https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathisfun04&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-17T00:30:11Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2006_UNCO_Math_Contest_II_Problems/Problem_4&diff=85405 2006 UNCO Math Contest II Problems/Problem 4 2017-04-23T18:42:42Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Determine all positive integers &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;n^2+3&lt;/math&gt; divides evenly (without remainder) into &lt;math&gt;n^4-3n^2+10&lt;/math&gt; ?<br /> <br /> <br /> By polynomial division, &lt;math&gt;\frac{n^4 - 3n^2 + 10}{n^2 + 3} = n^2 - 6 + \frac{28}{n^2 + 3}&lt;/math&gt;. By default, &lt;math&gt;n^2 - 6 \in \mathbb{N}&lt;/math&gt; when &lt;math&gt;n \in \mathbb{N}&lt;/math&gt;, so we must find all positive integral values of &lt;math&gt;n&lt;/math&gt; for which &lt;math&gt;\frac{28}{n^2 + 3}&lt;/math&gt; is also an integer. Listing out the factors of &lt;math&gt;28&lt;/math&gt;, we see that &lt;math&gt;n^2 + 3 \in \{1, 2, 4, 7, 14, 28\}&lt;/math&gt;. Looking for positive integral values that satisfy these conditions yields &lt;math&gt;\boxed{n \in \{1, 2, 5\}}&lt;/math&gt;, as desired. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{UNCO Math Contest box|n=II|year=2006|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1968_AHSME_Problems/Problem_21&diff=83438 1968 AHSME Problems/Problem 21 2017-02-11T19:03:12Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;S=1!+2!+3!+\cdots +99!&lt;/math&gt;, then the units' digit in the value of S is:<br /> <br /> &lt;math&gt;\text{(A) } 9\quad<br /> \text{(B) } 8\quad<br /> \text{(C) } 5\quad<br /> \text{(D) } 3\quad<br /> \text{(E) } 0&lt;/math&gt;<br /> <br /> == Solution ==<br /> Note that every factorial after &lt;math&gt;5!&lt;/math&gt; has a unit digit of &lt;math&gt;0&lt;/math&gt;, meaning that we can disregard them. Thus, we only need to find the units digit of &lt;math&gt;1! + 2! + 3! + 4!&lt;/math&gt;, and as &lt;math&gt;1! + 2! + 3! + 4! = 3 \equiv 10&lt;/math&gt;, which means that the unit digit is &lt;math&gt;3&lt;/math&gt;, we have our answer of &lt;math&gt;\fbox{D}&lt;/math&gt; as desired.<br /> <br /> == See also ==<br /> {{AHSME box|year=1968|num-b=20|num-a=22}} <br /> <br /> [[Category: Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2006_UNCO_Math_Contest_II_Problems/Problem_2&diff=82662 2006 UNCO Math Contest II Problems/Problem 2 2017-02-04T00:02:32Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;a,b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are positive integers, how many integers are strictly between the product &lt;math&gt;abc&lt;/math&gt;<br /> and &lt;math&gt;(a+1)(b+1)(c+1)&lt;/math&gt; ? For example, there are 35 integers strictly between &lt;math&gt;24=2*3*4&lt;/math&gt; and &lt;math&gt;60=3*4*5.&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> We start by expanding &lt;math&gt;(a + 1)(b + 1)(c + 1)&lt;/math&gt;, and in doing so we obtain &lt;math&gt;abc + ab + ac + a + bc + b + c + 1&lt;/math&gt;. We then subtract &lt;math&gt;abc&lt;/math&gt; from &lt;math&gt;abc + ab + ac + a + bc + b + c + 1&lt;/math&gt;, to get &lt;math&gt;ab + ac + a + bc + b + c + 1&lt;/math&gt;. We finally subtract &lt;math&gt;1&lt;/math&gt; from this value, as the problem asks for how many integers are strictly in between, and we get our answer of &lt;math&gt;\boxed{ab + ac + bc + a + b + c}&lt;/math&gt; as desired.<br /> <br /> ==See Also==<br /> {{UNCO Math Contest box|n=II|year=2006|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_9&diff=82583 2016 AIME I Problems/Problem 9 2017-01-26T22:15:46Z <p>Mathisfun04: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has &lt;math&gt;AB=40,AC=31,&lt;/math&gt; and &lt;math&gt;\sin{A}=\frac{1}{5}&lt;/math&gt;. This triangle is inscribed in rectangle &lt;math&gt;AQRS&lt;/math&gt; with &lt;math&gt;B&lt;/math&gt; on &lt;math&gt;\overline{QR}&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; on &lt;math&gt;\overline{RS}&lt;/math&gt;. Find the maximum possible area of &lt;math&gt;AQRS&lt;/math&gt;.<br /> ==Solution==<br /> <br /> ===Solution 1===<br /> Note that if angle &lt;math&gt;BAC&lt;/math&gt; is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where &lt;math&gt;A&lt;/math&gt; is obtuse. Therefore, angle A is acute. Let angle &lt;math&gt;CAS=n&lt;/math&gt; and angle &lt;math&gt;BAQ=m&lt;/math&gt;. Then, &lt;math&gt;\overline{AS}=31\cos(n)&lt;/math&gt; and &lt;math&gt;\overline{AQ}=40\cos(m)&lt;/math&gt;. Then the area of rectangle &lt;math&gt;AQRS&lt;/math&gt; is &lt;math&gt;1240\cos(m)\cos(n)&lt;/math&gt;. By product-to-sum, &lt;math&gt;\cos(m)\cos(n)=\frac{1}{2}(\cos(m+n)+\cos(m-n))&lt;/math&gt;. &lt;math&gt;\cos(m+n)=\sin(90-m-n)=\sin(BAC)=\frac{1}{5}&lt;/math&gt;. The maximum possible value of &lt;math&gt;\cos(m-n)&lt;/math&gt; is 1, which occurs when &lt;math&gt;m=n&lt;/math&gt;. Thus the maximum possible value of &lt;math&gt;\cos(m)\cos(n)&lt;/math&gt; is &lt;math&gt;\frac{1}{2}(\frac{1}{5}+1)=\frac{3}{5}&lt;/math&gt; so the maximum possible area of &lt;math&gt;AQRS&lt;/math&gt; is &lt;math&gt;1240\times{\frac{3}{5}}=\fbox{744}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> As above, we note that angle &lt;math&gt;A&lt;/math&gt; must be acute. Therefore, let &lt;math&gt;A&lt;/math&gt; be the origin, and suppose that &lt;math&gt;Q&lt;/math&gt; is on the positive &lt;math&gt;x&lt;/math&gt; axis and &lt;math&gt;S&lt;/math&gt; is on the positive &lt;math&gt;y&lt;/math&gt; axis. We approach this using complex numbers. Let &lt;math&gt;w=\text{cis} A&lt;/math&gt;, and let &lt;math&gt;z&lt;/math&gt; be a complex number with &lt;math&gt;|z|=1&lt;/math&gt;, &lt;math&gt;\text{Arg}(z)\ge 0^\circ&lt;/math&gt; and &lt;math&gt;\text{Arg}(zw)\le90^\circ&lt;/math&gt;. Then we represent &lt;math&gt;B&lt;/math&gt; by &lt;math&gt;40z&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; by &lt;math&gt;31zw&lt;/math&gt;. The coordinates of &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; depend on the real part of &lt;math&gt;40z&lt;/math&gt; and the imaginary part of &lt;math&gt;31zw&lt;/math&gt;. Thus<br /> &lt;cmath&gt;[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).&lt;/cmath&gt;<br /> We can expand this, using the fact that &lt;math&gt;z\overline{z}=|z|^2&lt;/math&gt;, finding<br /> &lt;cmath&gt;[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).&lt;/cmath&gt;<br /> Now as &lt;math&gt;w=\text{cis}A&lt;/math&gt;, we know that &lt;math&gt;\Im(w)=\frac15&lt;/math&gt;. Also, &lt;math&gt;|z^2w|=1&lt;/math&gt;, so the maximum possible imaginary part of &lt;math&gt;z^2w&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt;. This is clearly achievable under our conditions on &lt;math&gt;z&lt;/math&gt;. Therefore, the maximum possible area of &lt;math&gt;AQRS&lt;/math&gt; is &lt;math&gt;620(1+\tfrac15)=\boxed{744}&lt;/math&gt;.<br /> <br /> ===Solution 3 (With Calculus)===<br /> Let &lt;math&gt;\theta&lt;/math&gt; be the angle &lt;math&gt;\angle BAQ&lt;/math&gt;. The height of the rectangle then can be expressed as &lt;math&gt;h = 31 \sin (A+\theta)&lt;/math&gt;, and the length of the rectangle can be expressed as &lt;math&gt;l = 40\cos \theta&lt;/math&gt;. The area of the rectangle can then be written as a function of &lt;math&gt;\theta&lt;/math&gt;, &lt;math&gt;[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta&lt;/math&gt;. For now, we will ignore the &lt;math&gt;1240&lt;/math&gt; and focus on the function &lt;math&gt;f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta&lt;/math&gt;.<br /> <br /> Taking the derivative, &lt;math&gt;f'(\theta) = \sin A \cdot -2\cos \theta \sin \theta + \cos A \cos 2\theta = \cos A \cos 2\theta - \sin A \sin 2\theta = \cos(2\theta + A)&lt;/math&gt;. Setting this equal to &lt;math&gt;0&lt;/math&gt;, we get &lt;math&gt;\cos(2 \theta + A) = 0 \Rightarrow 2\theta +A = 90, 270 ^\circ&lt;/math&gt;. Since we know that &lt;math&gt;A+ \theta &lt; 90&lt;/math&gt;, the &lt;math&gt;270^\circ&lt;/math&gt; solution is extraneous. Thus, we get that &lt;math&gt;\theta = \frac{90 - A}{2} = 45 - \frac{A}{2}&lt;/math&gt;.<br /> <br /> Plugging this value into the original area equation, &lt;math&gt;a(45 - \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2})&lt;/math&gt;. Using a product-to-sum formula, we get that: &lt;cmath&gt;1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) = &lt;/cmath&gt;<br /> &lt;cmath&gt;1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2})))= &lt;/cmath&gt;<br /> &lt;cmath&gt;620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}&lt;/cmath&gt;.<br /> <br /> ===Note on Problem Validity===<br /> <br /> It has been noted that this answer won't actually work. Let angle &lt;math&gt;QAB = m&lt;/math&gt; and angle &lt;math&gt;CAS = n&lt;/math&gt; as in Solution 1. Since we know (through that solution) that &lt;math&gt;m = n&lt;/math&gt;, we can call them each &lt;math&gt;\theta&lt;/math&gt;. The height of the rectangle is &lt;math&gt;AS = 31\cos\theta&lt;/math&gt;, and the distance &lt;math&gt;BQ = 40\sin\theta&lt;/math&gt;. We know that, if the triangle is to be inscribed in a rectangle, &lt;math&gt;AS \geq BQ&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;AS \geq BQ&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;31\cos\theta \geq 40\sin\theta&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac{31}{40} \geq \tan\theta&lt;/cmath&gt;<br /> <br /> However, &lt;math&gt;\tan\theta = \tan(\frac{90-A}{2}) = \frac{\sin(90-A)}{\cos(90-A)+1} = \frac{\cos A}{\sin A + 1} = \frac{\frac{2\sqrt6}{5}}{\frac{6}{5}} = \frac{\sqrt6}{3} &gt; \frac{31}{40}&lt;/math&gt;, so the triangle does not actually fit in the rectangle: specifically, B is above R and thus in the line containing segment QR but not on the actual segment or in the rectangle.<br /> <br /> &lt;asy&gt;<br /> size(200);<br /> pair A,B,C,Q,R,S;<br /> real r = (pi/2 - asin(1/5))/2;<br /> A = (0,0);<br /> B = 40*dir(r*180/pi);<br /> C = 31*dir(90-r*180/pi);<br /> draw(A--B--C--cycle);<br /> Q = (40*cos(r),0);<br /> R = (40*cos(r),31*cos(r));<br /> S = (0, 31*cos(r));<br /> draw(A--Q--R--S--cycle);<br /> <br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,NE);<br /> label(&quot;$C$&quot;,C,N);<br /> label(&quot;$Q$&quot;,Q,SE);<br /> label(&quot;$R$&quot;,R,E);<br /> label(&quot;$S$&quot;,S,NW);<br /> &lt;/asy&gt;<br /> The actual answer is a radical near &lt;math&gt;728&lt;/math&gt; (letting the triangle be inside the rectangle). The CAMC, however, has decided to accept only the answer &lt;math&gt;744&lt;/math&gt; despite the invalid problem statement.<br /> <br /> =See Also=<br /> {{AIME box|year=2016|n=I|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=Arithmetic_Mean-Geometric_Mean_Inequality&diff=82307 Arithmetic Mean-Geometric Mean Inequality 2017-01-14T17:25:38Z <p>Mathisfun04: /* Proof */</p> <hr /> <div>The '''Arithmetic Mean-Geometric Mean Inequality''' ('''AM-GM''' or '''AMGM''') is an elementary [[inequality]], and is generally one of the first ones taught in inequality courses.<br /> <br /> == Theorem ==<br /> AM-GM states that for any [[set]] of [[nonnegative]] [[real number]]s, the [[arithmetic mean]] of the set is greater than or [[equal]] to the [[geometric mean]] of the set. Algebraically, this is expressed as follows.<br /> <br /> For a set of nonnegative real numbers &lt;math&gt;a_1,a_2,\ldots,a_n&lt;/math&gt;, the following always holds:<br /> &lt;cmath&gt; \frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n} &lt;/cmath&gt;<br /> Using the shorthand notation for [[summation]]s and [[product]]s:<br /> &lt;cmath&gt; \sum_{i=1}^{n}\frac{a_i}{n} \geq \prod\limits_{i=1}^{n}a_i^{\frac{1}{n}} . &lt;/cmath&gt;<br /> For example, for the set &lt;math&gt;\{9,12,54\}&lt;/math&gt;, the arithmetic mean, 25, is greater than the geometric mean, 18; AM-GM guarantees this is always the case. <br /> <br /> The [[equality condition]] of this [[inequality]] states that the arithmetic mean and geometric mean are equal [[iff|if and only if]] all members of the set are equal.<br /> <br /> AM-GM can be used fairly frequently to solve [[Olympiad]]-level inequality problems, such as those on the [[United States of America Mathematics Olympiad | USAMO]] and [[International Mathematics Olympiad | IMO]].<br /> <br /> === Proof ===<br /> <br /> See here: [[Proofs of AM-GM]].<br /> <br /> === Weighted Form ===<br /> <br /> The weighted form of AM-GM is given by using [[weighted average]]s. For example, the weighted arithmetic mean of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;3:1&lt;/math&gt; is &lt;math&gt;\frac{3x+1y}{3+1}&lt;/math&gt; and the geometric is &lt;math&gt;\sqrt[3+1]{x^3y}&lt;/math&gt;.<br /> <br /> AM-GM applies to weighted averages. Specifically, the '''weighted AM-GM Inequality''' states that if &lt;math&gt;a_1, a_2, \dotsc, a_n&lt;/math&gt; are nonnegative real numbers, and &lt;math&gt;\lambda_1, \lambda_2, \dotsc, \lambda_n&lt;/math&gt; are nonnegative real numbers (the &quot;weights&quot;) which sum to 1, then<br /> &lt;cmath&gt; \lambda_1 a_1 + \lambda_2 a_2 + \dotsb + \lambda_n a_n \ge a_1^{\lambda_1} a_2^{\lambda_2} \dotsm a_n^{\lambda_n}, &lt;/cmath&gt;<br /> or, in more compact notation,<br /> &lt;cmath&gt; \sum_{i=1}^n \lambda_i a_i \ge \prod_{i=1}^n a_i^{\lambda_i} . &lt;/cmath&gt;<br /> Equality holds if and only if &lt;math&gt;a_i = a_j&lt;/math&gt; for all integers &lt;math&gt;i, j&lt;/math&gt; such that &lt;math&gt;\lambda_i \neq 0&lt;/math&gt; and &lt;math&gt;\lambda_j \neq 0&lt;/math&gt;.<br /> We obtain the unweighted form of AM-GM by setting &lt;math&gt;\lambda_1 = \lambda_2 = \dotsb = \lambda_n = 1/n&lt;/math&gt;.<br /> <br /> ==Extensions==<br /> <br /> * The [[power mean inequality]] is a generalization of AM-GM which places the arithemetic and geometric means on a continuum of different means.<br /> * The [[root-square-mean arithmetic-mean geometric-mean harmonic-mean inequality]] is special case of the power mean inequality.<br /> * Kedlaya also extended it greatly doing some stuff like the arithmetic mean of the sequence of geometric means is at least the geometric mean of the sequence of arithmetic means or something like that.<br /> <br /> ==Problems==<br /> <br /> === Introductory ===<br /> * Demonstrate that if &lt;math&gt;a_1a_2\cdots a_n=1&lt;/math&gt; then &lt;math&gt;a_1+a_2+\cdots +a_n\ge n&lt;/math&gt;.<br /> <br /> === Intermediate ===<br /> * Find the minimum value of &lt;math&gt;\frac{9x^2\sin^2 x + 4}{x\sin x}&lt;/math&gt; for &lt;math&gt;0 &lt; x &lt; \pi&lt;/math&gt;.<br /> ([[1983 AIME Problems/Problem 9|Source]])<br /> === Olympiad ===<br /> <br /> * Let &lt;math&gt;a &lt;/math&gt;, &lt;math&gt;b &lt;/math&gt;, and &lt;math&gt;c &lt;/math&gt; be positive real numbers. Prove that<br /> &lt;cmath&gt; (a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3 . &lt;/cmath&gt;<br /> ([[2004 USAMO Problems/Problem 5|Source]])<br /> <br /> == See Also ==<br /> <br /> * [[RMS-AM-GM-HM]]<br /> * [[Algebra]]<br /> * [[Inequalities]]<br /> <br /> <br /> [[Category:Inequality]]<br /> [[Category:Theorems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2010_UNCO_Math_Contest_II_Problems/Problem_9&diff=81967 2010 UNCO Math Contest II Problems/Problem 9 2016-12-22T00:33:30Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> (a) Find integers &lt;math&gt;A, B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; so that &lt;math&gt;A^3+B^4=C^5.&lt;/math&gt; Express your answers in exponential form.<br /> <br /> (b) Find integers &lt;math&gt;A, B, C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; so that &lt;math&gt;A^3+B^4+C^5=3^D.&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2010|n=II|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_17&diff=81966 1986 AHSME Problems/Problem 17 2016-12-22T00:26:56Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A drawer in a darkened room contains &lt;math&gt;100&lt;/math&gt; red socks, &lt;math&gt;80&lt;/math&gt; green socks, &lt;math&gt;60&lt;/math&gt; blue socks and &lt;math&gt;40&lt;/math&gt; black socks. <br /> A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn. <br /> What is the smallest number of socks that must be selected to guarantee that the selection contains at least &lt;math&gt;10&lt;/math&gt; pairs? <br /> (A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ 21\qquad<br /> \textbf{(B)}\ 23\qquad<br /> \textbf{(C)}\ 24\qquad<br /> \textbf{(D)}\ 30\qquad<br /> \textbf{(E)}\ 50 &lt;/math&gt;<br /> <br /> ==Solution==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{AHSME box|year=1986|num-b=16|num-a=18}} <br /> <br /> [[Category: Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_25&diff=81965 2016 AMC 8 Problems/Problem 25 2016-12-22T00:08:41Z <p>Mathisfun04: /* Solution 2 */</p> <hr /> <div>A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> <br /> ==Solution 1==<br /> Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height &lt;math&gt;15&lt;/math&gt; and base &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The Pythagorean triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt; tells us that these triangles have hypotenuses of &lt;math&gt;17&lt;/math&gt;. <br /> <br /> Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be &lt;math&gt;r&lt;/math&gt;.<br /> <br /> The area of the entire isosceles triangle is &lt;math&gt;\frac{(16)(15)}{2} = 120&lt;/math&gt;, so the area of each of the two congruent right triangles it gets split into is &lt;math&gt;\frac{120}{2} = 60&lt;/math&gt;. We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is &lt;math&gt;\frac{17r}{2}&lt;/math&gt;. Thus we can write the equation &lt;math&gt;\frac{17r}{2} = 60&lt;/math&gt;, so &lt;math&gt;17r = 120&lt;/math&gt;, so &lt;math&gt;r = \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, &lt;math&gt;60&lt;/math&gt;. &lt;math&gt;\frac{60}{17}&lt;/math&gt; times &lt;math&gt;2&lt;/math&gt; results in the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is &lt;math&gt; \boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Solution 3: Similar Triangles==<br /> &lt;asy&gt; pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label(&quot;$A$&quot;,A,SW); label(&quot;$B$&quot;,B,SE); label(&quot;$C$&quot;,C,N); label(&quot;$D$&quot;,D,S); label(&quot;$E$&quot;,E,NW);&lt;/asy&gt;<br /> Let's call the triangle &lt;math&gt;\triangle ABC,&lt;/math&gt; where &lt;math&gt;AB=16&lt;/math&gt; and &lt;math&gt; AC=BC.&lt;/math&gt; Let's say that &lt;math&gt;D&lt;/math&gt; is the midpoint of &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; is the point where &lt;math&gt;AC&lt;/math&gt; is tangent to the semicircle. We could also use &lt;math&gt;BC&lt;/math&gt; instead of &lt;math&gt;AC&lt;/math&gt; because of symmetry.<br /> <br /> Notice that &lt;math&gt;\triangle ACD \cong \triangle BCD,&lt;/math&gt; and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by &lt;math&gt;AA&lt;/math&gt; similarity, &lt;math&gt;\triangle AED \sim \triangle ADC,&lt;/math&gt; with &lt;math&gt;\angle EAD \cong \angle DAC&lt;/math&gt; and &lt;math&gt; \angle CDA \cong \angle DEA.&lt;/math&gt; This similarity means that we can create a proportion: &lt;math&gt;\frac{AD}{AB}=\frac{DE}{CD}.&lt;/math&gt; We plug in &lt;math&gt;AD=\frac{AB}{2}=8, AC=17,&lt;/math&gt; and &lt;math&gt;CD=15.&lt;/math&gt; After we multiply both sides by &lt;math&gt;15,&lt;/math&gt; we get &lt;math&gt;DE=\frac{8}{17} \cdot 15= \boxed{\textbf{(B) }\frac{120}{17}}.&lt;/math&gt;<br /> <br /> (By the way, we could also use &lt;math&gt;\triangle DEC \sim \triangle ADC.&lt;/math&gt;)<br /> <br /> ==Solution 4: Inscribed Circle==<br /> <br /> &lt;asy&gt; pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(&quot;$B$&quot;,B,SW); label(&quot;$D$&quot;,D,SE); label(&quot;$A$&quot;,A,N); label(&quot;$M$&quot;,M,S); label(&quot;$C$&quot;,C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));&lt;/asy&gt;<br /> <br /> We'll call this triangle &lt;math&gt;\triangle ABD&lt;/math&gt;. Let the midpoint of base &lt;math&gt;BD&lt;/math&gt; be &lt;math&gt;M&lt;/math&gt;. Divide the triangle in half by drawing a line from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;. Half the base of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{16}{2} = 8&lt;/math&gt;. The height is &lt;math&gt;15&lt;/math&gt;, which is given in the question. Using the Pythagorean Triple &lt;math&gt;8&lt;/math&gt;-&lt;math&gt;15&lt;/math&gt;-&lt;math&gt;17&lt;/math&gt;, the length of each of the legs (&lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;DA&lt;/math&gt;) is 17.<br /> <br /> Reflect the triangle over its base. This will create an inscribed circle in a rhombus &lt;math&gt;ABCD&lt;/math&gt;. Because &lt;math&gt;AB \cong DA&lt;/math&gt;, &lt;math&gt;BC \cong CD&lt;/math&gt;. Therefore &lt;math&gt;AB = BC = CD = DA&lt;/math&gt;.<br /> <br /> The semiperimeter &lt;math&gt;s&lt;/math&gt; of the rhombus is &lt;math&gt;\frac{AB + BC + CD + DA}{2} = \frac{(17)(4)}{2} = 34&lt;/math&gt;. Since the area of &lt;math&gt;\triangle ABD&lt;/math&gt; is &lt;math&gt;\frac{bh}{2}&lt;/math&gt;, the area &lt;math&gt;[ABCD]&lt;/math&gt; of the rhombus is twice that, which is &lt;math&gt;bh = (16)(15) = 240&lt;/math&gt;.<br /> <br /> The [https://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle#Incircle Formula for the Incircle of a Quadrilateral] is &lt;math&gt;s&lt;/math&gt;&lt;math&gt;r&lt;/math&gt; = &lt;math&gt;[ABCD]&lt;/math&gt;. Substituting the semiperimeter and area into the equation, &lt;math&gt;34r = 240&lt;/math&gt;. Solving this, &lt;math&gt;r = \frac{240}{34}&lt;/math&gt; = &lt;math&gt;\boxed{\textbf{(B) }\frac{120}{17}}&lt;/math&gt;.<br /> <br /> ==Part Solution 5: Answer Choices==<br /> Part Solution by e_power_pi_times_i<br /> <br /> Notice that the radius must be smaller than half the base (which is &lt;math&gt;8&lt;/math&gt;). Therefore answer choices &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, and &lt;math&gt;E&lt;/math&gt; are eliminated.<br /> If you do simple math, the square root of &lt;math&gt;3&lt;/math&gt; is about &lt;math&gt;1.7&lt;/math&gt;, and if you multiply that by four, you get &lt;math&gt;6.8&lt;/math&gt;. If you divide &lt;math&gt;120&lt;/math&gt; by &lt;math&gt;17&lt;/math&gt;, you get approximately &lt;math&gt;7&lt;/math&gt;, so you have about &lt;math&gt;\dfrac{1}{2}&lt;/math&gt; of a chance of getting it right.<br /> {{AMC8 box|year=2016|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:AoPS_Community_Awards&diff=81850 AoPS Wiki:AoPS Community Awards 2016-12-11T20:14:20Z <p>Mathisfun04: /* Participants */</p> <hr /> <div>This '''AoPS Community Awards''' page is a celebration of the accomplishments of members of the [[AoPS]] community.<br /> <br /> <br /> == IMO Participants and Medalists ==<br /> This is a list of members of the AoPS community who have competed for their country at the [[International Mathematical Olympiad]].<br /> <br /> === Participants ===<br /> * Shyam Narayanan (2015)<br /> * Prafulla Susil Dhariwal(2011,2012)<br /> * Akashnil Dutta(2009,2010,2011)<br /> * Krishanu Roy Sankar (2008)<br /> * Zachary Abel (2006) (AoPS assistant instructor)<br /> * Marco Avila (2006)<br /> * John Berman (2009)<br /> * Zarathustra Brady (2006)<br /> * Robert Cordwell (2005)<br /> * Wenyu Cao (2009, 2011)<br /> * Sherry Gong (2002, 2003, 2004, 2005, 2007)<br /> * Elyot Grant (2005)<br /> * Darij Grinberg (2006)<br /> * Mahbubul Hasan (2005)<br /> * Daniel Kane (AoPS assistant instructor)<br /> * Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br /> * Viktoriya Krakovna (2006)<br /> * Nate Ince (2004) (AoPS assistant instructor)<br /> * Brian Lawrence (2005, 2007) ([[WOOT]] instructor)<br /> * Thomas Mildorf (2005) (AoPS assistant instructor)<br /> * Alison Miller (2004) (AoPS assistant instructor)<br /> * Richard Peng (2005, 2006)<br /> * Eric Price (2005)<br /> * David Rhee (2004, 2005, 2006)<br /> * Peng Shi (2004, 2005, 2006)<br /> * Arne Smeets (2003, 2004)<br /> * Bobby Shen (2012, 2013)<br /> * Arnav Tripathy (2006, 2007)<br /> * [[Naoki Sato]] (AoPS instructor)<br /> * Yi Sun (2006)<br /> * [[Valentin Vornicu]] (AoPS/MathLinks webmaster)<br /> * Victor Wang (2013)<br /> * Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br /> * Alex Zhai (2005, 2006, 2007, 2008)<br /> * Yufei Zhao (2004, 2005, 2006)<br /> * Tigran Sloyan(2003,2004,2005,2006,2007)<br /> * Marco Avila (2006)<br /> * Vipul Naik (2003,2004)<br /> * Bhargav Narayanan (2007)<br /> * Tigran Hakobyan (2007)<br /> * Carmela Lao (2009, 2010)<br /> * Calvin Deng (2010, 2012, 2013)<br /> * Allen Yuan (2010)<br /> * In-sung Na (2010)<br /> * Xiaoyu He (2010, 2011)<br /> * Kevin Sun (2013)<br /> <br /> ===Perfect Scorers===<br /> *Le Hung Viet Bao (2003)<br /> *Brian Lawrence (2005)<br /> *Alex Zhai (2008)<br /> *Jeck Lim (2012)<br /> *Alex Song (2015)<br /> *Allen Liu (2016)<br /> *Yuan Yao (2016)<br /> *Junghun Ju (2016)<br /> <br /> === Gold medalists ===<br /> * Shyam Narayanan (2015)<br /> * Akashnil Dutta(2011)<br /> * Zarathustra Brady (2006)<br /> * Robert Cordwell (2005)<br /> * Darij Grinberg (2006)<br /> * Kiran Kedlaya (1990, 1992) ([[Art of Problem Solving Foundation]] board member)<br /> * Brian Lawrence (2005) ([[WOOT]] instructor)<br /> * Thomas Mildorf (2005) (AoPS assistant instructor)<br /> * Mark Sellke (2013, 2014)<br /> * Alison Miller (2004) (AoPS assistant instructor)<br /> * Arnav Tripathy (2006)<br /> * Eric Price (2005)<br /> * Yufei Zhao (2005)<br /> * Alex Zhai (2007, 2008)<br /> *Sherry Gong (2007)<br /> *Krishanu Sankar (2008)<br /> * John Berman (2009)<br /> * Xiaoyu He (2010, 2011)<br /> * Wenyu Cao (2011)<br /> * Prafulla Susil Dhariwal (2012)<br /> * Bobby Shen (2012, 2013)<br /> * James Tao (2013, 2014)<br /> * Victor Wang (2013)<br /> * Alex Song (2011, 2012, 2013, 2014, 2015)<br /> * Allen Liu (2014, 2015, 2016)<br /> * Junghun Ju (2015, 2016)<br /> * Ryan Alweiss (2015)<br /> * Evan Chen (2014)<br /> <br /> === Silver medalists ===<br /> * Akashnil Dutta (2009,2010)<br /> * Zachary Abel (2006) (AoPS assistant instructor)<br /> * Sherry Gong (2004, 2005)<br /> * Nate Ince (2004) (AoPS assistant instructor)<br /> * Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br /> * Viktoriya Krakovna (2006)<br /> * Hyun Soo Kim (2005) (AoPS assistant instructor)<br /> * Richard Peng (2005)<br /> * David Rhee (2006)<br /> * Naoki Sato (AoPS instructor)<br /> * Peng Shi (2006)<br /> * Arne Smeets (2004)<br /> * Yi Sun (2006)<br /> * [[Sam Vandervelde]] (1989) ([[WOOT]] instructor)<br /> * Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br /> * Alex Zhai (2006)<br /> * Yufei Zhao (2006)<br /> * Tigran Sloyan (2006,2007)<br /> * Vipul Naik (2003,2004)<br /> * Delong Meng (2009)<br /> * Qinxuan Pan (2009)<br /> * Wenyu Cao (2009)<br /> * Carmela Lao (2010)<br /> * Jafar Jafarov (2006)<br /> * Ali Gurel (1996)<br /> <br /> === Bronze medalists ===<br /> * Debdyuti Banerjee (2011)<br /> * Sherry Gong (2003)<br /> * Elyot Grant (2005)<br /> * Richard Peng (2006)<br /> * [[Naoki Sato]] (AoPS instructor)<br /> * [[Valentin Vornicu]] (AoPS/[[MathLinks]] webmaster)<br /> * Yufei Zhao (2004)<br /> * Tigran Sloyan(2004;2005)<br /> * Tigran Hakobyan (2007)<br /> * Carmela Lao (2009)<br /> * Jafar Jafarov (2007)<br /> * Kevin Sun (2013)<br /> <br /> == IPhO Participants and Medalists ==<br /> This is a list of members of the AoPS community who have competed for their country at the [[International Physics Olympiad]].<br /> === Participants ===<br /> * Sherry Gong (2006)<br /> * Yi Sun (2004)<br /> * Arnav Tripathy (2006)<br /> * Marianna Mao (2009)<br /> * Anand Natarajan (2009)<br /> * Bowei Liu (2009)<br /> * Daniel Li (2010)<br /> <br /> === Gold Medalists ===<br /> * Yi Sun (2004)<br /> * Rahul Singh (2007)<br /> * Marianna Mao (2009)<br /> * Anand Natarajan (2009)<br /> * Bowei Liu (2009)<br /> * Daniel Li(2010)<br /> * Eric Schneider<br /> * Kevin Zhou<br /> <br /> === Silver Medalists ===<br /> * Sherry Gong (2006)<br /> <br /> == IOI Medalists ==<br /> This list is of AoPSers who have won medals at the [[International Olympiad in Informatics]].<br /> <br /> ===Gold Medalists ===<br /> * David Benjamin (2008)<br /> * Brian Hamrick (2009)<br /> * Neal Wu (2008, 2009, 2010)<br /> * Wenyu Cao (2010)<br /> * Scott Wu (2012, 2013, 2014)<br /> <br /> ===Silver Medalists ===<br /> * Akashnil Dutta(2011)<br /> * Brian Hamrick (2008)<br /> * Jacob Steinhardt (2008)<br /> * Wenyu Cao (2009)<br /> * Travis Hance (2009)<br /> <br /> ===Bronze Medalists ===<br /> * David Benjamin (2007)<br /> <br /> <br /> == USAMO ==<br /> The following AoPSers have won the [[United States of America Mathematical Olympiad]] (USAMO). (Note that the definition of &quot;winner&quot; has changed over the years -- currently it is the top 12 scores on the USAMO, but in the past it has been the top 6 or top 8 scores.)<br /> === Perfect Scorers ===<br /> * Daniel Kane (AoPS assistant instructor)<br /> * Kiran Kedlaya (1991) ([[Art of Problem Solving Foundation]] board member)<br /> * Brian Lawrence (2006) ([[WOOT]] instructor)<br /> * Alex Zhu (2012)<br /> * Allen Liu (2015,2016)<br /> <br /> === Winners ===<br /> * Yakov Berchenko-Kogan (2006)<br /> * Wenyu Cao (2009)<br /> * Calvin Deng (2009, 2012, 2013)<br /> * Sherry Gong (2006, 2007)<br /> * Yi Han (2006)<br /> * Adam Hesterberg (2007)<br /> * Daniel Kane (AoPS assistant instructor)<br /> * Kiran Kedlaya (1990, 1991, 1992) ([[Art of Problem Solving Foundation]] board member)<br /> * Brian Lawrence (2005, 2006, 2007) ([[WOOT]] instructor)<br /> * Tedrick Leung (2006, 2007)<br /> * Haitao Mao (2007)<br /> * Richard Mccutchen (2006)<br /> * Shyam Narayanan (2015)<br /> * Albert Ni (2005)<br /> * [[David Patrick]] (1988) (AoPS instructor)<br /> * David Rolnick (2008) (AoPS assistant instructor)<br /> * [[Richard Rusczyk]] (1989) (AoPS founder)<br /> * Krishanu Sankar (2007,2008)<br /> * Bobby Shen (2013)<br /> * Peng Shi (2006)<br /> * Jacob Steinhardt (2007)<br /> * Yi Sun (2006)<br /> * Arnav Tripathy (2006, 2007)<br /> * Victor Wang (2012, 2013)<br /> * [[Sam Vandervelde]] (1987, 1989) ([[WOOT]] instructor)<br /> * Melanie Wood (1998, 1999) ([[WOOT]] instructor)<br /> * David Yang (2009,2011,2012)<br /> * Alex Zhai (2006, 2007,2008)<br /> * Yufei Zhao (2006)<br /> * David Bay Rush (2009)<br /> * Alex Song (2012,2013,2014,2015)<br /> * Evan Chen (2014)<br /> * James Tao (2014)<br /> * Scott Wu (2014)<br /> * Allen Liu (2014,2015,2016)<br /> * Ankan Bhattacharya (2016)<br /> * Yuan Yao (2016)<br /> <br /> == Putnam Fellows ==<br /> The top 5 students (including ties) on the collegiate [[Putnam Exam|William Lowell Putnam Competition]] are named Putnam Fellows.<br /> * David Ash (1981, 1982, 1983)<br /> * Matthew Ince (2005) (AoPS assistant instructor)<br /> * Daniel Kane (2003, 2004, 2005) (AoPS assistant instructor)<br /> * Kiran Kedlaya (1994, 1995, 1996) ([[AoPS Foundation]] board member)<br /> * Brian Lawrence (2007, 2008, 2010, 2011)<br /> * Mitchell Lee (2012, 2013)<br /> * Evan O'Dorney (2011, 2012, 2013)<br /> * Alexander Schwartz (2000, 2002)<br /> * Mark Sellke (2014)<br /> * Bobby Shen (2013, 2014)<br /> * Jan Siwanowicz (2001) <br /> * Arnav Tripathy (2008)<br /> * Melanie Wood (2002) ([[WOOT]] instructor)<br /> * David Yang (2013, 2014, 2015)<br /> <br /> == Siemens Competition Winners ==<br /> The annual [[Siemens Competition]] (formerly Siemens-Westinghouse) is a scientific research competition.<br /> * Michael Viscardi (1st Individual, 2005)<br /> * Lucia Mocz (2nd Team, 2006)<br /> <br /> ==Intel STS Finalists==<br /> The annual [[Intel Science Talent Search]] is a science competition seeking to find and reward the most scientifically accomplished seniors.<br /> <br /> * Jonathan Li (2011)<br /> *Evan O'Dorney (2011)<br /> * Philip Mocz (2008)<br /> * Yihe Dong (2008)<br /> * Qiaochu Yuan (2008)<br /> * Greg Brockman (2007)<br /> <br /> == Clay Junior Fellows ==<br /> Each year since 2003, the [[Clay Mathematics Institute]] has selected 12 Junior Fellows.<br /> * Thomas Belulovich (2005) (AoPS assistant instructor)<br /> * Atoshi Chowdhury (2003) (AoPS assistant instructor)<br /> * Robert Cordwell (2005)<br /> * Eve Drucker (2003) (AoPS assistant instructor)<br /> * Matthew Ince (2004) (AoPS assistant instructor)<br /> * Nate Ince (2004) (AoPS assistant instructor)<br /> * Hyun Soo Kim (2005) (AoPS assistant instructor)<br /> * Raju Krishnamoorthy (2005)<br /> * Alison Miller (2003) (AoPS assistant instructor)<br /> * Brian Rice (2003) (AoPS assistant instructor)<br /> * Dmitry Taubinski (2005) (AoPS assistant instructor)<br /> * Ameya Velingker (2005)<br /> <br /> <br /> == Perfect AIME Scores ==<br /> Very few students have ever achieved a perfect score on the [[American Invitational Mathematics Examination]] (AIME)<br /> <br /> * Matthew Babbitt (2013)<br /> * David Benjamin (2006)<br /> * [[Mathew Crawford]] (1992) (AoPS instructor)<br /> * Lewis Chen (2014)<br /> * Calvin Deng (2011)<br /> * Sam Elder (2008)<br /> * [[Sandor Lehoczky]] (1990) (AoPS author)<br /> * Tedrick Leung (2006)<br /> * Tony Liu (2006)<br /> * Haitao Mao (2008)<br /> * Shyam Narayanan (2012, 2014)<br /> * Vinayak Kumar (2016)<br /> * Bobby Shen (2010, 2011, 2012, 2013)<br /> * [[Richard Rusczyk]] (1989) (AoPS founder)<br /> * [[Sam Vandervelde]] (1988) ([[WOOT]] instructor)<br /> * Alexander Whatley (2012)<br /> * Scott Wu (2012)<br /> <br /> == Perfect AMC Scores ==<br /> === Perfect AMC 12 Scores ===<br /> The [[AMC 12]] is a challenging examination for students in grades 12 and below administered by the [[American Mathematics Competitions]].<br /> * Zachary Abel (2005) (AoPS assistant instructor)<br /> * David Benjamin (2006)<br /> * Wenyu Cao (2009)<br /> * Lewis Chen (2014)<br /> * Sam Elder (2008)<br /> * Ruozhou (Joe) Jia (2003) (AoPS assistant instructor)<br /> * Joel Lewis (2003) <br /> * Daniel Li (2009)<br /> * Jonathan Li (2010)<br /> * Jonathan Lowd (2003) (AoPS assistant instructor)<br /> * Thomas Mildorf (2004) (AoPS assistant instructor)<br /> * Alison Miller (2004) (AoPS assistant instructor)<br /> * Shyam Narayanan (2015)<br /> * Albert Ni (2003) (AoPS instructor)<br /> * Ajay Sharma (2004)<br /> * Bobby Shen (2009, 2011)<br /> * Matt Superdock (2009)<br /> * Arnav Tripathy (2006, 2007)<br /> * Qiaochu Yuan (2008)<br /> * Alex Zhai (2007)<br /> <br /> ===Perfect AMC 10 scorers===<br /> The [[AMC 10]] is a challenging examination for students in grades 10 and below administered by the [[American Mathematics Competitions]].<br /> <br /> * Matthew Babbitt (2009)<br /> * Sergei Bernstein (2007)<br /> * Yifan Cao (2005)<br /> * Kevin Chen (2007, 2008, 2009)<br /> * Lewis Chen(2009)<br /> * Yongyi Chen (2009)<br /> * In Young Cho (2007)<br /> * Mario Choi (2007)<br /> * Calvin Deng (2008)<br /> * Billy Dorminy (2007)<br /> * Zhou Fan (2005)<br /> * Albert Gu (2007)<br /> * Robin He (2007)<br /> * Keone Hon (2005)<br /> * Susan Hu (2005)<br /> * Lyndon Ji (2008)<br /> * Sam Keller (2007)<br /> * Michael Kural (2013)<br /> * Vincent Le (2006)<br /> * Daniel Li (2007)<br /> * Jonathan Li (2009, 2007)<br /> * Kevin Li (2009)<br /> * Patricia Li (2005)<br /> * Carl Lian (2007)<br /> * Sam Lite (2009)<br /> * David Lu (2009)<br /> * Michael Ma (2009, 2010, 2011 in grades 3, 4, 5)<br /> * Thomas Mildorf (2002) (AoPS assistant instructor)<br /> * Anupa Murali (2008)<br /> * Shyam Narayanan (2013)<br /> * Graham O'Donnell (2015, 2016)<br /> * Edwin Peng (2013)<br /> * Max Rosett (2005, 2006) (AoPS assistant instructor)<br /> * Amrit Saxena (2009)<br /> * Eric Schneider (2009)<br /> * Maximilian Schindler (2009)<br /> * Bobby Shen (2009)<br /> * Jeffrey Shen (2008)<br /> * Lilly Shen (2009)<br /> * Richard Spence (2009)<br /> * Kyle Stankowski (2009)<br /> * Michael Tan (2009)<br /> * Michael Tang (2014, 2015)<br /> * Kevin Tian (2009)<br /> * Howard Tong (2005)<br /> * Sam Trabucco (2008)<br /> * Brent Woodhouse (2006, 2007)<br /> * Lawrence Wu (2009)<br /> * David Yang (2009)<br /> * Allen Yuan (2009)<br /> * Peijin Zhang (2009)<br /> * Jonathan Zhou (2007)<br /> * Alex Zhu (2009)<br /> <br /> === Perfect AHSME Scores ===<br /> The [[American High School Mathematics Examination]] (AHSME) was the predecessor of the AMC 12.<br /> * Christopher Chang (1994, 1995, 1996)<br /> * [[Mathew Crawford]] (1994, 1995) (AoPS instructor)<br /> * [[David Patrick]] (1988) (AoPS instructor)<br /> <br /> =='''Perfect USAMO Index'''==<br /> *Lewis Chen (2014)<br /> *Samuel Elder (2009)<br /> *Bobby Shen (2011)<br /> *Gabriel Caroll (1998, 1999, 2000, 2001)<br /> <br /> == MATHCOUNTS ==<br /> [[MathCounts]] is the premier middle school [[mathematics competition]] in the U.S.<br /> === National Champions ===<br /> * Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br /> * Albert Ni (2002) (AoPS instructor)<br /> * Adam Hesterberg (2003)<br /> * Neal Wu (2005)<br /> * Daesun Yim (2006)<br /> * Kevin Chen (2007)<br /> * Darryl Wu (2008)<br /> * Bobby Shen (2009)<br /> * Mark Sellke (2010)<br /> * Scott Wu (2011)<br /> * Chad Qian (2012)<br /> * Alec Sun (2013)<br /> * Swapnil Garg (2014)<br /> * Kevin Liu (2015)<br /> * Edward Wan (2016)<br /> <br /> === National Top 12 ===<br /> * Ashley Reiter Ahlin (1987) ([[WOOT]] instructor)<br /> * Andrew Ardito (2005, 2006)<br /> * David Benjamin (2004, 2005)<br /> * Nathan Benjamin (2005, 2006)<br /> * Wenyu Cao (2007)<br /> * Andrew Cai (2016)<br /> * Christopher Chang (1991, 1992)<br /> * Kevin Chen (2006, 2007)<br /> * Steven Chen (2009, 2010)<br /> * Andrew Chien (2003)<br /> * Peter Chien (2004)<br /> * Mario Choi (2007)<br /> * Joseph Chu (2004)<br /> * Alexander Clifton (2009)<br /> * [[Mathew Crawford]] (1990, 1991) (AoPS instructor)<br /> * Calvin Deng (2009)<br /> * Brian Hamrick (2006)<br /> * Frank Han (2015)<br /> * Adam Hesterberg (2002, 2003)<br /> * Jason Hyun (2008)<br /> * Ruozhou (Joe) Jia (2000) (AoPS assistant instructor)<br /> * Ben Kang (2016)<br /> * Sam Keller (2006)<br /> * Shaunak Kishore (2003, 2004)<br /> * Kiran Kota (2005)<br /> * Brian Lawrence (2003) ([[WOOT]] instructor)<br /> * Karlanna Lewis (2005)<br /> * Daniel Li (2006)<br /> * Patricia Li (2005)<br /> * Ray Li (2009)<br /> * Poh-Ling Loh (2000)<br /> * Brian Liu (2016)<br /> * David Lu (2008)<br /> * Albert Ni (2002) (AoPS assistant instructor)<br /> * Graham O'Donnell (2014)<br /> * Maximilian Schindler (2009)<br /> * Bobby Shen (2008, 2009)<br /> * Elizabeth Synge (2007)<br /> * Jason Trigg (2002)<br /> * [[Sam Vandervelde]] (1985) ([[WOOT]] instructor)<br /> * Edward Wan (2016)<br /> * Victor Wang (2009)<br /> * Alex Wei (2016)<br /> * Eric Wei (2016)<br /> * Ben Wright (2016)<br /> * Neal Wu (2005, 2006)<br /> * Rolland Wu (2006)<br /> * Xiaoyu He (2008)<br /> * Alex Xu (2016)<br /> * David Yang (2009)<br /> * Daesun Yim (2006)<br /> * Darren Yin (2002)<br /> * Justin Yu (2016)<br /> * Allen Yuan (2007)<br /> * Samuel Zbarsky (2008)<br /> * Alex Zhai (2004)<br /> * Mark Zhang (2005)<br /> * Alan Zhou (2009)<br /> * Mark Sellke (2009, 2010)<br /> * Eugene Chen (2010)<br /> * Lewis Chen (2010)<br /> * Shyam Narayanan (2010, 2011)<br /> * Franklyn Wang (2013, 2014)<br /> *Akshaj Kadaveru (2013, 2014)<br /> *Scott Wu (2010, 2011)<br /> *Walker Kroubalkian (2015)<br /> <br /> === Masters Round Champions ===<br /> * Christopher Chang (1991)<br /> * Brian Lawrence (2003) ([[WOOT]] instructor)<br /> * Sergei Bernstein (2005)<br /> * Daniel Li (2006)<br /> * Kevin Chen (2007)<br /> * Bobby Shen (2008)<br /> * Maximilian Schindler (2009)<br /> * Alex Song (2010)<br /> <br /> === National Test Champions ===<br /> * [[Mathew Crawford]] (1990) (AoPS instructor)<br /> * Adam Hesterberg (2003)<br /> * Sergei Bernstein (2005)<br /> * Neal Wu (2006)<br /> * Bobby Shen (2008)<br /> * David Yang (2009)<br /> * Mark Sellke (2010)<br /> *Shyam Narayanan (2011)<br /> * Kevin Liu (2014)<br /> * Andy Xu (2015)<br /> * Edward Wan (2016)<br /> <br /> === National Team Champions ===<br /> * Team Virginia: Divya Garg, Brian Hamrick, Daniel Li, Jimmy Clark (2006)<br /> * Team Massachusetts: Alec Sun, James Lin, Michael Ren, Matthew Lipman (2013)<br /> * Team California: Swapnil Garg, Harry Wang, Rajiv Movva, Jeffery Li (2014)<br /> <br /> == Harvard-MIT Math Tournament ==<br /> <br /> The [[HMMT]] 2007 winning team, the &quot;WOOTlings&quot;, consisted entirely of [[WOOT]]ers:<br /> <br /> * Wenyu Cao<br /> * Eric Chang<br /> * Jeremy Hahn<br /> * Alex Kandell<br /> * Adeel Khan<br /> * Sathish Nagappan<br /> * Krishanu Roy Sankar<br /> * Patrick Tenorio<br /> <br /> == ARML ==<br /> <br /> <br /> === ARML winners ===<br /> * Alex Song (2009, 2011)<br /> * Benjamin Gunby (2010)<br /> * Allen Liu (2012, 2013)<br /> <br /> === ARML Perfect Scorers ===<br /> * Alex Song (2011)<br /> * Shyam Narayanan (2012, 2014)<br /> <br /> === ARML Top 10 ===<br /> * Zachary Abel (2006) (AoPS assistant instructor)<br /> *Lewis Chen(2011,2012)<br /> * Benjamin Gunby (2010)<br /> * Bobby Shen (2010)<br /> * Michael Tang (2015)<br /> * Seva Tchernov (2007)<br /> * Arnav Tripathy (2007)<br /> * David Yang (2009)<br /> * Daesun Yim (2008)<br /> * Alex Song (2009,2011,2012)<br /> *Bailey Wang (2009)(WOOT Grader)<br /> * Shyam Narayanan (2011, 2012 (Perfect Score), 2014 (Perfect Score), 2015)<br /> <br /> == See also ==<br /> * [[Academic competitions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> * [[Academic scholarships]]<br /> <br /> <br /> <br /> [[Category:Art of Problem Solving]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2006_UNCO_Math_Contest_II_Problems/Problem_6&diff=81836 2006 UNCO Math Contest II Problems/Problem 6 2016-12-09T21:58:06Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> The sum of all of the positive integer divisors of &lt;math&gt;6^2=36&lt;/math&gt; is &lt;math&gt;1+2+3+4+6+9+12+18+36=91&lt;/math&gt;<br /> <br /> (a) Determine a nice closed formula (i.e. without dots or the summation symbol) for the sum of all positive divisors of &lt;math&gt;6^n&lt;/math&gt;.<br /> <br /> (b) Repeat for &lt;math&gt;12^n&lt;/math&gt;.<br /> <br /> (c) Generalize.<br /> <br /> ==Solution==<br /> {{solution}}<br /> <br /> ==See Also==<br /> {{UNCO Math Contest box|n=II|year=2006|num-b=5|num-a=7}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2006_UNCO_Math_Contest_II_Problems/Problem_5&diff=81835 2006 UNCO Math Contest II Problems/Problem 5 2016-12-09T21:54:53Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> In the figure &lt;math&gt;BD&lt;/math&gt; is parallel to &lt;math&gt;AE&lt;/math&gt; and also &lt;math&gt;BF&lt;/math&gt; is parallel to &lt;math&gt;DE&lt;/math&gt;. The area of the larger triangle &lt;math&gt;ACE&lt;/math&gt; is &lt;math&gt;128&lt;/math&gt;.<br /> The area of the trapezoid &lt;math&gt;BDEA&lt;/math&gt; is &lt;math&gt;78&lt;/math&gt;. Determine the area of triangle &lt;math&gt;ABF&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(1,2)--(4,0)--cycle,black);<br /> draw((1/2,1)--(2.5,1)--(2,0),black);<br /> MP(&quot;A&quot;,(4,0),SE);MP(&quot;C&quot;,(1,2),N);MP(&quot;E&quot;,(0,0),SW);<br /> MP(&quot;D&quot;,(.5,1),W);MP(&quot;B&quot;,(2.5,1),NE);MP(&quot;F&quot;,(2,0),S);<br /> &lt;/asy&gt;<br /> <br /> ==Solution==<br /> {{solution}}<br /> <br /> ==See Also==<br /> {{UNCO Math Contest box|n=II|year=2006|num-b=4|num-a=6}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2006_UNCO_Math_Contest_II_Problems/Problem_4&diff=81834 2006 UNCO Math Contest II Problems/Problem 4 2016-12-09T21:54:27Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Determine all positive integers &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;n^2+3&lt;/math&gt; divides evenly (without remainder) into &lt;math&gt;n^4-3n^2+10&lt;/math&gt; ?<br /> <br /> <br /> ==Solution==<br /> {{solution}}<br /> <br /> ==See Also==<br /> {{UNCO Math Contest box|n=II|year=2006|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2006_UNCO_Math_Contest_II_Problems/Problem_2&diff=81833 2006 UNCO Math Contest II Problems/Problem 2 2016-12-09T21:53:20Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;a,b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are positive integers, how many integers are strictly between the product &lt;math&gt;abc&lt;/math&gt;<br /> and &lt;math&gt;(a+1)(b+1)(c+1)&lt;/math&gt; ? For example, there are 35 integers strictly between &lt;math&gt;24=2*3*4&lt;/math&gt; and &lt;math&gt;60=3*4*5.&lt;/math&gt;<br /> <br /> ==Solution==<br /> {{solution}}<br /> <br /> ==See Also==<br /> {{UNCO Math Contest box|n=II|year=2006|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_7&diff=81806 2007 UNCO Math Contest II Problems/Problem 7 2016-12-06T01:05:10Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> (a) Express the infinite sum &lt;math&gt;S= 1+ \frac{1}{3}+\frac{1}{3^2}+ \frac{1}{3^3}+ \cdots&lt;/math&gt; as a reduced fraction.<br /> <br /> (b) Express the infinite sum &lt;math&gt;T=\frac{1}{5}+ \frac{1}{25}+ \frac{2}{125}+ \frac{3}{625}+ \frac{5}{3125}+ \cdots&lt;/math&gt;<br /> as a reduced fraction. Here the denominators are powers of &lt;math&gt;5&lt;/math&gt; and the numerators &lt;math&gt;1, 1, 2, 3, 5, \ldots&lt;/math&gt; are the Fibonacci numbers<br /> &lt;math&gt;F_n&lt;/math&gt; where &lt;math&gt;F_n=F_{n-1}+F_{n-2}&lt;/math&gt;.<br /> <br /> == Solution == <br /> <br /> Part A: Knowing that the formula for an infinite geometric series is &lt;math&gt;A/(1 - r)&lt;/math&gt;, where &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are the first term and common ratio respectively, we compute &lt;math&gt;1/(1 - 1/3) = 3/2&lt;/math&gt;, and we have our answer of &lt;math&gt;2/3&lt;/math&gt;. <br /> <br /> <br /> <br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_7&diff=81805 2007 UNCO Math Contest II Problems/Problem 7 2016-12-06T01:03:01Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> (a) Express the infinite sum &lt;math&gt;S= 1+ \frac{1}{3}+\frac{1}{3^2}+ \frac{1}{3^3}+ \cdots&lt;/math&gt; as a reduced fraction.<br /> <br /> (b) Express the infinite sum &lt;math&gt;T=\frac{1}{5}+ \frac{1}{25}+ \frac{2}{125}+ \frac{3}{625}+ \frac{5}{3125}+ \cdots&lt;/math&gt;<br /> as a reduced fraction. Here the denominators are powers of &lt;math&gt;5&lt;/math&gt; and the numerators &lt;math&gt;1, 1, 2, 3, 5, \ldots&lt;/math&gt; are the Fibonacci numbers<br /> &lt;math&gt;F_n&lt;/math&gt; where &lt;math&gt;F_n=F_{n-1}+F_{n-2}&lt;/math&gt;.<br /> <br /> == Solution == <br /> <br /> Part A: Knowing that the formula for an infinite geometric series is &lt;math&gt;A/(1 - r)&lt;/math&gt;, where &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;r&lt;/math&gt; are the first term and common ratio respectively, we compute &lt;math&gt;1/(1 - 1/3) = 3/2&lt;/math&gt;, and therefore, we have our answer of &lt;math&gt;2/3&lt;/math&gt;. <br /> <br /> <br /> <br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_10&diff=81804 2007 UNCO Math Contest II Problems/Problem 10 2016-12-06T00:57:41Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A quaternary “number” is an arrangement of digits, each of which is &lt;math&gt;0, 1, 2, 3.&lt;/math&gt;<br /> Some examples: &lt;math&gt;001, 3220, 022113.&lt;/math&gt;<br /> <br /> (a) How many &lt;math&gt;6&lt;/math&gt;-digit quaternary numbers are there in which each of &lt;math&gt;0, 1&lt;/math&gt; appear at least once?<br /> <br /> (b) How many &lt;math&gt;n&lt;/math&gt;-digit quaternary numbers are there in which each of &lt;math&gt;0, 1, 2,&lt;/math&gt; appear at least<br /> once? Test your answer with &lt;math&gt;n=3.&lt;/math&gt;<br /> <br /> (c) Generalize.<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=9|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_9&diff=81803 2007 UNCO Math Contest II Problems/Problem 9 2016-12-06T00:57:15Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A circle is inscribed in an equilateral triangle whose side<br /> length is &lt;math&gt;2&lt;/math&gt;. Then another circle is inscribed externally<br /> tangent to the first circle but inside the triangle as shown.<br /> And then another, and another. If this process continues<br /> forever what is the total area of all the circles? Express<br /> your answer as an exact multiple of &lt;math&gt;\pi&lt;/math&gt; (and not as a<br /> decimal approximation).<br /> <br /> &lt;asy&gt;<br /> path T=polygon(3);<br /> draw(unitcircle,black);<br /> draw(scale(2)*T,black);<br /> draw(shift(2/sqrt(3),-2/3)*scale(1/3)*unitcircle,black);<br /> draw(shift(2/sqrt(3)/3,-2/9)*shift(2/sqrt(3),-2/3)*scale(1/9)*unitcircle,black);<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_8&diff=81802 2007 UNCO Math Contest II Problems/Problem 8 2016-12-06T00:56:51Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A regular decagon &lt;math&gt;P_1P_2P_3\cdots P_{10}&lt;/math&gt; is drawn<br /> in the coordinate plane with &lt;math&gt;P_1&lt;/math&gt; at &lt;math&gt;(2,0)&lt;/math&gt;<br /> and &lt;math&gt;P_6&lt;/math&gt; at &lt;math&gt;(8,0)&lt;/math&gt;. If &lt;math&gt;P_n&lt;/math&gt; denotes the point<br /> &lt;math&gt;(x_n ,y_n )&lt;/math&gt;, compute the numerical value of<br /> the following product of complex numbers:<br /> &lt;math&gt;( x_1+iy_1)( x_2+iy_2)( x_3+iy_3) \cdots (x_{10} + iy_{10})&lt;/math&gt;<br /> where &lt;math&gt;i^2 = -1&lt;/math&gt; as usual.<br /> <br /> &lt;asy&gt;<br /> draw(polygon(10),dot);<br /> draw((-2,0)--(2,0),black);<br /> draw((-5/3,-2)--(-5/3,2),black);<br /> MP(&quot;P_1&quot;,(-1,0),NW);MP(&quot;(2,0)&quot;,(-.9,0),SW);<br /> MP(&quot;P_6&quot;,(1,0),NE);MP(&quot;(8,0)&quot;,(.9,0),SE);<br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_7&diff=81801 2007 UNCO Math Contest II Problems/Problem 7 2016-12-06T00:54:56Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> (a) Express the infinite sum &lt;math&gt;S= 1+ \frac{1}{3}+\frac{1}{3^2}+ \frac{1}{3^3}+ \cdots&lt;/math&gt; as a reduced fraction.<br /> <br /> (b) Express the infinite sum &lt;math&gt;T=\frac{1}{5}+ \frac{1}{25}+ \frac{2}{125}+ \frac{3}{625}+ \frac{5}{3125}+ \cdots&lt;/math&gt;<br /> as a reduced fraction. Here the denominators are powers of &lt;math&gt;5&lt;/math&gt; and the numerators &lt;math&gt;1, 1, 2, 3, 5, \ldots&lt;/math&gt; are the Fibonacci numbers<br /> &lt;math&gt;F_n&lt;/math&gt; where &lt;math&gt;F_n=F_{n-1}+F_{n-2}&lt;/math&gt;.<br /> <br /> == Solution == <br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_6&diff=81800 2007 UNCO Math Contest II Problems/Problem 6 2016-12-06T00:54:28Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> (a) Demonstrate that every odd number &lt;math&gt;2n+1&lt;/math&gt; can be expressed as a difference of two squares.<br /> <br /> (b) Demonstrate which even numbers can be expressed as a difference of two squares.<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_5&diff=81799 2007 UNCO Math Contest II Problems/Problem 5 2016-12-06T00:53:57Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Ten different playing cards have the numbers<br /> &lt;math&gt;1, 1, 2, 2, 3, 3, 4, 4, 5, 5&lt;/math&gt; written on them<br /> as shown. Three cards are selected at random<br /> without replacement. What is the<br /> probability that the sum of the<br /> numbers on the three cards is divisible by &lt;math&gt;7&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(200,100);<br /> path Card=(arc((.2,.2),.2,180,270)--(1.8,0)--arc((1.8,.2),.2,270,360)--(2,2.8)--arc((1.8,2.8),.2,0,90)--(.2,3)--arc((.2,2.8),.2,90,180)--(0,.2));<br /> draw(Card,black);<br /> draw(shift(3,0)*Card,black);<br /> draw(shift(6,0)*Card,black);<br /> draw(shift(9,0)*Card,black);<br /> draw(shift(12,0)*Card,black);<br /> draw(shift(0,-4)*Card,black);<br /> draw(shift(3,-4)*Card,black);<br /> draw(shift(6,-4)*Card,black);<br /> draw(shift(9,-4)*Card,black);<br /> draw(shift(12,-4)*Card,black);<br /> MP(&quot;1&quot;,(.25,2.1),N);MP(&quot;1&quot;,(1.75,-.1),N);<br /> MP(&quot;1&quot;,(.25,2.1-4),N);MP(&quot;1&quot;,(1.75,-4.1),N);<br /> MP(&quot;\spadesuit&quot;,(1,1),N);MP(&quot;\clubsuit&quot;,(1,1-4),N);<br /> <br /> MP(&quot;2&quot;,(3.25,2.1),N);MP(&quot;2&quot;,(4.75,-.1),N);<br /> MP(&quot;2&quot;,(3.25,2.1-4),N);MP(&quot;2&quot;,(4.75,-4.1),N);<br /> MP(&quot;\spadesuit&quot;,(4,1.75),N);MP(&quot;\clubsuit&quot;,(4,1.75-4),N);<br /> MP(180,&quot;\spadesuit&quot;,(4,.25),N);MP(180,&quot;\clubsuit&quot;,(4,.25-4),N);<br /> <br /> <br /> MP(&quot;3&quot;,(6.25,2.1),N);MP(&quot;3&quot;,(7.75,-.1),N);<br /> MP(&quot;3&quot;,(6.25,2.1-4),N);MP(&quot;3&quot;,(7.75,-4.1),N);<br /> MP(&quot;\spadesuit&quot;,(7,1.75),N);MP(&quot;\clubsuit&quot;,(7,1.85-4),N);<br /> MP(&quot;\spadesuit&quot;,(7,1),N);MP(&quot;\clubsuit&quot;,(7,1-4),N);<br /> MP(180,&quot;\spadesuit&quot;,(7,.25),N);MP(180,&quot;\clubsuit&quot;,(7,.25-4),N);<br /> <br /> MP(&quot;4&quot;,(9.25,2.1),N);MP(&quot;4&quot;,(10.75,-.1),N);<br /> MP(&quot;4&quot;,(9.25,2.1-4),N);MP(&quot;4&quot;,(10.75,-4.1),N);<br /> MP(&quot;\spadesuit&quot;,(10-.3,1.75),N);MP(&quot;\clubsuit&quot;,(10-.3,1.75-4),N);<br /> MP(180,&quot;\spadesuit&quot;,(10-.3,.25),N);MP(180,&quot;\clubsuit&quot;,(10-.3,.25-4),N);<br /> MP(&quot;\spadesuit&quot;,(10+.3,1.75),N);MP(&quot;\clubsuit&quot;,(10+.3,1.75-4),N);<br /> MP(180,&quot;\spadesuit&quot;,(10+.3,.25),N);MP(180,&quot;\clubsuit&quot;,(10+.3,.25-4),N);<br /> <br /> MP(&quot;5&quot;,(12.25,2.1),N);MP(&quot;5&quot;,(13.75,-.1),N);<br /> MP(&quot;5&quot;,(12.25,2.1-4),N);MP(&quot;5&quot;,(13.75,-4.1),N);<br /> MP(&quot;\spadesuit&quot;,(13-.3,1.75),N);MP(&quot;\clubsuit&quot;,(13-.3,1.75-4),N);<br /> MP(&quot;\spadesuit&quot;,(13,1),N);MP(&quot;\clubsuit&quot;,(13,1-4),N);<br /> MP(180,&quot;\spadesuit&quot;,(13-.3,.25),N);MP(180,&quot;\clubsuit&quot;,(13-.3,.25-4),N);<br /> MP(&quot;\spadesuit&quot;,(13+.3,1.75),N);MP(&quot;\clubsuit&quot;,(13+.3,1.75-4),N);<br /> MP(180,&quot;\spadesuit&quot;,(13+.3,.25),N);MP(180,&quot;\clubsuit&quot;,(13+.3,.25-4),N);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=4|num-a=6}}<br /> <br /> [[Category:Introductory Probability Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_4&diff=81798 2007 UNCO Math Contest II Problems/Problem 4 2016-12-06T00:53:33Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> If &lt;math&gt;x&lt;/math&gt; is a primitive cube root of one (this means that &lt;math&gt;x^3 =1&lt;/math&gt; but &lt;math&gt;x \ne 1&lt;/math&gt;) compute the value of<br /> &lt;cmath&gt;x^{2006}+\frac{1}{x^{2006}}+x^{2007}+\frac{1}{x^{2007}}.&lt;/cmath&gt;<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_3&diff=81797 2007 UNCO Math Contest II Problems/Problem 3 2016-12-06T00:53:13Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> State the general rule illustrated here and prove it:<br /> <br /> &lt;math&gt; 1 ,\quad \begin{tabular}{cc} 1&amp;1\\1&amp;2\end{tabular} ,\quad \begin{tabular}{ccc} 1&amp;1&amp;1\\1&amp;2&amp;2\\1&amp;2&amp;3\end{tabular},\quad \begin{tabular}{cccc} 1&amp;1&amp;1&amp;1\\1&amp;2&amp;2&amp;2\\1&amp;2&amp;3&amp;3\\1&amp;2&amp;3&amp;4 \end{tabular} ,\quad \begin{tabular}{ccccc} 1&amp;1&amp;1&amp;1&amp;1\\1&amp;2&amp;2&amp;2&amp;2\\1&amp;2&amp;3&amp;3&amp;3\\1&amp;2&amp;3&amp;4&amp;4\\1&amp;2&amp;3&amp;4&amp;5 \end{tabular} ,\quad \cdots&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2007_UNCO_Math_Contest_II_Problems/Problem_2&diff=81796 2007 UNCO Math Contest II Problems/Problem 2 2016-12-06T00:51:00Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In Grants Pass, Oregon &lt;math&gt;\frac{4}{5}&lt;/math&gt; of the men are married to &lt;math&gt;\frac{3}{7}&lt;/math&gt; of the women. <br /> What fraction of the adult population is married? Give a possible generalization.<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2007|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=Simon%27s_Favorite_Factoring_Trick&diff=81766 Simon's Favorite Factoring Trick 2016-12-04T15:54:22Z <p>Mathisfun04: /* The General Statement */</p> <hr /> <div><br /> ==About==<br /> '''Dr. Simon's Favorite Factoring Trick''' (abbreviated '''SFFT''') is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta | Simon Rubinstein-Salzedo]].<br /> <br /> ==The General Statement==<br /> The general statement of SFFT is: &lt;math&gt;{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)&lt;/math&gt;. Two special common cases are: &lt;math&gt;xy + x + y + 1 = (x+1)(y+1)&lt;/math&gt; and &lt;math&gt;xy - x - y +1 = (x-1)(y-1)&lt;/math&gt;.<br /> <br /> The act of adding &lt;math&gt;{jk}&lt;/math&gt; to &lt;math&gt;{xy}+{xk}+{yj}&lt;/math&gt; in order to be able to factor it could be called &quot;completing the rectangle&quot; in analogy to the more familiar &quot;completing the square.&quot;<br /> <br /> == Applications ==<br /> This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are variables and &lt;math&gt;j,k&lt;/math&gt; are known constants. Also, it is typically necessary to add the &lt;math&gt;jk&lt;/math&gt; term to both sides to perform the factorization.<br /> <br /> == Problems ==<br /> ===Introductory===<br /> *Two different [[prime number]]s between &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;18&lt;/math&gt; are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?<br /> <br /> &lt;math&gt; \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 } &lt;/math&gt;<br /> <br /> ([[2000 AMC 12/Problem 6|Source]])<br /> ===Intermediate===<br /> *&lt;math&gt;m, n&lt;/math&gt; are integers such that &lt;math&gt;m^2 + 3m^2n^2 = 30n^2 + 517&lt;/math&gt;. Find &lt;math&gt;3m^2n^2&lt;/math&gt;.<br /> <br /> ([[1987 AIME Problems/Problem 5|Source]])<br /> <br /> *The integer &lt;math&gt;N&lt;/math&gt; is positive. There are exactly &lt;math&gt;2005&lt;/math&gt; pairs &lt;math&gt;(x, y)&lt;/math&gt; of positive integers satisfying:<br /> <br /> &lt;cmath&gt;\frac 1x +\frac 1y = \frac 1N&lt;/cmath&gt;<br /> <br /> Prove that &lt;math&gt;N&lt;/math&gt; is a perfect square. (British Mathematical Olympiad Round 2, 2005)<br /> <br /> == See Also ==<br /> * [[Algebra]]<br /> * [[Factoring]]<br /> <br /> [[Category:Elementary algebra]]<br /> [[Category:Theorems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=User_talk:Mathisfun04&diff=81668 User talk:Mathisfun04 2016-11-26T21:34:09Z <p>Mathisfun04: Created page with &quot;Hi everyone! My name is Mathisfun04, and I'm a lively student who loves math! I first discovered Aops a couple months ago, and I just fell in love with the site! I think that...&quot;</p> <hr /> <div>Hi everyone!<br /> <br /> My name is Mathisfun04, and I'm a lively student who loves math! I first discovered Aops a couple months ago, and I just fell in love with the site! I think that it's the best out there. And ever since, I have been dedicated to making Aops the best it can be. I found AopsWiki a couple days ago, and decided that I should help cultivate and inspire learning in it. I hope to use Aops and its beautiful features for many years to come!<br /> <br /> And also, feel free to friend me! And if you see anything wrong in my edits, PM me and I'll fix it!<br /> <br /> Mathisfun04</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=81663 1985 AJHSME Problems/Problem 21 2016-11-26T19:46:28Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Mr. Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assume his salary is &lt;math&gt;100&lt;/math&gt; dollars. Then in the next year, he would have &lt;math&gt;110&lt;/math&gt; dollars, and in the next he would have &lt;math&gt;121&lt;/math&gt; dollars. The next year he would have &lt;math&gt;133.1&lt;/math&gt; dollars and in the final year, he would have &lt;math&gt;146.41&lt;/math&gt;. As the total increase is greater than &lt;math&gt;45\%&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt; <br /> <br /> Note that we could have generalized this for any integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=20|num-a=22}}<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=81662 1985 AJHSME Problems/Problem 21 2016-11-26T19:46:06Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Mr. Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assume his salary is &lt;math&gt;100&lt;/math&gt; dollars. Then in the next year, he would have &lt;math&gt;110&lt;/math&gt; dollars, and in the next he would have &lt;math&gt;121&lt;/math&gt; dollars. The next year he would have &lt;math&gt;133.1&lt;/math&gt; dollars and in the last year, he would have &lt;math&gt;146.41&lt;/math&gt;. As the total increase is greater than &lt;math&gt;45\%&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt; <br /> <br /> Note that we could have generalized this for any integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=20|num-a=22}}<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=81661 1985 AJHSME Problems/Problem 21 2016-11-26T19:45:46Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Mr. Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assume his salary is &lt;math&gt;100&lt;/math&gt; dollars. Then in the next year, he would have &lt;math&gt;110 dollars&lt;/math&gt;, and in the next he would have &lt;math&gt;121&lt;/math&gt; dollars. The next year he would have &lt;math&gt;133.1&lt;/math&gt; dollars and in the last year, he would have &lt;math&gt;146.41&lt;/math&gt;. As the total increase is greater than &lt;math&gt;45\%&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt; <br /> <br /> Note that we could have generalized this for any integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=20|num-a=22}}<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=81660 1985 AJHSME Problems/Problem 21 2016-11-26T19:43:46Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Mr. Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assume his salary is &lt;math&gt;100&lt;/math&gt; dollars. Then in the next year, he would have &lt;math&gt;110 dollars&lt;/math&gt;, and in the next he would have &lt;math&gt;121&lt;/math&gt; dollars. The next year he would have &lt;math&gt;133.1&lt;/math&gt; dollars and in the last year, he would have &lt;math&gt;146.41&lt;/math&gt;. As the total increase is greater than &lt;math&gt;45%&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt; <br /> <br /> Note that we could have generalized this for any integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=20|num-a=22}}<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=81659 1985 AJHSME Problems/Problem 21 2016-11-26T19:43:11Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Mr. Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assume his salary is &lt;math&gt;100&lt;/math&gt; dollars. Then in the next year, he would have &lt;math&gt;110 dollars&lt;/math&gt;, and in the next he would have &lt;math&gt;121&lt;/math&gt; dollars. The next year he would have &lt;math&gt;133.1&lt;/math&gt; dollars and in the last year, he would have &lt;math&gt;146.41&lt;/math&gt;. As the total increase is greater than &lt;math&gt;45%&lt;/math&gt;, the answer is &lt;math&gt;boxedE&lt;/math&gt;. <br /> <br /> Note that we could have generalized this for any integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=20|num-a=22}}<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_21&diff=81658 1985 AJHSME Problems/Problem 21 2016-11-26T19:42:27Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Mr. Green receives a &lt;math&gt;10\% &lt;/math&gt; raise every year. His salary after four such raises has gone up by what percent?<br /> <br /> &lt;math&gt;\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\% &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assume his salary is &lt;math&gt;100&lt;/math&gt; dollars. Then in the next year, he would have &lt;math&gt;110 dollars&lt;/math&gt;, and in the next he would have &lt;math&gt;121&lt;/math&gt; dollars. The next year he would have &lt;math&gt;133.1&lt;/math&gt; dollars and in the last year, he would have &lt;math&gt;146.41&lt;/math&gt;. As the total increase is greater than &lt;math&gt;45%&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\text{E}}&lt;/math&gt;. <br /> <br /> Note that we could have generalized this for any integer &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=20|num-a=22}}<br /> [[Category:Introductory Algebra Problems]]<br /> <br /> <br /> {{MAA Notice}}</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=1985_AJHSME_Problems/Problem_2&diff=81657 1985 AJHSME Problems/Problem 2 2016-11-26T19:38:38Z <p>Mathisfun04: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> &lt;math&gt;90+91+92+93+94+95+96+97+98+99=&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1===<br /> One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. <br /> We find a simpler problem in this problem, and simplify -&gt; &lt;math&gt;90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9&lt;/math&gt;<br /> <br /> We know &lt;math&gt;90 \times 10&lt;/math&gt;, that's easy - &lt;math&gt;900&lt;/math&gt;. So how do we find &lt;math&gt;1 + 2 + ... + 8 + 9&lt;/math&gt;?<br /> <br /> We rearrange the numbers to make &lt;math&gt;(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5&lt;/math&gt;. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. &lt;math&gt;4 \times 10 + 5 = 45&lt;/math&gt;. Adding that on to 900 makes 945.<br /> <br /> 945 is &lt;math&gt;\boxed{\text{B}}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Instead of breaking the sum and then rearranging, we can start by rearranging:<br /> &lt;cmath&gt;\begin{align*}<br /> 90+91+92+\cdots +98+99 &amp;= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\<br /> &amp;= 189+189+189+189+189 \\<br /> &amp;= 945\rightarrow \boxed{\text{B}} <br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ===Solution 3===<br /> <br /> We can use a formula. <br /> <br /> It is &lt;math&gt;\frac{n}{2}\times&lt;/math&gt; (First term+Last term) where &lt;math&gt;n&lt;/math&gt; is the number of terms in the sequence. <br /> <br /> Applying it here:<br /> <br /> &lt;math&gt;\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AJHSME box|year=1985|num-b=1|num-a=3}}<br /> <br /> <br /> {{MAA Notice}}<br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_10&diff=81656 2008 UNCO Math Contest II Problems/Problem 10 2016-11-26T19:34:01Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt;f(n,2)&lt;/math&gt; be the number of ways of splitting &lt;math&gt;2n&lt;/math&gt; people into &lt;math&gt;n&lt;/math&gt; groups, each of size &lt;math&gt;2&lt;/math&gt;. As an example,<br /> <br /> the &lt;math&gt;4&lt;/math&gt; people &lt;math&gt;A, B, C, D&lt;/math&gt; can be split into &lt;math&gt;3&lt;/math&gt; groups: &lt;math&gt;\fbox{AB} \ \fbox{CD} ; \fbox{AC} \ \fbox{BD} ;&lt;/math&gt;<br /> and &lt;math&gt;\fbox{AD} \ \fbox{BC}.&lt;/math&gt; <br /> <br /> Hence &lt;math&gt;f(2,2)= 3.&lt;/math&gt;<br /> <br /> (a) Compute &lt;math&gt;f(3,2)&lt;/math&gt; and &lt;math&gt;f(4,2).&lt;/math&gt;<br /> <br /> (b) Conjecture a formula for &lt;math&gt;f(n,2).&lt;/math&gt;<br /> <br /> (c) Let &lt;math&gt;f(n,3)&lt;/math&gt; be the number of ways of splitting &lt;math&gt;\left \{1, 2, 3,\ldots ,3n \right \}&lt;/math&gt; into &lt;math&gt;n&lt;/math&gt; subsets of size &lt;math&gt;3&lt;/math&gt;.<br /> Compute &lt;math&gt;f(2,3),f(3,3)&lt;/math&gt; and conjecture a formula for &lt;math&gt;f(n,3).&lt;/math&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|num-b=9|after=Last Question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_9&diff=81655 2008 UNCO Math Contest II Problems/Problem 9 2016-11-26T19:33:10Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;C_n = 1+10 +10^2 + \cdots + 10^{n-1}.&lt;/math&gt;<br /> <br /> (a) Prove that &lt;math&gt; 9C_n = 10^n -1.&lt;/math&gt;<br /> <br /> (b) Prove that &lt;math&gt;(3C_3+ 2)^2 =112225.&lt;/math&gt;<br /> <br /> (c) Prove that each term in the following sequence is a perfect square:<br /> &lt;cmath&gt;25, 1225, 112225, 11122225, 1111222225,\ldots &lt;/cmath&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_7&diff=81654 2008 UNCO Math Contest II Problems/Problem 7 2016-11-26T19:32:50Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Determine the value of &lt;math&gt;a&lt;/math&gt; so that the following fraction reduces to a quotient of two linear<br /> expressions: &lt;cmath&gt;\frac{x^3+(a-10)x^2-x+(a-6)}{x^3+(a-6)x^2-x+(a-10)}&lt;/cmath&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_6&diff=81653 2008 UNCO Math Contest II Problems/Problem 6 2016-11-26T19:32:33Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> <br /> Points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are on the same side of<br /> line &lt;math&gt;L&lt;/math&gt; in the plane. &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt; units away<br /> from &lt;math&gt;L, B&lt;/math&gt; is &lt;math&gt;9&lt;/math&gt; units away from &lt;math&gt;L&lt;/math&gt;.<br /> The distance between &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt;. For<br /> all points &lt;math&gt;P&lt;/math&gt; on &lt;math&gt;L&lt;/math&gt; what is the smallest<br /> value of the sum &lt;math&gt;AP + PB&lt;/math&gt; of the distances<br /> from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt; and from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; ?<br /> <br /> &lt;asy&gt;<br /> draw((-1,0)--(16,0),arrow=Arrow());<br /> draw((16,0)--(-1,0),arrow=Arrow());<br /> draw((2,0)--(2,5)--(2+sqrt(128),9)--(2+sqrt(128),0),black);<br /> draw((2,5)--(8,0)--(2+sqrt(128),9),dashed);<br /> dot((2,5));dot((8,0));dot((2+sqrt(128),9));<br /> MP(&quot;A&quot;,(2,5),W);MP(&quot;P&quot;,(8,0),S);MP(&quot;B&quot;,(2+sqrt(128),9),E);MP(&quot;L&quot;,(14,0),S);<br /> MP(&quot;5&quot;,(2,2.5),W);MP(&quot;12&quot;,(2+sqrt(128)/2,7),N);MP(&quot;9&quot;,(2+sqrt(128),4.5),E);<br /> &lt;/asy&gt;<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_5&diff=81652 2008 UNCO Math Contest II Problems/Problem 5 2016-11-26T19:32:13Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> 5. The sum of &lt;math&gt;400, 3, 500, 800&lt;/math&gt; and &lt;math&gt;305&lt;/math&gt; is &lt;math&gt;2008&lt;/math&gt; and the product of these five numbers is<br /> &lt;math&gt;146400000000 = 1464 \times 10^8.&lt;/math&gt;<br /> <br /> (a) Determine the largest number which is the product of positive integers whose sum is &lt;math&gt;2008&lt;/math&gt;.<br /> <br /> (b) Determine the largest number which is the product of positive integers whose sum is &lt;math&gt;n&lt;/math&gt;.<br /> <br /> <br /> == Solution == <br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|num-b=4|num-a=6}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_2&diff=81651 2008 UNCO Math Contest II Problems/Problem 2 2016-11-26T19:31:02Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;S = \left \{a,b,c,d \right \}&lt;/math&gt; be a set of four positive integers. If pairs of distinct elements of &lt;math&gt;S&lt;/math&gt;<br /> are added, the following six sums are obtained: &lt;math&gt;5, 10, 11, 13, 14, 19.&lt;/math&gt; Determine the values of &lt;math&gt;a,<br /> b, c&lt;/math&gt;, and &lt;math&gt;d.&lt;/math&gt; [Hint: there are two possibilities.]<br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2008_UNCO_Math_Contest_II_Problems/Problem_1&diff=81650 2008 UNCO Math Contest II Problems/Problem 1 2016-11-26T19:30:42Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Determine the number of &lt;math&gt;3 \times 3&lt;/math&gt; square arrays<br /> whose row and column sums are equal to &lt;math&gt;2&lt;/math&gt;,<br /> using &lt;math&gt;0, 1, 2&lt;/math&gt; as entries. Entries may be<br /> repeated, and not all of &lt;math&gt;0, 1, 2&lt;/math&gt; need be used as the<br /> two examples show.<br /> <br /> &lt;cmath&gt;\begin{tabular}{c c c c c c c c c}<br /> 1 &amp; 1 &amp; 0 &amp; &amp; &amp; &amp; 0 &amp; 1 &amp; 1 \\<br /> 0 &amp; 0 &amp; 2 &amp; &amp; &amp; &amp; 1 &amp; 1 &amp; 0 \\<br /> 1 &amp; 1 &amp; 0 &amp; &amp; &amp; &amp; 1 &amp; 0 &amp; 1 \\<br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> <br /> == Solution == <br /> {{solution}}<br /> <br /> == See Also ==<br /> {{UNCO Math Contest box|n=II|year=2008|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_10&diff=81642 2009 UNCO Math Contest II Problems/Problem 10 2016-11-26T13:49:52Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt;S=\left \{1,2,3,\ldots ,n\right \}&lt;/math&gt;. Determine the number of subsets &lt;math&gt;A&lt;/math&gt; of &lt;math&gt;S&lt;/math&gt; such that &lt;math&gt;A&lt;/math&gt; contains at least two<br /> elements and such that no two elements of &lt;math&gt;A&lt;/math&gt; differ by &lt;math&gt;1&lt;/math&gt; when<br /> <br /> (a) &lt;math&gt;n=10&lt;/math&gt;<br /> <br /> (b) &lt;math&gt;n=20&lt;/math&gt;<br /> <br /> (c) generalize for any &lt;math&gt;n&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_9&diff=81641 2009 UNCO Math Contest II Problems/Problem 9 2016-11-26T13:49:31Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A square is divided into three pieces of<br /> equal area by two parallel lines as shown.<br /> If the distance between the two parallel<br /> lines is &lt;math&gt;8&lt;/math&gt; what is the area of the square?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(1,0)--(1,1)--(0,1)--cycle,black);<br /> draw((1,0)--(0,2/3),black);<br /> draw((1,1/3)--(0,1),black);<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_8&diff=81640 2009 UNCO Math Contest II Problems/Problem 8 2016-11-26T13:48:05Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Two diagonals are drawn in the trapezoid<br /> forming four triangles. The areas of two of the<br /> triangles are &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;25&lt;/math&gt; as shown. What is the total<br /> area of the trapezoid?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(20,0)--(2,4)--(14,4)--(0,0),black);<br /> draw((0,0)--(2,4)--(14,4)--(20,0),black);<br /> MP(&quot;25&quot;,(9,.25),N);MP(&quot;9&quot;,(9,2.25),N);<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_7&diff=81639 2009 UNCO Math Contest II Problems/Problem 7 2016-11-26T13:47:48Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> A polynomial &lt;math&gt;P(x)&lt;/math&gt; has a remainder of &lt;math&gt;4&lt;/math&gt; when divided by &lt;math&gt;x+2&lt;/math&gt; and a remainder of &lt;math&gt;14&lt;/math&gt; when divided<br /> by &lt;math&gt;x-3.&lt;/math&gt; What is the remainder when &lt;math&gt;P(x)&lt;/math&gt; is divided by &lt;math&gt;(x+2)(x-3)&lt;/math&gt;?<br /> <br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_6&diff=81638 2009 UNCO Math Contest II Problems/Problem 6 2016-11-26T13:47:32Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let each of &lt;math&gt;m&lt;/math&gt; distinct points on the positive<br /> &lt;math&gt;x&lt;/math&gt;-axis be joined to each of &lt;math&gt;n&lt;/math&gt; distinct points on<br /> the positive &lt;math&gt;y&lt;/math&gt;-axis. Assume no three segments<br /> are concurrent (except at the axes). Obtain<br /> with proof a formula for the number of interior<br /> intersection points. The diagram shows that<br /> the answer is &lt;math&gt;3&lt;/math&gt; when &lt;math&gt;m=3&lt;/math&gt; and &lt;math&gt;n=2.&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,3),arrow=Arrow());<br /> draw((0,0)--(4,0),arrow=Arrow());<br /> for(int x=0;x&lt;4;++x){<br /> for(int y=0;y&lt;3;++y){<br /> D((x,0)--(0,y),black);<br /> }}<br /> dot(IP((2,0)--(0,1),(1,0)--(0,2)));<br /> dot(IP((3,0)--(0,1),(1,0)--(0,2)));<br /> dot(IP((3,0)--(0,1),(2,0)--(0,2)));<br /> &lt;/asy&gt;<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_5&diff=81637 2009 UNCO Math Contest II Problems/Problem 5 2016-11-26T13:47:14Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> The two large isosceles right triangles are congruent.<br /> If the area of the inscribed square &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;225&lt;/math&gt; square<br /> units, what is the area of the inscribed square &lt;math&gt;B&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(1,0)--(0,1)--cycle,black);<br /> draw((0,0)--(1/2,0)--(1/2,1/2)--(0,1/2)--cycle,black);<br /> MP(&quot;A&quot;,(1/4,1/8),N);<br /> draw((2,0)--(3,0)--(2,1)--cycle,black);<br /> draw((2+1/3,0)--(2+2/3,1/3)--(2+1/3,2/3)--(2,1/3)--cycle,black);<br /> MP(&quot;B&quot;,(2+1/3,1/8),N);<br /> &lt;/asy&gt;<br /> <br /> == Solution == <br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=4|num-a=6}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_3&diff=81636 2009 UNCO Math Contest II Problems/Problem 3 2016-11-26T13:46:07Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> An army of ants is organizing a march to the Obama inauguration. If they form columns of &lt;math&gt;10&lt;/math&gt; ants<br /> there are &lt;math&gt;8&lt;/math&gt; left over. If they form columns of &lt;math&gt;7, 11&lt;/math&gt; or &lt;math&gt;13&lt;/math&gt; ants there are &lt;math&gt;2&lt;/math&gt; left over. What is the<br /> smallest number of ants that could be in the army?<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_2&diff=81635 2009 UNCO Math Contest II Problems/Problem 2 2016-11-26T13:45:47Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> (a) Let &lt;math&gt;Q_n=1^n+2^n&lt;/math&gt;. For how many &lt;math&gt;n&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt; inclusive is &lt;math&gt;Q_n&lt;/math&gt; a multiple of &lt;math&gt;5&lt;/math&gt;?<br /> <br /> (b) For how many &lt;math&gt;n&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;100&lt;/math&gt; inclusive is &lt;math&gt;R_n=1^n+2^n+3^n+4^n&lt;/math&gt; a multiple of 5?<br /> <br /> <br /> == Solution ==<br /> {{solution}}<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=1|num-a=3}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_1&diff=81634 2009 UNCO Math Contest II Problems/Problem 1 2016-11-26T13:45:24Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> How many positive &lt;math&gt;3&lt;/math&gt;-digit numbers &lt;math&gt;abc&lt;/math&gt; are there such that &lt;math&gt;a+b=c&lt;/math&gt; For example, &lt;math&gt;202&lt;/math&gt; and &lt;math&gt;178&lt;/math&gt;<br /> have this property but &lt;math&gt;245&lt;/math&gt; and &lt;math&gt;317&lt;/math&gt; do not.<br /> <br /> <br /> == Solution ==<br /> <br /> Here, we need to find &lt;math&gt;a,b\in \Bbb N_0&lt;/math&gt; such that &lt;math&gt;1\le a\le 9&lt;/math&gt; , &lt;math&gt;0\le b\le 9&lt;/math&gt; and &lt;math&gt;a+b=c&lt;/math&gt; where &lt;math&gt;c\in \Bbb N, c\le 9&lt;/math&gt;.<br /> <br /> If we place &lt;math&gt;a=1&lt;/math&gt;, then we can place &lt;math&gt;0,1,2,3,4,5,6,7,8&lt;/math&gt; as &lt;math&gt;b&lt;/math&gt;, i.e. in &lt;math&gt;9&lt;/math&gt; ways.<br /> <br /> Similarly, if we place &lt;math&gt;a=2&lt;/math&gt;, we can place &lt;math&gt;b=0,1,2,3,4,5,6,7&lt;/math&gt; i.e. in &lt;math&gt;8&lt;/math&gt; ways.<br /> <br /> &lt;cmath&gt;\dots&lt;/cmath&gt;<br /> <br /> If, we place &lt;math&gt;a=9&lt;/math&gt;, we have the only choice &lt;math&gt;b=0&lt;/math&gt;, in &lt;math&gt;2&lt;/math&gt; ways.<br /> <br /> So, in order to get the number of possibilities, we have to add the no. of all the possibilities we got, i.e. the answer is &lt;cmath&gt;\color{red}{1+2+3+4+5+6+7+8+9=\frac {9\times 10}{2}}=\color{blue}{45}&lt;/cmath&gt;<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mathisfun04 https://artofproblemsolving.com/wiki/index.php?title=2009_UNCO_Math_Contest_II_Problems/Problem_4&diff=81627 2009 UNCO Math Contest II Problems/Problem 4 2016-11-26T03:01:31Z <p>Mathisfun04: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> How many perfect squares are divisors of the product &lt;math&gt;1!\cdot 2!\cdot 3!\cdot 4!\cdot 5!\cdot 6!\cdot 7!\cdot 8!&lt;/math&gt; ? (Here, for<br /> example, &lt;math&gt;4!&lt;/math&gt; means &lt;math&gt;4\cdot 3\cdot 2\cdot 1&lt;/math&gt;)<br /> <br /> == Solution == <br /> <br /> We first factorize the product as &lt;math&gt;2^7\cdot3^2\cdot5\cdot7&lt;/math&gt;. Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are four even powers of two that are less than or equal to &lt;math&gt;2^7&lt;/math&gt;, namely &lt;math&gt;2^0&lt;/math&gt;, &lt;math&gt;2^2&lt;/math&gt;, &lt;math&gt;2^4&lt;/math&gt;, &lt;math&gt;2^6&lt;/math&gt;, as &lt;math&gt;0&lt;/math&gt; is even. Repeating this process with three, five, and seven, we find that three has two even powers, &lt;math&gt;3^0&lt;/math&gt;, and &lt;math&gt;3^2&lt;/math&gt;, and that five and seven have only one even power, &lt;math&gt;5^0&lt;/math&gt; and &lt;math&gt;7^0&lt;/math&gt; respectively. Multiplying this we have &lt;math&gt;4\cdot2\cdot1\cdot1 = 8&lt;/math&gt;. Therefore, our answer is &lt;math&gt;\boxed8&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{UNCO Math Contest box|year=2009|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Mathisfun04