https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Obtuse&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-05T19:06:12Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality&diff=122616 Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality 2020-05-19T01:09:17Z <p>Obtuse: Undo revision 122614 by Severusslouch (talk)</p> <hr /> <div>The '''Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality''' (RMS-AM-GM-HM), is an [[inequality]] of the [[root-mean square]], [[arithmetic mean]], [[geometric mean]], and [[harmonic mean]] of a set of [[positive]] [[real number]]s &lt;math&gt;x_1,\ldots,x_n&lt;/math&gt; that says:<br /> <br /> &lt;cmath&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/cmath&gt;<br /> <br /> with equality if and only if &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;. This inequality can be expanded to the [[power mean inequality]].<br /> <br /> As a consequence we can have the following inequality:<br /> If &lt;math&gt;x_1,x_2,\cdots,x_n&lt;/math&gt; are positive reals, then <br /> &lt;cmath&gt;(x_1+x_2+\cdots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots \frac{1}{x_n}\right) \geq n^2&lt;/cmath&gt;<br /> with equality if and only if &lt;math&gt;x_1=x_2=\cdots=x_n&lt;/math&gt;; which follows directly by cross multiplication from the AM-HM inequality. This is extremely useful in problem solving.<br /> <br /> The Root Mean Square is also known as the [[quadratic mean]], and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality. <br /> <br /> == Proof ==<br /> <br /> The inequality &lt;math&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}&lt;/math&gt; is a direct consequence of the [[Cauchy-Schwarz Inequality]];<br /> &lt;cmath&gt;(x_1^2+x_2^2+\cdots +x_n^2)(1+1+\cdots +1)\geq (x_1+x_2+\cdots +x_n)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}\geq \left(\frac{x_1+x_2+\cdots +x_n}{n}\right)^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n}&lt;/cmath&gt;<br /> Alternatively, the RMS-AM can be proved using Jensen's inequality:<br /> Suppose we let &lt;math&gt;F(x)=x^2&lt;/math&gt; (We know that &lt;math&gt;F(x)&lt;/math&gt; is convex because &lt;math&gt;F'(x)=2x&lt;/math&gt; and therefore &lt;math&gt;F''(x)=2&gt;0&lt;/math&gt;). <br /> We have:<br /> &lt;cmath&gt;F\left(\frac{x_1}{n}+\cdots+\frac{x_n}{n}\right)\le \frac{F(x_1)}{n}+\cdots+\frac{F(x_n)}{n}&lt;/cmath&gt;<br /> Factoring out the &lt;math&gt;\frac{1}{n}&lt;/math&gt; yields:<br /> &lt;cmath&gt;F\left(\frac{x_1+\cdots+x_n}{n}\right)\le \frac {F(x_1)+\cdots+F(x_n)}{n}&lt;/cmath&gt;<br /> &lt;cmath&gt;\left(\frac{x_1+\cdots+x_n}{n}\right)^2 \le \frac{x_1^2+\cdots+x_n^2}{n}&lt;/cmath&gt;<br /> Taking the square root to both sides (remember that both are positive):<br /> &lt;cmath&gt;\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}} \ge\frac{x_1+\cdots+x_n}{n} \blacksquare.&lt;/cmath&gt;<br /> <br /> <br /> <br /> The inequality &lt;math&gt;\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}&lt;/math&gt; is called the AM-GM inequality, and proofs can be found [[Proofs of AM-GM|here]].<br /> <br /> <br /> The inequality &lt;math&gt;\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt; is a direct consequence of AM-GM; &lt;math&gt;\frac{\sum_{i=1}^{n}\sqrt[n]{\frac{x_1x_2\cdots x_n}{x_i^n}}}{n}\geq 1&lt;/math&gt;, so &lt;math&gt;\sqrt[n]{x_1x_2\cdots x_n}\frac{\sum_{i=1}^{n}\frac{1}{x_i}}{n}\geq 1&lt;/math&gt;, so &lt;math&gt;\sqrt[n]{x_1\cdots x_n}\ge\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}&lt;/math&gt;.<br /> <br /> Therefore the original inequality is true.<br /> <br /> ===Geometric Proof===<br /> <br /> &lt;asy&gt;size(250);<br /> pair O=(0,0),A=(-1,0),B=(0,1),C=(1,0),P=(1/2,0),Q=(1/2,sqrt(3)/2),R=foot(P,Q,O);<br /> draw(B--O--C--arc(O,C,A)--O--R--P); rightanglemark(O,P,R);<br /> draw(O--B,red);<br /> draw(P--Q,blue);<br /> draw(B--P,green);<br /> draw(R--Q,magenta);<br /> draw((A-(0,0.05))--(P-(0,0.05)),Arrows);<br /> draw((P-(0,0.05))--(C-(0,0.05)),Arrows);<br /> label(&quot;AM&quot;,(O+B)/2,W,red);<br /> label(&quot;GM&quot;,(P+Q)/2,E,blue);<br /> label(&quot;HM&quot;,(R+Q)/2,unit(P-R),magenta);<br /> label(&quot;RMS&quot;,(3B+P)/4,unit(foot(O,B,P)),green);<br /> label(&quot;$a$&quot;,(A+P)/2,3*S);<br /> label(&quot;$b$&quot;,(P+C)/2,3*S);&lt;/asy&gt;<br /> <br /> The inequality is clearly shown in this diagram for &lt;math&gt;n=2&lt;/math&gt;<br /> <br /> {{stub}}<br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_6&diff=96501 2005 AIME II Problems/Problem 6 2018-07-26T04:31:51Z <p>Obtuse: </p> <hr /> <div>== Problem ==<br /> The cards in a stack of &lt;math&gt; 2n &lt;/math&gt; cards are numbered [[consecutive]]ly from 1 through &lt;math&gt; 2n &lt;/math&gt; from top to bottom. The top &lt;math&gt; n &lt;/math&gt; cards are removed, kept in order, and form pile &lt;math&gt; A. &lt;/math&gt; The remaining cards form pile &lt;math&gt; B. &lt;/math&gt; The cards are then restacked by taking cards alternately from the tops of pile &lt;math&gt; B &lt;/math&gt; and &lt;math&gt; A, &lt;/math&gt; respectively. In this process, card number &lt;math&gt; (n+1) &lt;/math&gt; becomes the bottom card of the new stack, card number 1 is on top of this card, and so on, until piles &lt;math&gt; A &lt;/math&gt; and &lt;math&gt; B &lt;/math&gt; are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named magical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical stack in which card number 131 retains its original position.<br /> <br /> == Solution 1 ==<br /> Since a card from B is placed on the bottom of the new stack, notice that cards from pile B will be marked as an even number in the new pile, while cards from pile A will be marked as odd in the new pile. Since 131 is odd and retains its original position in the stack, it must be in pile A. Also to retain its original position, exactly &lt;math&gt;131 - 1 = 130&lt;/math&gt; numbers must be in front of it. There are &lt;math&gt;\frac{130}{2} = 65&lt;/math&gt; cards from each of piles A, B in front of card 131. This suggests that &lt;math&gt;n = 131 + 65 = 196&lt;/math&gt;; the total number of cards is &lt;math&gt;196 \cdot 2 = \boxed{392}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> If you index the final stack &lt;math&gt;1,2,\dots,2n&lt;/math&gt;, you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers in pile A are the consecutive integers &lt;math&gt;1,2,\dots, n&lt;/math&gt;. This all leads us to the following equation.<br /> &lt;cmath&gt;<br /> 131=2n-2(131)+1\implies2n=\boxed{392}<br /> &lt;/cmath&gt;<br /> == See also ==<br /> {{AIME box|year=2005|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=87161 2016 AMC 10A Problems 2017-08-23T01:30:01Z <p>Obtuse: </p> <hr /> <div>==Problem 1==<br /> What is the value of &lt;math&gt;\dfrac{11!-10!}{9!}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> For what value of &lt;math&gt;x&lt;/math&gt; does &lt;math&gt;10^{x}\cdot 100^{2x}=1000^{5}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> For every dollar Ben spent on bagels, David spent &lt;math&gt;25&lt;/math&gt; cents less. Ben paid &lt;math&gt;\$12.50&lt;/math&gt; more than David. How much did they spend in the bagel store together?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> The remainder can be defined for all real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;y \neq 0&lt;/math&gt; by &lt;cmath&gt;\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor&lt;/cmath&gt;where &lt;math&gt;\left \lfloor \tfrac{x}{y} \right \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;. What is the value of &lt;math&gt;\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> A rectangular box has integer side lengths in the ratio &lt;math&gt;1: 3: 4&lt;/math&gt;. Which of the following could be the volume of the box?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ximena lists the whole numbers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;30&lt;/math&gt; once. Emilio copies Ximena's numbers, replacing each occurrence of the digit &lt;math&gt;2&lt;/math&gt; by the digit &lt;math&gt;1&lt;/math&gt;. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The mean, median, and mode of the &lt;math&gt;7&lt;/math&gt; data values &lt;math&gt;60, 100, x, 40, 50, 200, 90&lt;/math&gt; are all equal to &lt;math&gt;x&lt;/math&gt;. What is the value of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays &lt;math&gt;40&lt;/math&gt; coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> A triangular array of &lt;math&gt;2016&lt;/math&gt; coins has &lt;math&gt;1&lt;/math&gt; coin in the first row, &lt;math&gt;2&lt;/math&gt; coins in the second row, &lt;math&gt;3&lt;/math&gt; coins in the third row, and so on up to &lt;math&gt;N&lt;/math&gt; coins in the &lt;math&gt;N&lt;/math&gt;th row. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<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 /> [[2016 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is &lt;math&gt;1&lt;/math&gt; foot wide on all four sides. What is the length in feet of the inner rectangle?<br /> &lt;asy&gt;<br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> path rectangle(pair X, pair Y){<br /> return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br /> }<br /> filldraw(rectangle((0,0),(7,5)),gray(0.5));<br /> filldraw(rectangle((1,1),(6,4)),gray(0.75));<br /> filldraw(rectangle((2,2),(5,3)),white);<br /> <br /> label(&quot;$1$&quot;,(0.5,2.5));<br /> draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br /> draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(1.5,2.5));<br /> draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br /> draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.5,2.5));<br /> draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br /> draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(4.1,1.5));<br /> draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br /> draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br /> <br /> label(&quot;$1$&quot;,(3.7,0.5));<br /> draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br /> draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br /> &lt;/asy&gt;<br /> <br /> &lt;cmath&gt;\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8&lt;/cmath&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> What is the area of the shaded region of the given &lt;math&gt;8 \times 5&lt;/math&gt; rectangle?<br /> <br /> &lt;asy&gt;<br /> <br /> size(6cm);<br /> defaultpen(fontsize(9pt));<br /> draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br /> filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br /> <br /> label(&quot;$1$&quot;,(1/2,5),dir(90));<br /> label(&quot;$7$&quot;,(9/2,5),dir(90));<br /> <br /> label(&quot;$1$&quot;,(8,1/2),dir(0));<br /> label(&quot;$4$&quot;,(8,3),dir(0));<br /> <br /> label(&quot;$1$&quot;,(15/2,0),dir(270));<br /> label(&quot;$7$&quot;,(7/2,0),dir(270));<br /> <br /> label(&quot;$1$&quot;,(0,9/2),dir(180));<br /> label(&quot;$4$&quot;,(0,2),dir(180));<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Three distinct integers are selected at random between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2016&lt;/math&gt;, inclusive. Which of the following is a correct statement about the probability &lt;math&gt;p&lt;/math&gt; that the product of the three integers is odd?<br /> <br /> &lt;math&gt;\textbf{(A)}\ p&lt;\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}&lt;p&lt;\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p&gt;\dfrac{1}{3}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Five friends sat in a movie theater in a row containing &lt;math&gt;5&lt;/math&gt; seats, numbered &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; from left to right. (The directions &quot;left&quot; and &quot;right&quot; are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> How many ways are there to write &lt;math&gt;2016&lt;/math&gt; as the sum of twos and threes, ignoring order? (For example, &lt;math&gt;1008\cdot 2 + 0\cdot 3&lt;/math&gt; and &lt;math&gt;402\cdot 2 + 404\cdot 3&lt;/math&gt; are two such ways.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Seven cookies of radius &lt;math&gt;1&lt;/math&gt; inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br /> <br /> &lt;asy&gt;<br /> draw(circle((0,0),3));<br /> draw(circle((0,0),1));<br /> draw(circle((1,sqrt(3)),1));<br /> draw(circle((-1,sqrt(3)),1)); <br /> draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br /> draw(circle((2,0),1)); draw(circle((-2,0),1)); &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> A triangle with vertices &lt;math&gt;A(0, 2)&lt;/math&gt;, &lt;math&gt;B(-3, 2)&lt;/math&gt;, and &lt;math&gt;C(-3, 0)&lt;/math&gt; is reflected about the &lt;math&gt;x&lt;/math&gt;-axis, then the image &lt;math&gt;\triangle A'B'C'&lt;/math&gt; is rotated counterclockwise about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt; to produce &lt;math&gt;\triangle A''B''C''&lt;/math&gt;. Which of the following transformations will return &lt;math&gt;\triangle A''B''C''&lt;/math&gt; to &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; counterclockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; clockwise rotation about the origin by &lt;math&gt;90^{\circ}&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; reflection about the &lt;math&gt;x&lt;/math&gt;-axis <br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; reflection about the line &lt;math&gt;y = x&lt;/math&gt; <br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; reflection about the &lt;math&gt;y&lt;/math&gt;-axis.<br /> <br /> [[2016 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;N&lt;/math&gt; be a positive multiple of &lt;math&gt;5&lt;/math&gt;. One red ball and &lt;math&gt;N&lt;/math&gt; green balls are arranged in a line in random order. Let &lt;math&gt;P(N)&lt;/math&gt; be the probability that at least &lt;math&gt;\tfrac{3}{5}&lt;/math&gt; of the green balls are on the same side of the red ball. Observe that &lt;math&gt;P(5)=1&lt;/math&gt; and that &lt;math&gt;P(N)&lt;/math&gt; approaches &lt;math&gt;\tfrac{4}{5}&lt;/math&gt; as &lt;math&gt;N&lt;/math&gt; grows large. What is the sum of the digits of the least value of &lt;math&gt;N&lt;/math&gt; such that &lt;math&gt;P(N) &lt; \tfrac{321}{400}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> In rectangle &lt;math&gt;ABCD,&lt;/math&gt; &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;BC=3&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; between &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, and point &lt;math&gt;F&lt;/math&gt; between &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; are such that &lt;math&gt;BE=EF=FC&lt;/math&gt;. Segments &lt;math&gt;\overline{AE}&lt;/math&gt; and &lt;math&gt;\overline{AF}&lt;/math&gt; intersect &lt;math&gt;\overline{BD}&lt;/math&gt; at &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt;, respectively. The ratio &lt;math&gt;BP:PQ:QD&lt;/math&gt; can be written as &lt;math&gt;r:s:t&lt;/math&gt; where the greatest common factor of &lt;math&gt;r,s&lt;/math&gt; and &lt;math&gt;t&lt;/math&gt; is 1. What is &lt;math&gt;r+s+t&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20&lt;/math&gt;<br /> <br /> <br /> [[2016 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> For some particular value of &lt;math&gt;N&lt;/math&gt;, when &lt;math&gt;(a+b+c+d+1)^N&lt;/math&gt; is expanded and like terms are combined, the resulting expression contains exactly &lt;math&gt;1001&lt;/math&gt; terms that include all four variables &lt;math&gt;a, b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt;, each to some positive power. What is &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 20|Solution]]<br /> ==Problem 21==<br /> Circles with centers &lt;math&gt;P, Q&lt;/math&gt; and &lt;math&gt;R&lt;/math&gt;, having radii &lt;math&gt;1, 2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;, respectively, lie on the same side of line &lt;math&gt;l&lt;/math&gt; and are tangent to &lt;math&gt;l&lt;/math&gt; at &lt;math&gt;P', Q'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;, respectively, with &lt;math&gt;Q'&lt;/math&gt; between &lt;math&gt;P'&lt;/math&gt; and &lt;math&gt;R'&lt;/math&gt;. The circle with center &lt;math&gt;Q&lt;/math&gt; is externally tangent to each of the other two circles. What is the area of triangle &lt;math&gt;PQR&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> A binary operation &lt;math&gt;\diamondsuit&lt;/math&gt; has the properties that &lt;math&gt;a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c&lt;/math&gt; and that &lt;math&gt;a\,\diamondsuit \,a=1&lt;/math&gt; for all nonzero real numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. (Here &lt;math&gt;\cdot&lt;/math&gt; represents multiplication). The solution to the equation &lt;math&gt;2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100&lt;/math&gt; can be written as &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A quadrilateral is inscribed in a circle of radius &lt;math&gt;200\sqrt{2}&lt;/math&gt;. Three of the sides of this quadrilateral have length &lt;math&gt;200&lt;/math&gt;. What is the length of the fourth side?<br /> <br /> &lt;math&gt;\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> <br /> ==Problem 25==<br /> How many ordered triples &lt;math&gt;(x,y,z)&lt;/math&gt; of positive integers satisfy &lt;math&gt;\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600&lt;/math&gt; and &lt;math&gt;\text{lcm}(y,z)=900&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64&lt;/math&gt;<br /> <br /> [[2016 AMC 10A Problems/Problem 25|Solution]]</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_15&diff=87094 2015 AMC 12A Problems/Problem 15 2017-08-18T18:47:45Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> What is the minimum number of digits to the right of the decimal point needed to express the fraction &lt;math&gt;\frac{123456789}{2^{26}\cdot 5^4}&lt;/math&gt; as a decimal?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We can rewrite the fraction as &lt;math&gt;\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}&lt;/math&gt;. Since the last digit of the numerator is odd, a &lt;math&gt;5&lt;/math&gt; is added to the right if the numerator is divided by &lt;math&gt;2&lt;/math&gt;, and this will continuously happen because &lt;math&gt;5&lt;/math&gt;, itself, is odd. Indeed, this happens twenty-two times since we divide by &lt;math&gt;2&lt;/math&gt; twenty-two times, so we will need &lt;math&gt;22&lt;/math&gt; more digits. Hence, the answer is &lt;math&gt;4 + 22 = \boxed{\textbf{(C)}\ 26}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Multiply the numerator and denominator of the fraction by &lt;math&gt;5^{22}&lt;/math&gt; (which is the same as multiplying by 1) to give &lt;math&gt;\frac{5^{22} \cdot 123456789}{10^{26}}&lt;/math&gt;. Now, instead of thinking about this as a fraction, think of it as the division calculation &lt;math&gt;(5^{22} \cdot 123456789) \div 10^{26}&lt;/math&gt; . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, <br /> &lt;math&gt;\boxed{\textbf{(C)}\ 26}&lt;/math&gt; is the minimum number of digits to the right of the decimal point needed.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=14|num-a=16}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_15&diff=87093 2015 AMC 12A Problems/Problem 15 2017-08-18T18:46:20Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> What is the minimum number of digits to the right of the decimal point needed to express the fraction &lt;math&gt;\frac{123456789}{2^{26}\cdot 5^4}&lt;/math&gt; as a decimal?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We can rewrite the fraction as &lt;math&gt;\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}&lt;/math&gt;. Since the last digit of the numerator is odd, a &lt;math&gt;5&lt;/math&gt; is added to the right if the numerator is divided by &lt;math&gt;2&lt;/math&gt;, and this will continuously happen because &lt;math&gt;5&lt;/math&gt;, itself, is odd. Indeed, this happens twenty-two times since we divide by &lt;math&gt;2&lt;/math&gt; twenty-two times, so we will need &lt;math&gt;22&lt;/math&gt; more digits. Hence, the answer is &lt;math&gt;4 + 22 = 26 \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Multiply the numerator and denominator of the fraction by &lt;math&gt;5^{22}&lt;/math&gt; (which is the same as multiplying by 1) to give &lt;math&gt;\frac{5^{22} \cdot 123456789}{10^{26}}&lt;/math&gt;. Now, instead of thinking about this as a fraction, think of it as the division calculation &lt;math&gt;(5^{22} \cdot 123456789) \div 10^{26}&lt;/math&gt; . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, <br /> &lt;math&gt;\boxed{\textbf{(C)}\ 26}&lt;/math&gt; is the minimum number of digits to the right of the decimal point needed.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=14|num-a=16}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_15&diff=87092 2015 AMC 12A Problems/Problem 15 2017-08-18T18:45:13Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> What is the minimum number of digits to the right of the decimal point needed to express the fraction &lt;math&gt;\frac{123456789}{2^{26}\cdot 5^4}&lt;/math&gt; as a decimal?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We can rewrite the fraction as &lt;math&gt;\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}&lt;/math&gt;. Since the last digit of the numerator is odd, a &lt;math&gt;5&lt;/math&gt; is added to the right if the numerator is divided by &lt;math&gt;2&lt;/math&gt;, and this will continuously happen because &lt;math&gt;5&lt;/math&gt;, itself, is odd. Indeed, this happens twenty-two times since we divide by &lt;math&gt;2&lt;/math&gt; twenty-two times, so we will need &lt;math&gt;22&lt;/math&gt; more digits. Hence, the answer is &lt;math&gt;4 + 22 = 26 \textbf{ (C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Multiply the numerator and denominator of the fraction by &lt;math&gt;5^{22}&lt;/math&gt; (which is the same as multiplying by 1) to give &lt;math&gt;\frac{5^{22} \cdot 123456789}{10^{26}}&lt;/math&gt;. Now, instead of thinking about this as a fraction, think of it as the division calculation &lt;math&gt;(5^{22} \cdot 123456789) \div 10^{26}&lt;/math&gt; . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit further to the right of the decimal point. Thus, <br /> &lt;math&gt;\boxed{\textbf{(C)}\ 26}&lt;/math&gt; is the minimum number of digits to the right of the decimal point needed.<br /> <br /> == See Also ==<br /> {{AMC12 box|year=2015|ab=A|num-b=14|num-a=16}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=86956 2016 AMC 10A Problems/Problem 22 2017-08-09T17:13:32Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 11&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> The number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes. This tells us that none of these factors can be 1. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without &lt;math&gt;1&lt;/math&gt;s is &lt;math&gt;2\cdot 5\cdot 11&lt;/math&gt;.<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_15&diff=86882 2016 AMC 12B Problems/Problem 15 2017-08-06T17:37:38Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First assign each face the letters &lt;math&gt;a,b,c,d,e,f&lt;/math&gt;. The sum of the product of the faces is &lt;math&gt;abc+acd+ade+aeb+fbc+fcd+fde+feb&lt;/math&gt;. We can factor this into &lt;math&gt;(a+f)(b+c)(d+e)&lt;/math&gt; which is the product of the sum of each pair of opposite faces. In order to maximize &lt;math&gt;(a+f)(b+c)(d+e)&lt;/math&gt; we use the numbers &lt;math&gt;(7+2)(6+3)(5+4)&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(D)}\ 729 }&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> We proceed from the factorization in the above solution. By the AM-GM inequality,<br /> <br /> &lt;cmath&gt;\frac{a_1+a_2++a_3}{3}\geq\sqrt{a_1a_2a_3}&lt;/cmath&gt;<br /> <br /> Cubing both sides,<br /> <br /> &lt;cmath&gt;\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}&lt;/cmath&gt;<br /> <br /> Let &lt;math&gt;a_1=(a+f)&lt;/math&gt;, &lt;math&gt;a_2=(b+c)&lt;/math&gt;, and &lt;math&gt;a_3=(d+e)&lt;/math&gt;. Let's substitute in these values.<br /> <br /> &lt;cmath&gt;\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;a+b+c+d+e+f&lt;/math&gt; is fixed at 27.<br /> <br /> &lt;cmath&gt;\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\boxed{\textbf{(D)}\ 729 }\geq{(a+f)(b+c)(d+e)}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_15&diff=86881 2016 AMC 12B Problems/Problem 15 2017-08-06T17:36:59Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad<br /> \textbf{(B)}\ 343 \qquad<br /> \textbf{(C)}\ 625 \qquad<br /> \textbf{(D)}\ 729 \qquad<br /> \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First assign each face the letters &lt;math&gt;a,b,c,d,e,f&lt;/math&gt;. The sum of the product of the faces is &lt;math&gt;abc+acd+ade+aeb+fbc+fcd+fde+feb&lt;/math&gt;. We can factor this into &lt;math&gt;(a+f)(b+c)(d+e)&lt;/math&gt; which is the product of the sum of each pair of opposite faces. In order to maximize &lt;math&gt;(a+f)(b+c)(d+e)&lt;/math&gt; we use the numbers &lt;math&gt;(7+2)(6+3)(5+4)&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(D)}\ 729 }&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86870 2016 AMC 12B Problems/Problem 14 2017-08-06T04:42:24Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then doing some calculus}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=S'&lt;/math&gt; and &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;\frac{2}{(a-1)^3}=S''&lt;/math&gt; and &lt;math&gt;\frac{2}{(0-1)^3}&lt;/math&gt; is negative (implying a relative maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;\frac{2}{(2-1)^3}&lt;/math&gt; is positive (implying a relative minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=4&lt;/math&gt;. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt; and that our answer is indeed &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; However, to be sure of this outside of this cop-out, one can analyze the end behavior of &lt;math&gt;S&lt;/math&gt;, how &lt;math&gt;S&lt;/math&gt; behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being clever}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86869 2016 AMC 12B Problems/Problem 14 2017-08-06T04:42:02Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then doing some calculus}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=S'&lt;/math&gt; and &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;\frac{2}{(a-1)^3}=S''&lt;/math&gt; and &lt;math&gt;\frac{2}{(0-1)^3}&lt;/math&gt; is negative (implying a relative maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;\frac{2}{(2-1)^3}&lt;/math&gt; is positive (implying a relative minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=4&lt;/math&gt;. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt; and that our answer is indeed &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; However, to be sure of this outside of this cop-out, one can analyze the end behavior of &lt;math&gt;S&lt;/math&gt;, how &lt;math&gt;S&lt;/math&gt; behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being clever}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86868 2016 AMC 12B Problems/Problem 14 2017-08-06T04:37:27Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then doing some calculus}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=S'&lt;/math&gt; and &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;\frac{2}{(a-1)^3}=S''&lt;/math&gt; and &lt;math&gt;\frac{2}{(0-1)^3}&lt;/math&gt; is negative (implying a relative maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;\frac{2}{(2-1)^3}&lt;/math&gt; is positive (implying a relative minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=4&lt;/math&gt;. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt; and that our answer is indeed &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; However, to be sure of this outside of this cop-out, one can analyze the end behavior of &lt;math&gt;S&lt;/math&gt;, how &lt;math&gt;S&lt;/math&gt; behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being clever}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86867 2016 AMC 12B Problems/Problem 14 2017-08-06T04:32:48Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then doing some lame calculus}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=S'&lt;/math&gt; and &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;\frac{2}{(a-1)^3}=S''&lt;/math&gt; and &lt;math&gt;\frac{2}{(0-1)^3}&lt;/math&gt; is negative (implying a relative maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;\frac{2}{(2-1)^3}&lt;/math&gt; is positive (implying a relative minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=4&lt;/math&gt;. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt; and that our answer is indeed &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; However, to be sure of this outside of this cop-out, one can analyze the end behavior of &lt;math&gt;S&lt;/math&gt;, how &lt;math&gt;S&lt;/math&gt; behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being clever}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86866 2016 AMC 12B Problems/Problem 14 2017-08-06T04:31:13Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=S'&lt;/math&gt; and &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;\frac{2}{(a-1)^3}=S''&lt;/math&gt; and &lt;math&gt;\frac{2}{(0-1)^3}&lt;/math&gt; is negative (implying a relative maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;\frac{2}{(2-1)^3}&lt;/math&gt; is positive (implying a relative minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=4&lt;/math&gt;. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt; and that our answer is indeed &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; However, to be sure of this outside of this cop-out, one can analyze the end behavior of &lt;math&gt;S&lt;/math&gt;, how &lt;math&gt;S&lt;/math&gt; behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the &quot;positive parts&quot; of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being a clever cow}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86865 2016 AMC 12B Problems/Problem 14 2017-08-06T02:50:21Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=S'&lt;/math&gt; and &lt;math&gt;\frac{(a-2)a}{(a-1)^2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;\frac{2}{(a-1)^3}=S''&lt;/math&gt; and &lt;math&gt;\frac{2}{(0-1)^3}&lt;/math&gt; is negative (implying a maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;\frac{2}{(2-1)^3}&lt;/math&gt; is positive (implying a minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; This is the only positive minimum &lt;math&gt;S&lt;/math&gt; has, so it must be the answer.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being a clever cow}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86864 2016 AMC 12B Problems/Problem 14 2017-08-06T02:38:54Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;2a(a-1)^{-1}-a^2(a-1)^{-2}=S'&lt;/math&gt; and &lt;math&gt;2a(a-1)^{-1}-a^2(a-1)^{-2}=0&lt;/math&gt; at &lt;math&gt;a=0&lt;/math&gt; and &lt;math&gt;a=2&lt;/math&gt;. &lt;math&gt;2(a-1)^{-3}=S''&lt;/math&gt; and &lt;math&gt;2(0-1)^{-3}&lt;/math&gt; is negative (implying a maximum occurs at &lt;math&gt;a=0&lt;/math&gt;) and &lt;math&gt;2(2-1)^{-3}&lt;/math&gt; is positive (implying a minimum occurs at &lt;math&gt;a=2&lt;/math&gt;). At &lt;math&gt;a=2&lt;/math&gt;, &lt;math&gt;S=\boxed{\textbf{(E)}\ 4}.&lt;/math&gt; This is the only positive minimum &lt;math&gt;S&lt;/math&gt; has, so it must be the answer.<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then being a clever cow}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of &lt;math&gt;S&lt;/math&gt;. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at &lt;math&gt;x=\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86828 2016 AMC 12B Problems/Problem 14 2017-08-05T18:19:03Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. I'll be back soon to finish this solution. Please don't delete it. :(<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. I'll be back soon to finish this solution. Please don't delete it. :(<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86827 2016 AMC 12B Problems/Problem 14 2017-08-05T18:18:12Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. We proceed by graphing in the &lt;math&gt;aS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then graphing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. We proceed by graphing in the &lt;math&gt;rS&lt;/math&gt; plane and find the answer is &lt;math&gt;\boxed{\textbf{(E)}\ 4}.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{a} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{a^2}{a-1}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S. <br /> <br /> &lt;math&gt;\textbf{Solving in terms of \textit{r} then calculus-ing}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;S=\frac{1}{-r^2+r}&lt;/cmath&gt;<br /> <br /> We seek the smallest positive value of S.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86826 2016 AMC 12B Problems/Problem 14 2017-08-05T17:54:57Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> A geometric sequence always looks like<br /> <br /> &lt;cmath&gt;a,ar,ar^2,ar^3,\dots&lt;/cmath&gt;<br /> <br /> and they say that the second term &lt;math&gt;ar=1&lt;/math&gt;. You should know that the sum of an infinite geometric series (denoted by &lt;math&gt;S&lt;/math&gt; here) is &lt;math&gt;\frac{a}{1-r}&lt;/math&gt;. We now have a system of equations which allows us to find &lt;math&gt;S&lt;/math&gt; in one variable.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> ar&amp;=1 \\<br /> S&amp;=\frac{a}{1-r}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86824 2016 AMC 12B Problems/Problem 14 2017-08-05T17:39:26Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> <br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86823 2016 AMC 12B Problems/Problem 14 2017-08-05T17:39:11Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> &lt;math&gt;\textbf{Calculus}&lt;/math&gt;<br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{AM-GM}&lt;/math&gt;<br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&diff=86822 2016 AMC 12B Problems/Problem 14 2017-08-05T17:38:51Z <p>Obtuse: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ \sqrt{5} \qquad<br /> \textbf{(D)}\ 3 \qquad<br /> \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> ==Solution==<br /> The second term in a geometric series is &lt;math&gt;a_2 = a \cdot r&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the common ratio for the series and &lt;math&gt;a&lt;/math&gt; is the first term of the series. So we know that &lt;math&gt;a\cdot r = 1&lt;/math&gt; and we wish to find the minimum value of the infinite sum of the series. We know that: &lt;math&gt;S_\infty = \frac{a}{1-r}&lt;/math&gt; and substituting in &lt;math&gt;a=\frac{1}{r}&lt;/math&gt;, we get that &lt;math&gt;S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}&lt;/math&gt;. From here, you can either use calculus or AM-GM.<br /> <br /> \textbf{Calculus}<br /> Let &lt;math&gt;f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}&lt;/math&gt;, then &lt;math&gt;f'(x) = -(x-x^2)^{-2}\cdot (1-2x)&lt;/math&gt;. Since &lt;math&gt;f(0)&lt;/math&gt; and &lt;math&gt;f(1)&lt;/math&gt; are undefined &lt;math&gt;x \neq 0,1&lt;/math&gt;. This means that we only need to find where the derivative equals &lt;math&gt;0&lt;/math&gt;, meaning &lt;math&gt;1-2x = 0 \Rightarrow x =\frac{1}{2}&lt;/math&gt;. So &lt;math&gt; r = \frac{1}{2}&lt;/math&gt;, meaning that &lt;math&gt;S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4&lt;/math&gt;<br /> <br /> \textbf{AM-GM}<br /> For 2 positive real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;\frac{a+b}{2} \geq \sqrt{ab}&lt;/math&gt;. Let &lt;math&gt;a = \frac{1}{r}&lt;/math&gt; and &lt;math&gt;b = \frac{1}{1-r}&lt;/math&gt;. Then: &lt;math&gt;\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}&lt;/math&gt;. This implies that &lt;math&gt;\frac{S_\infty}{2} \geq \sqrt{S_\infty}&lt;/math&gt;. or &lt;math&gt;S_\infty^2 \geq 4 \cdot S_\infty&lt;/math&gt;. Rearranging : &lt;math&gt;(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4&lt;/math&gt;. Thus, the smallest value is &lt;math&gt;S_\infty = 4&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=Number&diff=86821 Number 2017-08-05T17:35:11Z <p>Obtuse: </p> <hr /> <div>A '''number''' is an entity, abstract in nature, that is used to describe another object. Numbers are most often used for counting and determining the [[magnitude]] of something. <br /> <br /> <br /> == Common [[set]]s of numbers ==<br /> <br /> * [[Algebraic number]]s<br /> * [[Complex number]]s<br /> * [[Counting number]]s<br /> * [[Integer]]s<br /> * [[Irrational number]]s<br /> * [[Rational number]]s<br /> * [[Real number]]s<br /> * [[Transcendental number]]s<br /> <br /> <br /> == See also ==<br /> * [[Number theory]]<br /> * [[Constant]]<br /> <br /> {{stub}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_12&diff=86815 2016 AMC 12B Problems/Problem 12 2017-08-04T23:08:09Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> All the numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9&lt;/math&gt; are written in a &lt;math&gt;3\times3&lt;/math&gt; array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to &lt;math&gt;18&lt;/math&gt;. What is the number in the center?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Solution by Mlux:<br /> Draw a &lt;math&gt;3\times3&lt;/math&gt; matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to &lt;math&gt;18&lt;/math&gt;. Trying &lt;math&gt;1+3+5+9 = 18&lt;/math&gt; works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a &lt;math&gt;2&lt;/math&gt; in between the &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt;. There is a &lt;math&gt;4&lt;/math&gt; between &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. The final grid should similar to this.<br /> <br /> &lt;math&gt;\newline<br /> 1, 2, 3\newline<br /> 8, 7, 4\newline<br /> 9, 6, 5&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(C)}7&lt;/math&gt; is in the middle.<br /> ==Solution 2==<br /> If we color the square like a chessboard, since the numbers alternate between even and odd, and there are five odd numbers and four even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges. Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be &lt;math&gt;25 - 18 = 7&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=11|num-a=13}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_6&diff=86814 2016 AMC 12B Problems/Problem 6 2017-08-04T22:47:53Z <p>Obtuse: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> All three vertices of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; lie on the parabola defined by &lt;math&gt;y=x^2&lt;/math&gt;, with &lt;math&gt;A&lt;/math&gt; at the origin and &lt;math&gt;\overline{BC}&lt;/math&gt; parallel to the &lt;math&gt;x&lt;/math&gt;-axis. The area of the triangle is &lt;math&gt;64&lt;/math&gt;. What is the length of &lt;math&gt;BC&lt;/math&gt;? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16&lt;/math&gt;<br /> <br /> ==Solution==<br /> By: Albert471<br /> <br /> Plotting points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on the graph shows that they are at &lt;math&gt;\left( -x,x^2\right)&lt;/math&gt; and &lt;math&gt;\left( x,x^2\right)&lt;/math&gt;, which is isosceles. By setting up the triangle area formula you get: &lt;math&gt;64=\frac{1}{2}*2x*x^2 = 64=x^3&lt;/math&gt; Making x=4, and the length of &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;2x&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C)}\ 8}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_19&diff=86804 2016 AMC 12A Problems/Problem 19 2017-08-04T05:23:22Z <p>Obtuse: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Jerry starts at &lt;math&gt;0&lt;/math&gt; on the real number line. He tosses a fair coin &lt;math&gt;8&lt;/math&gt; times. When he gets heads, he moves &lt;math&gt;1&lt;/math&gt; unit in the positive direction; when he gets tails, he moves &lt;math&gt;1&lt;/math&gt; unit in the negative direction. The probability that he reaches &lt;math&gt;4&lt;/math&gt; at some time during this process &lt;math&gt;\frac{a}{b},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b?&lt;/math&gt; (For example, he succeeds if his sequence of tosses is &lt;math&gt;HTHHHHHH.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;6&lt;/math&gt; to &lt;math&gt;8&lt;/math&gt; heads, we are guaranteed to hit &lt;math&gt;4&lt;/math&gt; heads, so the sum here is &lt;math&gt;\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37&lt;/math&gt;.<br /> <br /> For &lt;math&gt;4&lt;/math&gt; heads, you have to hit the &lt;math&gt;4&lt;/math&gt; heads at the start so there's only one way, &lt;math&gt;1&lt;/math&gt;.<br /> <br /> For &lt;math&gt;5&lt;/math&gt; heads, we either start off with &lt;math&gt;4&lt;/math&gt; heads, which gives us &lt;math&gt;4\textbf{C}1=4&lt;/math&gt; ways to arrange the other flips, or we start off with five heads and one tail, which has &lt;math&gt;6&lt;/math&gt; ways minus the &lt;math&gt;2&lt;/math&gt; overlapping cases, &lt;math&gt;\text{HHHHHTTT}&lt;/math&gt; and &lt;math&gt;\text{HHHHTHTT}&lt;/math&gt;. Total ways: &lt;math&gt;8&lt;/math&gt;.<br /> <br /> Then we sum to get &lt;math&gt;46&lt;/math&gt;. There are a total of &lt;math&gt;2^8=256&lt;/math&gt; possible sequences of &lt;math&gt;8&lt;/math&gt; coin flips, so the probability is &lt;math&gt;\frac{46}{256}=\frac{23}{128}&lt;/math&gt;. Summing, we get &lt;math&gt;23+128=\boxed{\textbf{(B) }151}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases: <br /> <br /> (Case 1): The first four flips are heads. Then, the last four flips can be anything so &lt;math&gt;2^4=16&lt;/math&gt; possibilities work. <br /> <br /> (Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so &lt;math&gt;2^2=4&lt;/math&gt; flips work. &lt;math&gt;4*4=16&lt;/math&gt;. <br /> <br /> (Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips. <br /> <br /> (1 tail in first four flips). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are &lt;math&gt;4*2=8&lt;/math&gt; possibilities. <br /> <br /> (2 tails in first four flips). In this case, the tails can be in &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; positions. <br /> <br /> Adding these cases up and taking the total out of &lt;math&gt;2^8=256&lt;/math&gt; yields &lt;math&gt;\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}&lt;/math&gt;. This means the answer is &lt;math&gt;23+128=\boxed{\textbf{(B) }151}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> (Inspired by https://www.youtube.com/watch?v=EZYzjmd_f2g by Walt S)<br /> <br /> Draw a triangle of dots as seen in the video. Now, we'll do something different than he did. Draw a dotted line between dots that correspond to 3 on the number line and dots that correspond to 4 on the number line. This is the threshold that must be passed to have a successful sequence. We consider the three entry points to the immediate right of this line. Doing a tiny bit of fun bashing as we slide down this triangle, we find there is 1 way to cross the line for the first time through the uppermost entry point, 4 ways to cross the line for the first time through the middle entry point, and 14 ways to cross the line for the first time through the lowermost entry point.<br /> <br /> Superimpose Pascal's triangle over the dots. Notice that the number of ways to get to a dot is equal to its corresponding number on Pascal's triangle. Rows on Pascal's triangle sum to powers of 2. There are &lt;math&gt;2^4=16&lt;/math&gt; ways to get to the 4th row (remember that when working with Pascal's triangle, we start counting with a 0th row), and we found that 1 of them makes us cross the threshold for the first time. There are &lt;math&gt;2^6=64&lt;/math&gt; ways to get to the 6th row, and we found that 4 of them make us cross the threshold for the first time. There are &lt;math&gt;2^8=256&lt;/math&gt; ways to get to the 8th row, and we found that 14 of them make us cross the threshold for the first time.<br /> <br /> Adding up these probabilities &lt;cmath&gt;\frac{1}{16}+\frac{4}{64}+\frac{14}{256}=\frac{23}{128}&lt;/cmath&gt; and summing &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;cmath&gt;a+b=23+128=\boxed{\textbf{(B)}\ 151}&lt;/cmath&gt; we get our answer.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_16&diff=86803 2016 AMC 12B Problems/Problem 16 2017-08-04T05:03:21Z <p>Obtuse: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> <br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.<br /> <br /> For the first case, we can cleverly choose the convenient form of our sequence to be<br /> &lt;cmath&gt;a-n,\cdots, a-1, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> <br /> because then our sum will just be &lt;math&gt;(2n+1)a&lt;/math&gt;. We now have <br /> &lt;cmath&gt;(2n+1)a = 345&lt;/cmath&gt;<br /> and &lt;math&gt;a&lt;/math&gt; will have a solution when &lt;math&gt;\frac{345}{2n+1}&lt;/math&gt; is an integer, namely when &lt;math&gt;2n+1&lt;/math&gt; is a divisor of 345. We check that <br /> &lt;cmath&gt;2n+1 = 3, 5, 15, 23&lt;/cmath&gt;<br /> work, and no more, because &lt;math&gt;2n+1=1&lt;/math&gt; does not satisfy the requirements of two or more consecutive integers, and when &lt;math&gt;2n+1&lt;/math&gt; equals the next biggest factor, &lt;math&gt;69&lt;/math&gt;, there must be negative integers in the sequence. Our solutions are &lt;math&gt;\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}&lt;/math&gt;.<br /> <br /> For the even cases, we choose our sequence to be of the form:<br /> &lt;cmath&gt;a-(n-1), \cdots, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> so the sum is &lt;math&gt;\frac{(2n)(2a+1)}{2} = n(2a+1)&lt;/math&gt;. In this case, we find our solutions to be &lt;math&gt;\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}&lt;/math&gt;.<br /> <br /> We have found all 7 solutions and our answer is &lt;math&gt;\boxed{\textbf{(E)} \, 7}&lt;/math&gt;.<br /> <br /> ==Solution 2== <br /> <br /> The sum from &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and &lt;math&gt;a&gt;b &lt;/math&gt; is <br /> <br /> &lt;math&gt;S=\dfrac{(a-b+1)(a+b)}{2}&lt;/math&gt; <br /> <br /> &lt;math&gt;345=\dfrac{(a-b+1)(a+b)}{2}&lt;/math&gt; <br /> <br /> &lt;math&gt;2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)&lt;/math&gt;<br /> <br /> Let &lt;math&gt;c=a-b+1&lt;/math&gt; and &lt;math&gt;d=a+b&lt;/math&gt; <br /> <br /> &lt;math&gt;2\cdot 3\cdot 5\cdot 23=c\cdot d&lt;/math&gt;<br /> <br /> If we factor &lt;math&gt;690&lt;/math&gt; into all of its factor groups &lt;math&gt;(\text{exg}~ (10,69) ~\text{or} ~(15,46))&lt;/math&gt; we will have several ordered pairs &lt;math&gt;(c,d)&lt;/math&gt; where &lt;math&gt;c&lt;d&lt;/math&gt;<br /> <br /> The number of possible values for &lt;math&gt;c&lt;/math&gt; is half the number of factors of &lt;math&gt;690&lt;/math&gt; which is &lt;math&gt;\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8&lt;/math&gt; <br /> <br /> However, we have one extraneous case of &lt;math&gt;(1,690)&lt;/math&gt; because here, &lt;math&gt;a=b&lt;/math&gt; and we have the sum of one consecutive number which is not allowed by the question.<br /> <br /> Thus the answer is &lt;math&gt;8-1=7&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(E)} \,7}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> There is a handy formula for this problem: The number of odd factors of &lt;math&gt;345&lt;/math&gt;<br /> <br /> &lt;math&gt;345 = 5*3*23&lt;/math&gt;<br /> <br /> &lt;math&gt;2^3 = 8&lt;/math&gt;<br /> <br /> There are 8 ways to have an increasing sum of positive integers that add to 345. However, we have to subtract one for the case where it is just &lt;math&gt;345&lt;/math&gt;. The problem wants two or more consecutive integers.<br /> <br /> Therefore, &lt;math&gt;8-1=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(E)} \,7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> We're dealing with an increasing arithmetic progression of common difference 1. Let &lt;math&gt;x&lt;/math&gt; be the number of terms in a summation. Let &lt;math&gt;y&lt;/math&gt; be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x \cdot \frac{y+y+[(x-1)1]}{2}&amp;=345 \\<br /> 2xy+x^2-x&amp;=690<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that &lt;math&gt;x&lt;/math&gt;... voilà!<br /> <br /> &lt;cmath&gt;(x)(x+2y-1)=690&lt;/cmath&gt;<br /> <br /> There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is &lt;math&gt;2y-1&lt;/math&gt; more than the expression in the parenthesis on the left. &lt;math&gt;y&lt;/math&gt; is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs. The problem also says the &quot;increasing sequence&quot; has to have &quot;two or more&quot; terms, so &lt;math&gt;x \geqslant 2&lt;/math&gt;. This eliminates the factor pair &lt;math&gt;1 \cdot 690&lt;/math&gt;. With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found &lt;math&gt;\boxed{\textbf{(E)}\ 7}&lt;/math&gt; ways to write 345 in the silly way outlined by the problem.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_16&diff=86802 2016 AMC 12B Problems/Problem 16 2017-08-04T04:55:38Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.<br /> <br /> For the first case, we can cleverly choose the convenient form of our sequence to be<br /> &lt;cmath&gt;a-n,\cdots, a-1, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> <br /> because then our sum will just be &lt;math&gt;(2n+1)a&lt;/math&gt;. We now have <br /> &lt;cmath&gt;(2n+1)a = 345&lt;/cmath&gt;<br /> and &lt;math&gt;a&lt;/math&gt; will have a solution when &lt;math&gt;\frac{345}{2n+1}&lt;/math&gt; is an integer, namely when &lt;math&gt;2n+1&lt;/math&gt; is a divisor of 345. We check that <br /> &lt;cmath&gt;2n+1 = 3, 5, 15, 23&lt;/cmath&gt;<br /> work, and no more, because &lt;math&gt;2n+1=1&lt;/math&gt; does not satisfy the requirements of two or more consecutive integers, and when &lt;math&gt;2n+1&lt;/math&gt; equals the next biggest factor, &lt;math&gt;69&lt;/math&gt;, there must be negative integers in the sequence. Our solutions are &lt;math&gt;\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}&lt;/math&gt;.<br /> <br /> For the even cases, we choose our sequence to be of the form:<br /> &lt;cmath&gt;a-(n-1), \cdots, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> so the sum is &lt;math&gt;\frac{(2n)(2a+1)}{2} = n(2a+1)&lt;/math&gt;. In this case, we find our solutions to be &lt;math&gt;\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}&lt;/math&gt;.<br /> <br /> We have found all 7 solutions and our answer is &lt;math&gt;\boxed{\textbf{(E)} \, 7}&lt;/math&gt;.<br /> <br /> ==Solution 2== <br /> <br /> The sum from &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and &lt;math&gt;a&gt;b &lt;/math&gt; is <br /> <br /> &lt;math&gt;S=\dfrac{(a-b+1)(a+b)}{2}&lt;/math&gt; <br /> <br /> &lt;math&gt;345=\dfrac{(a-b+1)(a+b)}{2}&lt;/math&gt; <br /> <br /> &lt;math&gt;2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)&lt;/math&gt;<br /> <br /> Let &lt;math&gt;c=a-b+1&lt;/math&gt; and &lt;math&gt;d=a+b&lt;/math&gt; <br /> <br /> &lt;math&gt;2\cdot 3\cdot 5\cdot 23=c\cdot d&lt;/math&gt;<br /> <br /> If we factor &lt;math&gt;690&lt;/math&gt; into all of its factor groups &lt;math&gt;(\text{exg}~ (10,69) ~\text{or} ~(15,46))&lt;/math&gt; we will have several ordered pairs &lt;math&gt;(c,d)&lt;/math&gt; where &lt;math&gt;c&lt;d&lt;/math&gt;<br /> <br /> The number of possible values for &lt;math&gt;c&lt;/math&gt; is half the number of factors of &lt;math&gt;690&lt;/math&gt; which is &lt;math&gt;\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8&lt;/math&gt; <br /> <br /> However, we have one extraneous case of &lt;math&gt;(1,690)&lt;/math&gt; because here, &lt;math&gt;a=b&lt;/math&gt; and we have the sum of one consecutive number which is not allowed by the question.<br /> <br /> Thus the answer is &lt;math&gt;8-1=7&lt;/math&gt;<br /> <br /> &lt;math&gt;\boxed{\textbf{(E)} \,7}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> There is a handy formula for this problem: The number of odd factors of &lt;math&gt;345&lt;/math&gt;<br /> <br /> &lt;math&gt;345 = 5*3*23&lt;/math&gt;<br /> <br /> &lt;math&gt;2^3 = 8&lt;/math&gt;<br /> <br /> There are 8 ways to have an increasing sum of positive integers that add to 345. However, we have to subtract one for the case where it is just &lt;math&gt;345&lt;/math&gt;. The problem wants two or more consecutive integers.<br /> <br /> Therefore, &lt;math&gt;8-1=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(E)} \,7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> We're dealing with an increasing arithmetic progression of common difference 1. Let &lt;math&gt;x&lt;/math&gt; be the number of terms in a summation. Let &lt;math&gt;y&lt;/math&gt; be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> x \cdot \frac{y+y+[(x-1)1]}{2}&amp;=345 \\<br /> 2xy+x^2-x&amp;=690<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that &lt;math&gt;x&lt;/math&gt;... voila!<br /> <br /> &lt;cmath&gt;(x)(x+2y-1)=690&lt;/cmath&gt;<br /> <br /> There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is &lt;math&gt;2y-1&lt;/math&gt; more than the expression in the parenthesis on the left. &lt;math&gt;y&lt;/math&gt; is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs that were just mirrors of the other 8. The problem also says the &quot;increasing sequence&quot; has to have &quot;two or more&quot; terms, so &lt;math&gt;x \geqslant 2&lt;/math&gt;. This eliminates the factor pair &lt;math&gt;1 \cdot 690&lt;/math&gt;. With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found &lt;math&gt;\boxed{\textbf{(E)}\ 7}&lt;/math&gt; ways to write 345 in the silly way outlined by the problem.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_20&diff=86642 2011 AMC 12A Problems/Problem 20 2017-07-27T01:58:36Z <p>Obtuse: </p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;f(x)=ax^2+bx+c&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; are integers. Suppose that &lt;math&gt;f(1)=0&lt;/math&gt;, &lt;math&gt;50&lt;f(7)&lt;60&lt;/math&gt;, &lt;math&gt;70&lt;f(8)&lt;80&lt;/math&gt;, &lt;math&gt;5000k&lt;f(100)&lt;5000(k+1)&lt;/math&gt; for some integer &lt;math&gt;k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 1 \qquad<br /> \textbf{(B)}\ 2 \qquad<br /> \textbf{(C)}\ 3 \qquad<br /> \textbf{(D)}\ 4 \qquad<br /> \textbf{(E)}\ 5 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> From &lt;math&gt;f(1) = 0&lt;/math&gt;, we know that &lt;math&gt;a+b+c = 0&lt;/math&gt;.<br /> <br /> From the first inequality, we get &lt;math&gt;50 &lt; 49a+7b+c &lt; 60&lt;/math&gt;. Subtracting &lt;math&gt;a+b+c = 0&lt;/math&gt; from this gives us &lt;math&gt;50 &lt; 48a+6b &lt; 60&lt;/math&gt;, and thus &lt;math&gt;\frac{25}{3} &lt; 8a+b &lt; 10&lt;/math&gt;. Since &lt;math&gt;8a+b&lt;/math&gt; must be an integer, it follows that &lt;math&gt;8a+b = 9&lt;/math&gt;.<br /> <br /> Similarly, from the second inequality, we get &lt;math&gt;70 &lt; 64a+8b+c &lt; 80&lt;/math&gt;. Again subtracting &lt;math&gt;a+b+c = 0&lt;/math&gt; from this gives us &lt;math&gt;70 &lt; 63a+7b &lt; 80&lt;/math&gt;, or &lt;math&gt;10 &lt; 9a+b &lt; \frac{80}{7}&lt;/math&gt;. It follows from this that &lt;math&gt;9a+b = 11&lt;/math&gt;.<br /> <br /> We now have a system of three equations: &lt;math&gt;a+b+c = 0&lt;/math&gt;, &lt;math&gt;8a+b = 9&lt;/math&gt;, and &lt;math&gt;9a+b = 11&lt;/math&gt;. Solving gives us &lt;math&gt;(a, b, c) = (2, -7, 5)&lt;/math&gt; and from this we find that &lt;math&gt;f(100) = 2(100)^2-7(100)+5 = 19305&lt;/math&gt;<br /> <br /> Since &lt;math&gt;15000 &lt; 19305 &lt; 20000 \to 5000(3) &lt; 19305 &lt; 5000(4)&lt;/math&gt;, we find that &lt;math&gt;k = 3 \rightarrow \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> &lt;math&gt;f(x)&lt;/math&gt; is some non-monic quadratic with a root at &lt;math&gt;x=1&lt;/math&gt;. Knowing this, we'll forget their silly &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; and instead write it as &lt;math&gt;f(x)=p(x-1)(x-r)&lt;/math&gt;.<br /> <br /> &lt;math&gt;f(7)=6p(7-r)&lt;/math&gt;, so &lt;math&gt;f(7)&lt;/math&gt; is a multiple of 6. They say &lt;math&gt;f(7)&lt;/math&gt; is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, &lt;math&gt;f(7)=6p(7-r)=54&lt;/math&gt;.<br /> <br /> &lt;math&gt;f(8)=7p(8-r)&lt;/math&gt;, so &lt;math&gt;f(8)&lt;/math&gt; is a multiple of 7. They say &lt;math&gt;f(8)&lt;/math&gt; is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, &lt;math&gt;f(8)=7p(8-r)=77&lt;/math&gt;.<br /> <br /> Now, we solve a system of equations in two variables. <br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 6p(7-r)&amp;=54 \\<br /> 7p(8-r)&amp;=77 \\<br /> \\<br /> p(7-r)&amp;=9 \\<br /> p(8-r)&amp;=11 \\<br /> \\<br /> 7p-pr&amp;=9 \\<br /> 8p-pr&amp;=11 \\<br /> \\<br /> (8p-pr)-(7p-pr)&amp;=11-9 \\<br /> \\<br /> p&amp;=2 \\<br /> \\<br /> 2(7-r)&amp;=9 \\<br /> \\<br /> r&amp;=2.5<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> &lt;math&gt;f(100)=2(100-1)(100-2.5)=19305 \implies k=3 \implies \boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_19&diff=86550 2016 AMC 12A Problems/Problem 19 2017-07-24T16:19:54Z <p>Obtuse: </p> <hr /> <div>==Problem==<br /> <br /> Jerry starts at &lt;math&gt;0&lt;/math&gt; on the real number line. He tosses a fair coin &lt;math&gt;8&lt;/math&gt; times. When he gets heads, he moves &lt;math&gt;1&lt;/math&gt; unit in the positive direction; when he gets tails, he moves &lt;math&gt;1&lt;/math&gt; unit in the negative direction. The probability that he reaches &lt;math&gt;4&lt;/math&gt; at some time during this process &lt;math&gt;\frac{a}{b},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b?&lt;/math&gt; (For example, he succeeds if his sequence of tosses is &lt;math&gt;HTHHHHHH.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> For &lt;math&gt;6&lt;/math&gt; to &lt;math&gt;8&lt;/math&gt; heads, we are guaranteed to hit &lt;math&gt;4&lt;/math&gt; heads, so the sum here is &lt;math&gt;\binom{8}{2}+\binom{8}{1}+\binom{8}{0}=28+8+1=37&lt;/math&gt;.<br /> <br /> For &lt;math&gt;4&lt;/math&gt; heads, you have to hit the &lt;math&gt;4&lt;/math&gt; heads at the start so there's only one way, &lt;math&gt;1&lt;/math&gt;.<br /> <br /> For &lt;math&gt;5&lt;/math&gt; heads, we either start off with &lt;math&gt;4&lt;/math&gt; heads, which gives us &lt;math&gt;4\textbf{C}1=4&lt;/math&gt; ways to arrange the other flips, or we start off with five heads and one tail, which has &lt;math&gt;6&lt;/math&gt; ways minus the &lt;math&gt;2&lt;/math&gt; overlapping cases, &lt;math&gt;\text{HHHHHTTT}&lt;/math&gt; and &lt;math&gt;\text{HHHHTHTT}&lt;/math&gt;. Total ways: &lt;math&gt;8&lt;/math&gt;.<br /> <br /> Then we sum to get &lt;math&gt;46&lt;/math&gt;. There are a total of &lt;math&gt;2^8=256&lt;/math&gt; possible sequences of &lt;math&gt;8&lt;/math&gt; coin flips, so the probability is &lt;math&gt;\frac{46}{256}=\frac{23}{128}&lt;/math&gt;. Summing, we get &lt;math&gt;23+128=\boxed{\textbf{(B) }151}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases: <br /> <br /> (Case 1): The first four flips are heads. Then, the last four flips can be anything so &lt;math&gt;2^4=16&lt;/math&gt; possibilities work. <br /> <br /> (Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so &lt;math&gt;2^2=4&lt;/math&gt; flips work. &lt;math&gt;4*4=16&lt;/math&gt;. <br /> <br /> (Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips. <br /> <br /> (1 tail in first four flips). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are &lt;math&gt;4*2=8&lt;/math&gt; possibilities. <br /> <br /> (2 tails in first four flips). In this case, the tails can be in &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; positions. <br /> <br /> Adding these cases up and taking the total out of &lt;math&gt;2^8=256&lt;/math&gt; yields &lt;math&gt;\frac{16+16+8+6}{256}=\frac{46}{256}=\frac{23}{128}&lt;/math&gt;. This means the answer is &lt;math&gt;23+128=\boxed{\textbf{(B) }151}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> (Inspired by https://www.youtube.com/watch?v=EZYzjmd_f2g by Walt S)<br /> <br /> Draw a triangle of dots as seen in the video. Now, we'll do something different than he did. Draw a dotted line between dots that correspond to 3 on the number line and dots that correspond to 4 on the number line. This is the threshold that must be passed to have a successful sequence. We consider the three entry points to the immediate right of this line. Doing a tiny bit of fun bashing as we slide down this triangle, we find there is 1 way to cross the line for the first time through the uppermost entry point, 4 ways to cross the line for the first time through the middle entry point, and 14 ways to cross the line for the first time through the lowermost entry point.<br /> <br /> Superimpose Pascal's triangle over the dots. Notice that the number of ways to get to a dot is equal to its corresponding number on Pascal's triangle. Rows on Pascal's triangle sum to powers of 2. There are &lt;math&gt;2^4=16&lt;/math&gt; ways to get to the 4th row (remember that we start with the 0th row), and we found that 1 of them makes us cross the threshold for the first time. There are &lt;math&gt;2^6=64&lt;/math&gt; ways to get to the 6th row, and we found that 4 of them make us cross the threshold for the first time. There are &lt;math&gt;2^8=256&lt;/math&gt; ways to get to the 8th row, and we found that 14 of them make us cross the threshold for the first time.<br /> <br /> Adding up these probabilities &lt;cmath&gt;\frac{1}{16}+\frac{4}{64}+\frac{14}{256}=\frac{23}{128}&lt;/cmath&gt; and summing &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, &lt;cmath&gt;a+b=23+128=\boxed{\textbf{(B)}\ 151}&lt;/cmath&gt; we get our answer.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=86535 2016 AMC 10A Problems/Problem 22 2017-07-24T02:42:08Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 11&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> The number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes. This tells us that the product cannot contain any &lt;math&gt;1&lt;/math&gt;'s. The number of factors is given as 110. The only way to write 110 as a product of at least three factors without &lt;math&gt;1&lt;/math&gt;s is &lt;math&gt;2\cdot 5\cdot 11&lt;/math&gt;.<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=86534 2016 AMC 10A Problems/Problem 22 2017-07-24T02:40:30Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 11&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> The number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes. This tells us that the product cannot contain any &lt;math&gt;1&lt;/math&gt;'s. The only way to write 110 as a product of at least three factors without &lt;math&gt;1&lt;/math&gt;s is &lt;math&gt;2\cdot 5\cdot 11&lt;/math&gt;.<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=86533 2016 AMC 10A Problems/Problem 22 2017-07-24T02:40:04Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 11&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> The number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes. This tells us that the product cannot contain any &lt;math&gt;1&lt;/math&gt;s. The only way to write 110 as a product of at least three factors without &lt;math&gt;1&lt;/math&gt;s is &lt;math&gt;2\cdot 5\cdot 11&lt;/math&gt;.<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=86532 2016 AMC 10A Problems/Problem 22 2017-07-24T02:37:06Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 11&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> Furthermore, since the number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes, we see that there can be at most three distinct prime factors for a number with 110 factors. (This product cannot contain any &lt;math&gt;1&lt;/math&gt;s, so the only way to write 110 as a product of at least three factors is &lt;math&gt;2\cdot 5\cdot 11&lt;/math&gt;.)<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_22&diff=86531 2016 AMC 10A Problems/Problem 22 2017-07-24T02:36:20Z <p>Obtuse: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> For some positive integer &lt;math&gt;n&lt;/math&gt;, the number &lt;math&gt;110n^3&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; positive integer divisors, including &lt;math&gt;1&lt;/math&gt; and the number &lt;math&gt;110n^3&lt;/math&gt;. How many positive integer divisors does the number &lt;math&gt;81n^4&lt;/math&gt; have?<br /> <br /> &lt;math&gt;\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the prime factorization of &lt;math&gt;110&lt;/math&gt; is &lt;math&gt;2 \cdot 5 \cdot 11&lt;/math&gt;, we have that the number is equal to &lt;math&gt;2 \cdot 5 \cdot 11 \cdot n^3&lt;/math&gt;. This has &lt;math&gt;2 \cdot 2 \cdot 2=8&lt;/math&gt; factors when &lt;math&gt;n=1&lt;/math&gt;. This needs a multiple of 11 factors, which we can achieve by setting &lt;math&gt;n=2^3&lt;/math&gt;, so we have &lt;math&gt;2^{10} \cdot 5 \cdot 11&lt;/math&gt; has &lt;math&gt;44&lt;/math&gt; factors. To achieve the desired &lt;math&gt;110&lt;/math&gt; factors, we need the number of factors to also be divisible by &lt;math&gt;5&lt;/math&gt;, so we can set &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;, so &lt;math&gt;2^{10} \cdot 5^4 \cdot 11&lt;/math&gt; has &lt;math&gt;110&lt;/math&gt; factors. Therefore, &lt;math&gt;n=2^3 \cdot 5&lt;/math&gt;. In order to find the number of factors of &lt;math&gt;81n^4&lt;/math&gt;, we raise this to the fourth power and multiply it by &lt;math&gt;81&lt;/math&gt;, and find the factors of that number. We have &lt;math&gt;3^4 \cdot 2^{12} \cdot 5^4&lt;/math&gt;, and this has &lt;math&gt;5 \cdot 13 \cdot 5=\boxed{\textbf{(D) }325}&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> &lt;math&gt;110n^3&lt;/math&gt; clearly has at least three distinct prime factors, namely 2, 5, and 11.<br /> <br /> Furthermore, since the number of factors of &lt;math&gt;p_1^{n_1}\cdots p_k^{n_k}&lt;/math&gt; is &lt;math&gt;(n_1+1)\cdots(n_k+1)&lt;/math&gt; when the &lt;math&gt;p&lt;/math&gt;'s are distinct primes, we see that there can be at most three distinct prime factors for a number with 110 factors. (This product cannot contain any 1s, so the only way to write 110 as a product of at least three factors is &lt;math&gt;2\cdot 5\cdot 11&lt;/math&gt;.<br /> <br /> We conclude that &lt;math&gt;110n^3&lt;/math&gt; has only the three prime factors 2, 5, and 11 and that the multiplicities are 1, 4, and 10 in some order. I.e., there are six different possible values of &lt;math&gt;n&lt;/math&gt; all of the form &lt;math&gt;n=p_1\cdot p_2^3&lt;/math&gt;.<br /> <br /> &lt;math&gt;81n^4&lt;/math&gt; thus has prime factorization &lt;math&gt;81n^4=3^4\cdot p_1^4\cdot p_2^{12}&lt;/math&gt; and a factor count of &lt;math&gt;5\cdot5\cdot13=\boxed{\textbf{(D) }325}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_9&diff=86403 2017 AMC 12A Problems/Problem 9 2017-07-16T00:45:54Z <p>Obtuse: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x+2&lt;/math&gt;, and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\leqslant 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y\leqslant 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x\leqslant 1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x\leqslant 1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\leqslant y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value (using &lt;math&gt;x-2&lt;/math&gt; as the common value works as well). Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\geqslant 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=x+6&lt;/math&gt; where &lt;math&gt;y\geqslant 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;S&lt;/math&gt; is made up of three rays with common endpoint &lt;math&gt;(1, 7)&lt;/math&gt;, the answer is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Solution by TheMathematicsTiger7<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_9&diff=86402 2017 AMC 12A Problems/Problem 9 2017-07-16T00:44:29Z <p>Obtuse: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x+2&lt;/math&gt;, and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\leqslant 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y\leqslant 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x\leqslant 1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x\leqslant 1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\leqslant y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value (using &lt;math&gt;x-2&lt;/math&gt; as the common value works as well). Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y\geqslant 7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=x+6&lt;/math&gt; where &lt;math&gt;y\geqslant 7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;S&lt;/math&gt; is made up of three rays with common endpoint &lt;math&gt;(1, 7)&lt;/math&gt;, the answer is &lt;math&gt;\boxed{E}&lt;/math&gt;<br /> <br /> Solution by TheMathematicsTiger7<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_9&diff=86401 2017 AMC 12A Problems/Problem 9 2017-07-16T00:42:54Z <p>Obtuse: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x+2&lt;/math&gt;, and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y&lt;7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y&lt;7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x&lt;1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x&lt;1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\leqslant y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value (using &lt;math&gt;x-2&lt;/math&gt; as the common value works as well). Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y&gt;7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=x+6&lt;/math&gt; where &lt;math&gt;y&gt;7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;S&lt;/math&gt; is made up of three rays with common endpoint &lt;math&gt;(1, 7)&lt;/math&gt;, the answer is &lt;math&gt;\boxed{E}&lt;/math&gt;<br /> <br /> Solution by TheMathematicsTiger7<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_9&diff=86400 2017 AMC 12A Problems/Problem 9 2017-07-16T00:42:31Z <p>Obtuse: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;x+2&lt;/math&gt;, and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\ \qquad\textbf{(C)}\ \text{three lines whose pairwise intersections are three distinct points} \\ \qquad\textbf{(D)}\ \text{a triangle}\qquad\textbf{(E)}\ \text{three rays with a common point} &lt;/math&gt;<br /> <br /> ==Solution==<br /> If the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;x+2&lt;/math&gt;, then &lt;math&gt;x=1&lt;/math&gt;. Also, &lt;math&gt;y-4\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y&lt;7&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;x=1&lt;/math&gt; where &lt;math&gt;y&lt;7&lt;/math&gt; is part of &lt;math&gt;S&lt;/math&gt;. This is a ray with an endpoint of &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> Similar to the process above, we assume that the two equal values are &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt;. Solving the equation &lt;math&gt;3=y-4&lt;/math&gt; then &lt;math&gt;y=7&lt;/math&gt;. Also, &lt;math&gt;x+2\leqslant 3&lt;/math&gt; because 3 is the common value. Solving for &lt;math&gt;x&lt;/math&gt;, we get &lt;math&gt;x&lt;1&lt;/math&gt;. Therefore the portion of the line &lt;math&gt;y=7&lt;/math&gt; where &lt;math&gt;x&lt;1&lt;/math&gt; is also part of &lt;math&gt;S&lt;/math&gt;. This is another ray with the same endpoint as the above ray: &lt;math&gt;(1, 7)&lt;/math&gt;.<br /> <br /> If &lt;math&gt;x+2&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are the two equal values, then &lt;math&gt;x+2=y-4&lt;/math&gt;. Solving the equation for &lt;math&gt;y&lt;/math&gt;, we get &lt;math&gt;y=x+6&lt;/math&gt;. Also &lt;math&gt;3\leqslant y-4&lt;/math&gt; because &lt;math&gt;y-4&lt;/math&gt; is one way to express the common value (&lt;math&gt;Using x-2 as the common value works as well). Solving for &lt;/math&gt;y&lt;math&gt;, we get &lt;/math&gt;y&gt;7&lt;math&gt;. Therefore the portion of the line &lt;/math&gt;y=x+6&lt;math&gt; where &lt;/math&gt;y&gt;7&lt;math&gt; is part of &lt;/math&gt;S&lt;math&gt; like the other two rays. The lowest possible value that can be achieved is also &lt;/math&gt;(1, 7)&lt;math&gt;.<br /> <br /> Since &lt;/math&gt;S&lt;math&gt; is made up of three rays with common endpoint &lt;/math&gt;(1, 7)&lt;math&gt;, the answer is &lt;/math&gt;\boxed{E}$<br /> <br /> Solution by TheMathematicsTiger7<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2017|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Obtuse https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems&diff=86125 2016 AMC 10B Problems 2017-06-22T16:34:01Z <p>Obtuse: /* Problem 7 */</p> <hr /> <div>==Problem 1== <br /> What is the value of &lt;math&gt;\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}&lt;/math&gt; when &lt;math&gt;a= \frac{1}{2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> If &lt;math&gt;n\heartsuit m=n^3m^2&lt;/math&gt;, what is &lt;math&gt;\frac{2\heartsuit 4}{4\heartsuit 2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Let &lt;math&gt;x=-2016&lt;/math&gt;. What is the value of &lt;math&gt;\bigg|&lt;/math&gt; &lt;math&gt;||x|-x|-|x|&lt;/math&gt; &lt;math&gt;\bigg|&lt;/math&gt; &lt;math&gt;-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Zoey read &lt;math&gt;15&lt;/math&gt; books, one at a time. The first book took her &lt;math&gt;1&lt;/math&gt; day to read, the second book took her &lt;math&gt;2&lt;/math&gt; days to read, the third book took her &lt;math&gt;3&lt;/math&gt; days to read, and so on, with each book taking her &lt;math&gt;1&lt;/math&gt; more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day the week did she finish her &lt;math&gt;15&lt;/math&gt;th book?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Sunday}\qquad\textbf{(B)}\ \text{Monday}\qquad\textbf{(C)}\ \text{Wednesday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The mean age of Amanda's &lt;math&gt;4&lt;/math&gt; cousins is &lt;math&gt;8&lt;/math&gt;, and their median age is &lt;math&gt;5&lt;/math&gt;. What is the sum of the ages of Amanda's youngest and oldest cousins?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number &lt;math&gt;S&lt;/math&gt;. What is the smallest possible value for the sum of the digits of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The ratio of the measures of two acute angles is &lt;math&gt;5:4&lt;/math&gt;, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> What is the tens digit of &lt;math&gt;2015^{2016}-2017?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> All three vertices of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; are lying on the parabola defined by &lt;math&gt;y=x^2&lt;/math&gt;, with &lt;math&gt;A&lt;/math&gt; at the origin and &lt;math&gt;\overline{BC}&lt;/math&gt; parallel to the &lt;math&gt;x&lt;/math&gt;-axis. The area of the triangle is &lt;math&gt;64&lt;/math&gt;. What is the length of &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A thin piece of wood of uniform density in the shape of an equilateral triangle with side length &lt;math&gt;3&lt;/math&gt; inches weighs &lt;math&gt;12&lt;/math&gt; ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of &lt;math&gt;5&lt;/math&gt; inches. Which of the following is closest to the weight, in ounces, of the second piece?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Carl decided to fence in his rectangular garden. He bought &lt;math&gt;20&lt;/math&gt; fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly &lt;math&gt;4&lt;/math&gt; yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Two different numbers are selected at random from &lt;math&gt;( 1, 2, 3, 4, 5)&lt;/math&gt; and multiplied together. What is the probability that the product is even?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for &lt;math&gt;1000&lt;/math&gt; of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these &lt;math&gt;1000&lt;/math&gt; babies were in sets of quadruplets?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line &lt;math&gt;y=\pi x&lt;/math&gt;, the line &lt;math&gt;y=-0.1&lt;/math&gt; and the line &lt;math&gt;x=5.1?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> All the numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9&lt;/math&gt; are written in a &lt;math&gt;3\times3&lt;/math&gt; array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to &lt;math&gt;18&lt;/math&gt;. What is the number in the center?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5&lt;/math&gt; and &lt;math&gt;BC=4&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;EB=1&lt;/math&gt;, point &lt;math&gt;G&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;CG=1&lt;/math&gt;. and point &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\overline{CD}&lt;/math&gt; so that &lt;math&gt;DF=2&lt;/math&gt;. Segments &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; intersect &lt;math&gt;\overline{EF}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, respectively. What is the value of &lt;math&gt;\frac{PQ}{EF}&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt; pair A1=(2,0),A2=(4,4);<br /> pair B1=(0,4),B2=(5,1);<br /> pair C1=(5,0),C2=(0,4); <br /> draw(A1--A2);<br /> draw(B1--B2);<br /> draw(C1--C2);<br /> draw((0,0)--B1--(5,4)--C1--cycle);<br /> dot((20/7,12/7));<br /> dot((3.07692307692,2.15384615384));<br /> label(&quot;$Q$&quot;,(3.07692307692,2.15384615384),N);<br /> label(&quot;$P$&quot;,(20/7,12/7),W);<br /> label(&quot;$A$&quot;,(0,4), NW);<br /> label(&quot;$B$&quot;,(5,4), NE);<br /> label(&quot;$C$&quot;,(5,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$F$&quot;,(2,0),S); label(&quot;$G$&quot;,(5,1),E);<br /> label(&quot;$E\$&quot;,(4,4),N);<br /> <br /> dot(A1); dot(A2);<br /> dot(B1); dot(B2);<br /> dot(C1); dot(C2);<br /> dot((0,0)); dot((5,4));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad<br /> \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad<br /> \textbf{(C)}~\frac{9}{82} \qquad<br /> \textbf{(D)}~\frac{10}{91}\qquad<br /> \textbf{(E)}~\frac19&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A(2,2)&lt;/math&gt; to the circle of radius &lt;math&gt;3&lt;/math&gt; centered at &lt;math&gt;A’(5,6)&lt;/math&gt;. What distance does the origin &lt;math&gt;O(0,0)&lt;/math&gt;, move under this transformation?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> What is the area of the region enclosed by the graph of the equation &lt;math&gt;x^2+y^2=|x|+|y|?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<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 \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt;, points &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are chosen on sides &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}&lt;/math&gt;, and &lt;math&gt;\overline{FA}&lt;/math&gt; respectively, so lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;ZW&lt;/math&gt;, &lt;math&gt;YX&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; are parallel and equally spaced. What is the ratio of the area of hexagon &lt;math&gt;WCXYFZ&lt;/math&gt; to the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> How many four-digit integers &lt;math&gt;abcd&lt;/math&gt;, with &lt;math&gt;a \neq 0&lt;/math&gt;, have the property that the three two-digit integers &lt;math&gt;ab&lt;bc&lt;cd&lt;/math&gt; form an increasing arithmetic sequence? One such number is &lt;math&gt;4692&lt;/math&gt;, where &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=6&lt;/math&gt;, &lt;math&gt;c=9&lt;/math&gt;, and &lt;math&gt;d=2&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;r&lt;/math&gt;. How many distinct values does &lt;math&gt;f(x)&lt;/math&gt; assume for &lt;math&gt;x \ge 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> <br /> ==See also==<br /> {{AMC10 box|year=2016|ab=B|before=[[2016 AMC 10A Problems]]|after=2017 AMC 10 A}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2016 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Obtuse