&lt;/math&gt;, but &lt;math&gt;N&gt;2M&lt;/math&gt; so this will not work.<br /> <br /> ===Solution 2===<br /> &lt;math&gt;76=3N+4M &gt; 10M&lt;/math&gt;, giving &lt;math&gt;M \le 7&lt;/math&gt;.<br /> Since &lt;math&gt;M&gt;4&lt;/math&gt;, we have &lt;math&gt;M=5,6,7&lt;/math&gt;.<br /> Since &lt;math&gt;4M&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\pmod{3}&lt;/math&gt;, we must have &lt;math&gt;M&lt;/math&gt; equal to &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\pmod{3}&lt;/math&gt;, so &lt;math&gt;M=7&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;3N=48&lt;/math&gt;, as desired. The answer is &lt;math&gt;\boxed{\textbf{(B)}~48}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_24&diff=105999 2015 AMC 8 Problems/Problem 24 2019-05-29T16:50:34Z <p>Ggs: /* Solution 1 */</p> <hr /> <div>A baseball league consists of two four-team divisions. Each team plays every other team in its division &lt;math&gt;N&lt;/math&gt; games. Each team plays every team in the other division &lt;math&gt;M&lt;/math&gt; games with &lt;math&gt;N&gt;2M&lt;/math&gt; and &lt;math&gt;M&gt;4&lt;/math&gt;. Each team plays a 76 game schedule. How many games does a team play within its own division?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 36 \qquad<br /> \textbf{(B) } 48 \qquad<br /> \textbf{(C) } 54 \qquad<br /> \textbf{(D) } 60 \qquad<br /> \textbf{(E) } 72<br /> &lt;/math&gt;<br /> <br /> ====Solution 1====<br /> On one team they play &lt;math&gt;\binom{3}{2}N&lt;/math&gt; games in their division and &lt;math&gt;4(M)&lt;/math&gt; games in the other. This gives &lt;math&gt;3N+4M=76&lt;/math&gt; <br /> <br /> Since &lt;math&gt;M&gt;4&lt;/math&gt; we start by trying &lt;math&gt;M=5&lt;/math&gt;. This doesn't work because &lt;math&gt;56&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;.<br /> <br /> Next, &lt;math&gt;M=6&lt;/math&gt; does not work because &lt;math&gt;52&lt;/math&gt; is not divisible by &lt;math&gt;3&lt;/math&gt;<br /> <br /> We try &lt;math&gt;M=7&lt;/math&gt; this does work giving &lt;math&gt;N=16,~M=7&lt;/math&gt; and thus &lt;math&gt;3\times 16=\boxed{\textbf{(B)}~48}&lt;/math&gt; games in their division.<br /> <br /> &lt;math&gt;M=10&lt;/math&gt; seems to work,until we realize this gives &lt;math&gt;N=12&lt;/math&gt;, but &lt;math&gt;N&gt;2M&lt;/math&gt; so this will not work.<br /> <br /> ===Solution 2===<br /> &lt;math&gt;76=3N+4M &gt; 10M&lt;/math&gt;, giving &lt;math&gt;M \le 7&lt;/math&gt;.<br /> Since &lt;math&gt;M&gt;4&lt;/math&gt;, we have &lt;math&gt;M=5,6,7&lt;/math&gt;.<br /> Since &lt;math&gt;4M&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\pmod{3}&lt;/math&gt;, we must have &lt;math&gt;M&lt;/math&gt; equal to &lt;math&gt;1&lt;/math&gt; &lt;math&gt;\pmod{3}&lt;/math&gt;, so &lt;math&gt;M=7&lt;/math&gt;.<br /> <br /> This gives &lt;math&gt;3N=48&lt;/math&gt;, as desired. The answer is &lt;math&gt;\boxed{\textbf{(B)}~48}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=105997 2016 AMC 8 Problems/Problem 23 2019-05-29T01:19:32Z <p>Ggs: /* Solution 1 */</p> <hr /> <div>Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing the diagram:<br /> <br /> &lt;asy<br /> <br /> label(&quot;&lt;math&gt;D&lt;/math&gt;&quot;, D, SE);<br /> label(&quot;&lt;math&gt;E&lt;/math&gt;&quot;, E, N);<br /> &lt;/asy&gt;%<br /> <br /> we see that &lt;math&gt;\triangle EAB&lt;/math&gt; is equilateral as each side is the radius of one of the two circles. Therefore, &lt;math&gt;\overarc{EB}=m\angle EAB=60^\circ&lt;/math&gt;. Therefore, since it is an inscribed angle<br /> <br /> ==Solution 2==<br /> As in Solution 1, observe that &lt;math&gt;\triangle{EAB}&lt;/math&gt; is equilateral. Therefore, &lt;math&gt;m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}&lt;/math&gt;. Since &lt;math&gt;CD&lt;/math&gt; is a straight line, we conclude that &lt;math&gt;m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;. Since &lt;math&gt;BE=BD&lt;/math&gt; (both are radii of the same circle), &lt;math&gt;\triangle{BED}&lt;/math&gt; is isosceles, meaning that &lt;math&gt;m\angle{BED}=m\angle{BDE}=30^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;m\angle{AEC}=m\angle{ACE}=30^{\circ}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{AMC8 box|year=2016|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=105996 2016 AMC 8 Problems/Problem 23 2019-05-29T01:19:09Z <p>Ggs: /* Solution 1 */</p> <hr /> <div>Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing the diagram:<br /> <br /> &lt;math&gt;&lt;asy<br /> <br /> label(&quot;&lt;/math&gt;D&lt;math&gt;&quot;, D, SE);<br /> label(&quot;&lt;/math&gt;E&lt;math&gt;&quot;, E, N);<br /> &lt;/asy&gt;%&lt;/math&gt;<br /> <br /> we see that &lt;math&gt;\triangle EAB&lt;/math&gt; is equilateral as each side is the radius of one of the two circles. Therefore, &lt;math&gt;\overarc{EB}=m\angle EAB=60^\circ&lt;/math&gt;. Therefore, since it is an inscribed angle<br /> <br /> ==Solution 2==<br /> As in Solution 1, observe that &lt;math&gt;\triangle{EAB}&lt;/math&gt; is equilateral. Therefore, &lt;math&gt;m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}&lt;/math&gt;. Since &lt;math&gt;CD&lt;/math&gt; is a straight line, we conclude that &lt;math&gt;m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;. Since &lt;math&gt;BE=BD&lt;/math&gt; (both are radii of the same circle), &lt;math&gt;\triangle{BED}&lt;/math&gt; is isosceles, meaning that &lt;math&gt;m\angle{BED}=m\angle{BDE}=30^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;m\angle{AEC}=m\angle{ACE}=30^{\circ}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{AMC8 box|year=2016|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_23&diff=105995 2016 AMC 8 Problems/Problem 23 2019-05-29T01:18:36Z <p>Ggs: /* Solution 1 */</p> <hr /> <div>Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Drawing the diagram:<br /> <br /> &lt;math&gt;&lt;asy<br /> <br /> label(&quot;&lt;/math&gt;D&lt;math&gt;&quot;, D, SE);<br /> label(&quot;&lt;/math&gt;E&lt;math&gt;&quot;, E, N);<br /> &lt;/asy&gt;%<br /> <br /> we see that &lt;/math&gt;\triangle EAB&lt;math&gt; is equilateral as each side is the radius of one of the two circles. Therefore, &lt;/math&gt;\overarc{EB}=m\angle EAB=60^\circ\$. Therefore, since it is an inscribed angle<br /> <br /> ==Solution 2==<br /> As in Solution 1, observe that &lt;math&gt;\triangle{EAB}&lt;/math&gt; is equilateral. Therefore, &lt;math&gt;m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}&lt;/math&gt;. Since &lt;math&gt;CD&lt;/math&gt; is a straight line, we conclude that &lt;math&gt;m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}&lt;/math&gt;. Since &lt;math&gt;BE=BD&lt;/math&gt; (both are radii of the same circle), &lt;math&gt;\triangle{BED}&lt;/math&gt; is isosceles, meaning that &lt;math&gt;m\angle{BED}=m\angle{BDE}=30^{\circ}&lt;/math&gt;. Similarly, &lt;math&gt;m\angle{AEC}=m\angle{ACE}=30^{\circ}&lt;/math&gt;. <br /> <br /> Now, &lt;math&gt;\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(C) }\ 120}&lt;/math&gt;.<br /> <br /> <br /> <br /> {{AMC8 box|year=2016|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Ggs https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_3&diff=105979 2017 AMC 8 Problems/Problem 3 2019-05-27T16:28:22Z <p>Ggs: /* Solution */</p> <hr /> <div>==Problem 3==<br /> <br /> What is the value of the expression &lt;math&gt;\sqrt{16\sqrt{8\sqrt{4}}}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;\sqrt{16\sqrt{8\sqrt{4}}}&lt;/math&gt; = &lt;math&gt;\sqrt{16\sqrt{8\cdot 2}}&lt;/math&gt; = &lt;math&gt;\sqrt{16\sqrt{16}}&lt;/math&gt; = &lt;math&gt;\sqrt{16\cdot 4}&lt;/math&gt; = &lt;math&gt;\sqrt{64}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(C)}\ 8}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=2|num-a=4}}<br /> <br /> {{MAA Notice}}</div> Ggs