https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Emerald+block&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-22T02:35:36Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_2&diff=151214 2019 USAJMO Problems/Problem 2 2021-04-07T22:53:22Z <p>Emerald block: Add new solution, fix incorrect example of f and g in Solution 1</p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;\mathbb Z&lt;/math&gt; be the set of all integers. Find all pairs of integers &lt;math&gt;(a,b)&lt;/math&gt; for which there exist functions &lt;math&gt;f:\mathbb Z\rightarrow\mathbb Z&lt;/math&gt; and &lt;math&gt;g:\mathbb Z\rightarrow\mathbb Z&lt;/math&gt; satisfying &lt;cmath&gt;f(g(x))=x+a\quad\text{and}\quad g(f(x))=x+b&lt;/cmath&gt; for all integers &lt;math&gt;x&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> We claim that the answer is &lt;math&gt;|a|=|b|&lt;/math&gt;.<br /> <br /> Proof:<br /> &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; are surjective because &lt;math&gt;x+a&lt;/math&gt; and &lt;math&gt;x+b&lt;/math&gt; can take on any integral value, and by evaluating the parentheses in different order, we find &lt;math&gt;f(g(f(x)))=f(x+b)=f(x)+a&lt;/math&gt; and &lt;math&gt;g(f(g(x)))=g(x+a)=g(x)+b&lt;/math&gt;. We see that if &lt;math&gt;a=0&lt;/math&gt; then &lt;math&gt;g(x)=g(x)+b&lt;/math&gt; to &lt;math&gt;b=0&lt;/math&gt; as well, so similarly if &lt;math&gt;b=0&lt;/math&gt; then &lt;math&gt;a=0&lt;/math&gt;, so now assume &lt;math&gt;a, b\ne 0&lt;/math&gt;. <br /> <br /> We see that if &lt;math&gt;x=|b|n&lt;/math&gt; then &lt;math&gt;f(x)\equiv f(0) \pmod{|a|}&lt;/math&gt;, if &lt;math&gt;x=|b|n+1&lt;/math&gt; then &lt;math&gt;f(x)\equiv f(1)\pmod{|a|}&lt;/math&gt;, if &lt;math&gt;x=|b|n+2&lt;/math&gt; then &lt;math&gt;f(x)\equiv f(2)\pmod{|a|}&lt;/math&gt;... if &lt;math&gt;x=|b|(n+1)-1&lt;/math&gt; then &lt;math&gt;f(x)\equiv f(|b|-1)\pmod{|a|}&lt;/math&gt;. This means that the &lt;math&gt;b&lt;/math&gt;-element collection &lt;math&gt;\left\{f(0), f(1), f(2), ... ,f(|b|-1)\right\}&lt;/math&gt; contains all &lt;math&gt;|a|&lt;/math&gt; residues mod &lt;math&gt;|a|&lt;/math&gt; since &lt;math&gt;f&lt;/math&gt; is surjective, so &lt;math&gt;|b|\ge |a|&lt;/math&gt;. Doing the same to &lt;math&gt;g&lt;/math&gt; yields that &lt;math&gt;|a|\ge |b|&lt;/math&gt;, so this means that only &lt;math&gt;|a|=|b|&lt;/math&gt; can work.<br /> <br /> For &lt;math&gt;a=b&lt;/math&gt; let &lt;math&gt;f(x)=x&lt;/math&gt; and &lt;math&gt;g(x)=x+a&lt;/math&gt;, and for &lt;math&gt;a=-b&lt;/math&gt; let &lt;math&gt;f(x)=-x&lt;/math&gt; and &lt;math&gt;g(x)=-x-a&lt;/math&gt;, so &lt;math&gt;|a|=|b|&lt;/math&gt; does work and are the only solutions, as desired.<br /> <br /> -Stormersyle<br /> <br /> ==Solution 2==<br /> <br /> We claim that &lt;math&gt;f&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt; exist if and only if &lt;math&gt;|a|=|b|&lt;/math&gt;.<br /> <br /> &lt;b&gt;Only If:&lt;/b&gt;<br /> <br /> For some fixed &lt;math&gt;j&lt;/math&gt;, let &lt;math&gt;f(j)=k&lt;/math&gt;.<br /> <br /> If &lt;math&gt;b=0&lt;/math&gt;, then &lt;math&gt;g(k)=j&lt;/math&gt;. Suppose &lt;math&gt;a\ne 0&lt;/math&gt;. Then &lt;math&gt;f(j)=f(g(k))=k+a\ne k&lt;/math&gt;, a contradiction. Thus, &lt;math&gt;a=0&lt;/math&gt;. Similarly, if &lt;math&gt;a=0&lt;/math&gt;, then &lt;math&gt;b=0&lt;/math&gt;, satisfying &lt;math&gt;|a|=|b|&lt;/math&gt;.<br /> <br /> Otherwise, &lt;math&gt;a,b\ne 0&lt;/math&gt;. We know that &lt;math&gt;g(k)=g(f(j))=j+b&lt;/math&gt;, &lt;math&gt;f(j+b)=f(g(k))=k+a&lt;/math&gt;, &lt;math&gt;g(k+a)=j+2b&lt;/math&gt;, and so on: &lt;math&gt;f(j+nb)=k+na&lt;/math&gt; and &lt;math&gt;g(k+na)=j+(n+1)b&lt;/math&gt; for &lt;math&gt;n\ge 0&lt;/math&gt;.<br /> <br /> Consider the value of &lt;math&gt;g(k-a)&lt;/math&gt;. Suppose &lt;math&gt;g(k-a)=j'\ne j&lt;/math&gt;. Then &lt;math&gt;f(j')=k&lt;/math&gt; and &lt;math&gt;g(f(j'))=j+b\ne j'+b&lt;/math&gt;, a contradiction. Thus, &lt;math&gt;g(k-a)=j&lt;/math&gt;. We repeat with &lt;math&gt;f(j-b)&lt;/math&gt;. Suppose &lt;math&gt;f(j-b)=k'-b\ne k-b&lt;/math&gt;. Then &lt;math&gt;g(k'-b)=j&lt;/math&gt; and &lt;math&gt;f(g(k'-b))=k\ne k'&lt;/math&gt;, a contradition. Thus, &lt;math&gt;f(j-b)=k-b&lt;/math&gt;. Continuing, &lt;math&gt;g(k-2a)=j-a&lt;/math&gt;, and so on: &lt;math&gt;f(j+nb)=k+na&lt;/math&gt; and &lt;math&gt;g(k+na)=j+(n+1)b&lt;/math&gt; now for all &lt;math&gt;n&lt;/math&gt;.<br /> <br /> This defines &lt;math&gt;f(x)&lt;/math&gt; for all &lt;math&gt;x\equiv j\pmod{|b|}&lt;/math&gt; and &lt;math&gt;g(x)&lt;/math&gt; for all &lt;math&gt;x\equiv f(j)\pmod{|a|}&lt;/math&gt;.<br /> <br /> This means that &lt;math&gt;x\equiv j\pmod{|b|}\implies f(x)\equiv f(j)\pmod{|a|}&lt;/math&gt;, and &lt;math&gt;y\equiv f(j)\pmod{|a|}\implies g(y)\equiv j\pmod{|b|}&lt;/math&gt; which implies &lt;math&gt;f(x)\equiv f(j)\pmod{|a|}\implies x\equiv j\pmod{|b|}&lt;/math&gt;.<br /> <br /> As a result, &lt;math&gt;f(x)&lt;/math&gt; maps each residue mod &lt;math&gt;|b|&lt;/math&gt; to a unique residue mod &lt;math&gt;|a|&lt;/math&gt;, so &lt;math&gt;|a|\ge|b|&lt;/math&gt;. Similarly, &lt;math&gt;g(x)&lt;/math&gt; maps each residue mod &lt;math&gt;|a|&lt;/math&gt; to a unique residue mod &lt;math&gt;|b|&lt;/math&gt;, so &lt;math&gt;|b|\ge|a|&lt;/math&gt;. Therefore, &lt;math&gt;|a|=|b|&lt;/math&gt;.<br /> <br /> &lt;b&gt;If:&lt;/b&gt;<br /> <br /> &lt;math&gt;|a|=|b|&lt;/math&gt; means that either &lt;math&gt;a=b&lt;/math&gt; or &lt;math&gt;a=-b&lt;/math&gt;. &lt;math&gt;f(x)=x,g(x)=x+a&lt;/math&gt; works for the former and &lt;math&gt;f(x)=-x,g(x)=-x-a&lt;/math&gt; works for the latter, and we are done.<br /> <br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2019|num-b=1|num-a=3}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2017_IMO_Problems/Problem_2&diff=139420 2017 IMO Problems/Problem 2 2020-12-10T22:13:53Z <p>Emerald block: Undo incomplete solution (only accounts for integers)</p> <hr /> <div>Let &lt;math&gt;\mathbb{R}&lt;/math&gt; be the set of real numbers , determine all functions <br /> &lt;math&gt;f:\mathbb{R}\rightarrow\mathbb{R}&lt;/math&gt; such that for any real numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; &lt;math&gt;{f(f(x)f(y)) + f(x+y)}&lt;/math&gt; =&lt;math&gt;f(xy)&lt;/math&gt;</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_2&diff=124538 2009 AIME I Problems/Problem 2 2020-06-09T00:52:39Z <p>Emerald block: new solution</p> <hr /> <div>== Problem ==<br /> <br /> There is a complex number &lt;math&gt;z&lt;/math&gt; with imaginary part &lt;math&gt;164&lt;/math&gt; and a positive integer &lt;math&gt;n&lt;/math&gt; such that<br /> <br /> &lt;cmath&gt;\frac {z}{z + n} = 4i.&lt;/cmath&gt;<br /> <br /> Find &lt;math&gt;n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;z = a + 164i&lt;/math&gt;.<br /> <br /> Then<br /> &lt;cmath&gt;\frac {a + 164i}{a + 164i + n} = 4i&lt;/cmath&gt;<br /> and<br /> &lt;cmath&gt;a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.&lt;/cmath&gt;<br /> <br /> By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation, <br /> <br /> we conclude that<br /> &lt;cmath&gt;a = -656.&lt;/cmath&gt;<br /> <br /> By equating the imaginary terms on each side of the equation,<br /> <br /> we conclude that<br /> &lt;cmath&gt;164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).&lt;/cmath&gt;<br /> <br /> We now have an equation for &lt;math&gt;n&lt;/math&gt;: &lt;cmath&gt;4i \left (-656 + n \right ) = 164i,&lt;/cmath&gt;<br /> <br /> and this equation shows that &lt;math&gt;n = \boxed{697}.&lt;/math&gt;<br /> <br /> <br /> ==Solution 2==<br /> <br /> &lt;cmath&gt;\frac {z}{z+n}=4i&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1-\frac {n}{z+n}=4i&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;1-4i=\frac {n}{z+n}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac {1}{1-4i}=\frac {z+n}{n}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\frac {1+4i}{17}=\frac {z}{n}+1&lt;/cmath&gt;<br /> <br /> Since their imaginary part has to be equal,<br /> <br /> &lt;cmath&gt;\frac {4i}{17}=\frac {164i}{n}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n=\frac {(164)(17)}{4}=697&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;n = \boxed{697}.&lt;/cmath&gt;<br /> <br /> ==Solution 3 (Highly not recommended)==<br /> Below is an image of the complex plane. Let &lt;math&gt;\operatorname{Im}(z)&lt;/math&gt; denote the imaginary part of a complex number &lt;math&gt;z&lt;/math&gt;.<br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> <br /> <br /> xaxis(&quot;Re&quot;,Arrows);<br /> yaxis(&quot;Im&quot;,Arrows);<br /> <br /> real f(real x) {return 164;}<br /> pair F(real x) {return (x,f(x));}<br /> <br /> draw(graph(f,-700,100),red,Arrows);<br /> <br /> label(&quot;Im$(z)=164$&quot;,F(-330),N);<br /> <br /> <br /> pair z = (-656,164);<br /> dot(Label(&quot;$z$&quot;,align=N),z);<br /> dot(Label(&quot;$z+n$&quot;,align=N),z+(697,0));<br /> <br /> draw(Label(&quot;$4x$&quot;),z--(0,0));<br /> draw(Label(&quot;$x$&quot;),(0,0)--z+(697,0));<br /> <br /> markscalefactor=2;<br /> draw(rightanglemark(z,(0,0),z+(697,0)));<br /> &lt;/asy&gt;<br /> &lt;math&gt;z&lt;/math&gt; must lie on the line &lt;math&gt;\operatorname{Im}(z)=164&lt;/math&gt;. &lt;math&gt;z+n&lt;/math&gt; must also lie on the same line, since &lt;math&gt;n&lt;/math&gt; is real and does not affect the imaginary part of &lt;math&gt;z&lt;/math&gt;.<br /> <br /> Consider &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;z+n&lt;/math&gt; in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have &lt;math&gt;z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)&lt;/math&gt; and &lt;math&gt;\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the magnitude and &lt;math&gt;\theta&lt;/math&gt; is the phase, and &lt;math&gt;z_n=r_n\angle\theta_n&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;4i&lt;/math&gt; has magnitude &lt;math&gt;4&lt;/math&gt; and phase &lt;math&gt;90^\circ&lt;/math&gt; (since the positive imaginary axis points in a direction &lt;math&gt;90^\circ&lt;/math&gt; counterclockwise from the positive real axis), &lt;math&gt;z&lt;/math&gt; must have a magnitude &lt;math&gt;4&lt;/math&gt; times that of &lt;math&gt;z+n&lt;/math&gt;. We denote the length from the origin to &lt;math&gt;z+n&lt;/math&gt; with the value &lt;math&gt;x&lt;/math&gt; and the length from the origin to &lt;math&gt;z&lt;/math&gt; with the value &lt;math&gt;4x&lt;/math&gt;. Additionally, &lt;math&gt;z&lt;/math&gt;, the origin, and &lt;math&gt;z+n&lt;/math&gt; must form a right angle, with &lt;math&gt;z&lt;/math&gt; counterclockwise from &lt;math&gt;z+n&lt;/math&gt;.<br /> <br /> This means that &lt;math&gt;z&lt;/math&gt;, the origin, and &lt;math&gt;z+n&lt;/math&gt; form a right triangle. The hypotenuse is the length from &lt;math&gt;z&lt;/math&gt; to &lt;math&gt;z+n&lt;/math&gt; and has length &lt;math&gt;n&lt;/math&gt;, since &lt;math&gt;n&lt;/math&gt; is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as &lt;math&gt;\frac{x \cdot 4x}{2}&lt;/math&gt;, or using the hypotenuse and its corresponding altitude, as &lt;math&gt;\frac{164n}{2}&lt;/math&gt;, so &lt;math&gt;\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n&lt;/math&gt;. By Pythagorean Theorem, &lt;math&gt;x^2+(4x)^2 = n^2 \implies 17x^2 = n^2&lt;/math&gt;. Substituting out &lt;math&gt;x^2&lt;/math&gt; using the earlier equation, we get &lt;math&gt;17\cdot41n = n^2 \implies n = \boxed{697}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/NL79UexadzE<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AIME box|year=2009|n=I|num-b=1|num-a=3}}<br /> <br /> [[Category:Complex numbers]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=124215 2019 AIME I Problems/Problem 14 2020-06-07T02:44:44Z <p>Emerald block: new solution</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(k)&lt;/math&gt; is the Euler Totient Function of integer &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;\phi(k)&lt;/math&gt; is the number of positive integers less than &lt;math&gt;k&lt;/math&gt; relatively prime to &lt;math&gt;k&lt;/math&gt;. Define the numbers &lt;math&gt;k_1,k_2,k_3,\cdots,k_n&lt;/math&gt; to be the prime factors of &lt;math&gt;k&lt;/math&gt;. Then, we have&lt;cmath&gt;\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).&lt;/cmath&gt;A property of the Totient function is that, for any prime &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;\phi(p)=p-1&lt;/math&gt;.<br /> <br /> Euler's Totient Theorem states that&lt;cmath&gt;a^{\phi(k)} \equiv 1\pmod k&lt;/cmath&gt;if &lt;math&gt;\gcd(a,k)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Solution 2 (Basic Modular Arithmetic)==<br /> <br /> In this solution, &lt;math&gt;k&lt;/math&gt; will represent an arbitrary nonnegative integer.<br /> <br /> We will show that any potential prime &lt;math&gt;p&lt;/math&gt; must be of the form &lt;math&gt;16k+1&lt;/math&gt; through a proof by contradiction. Suppose that there exists some prime &lt;math&gt;p&lt;/math&gt; that can not be expressed in the form &lt;math&gt;16k+1&lt;/math&gt; that is a divisor of &lt;math&gt;2019^8+1&lt;/math&gt;. First, note that if the prime &lt;math&gt;p&lt;/math&gt; is a divisor of &lt;math&gt;2019&lt;/math&gt;, then &lt;math&gt;2019^8&lt;/math&gt; is a multiple of &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;2019^8+1&lt;/math&gt; is not. Thus, &lt;math&gt;p&lt;/math&gt; is not a divisor of &lt;math&gt;2019&lt;/math&gt;.<br /> <br /> Because &lt;math&gt;2019^8+1&lt;/math&gt; is a multiple of &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;2019^8+1\equiv0\pmod{p}&lt;/math&gt;. This means that &lt;math&gt;2019^8\equiv-1\pmod{p}&lt;/math&gt;, and by raising both sides to an arbitrary odd positive integer, we have that &lt;math&gt;2019^{16k+8}\equiv-1\pmod{p}&lt;/math&gt;.<br /> <br /> Then, since the problem requires an odd prime, &lt;math&gt;p&lt;/math&gt; can be expressed as &lt;math&gt;16k+m&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; is an odd integer ranging from &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;15&lt;/math&gt;, inclusive. By Fermat's Little Theorem, &lt;math&gt;2019^{p-1}\equiv1\pmod{p}&lt;/math&gt;, and plugging in values, we get &lt;math&gt;2019^{16k+n}\equiv1\pmod{p}&lt;/math&gt;, where &lt;math&gt;n=m-1&lt;/math&gt; and is thus an even integer ranging from &lt;math&gt;2&lt;/math&gt; to &lt;math&gt;14&lt;/math&gt;, inclusive.<br /> <br /> If &lt;math&gt;n=8&lt;/math&gt;, then &lt;math&gt;2019^{16k+8}\equiv1\pmod{p}&lt;/math&gt;, which creates a contradiction. If &lt;math&gt;n&lt;/math&gt; is not a multiple of &lt;math&gt;8&lt;/math&gt; but is a multiple of &lt;math&gt;4&lt;/math&gt;, squaring both sides of &lt;math&gt;2019^{16k+n}\equiv1\pmod{p}&lt;/math&gt; also results in the same contradictory equivalence. For all remaining &lt;math&gt;n&lt;/math&gt;, raising both sides of &lt;math&gt;2019^{16k+n}\equiv1\pmod{p}&lt;/math&gt; to the &lt;math&gt;4&lt;/math&gt;th power creates the same contradiction. (Note that &lt;math&gt;32k&lt;/math&gt; and &lt;math&gt;64k&lt;/math&gt; can both be expressed in the form &lt;math&gt;16k&lt;/math&gt;.)<br /> <br /> Since we have proved that no value of &lt;math&gt;n&lt;/math&gt; can work, this means that a prime must be of the form &lt;math&gt;16k+1&lt;/math&gt; in order to be a factor of &lt;math&gt;2019^8+1&lt;/math&gt;. The smallest prime of this form is &lt;math&gt;17&lt;/math&gt;, and testing it, we get<br /> &lt;cmath&gt;2019^8+1\equiv13^8+1\equiv169^4+1\equiv(-1)^4+1\equiv1+1\equiv2\pmod{17},&lt;/cmath&gt;<br /> so it does not work. The next smallest prime of the required form is &lt;math&gt;97&lt;/math&gt;, and testing it, we get<br /> &lt;cmath&gt;2019^8+1\equiv(-18)^8+1\equiv324^4+1\equiv33^4+1\equiv1089^2+1\equiv22^2+1\equiv484+1\equiv-1+1\equiv0\pmod{97},&lt;/cmath&gt;<br /> so it works. Thus, the answer is &lt;math&gt;\boxed{097}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_14&diff=123982 2019 AIME II Problems/Problem 14 2020-06-05T23:22:01Z <p>Emerald block: fixed small grammar mistake</p> <hr /> <div>==Problem==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that, given an unlimited supply of stamps of denominations &lt;math&gt;5,n,&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; cents, &lt;math&gt;91&lt;/math&gt; cents is the greatest postage that cannot be formed.<br /> <br /> ==Solution 1==<br /> <br /> By the Chicken McNugget theorem, the least possible value of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;91&lt;/math&gt; cents cannot be formed satisfies &lt;math&gt;5n - (5 + n) = 91 \implies n = 24&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be at least &lt;math&gt;24&lt;/math&gt;.<br /> <br /> For a value of &lt;math&gt;n&lt;/math&gt; to work, we must not only be unable to form the value &lt;math&gt;91&lt;/math&gt;, but we must also be able to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;96&lt;/math&gt;, as with these five values, we can form any value greater than &lt;math&gt;96&lt;/math&gt; by using additional &lt;math&gt;5&lt;/math&gt; cent stamps.<br /> <br /> Notice that we must form the value &lt;math&gt;96&lt;/math&gt; without forming the value &lt;math&gt;91&lt;/math&gt;. If we use any &lt;math&gt;5&lt;/math&gt; cent stamps when forming &lt;math&gt;96&lt;/math&gt;, we could simply remove one to get &lt;math&gt;91&lt;/math&gt;. This means that we must obtain the value &lt;math&gt;96&lt;/math&gt; using only stamps of denominations &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Recalling that &lt;math&gt;n \geq 24&lt;/math&gt;, we can easily figure out the working &lt;math&gt;(n,n+1)&lt;/math&gt; pairs that can used to obtain &lt;math&gt;96&lt;/math&gt;, as we can use at most &lt;math&gt;\frac{96}{24}=4&lt;/math&gt; stamps without going over. The potential sets are &lt;math&gt;(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)&lt;/math&gt;, and &lt;math&gt;(96, 97)&lt;/math&gt;.<br /> <br /> The last two obviously do not work, since they are too large to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;94&lt;/math&gt;, and by a little testing, only &lt;math&gt;(24, 25)&lt;/math&gt; and &lt;math&gt;(47, 48)&lt;/math&gt; can form the necessary values, so &lt;math&gt;n \in \{24, 47\}&lt;/math&gt;. &lt;math&gt;24 + 47 = \boxed{071}&lt;/math&gt;.<br /> <br /> ~Revision by [[User:emerald_block|emerald_block]]<br /> <br /> ===Note on finding and testing potential pairs===<br /> <br /> In order to find potential &lt;math&gt;(n,n+1)&lt;/math&gt; pairs, we simply test all combinations of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; that sum to less than &lt;math&gt;4n&lt;/math&gt; (so that &lt;math&gt;n\ge24&lt;/math&gt;) to see if they produce an integer value of &lt;math&gt;n&lt;/math&gt; when their sum is set to &lt;math&gt;96&lt;/math&gt;. Note that, since &lt;math&gt;96&lt;/math&gt; is divisible by &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;, we must either use only &lt;math&gt;n&lt;/math&gt; or only &lt;math&gt;n+1&lt;/math&gt;, as otherwise, the sum is guaranteed to not be divisible by one of the numbers &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c}<br /> \text{Combination} &amp; \text{Sum} &amp; n\text{-value} \\ \hline<br /> n,n,n,n &amp; 4n &amp; 24 \\<br /> n+1,n+1,n+1 &amp; 3n+3 &amp; 31 \\<br /> n,n,n &amp; 3n &amp; 32 \\<br /> n+1,n+1 &amp; 2n+2 &amp; 47 \\<br /> n,n &amp; 2n &amp; 48 \\<br /> n+1 &amp; n+1 &amp; 95 \\<br /> n &amp; n &amp; 96 \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> To test whether a pair works, we simply check that, using the number &lt;math&gt;5&lt;/math&gt; and the two numbers in the pair, it is impossible to form a sum of &lt;math&gt;91&lt;/math&gt;, and it is possible to form sums of &lt;math&gt;92&lt;/math&gt;, &lt;math&gt;93&lt;/math&gt;, and &lt;math&gt;94&lt;/math&gt;. (&lt;math&gt;95&lt;/math&gt; can always be formed using only &lt;math&gt;5&lt;/math&gt;s, and the pair is already able to form &lt;math&gt;96&lt;/math&gt; because that was how it was found.) We simply need to reach the residues &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;&lt;math&gt;\pmod{5}&lt;/math&gt; using only &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; without going over the number we are trying to form, while being unable to do so with the residue &lt;math&gt;1&lt;/math&gt;. As stated in the above solution, the last two pairs are clearly too large to work.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c|c|c}<br /> \text{Pair} &amp; \text{Not }91 &amp; 92 &amp; 93 &amp; 94 \\ \hline<br /> 24,25 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 31,32 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 32,33 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 47,48 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 48,49 &amp; \checkmark &amp; \times &amp; \checkmark &amp; \checkmark \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> (Note that if a pair is unable to fulfill a single requirement, there is no need to check the rest.)<br /> <br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> Notice that once we hit all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we'd be able to get any number greater (since we can continually add &lt;math&gt;5&lt;/math&gt; to each residue). Furthermore, &lt;math&gt;n\not\equiv 0,1\pmod{5}&lt;/math&gt; since otherwise &lt;math&gt;91&lt;/math&gt; is obtainable (by repeatedly adding &lt;math&gt;5&lt;/math&gt; to either &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n+1&lt;/math&gt;) Since the given numbers are &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n+1&lt;/math&gt;, we consider two cases: when &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt; and when &lt;math&gt;n&lt;/math&gt; is not that. <br /> <br /> When &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;, we can only hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; once we get to &lt;math&gt;4n&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; only contribute &lt;math&gt;1&lt;/math&gt; more residue &lt;math&gt;\bmod 5&lt;/math&gt;). Looking at multiples of &lt;math&gt;4&lt;/math&gt; greater than &lt;math&gt;91&lt;/math&gt; with &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, we get &lt;math&gt;n=24&lt;/math&gt;. It's easy to check that this works. Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem). <br /> <br /> Now, if &lt;math&gt;n\equiv 2,3\pmod{5}&lt;/math&gt;, then we'd need to go up to &lt;math&gt;2(n+1)=2n+2&lt;/math&gt; until we can hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; create &lt;math&gt;2&lt;/math&gt; distinct residues &lt;math&gt;\bmod{5}&lt;/math&gt;. Checking for such &lt;math&gt;n&lt;/math&gt; gives &lt;math&gt;n=47&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. It's easy to check that &lt;math&gt;n=47&lt;/math&gt; works, but &lt;math&gt;n=48&lt;/math&gt; does not (since &lt;math&gt;92&lt;/math&gt; is unobtainable). Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the &lt;math&gt;3 \pmod{5}&lt;/math&gt; case, the residue &lt;math&gt;2 \pmod{5}&lt;/math&gt; has will not be produced until &lt;math&gt;3(n+1)&lt;/math&gt; while the &lt;math&gt;1\pmod5&lt;/math&gt; case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to &lt;math&gt;1 \pmod5&lt;/math&gt;)<br /> <br /> Since we've checked all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we can be sure that these are all the possible values of &lt;math&gt;n&lt;/math&gt;. Hence, the answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;. - ktong<br /> <br /> ==Solution 3==<br /> Obviously &lt;math&gt;n\le 90&lt;/math&gt;. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If &lt;math&gt;n\equiv 0\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n+1&lt;/math&gt; and some 5's, so there are no solutions for this case. If &lt;math&gt;n\equiv 1\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n&lt;/math&gt; and some 5's, so there are no solutions for this case either.<br /> <br /> For &lt;math&gt;n\equiv 2\pmod{5}&lt;/math&gt;, &lt;math&gt;2n+2&lt;/math&gt; is the smallest value that can be formed which is 1 mod 5, so &lt;math&gt;2n+2=96&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;. We see that &lt;math&gt;92=45+47&lt;/math&gt;, &lt;math&gt;93=48+45&lt;/math&gt;, and &lt;math&gt;94=47+47&lt;/math&gt;, so &lt;math&gt;n=47&lt;/math&gt; does work. If &lt;math&gt;n\equiv 3\pmod{5}&lt;/math&gt;, then the smallest value that can be formed which is 1 mod 5 is &lt;math&gt;2n&lt;/math&gt;, so &lt;math&gt;2n=96&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. We see that &lt;math&gt;94=49+45&lt;/math&gt; and &lt;math&gt;93=48+45&lt;/math&gt;, but 92 cannot be formed, so there are no solutions for this case. If &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, then we can just ignore &lt;math&gt;n+1&lt;/math&gt; since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that &lt;math&gt;5n-n-5=91&lt;/math&gt; meaning &lt;math&gt;4n=96&lt;/math&gt; and &lt;math&gt;n=24&lt;/math&gt;.<br /> Hence, the only two &lt;math&gt;n&lt;/math&gt; that work are &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;, so our answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;.<br /> -Stormersyle<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_14&diff=123981 2019 AIME II Problems/Problem 14 2020-06-05T23:20:40Z <p>Emerald block: fixed mistake in text</p> <hr /> <div>==Problem==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that, given an unlimited supply of stamps of denominations &lt;math&gt;5,n,&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; cents, &lt;math&gt;91&lt;/math&gt; cents is the greatest postage that cannot be formed.<br /> <br /> ==Solution 1==<br /> <br /> By the Chicken McNugget theorem, the least possible value of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;91&lt;/math&gt; cents cannot be formed satisfies &lt;math&gt;5n - (5 + n) = 91 \implies n = 24&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be at least &lt;math&gt;24&lt;/math&gt;.<br /> <br /> For a value of &lt;math&gt;n&lt;/math&gt; to work, we must not only be unable to form the value &lt;math&gt;91&lt;/math&gt;, but we must also be able to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;96&lt;/math&gt;, as with these five values, we can form any value greater than &lt;math&gt;96&lt;/math&gt; by using additional &lt;math&gt;5&lt;/math&gt; cent stamps.<br /> <br /> Notice that we must form the value &lt;math&gt;96&lt;/math&gt; without forming the value &lt;math&gt;91&lt;/math&gt;. If we use any &lt;math&gt;5&lt;/math&gt; cent stamps when forming &lt;math&gt;96&lt;/math&gt;, we could simply remove one to get &lt;math&gt;91&lt;/math&gt;. This means that we must obtain the value &lt;math&gt;96&lt;/math&gt; using only stamps of denominations &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Recalling that &lt;math&gt;n \geq 24&lt;/math&gt;, we can easily figure out the working &lt;math&gt;(n,n+1)&lt;/math&gt; pairs that can used to obtain &lt;math&gt;96&lt;/math&gt;, as we can use at most &lt;math&gt;\frac{96}{24}=4&lt;/math&gt; stamps without going over. The potential sets are &lt;math&gt;(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)&lt;/math&gt;, and &lt;math&gt;(96, 97)&lt;/math&gt;.<br /> <br /> The last two obviously do not work, since they are too large to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;94&lt;/math&gt;, and by a little testing, only &lt;math&gt;(24, 25)&lt;/math&gt; and &lt;math&gt;(47, 48)&lt;/math&gt; can form the necessary values, so &lt;math&gt;n \in \{24, 47\}&lt;/math&gt;. &lt;math&gt;24 + 47 = \boxed{071}&lt;/math&gt;.<br /> <br /> ~Revision by [[User:emerald_block|emerald_block]]<br /> <br /> ===Note on finding and testing potential pairs===<br /> <br /> In order to find potential &lt;math&gt;(n,n+1)&lt;/math&gt; pairs, we simply test all combinations of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; that sum to less than &lt;math&gt;4n&lt;/math&gt; (so that &lt;math&gt;n\ge24&lt;/math&gt;) to see if they produce an integer value of &lt;math&gt;n&lt;/math&gt; when their sum is set to &lt;math&gt;96&lt;/math&gt;. Note that, since &lt;math&gt;96&lt;/math&gt; is divisible by &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;, we must either use only &lt;math&gt;n&lt;/math&gt; or only &lt;math&gt;n+1&lt;/math&gt;, as otherwise, the sum is guaranteed to not be divisible by one of the numbers &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c}<br /> \text{Combination} &amp; \text{Sum} &amp; n\text{-value} \\ \hline<br /> n,n,n,n &amp; 4n &amp; 24 \\<br /> n+1,n+1,n+1 &amp; 3n+3 &amp; 31 \\<br /> n,n,n &amp; 3n &amp; 32 \\<br /> n+1,n+1 &amp; 2n+2 &amp; 47 \\<br /> n,n &amp; 2n &amp; 48 \\<br /> n+1 &amp; n+1 &amp; 95 \\<br /> n &amp; n &amp; 96 \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> To test whether a pair works, we simply check that, using the number &lt;math&gt;5&lt;/math&gt; and the two numbers in the pair, it is impossible to form a sum of &lt;math&gt;91&lt;/math&gt;, and that it is possible to form sums of &lt;math&gt;92&lt;/math&gt;, &lt;math&gt;93&lt;/math&gt;, and &lt;math&gt;94&lt;/math&gt;. (&lt;math&gt;95&lt;/math&gt; can always be formed using only &lt;math&gt;5&lt;/math&gt;s, and the pair is already able to form &lt;math&gt;96&lt;/math&gt; because that was how it was found.) We simply need to reach the residues &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;&lt;math&gt;\pmod{5}&lt;/math&gt; using only &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; without going over the number we are trying to form, while being unable to do so with the residue &lt;math&gt;1&lt;/math&gt;. As stated in the above solution, the last two pairs are clearly too large to work.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c|c|c}<br /> \text{Pair} &amp; \text{Not }91 &amp; 92 &amp; 93 &amp; 94 \\ \hline<br /> 24,25 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 31,32 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 32,33 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 47,48 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 48,49 &amp; \checkmark &amp; \times &amp; \checkmark &amp; \checkmark \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> (Note that if a pair is unable to fulfill a single requirement, there is no need to check the rest.)<br /> <br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> Notice that once we hit all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we'd be able to get any number greater (since we can continually add &lt;math&gt;5&lt;/math&gt; to each residue). Furthermore, &lt;math&gt;n\not\equiv 0,1\pmod{5}&lt;/math&gt; since otherwise &lt;math&gt;91&lt;/math&gt; is obtainable (by repeatedly adding &lt;math&gt;5&lt;/math&gt; to either &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n+1&lt;/math&gt;) Since the given numbers are &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n+1&lt;/math&gt;, we consider two cases: when &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt; and when &lt;math&gt;n&lt;/math&gt; is not that. <br /> <br /> When &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;, we can only hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; once we get to &lt;math&gt;4n&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; only contribute &lt;math&gt;1&lt;/math&gt; more residue &lt;math&gt;\bmod 5&lt;/math&gt;). Looking at multiples of &lt;math&gt;4&lt;/math&gt; greater than &lt;math&gt;91&lt;/math&gt; with &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, we get &lt;math&gt;n=24&lt;/math&gt;. It's easy to check that this works. Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem). <br /> <br /> Now, if &lt;math&gt;n\equiv 2,3\pmod{5}&lt;/math&gt;, then we'd need to go up to &lt;math&gt;2(n+1)=2n+2&lt;/math&gt; until we can hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; create &lt;math&gt;2&lt;/math&gt; distinct residues &lt;math&gt;\bmod{5}&lt;/math&gt;. Checking for such &lt;math&gt;n&lt;/math&gt; gives &lt;math&gt;n=47&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. It's easy to check that &lt;math&gt;n=47&lt;/math&gt; works, but &lt;math&gt;n=48&lt;/math&gt; does not (since &lt;math&gt;92&lt;/math&gt; is unobtainable). Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the &lt;math&gt;3 \pmod{5}&lt;/math&gt; case, the residue &lt;math&gt;2 \pmod{5}&lt;/math&gt; has will not be produced until &lt;math&gt;3(n+1)&lt;/math&gt; while the &lt;math&gt;1\pmod5&lt;/math&gt; case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to &lt;math&gt;1 \pmod5&lt;/math&gt;)<br /> <br /> Since we've checked all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we can be sure that these are all the possible values of &lt;math&gt;n&lt;/math&gt;. Hence, the answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;. - ktong<br /> <br /> ==Solution 3==<br /> Obviously &lt;math&gt;n\le 90&lt;/math&gt;. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If &lt;math&gt;n\equiv 0\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n+1&lt;/math&gt; and some 5's, so there are no solutions for this case. If &lt;math&gt;n\equiv 1\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n&lt;/math&gt; and some 5's, so there are no solutions for this case either.<br /> <br /> For &lt;math&gt;n\equiv 2\pmod{5}&lt;/math&gt;, &lt;math&gt;2n+2&lt;/math&gt; is the smallest value that can be formed which is 1 mod 5, so &lt;math&gt;2n+2=96&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;. We see that &lt;math&gt;92=45+47&lt;/math&gt;, &lt;math&gt;93=48+45&lt;/math&gt;, and &lt;math&gt;94=47+47&lt;/math&gt;, so &lt;math&gt;n=47&lt;/math&gt; does work. If &lt;math&gt;n\equiv 3\pmod{5}&lt;/math&gt;, then the smallest value that can be formed which is 1 mod 5 is &lt;math&gt;2n&lt;/math&gt;, so &lt;math&gt;2n=96&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. We see that &lt;math&gt;94=49+45&lt;/math&gt; and &lt;math&gt;93=48+45&lt;/math&gt;, but 92 cannot be formed, so there are no solutions for this case. If &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, then we can just ignore &lt;math&gt;n+1&lt;/math&gt; since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that &lt;math&gt;5n-n-5=91&lt;/math&gt; meaning &lt;math&gt;4n=96&lt;/math&gt; and &lt;math&gt;n=24&lt;/math&gt;.<br /> Hence, the only two &lt;math&gt;n&lt;/math&gt; that work are &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;, so our answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;.<br /> -Stormersyle<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_14&diff=123952 2019 AIME II Problems/Problem 14 2020-06-05T22:58:25Z <p>Emerald block: fixed small formatting error</p> <hr /> <div>==Problem==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that, given an unlimited supply of stamps of denominations &lt;math&gt;5,n,&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; cents, &lt;math&gt;91&lt;/math&gt; cents is the greatest postage that cannot be formed.<br /> <br /> ==Solution 1==<br /> <br /> By the Chicken McNugget theorem, the least possible value of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;91&lt;/math&gt; cents cannot be formed satisfies &lt;math&gt;5n - (5 + n) = 91 \implies n = 24&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be at least &lt;math&gt;24&lt;/math&gt;.<br /> <br /> For a value of &lt;math&gt;n&lt;/math&gt; to work, we must not only be unable to form the value &lt;math&gt;91&lt;/math&gt;, but we must also be able to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;96&lt;/math&gt;, as with these five values, we can form any value greater than &lt;math&gt;96&lt;/math&gt; by using additional &lt;math&gt;5&lt;/math&gt; cent stamps.<br /> <br /> Notice that we must form the value &lt;math&gt;96&lt;/math&gt; without forming the value &lt;math&gt;91&lt;/math&gt;. If we use any &lt;math&gt;5&lt;/math&gt; cent stamps when forming &lt;math&gt;96&lt;/math&gt;, we could simply remove one to get &lt;math&gt;91&lt;/math&gt;. This means that we must obtain the value &lt;math&gt;96&lt;/math&gt; using only stamps of denominations &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Recalling that &lt;math&gt;n \geq 24&lt;/math&gt;, we can easily figure out the working &lt;math&gt;(n,n+1)&lt;/math&gt; pairs that can used to obtain &lt;math&gt;96&lt;/math&gt;, as we can use at most &lt;math&gt;\frac{96}{24}=4&lt;/math&gt; stamps without going over. The potential sets are &lt;math&gt;(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)&lt;/math&gt;, and &lt;math&gt;(96, 97)&lt;/math&gt;.<br /> <br /> The last two obviously do not work, since they are too large to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;94&lt;/math&gt;, and by a little testing, only &lt;math&gt;(24, 25)&lt;/math&gt; and &lt;math&gt;(47, 48)&lt;/math&gt; can form the necessary values, so &lt;math&gt;n \in \{24, 47\}&lt;/math&gt;. &lt;math&gt;24 + 47 = \boxed{071}&lt;/math&gt;.<br /> <br /> ~Revision by [[User:emerald_block|emerald_block]]<br /> <br /> ===Note on finding and testing potential pairs===<br /> <br /> In order to find potential &lt;math&gt;(n,n+1)&lt;/math&gt; pairs, we simply test all combinations of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; that sum to less than &lt;math&gt;4n&lt;/math&gt; (so that &lt;math&gt;n\ge24&lt;/math&gt;) to see if they produce an integer value of &lt;math&gt;n&lt;/math&gt; when their sum is set to &lt;math&gt;96&lt;/math&gt;. Note that, since &lt;math&gt;96&lt;/math&gt; is divisible by &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;, we must either use only &lt;math&gt;n&lt;/math&gt; or only &lt;math&gt;n+1&lt;/math&gt;, as otherwise, the sum is guaranteed to not be divisible by one of the numbers &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c}<br /> \text{Combination} &amp; \text{Sum} &amp; n\text{-value} \\ \hline<br /> n,n,n,n &amp; 4n &amp; 24 \\<br /> n+1,n+1,n+1 &amp; 3n+3 &amp; 31 \\<br /> n,n,n &amp; 3n &amp; 32 \\<br /> n+1,n+1 &amp; 2n+2 &amp; 47 \\<br /> n,n &amp; 2n &amp; 48 \\<br /> n+1 &amp; n+1 &amp; 95 \\<br /> n &amp; n &amp; 96 \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> To test whether a pair works, we simply check that it cannot form &lt;math&gt;91&lt;/math&gt;, and that it can form &lt;math&gt;92&lt;/math&gt;, &lt;math&gt;93&lt;/math&gt;, and &lt;math&gt;94&lt;/math&gt;. (&lt;math&gt;95&lt;/math&gt; can always be formed using only &lt;math&gt;5&lt;/math&gt;s, and the pair is already able to form &lt;math&gt;96&lt;/math&gt; because that was how it was found.) We simply need to reach the residues &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;&lt;math&gt;\pmod{5}&lt;/math&gt; using only &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; without going over the number we are trying to form, while being unable to do so with the residue &lt;math&gt;1&lt;/math&gt;. As stated in the above solution, the last two pairs are clearly too large to work.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c|c|c}<br /> \text{Pair} &amp; \text{Not }91 &amp; 92 &amp; 93 &amp; 94 \\ \hline<br /> 24,25 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 31,32 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 32,33 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 47,48 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 48,49 &amp; \checkmark &amp; \times &amp; \checkmark &amp; \checkmark \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> (Note that if a pair is unable to fulfill a single requirement, there is no need to check the rest.)<br /> <br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> Notice that once we hit all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we'd be able to get any number greater (since we can continually add &lt;math&gt;5&lt;/math&gt; to each residue). Furthermore, &lt;math&gt;n\not\equiv 0,1\pmod{5}&lt;/math&gt; since otherwise &lt;math&gt;91&lt;/math&gt; is obtainable (by repeatedly adding &lt;math&gt;5&lt;/math&gt; to either &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n+1&lt;/math&gt;) Since the given numbers are &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n+1&lt;/math&gt;, we consider two cases: when &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt; and when &lt;math&gt;n&lt;/math&gt; is not that. <br /> <br /> When &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;, we can only hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; once we get to &lt;math&gt;4n&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; only contribute &lt;math&gt;1&lt;/math&gt; more residue &lt;math&gt;\bmod 5&lt;/math&gt;). Looking at multiples of &lt;math&gt;4&lt;/math&gt; greater than &lt;math&gt;91&lt;/math&gt; with &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, we get &lt;math&gt;n=24&lt;/math&gt;. It's easy to check that this works. Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem). <br /> <br /> Now, if &lt;math&gt;n\equiv 2,3\pmod{5}&lt;/math&gt;, then we'd need to go up to &lt;math&gt;2(n+1)=2n+2&lt;/math&gt; until we can hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; create &lt;math&gt;2&lt;/math&gt; distinct residues &lt;math&gt;\bmod{5}&lt;/math&gt;. Checking for such &lt;math&gt;n&lt;/math&gt; gives &lt;math&gt;n=47&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. It's easy to check that &lt;math&gt;n=47&lt;/math&gt; works, but &lt;math&gt;n=48&lt;/math&gt; does not (since &lt;math&gt;92&lt;/math&gt; is unobtainable). Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the &lt;math&gt;3 \pmod{5}&lt;/math&gt; case, the residue &lt;math&gt;2 \pmod{5}&lt;/math&gt; has will not be produced until &lt;math&gt;3(n+1)&lt;/math&gt; while the &lt;math&gt;1\pmod5&lt;/math&gt; case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to &lt;math&gt;1 \pmod5&lt;/math&gt;)<br /> <br /> Since we've checked all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we can be sure that these are all the possible values of &lt;math&gt;n&lt;/math&gt;. Hence, the answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;. - ktong<br /> <br /> ==Solution 3==<br /> Obviously &lt;math&gt;n\le 90&lt;/math&gt;. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If &lt;math&gt;n\equiv 0\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n+1&lt;/math&gt; and some 5's, so there are no solutions for this case. If &lt;math&gt;n\equiv 1\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n&lt;/math&gt; and some 5's, so there are no solutions for this case either.<br /> <br /> For &lt;math&gt;n\equiv 2\pmod{5}&lt;/math&gt;, &lt;math&gt;2n+2&lt;/math&gt; is the smallest value that can be formed which is 1 mod 5, so &lt;math&gt;2n+2=96&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;. We see that &lt;math&gt;92=45+47&lt;/math&gt;, &lt;math&gt;93=48+45&lt;/math&gt;, and &lt;math&gt;94=47+47&lt;/math&gt;, so &lt;math&gt;n=47&lt;/math&gt; does work. If &lt;math&gt;n\equiv 3\pmod{5}&lt;/math&gt;, then the smallest value that can be formed which is 1 mod 5 is &lt;math&gt;2n&lt;/math&gt;, so &lt;math&gt;2n=96&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. We see that &lt;math&gt;94=49+45&lt;/math&gt; and &lt;math&gt;93=48+45&lt;/math&gt;, but 92 cannot be formed, so there are no solutions for this case. If &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, then we can just ignore &lt;math&gt;n+1&lt;/math&gt; since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that &lt;math&gt;5n-n-5=91&lt;/math&gt; meaning &lt;math&gt;4n=96&lt;/math&gt; and &lt;math&gt;n=24&lt;/math&gt;.<br /> Hence, the only two &lt;math&gt;n&lt;/math&gt; that work are &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;, so our answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;.<br /> -Stormersyle<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_14&diff=123944 2019 AIME II Problems/Problem 14 2020-06-05T22:51:09Z <p>Emerald block: added additional info</p> <hr /> <div>==Problem==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that, given an unlimited supply of stamps of denominations &lt;math&gt;5,n,&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; cents, &lt;math&gt;91&lt;/math&gt; cents is the greatest postage that cannot be formed.<br /> <br /> ==Solution 1==<br /> <br /> By the Chicken McNugget theorem, the least possible value of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;91&lt;/math&gt; cents cannot be formed satisfies &lt;math&gt;5n - (5 + n) = 91 \implies n = 24&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be at least &lt;math&gt;24&lt;/math&gt;.<br /> <br /> For a value of &lt;math&gt;n&lt;/math&gt; to work, we must not only be unable to form the value &lt;math&gt;91&lt;/math&gt;, but we must also be able to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;96&lt;/math&gt;, as with these five values, we can form any value greater than &lt;math&gt;96&lt;/math&gt; by using additional &lt;math&gt;5&lt;/math&gt; cent stamps.<br /> <br /> Notice that we must form the value &lt;math&gt;96&lt;/math&gt; without forming the value &lt;math&gt;91&lt;/math&gt;. If we use any &lt;math&gt;5&lt;/math&gt; cent stamps when forming &lt;math&gt;96&lt;/math&gt;, we could simply remove one to get &lt;math&gt;91&lt;/math&gt;. This means that we must obtain the value &lt;math&gt;96&lt;/math&gt; using only stamps of denominations &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Recalling that &lt;math&gt;n \geq 24&lt;/math&gt;, we can easily figure out the working &lt;math&gt;(n,n+1)&lt;/math&gt; pairs that can used to obtain &lt;math&gt;96&lt;/math&gt;, as we can use at most &lt;math&gt;\frac{96}{24}=4&lt;/math&gt; stamps without going over. The potential sets are &lt;math&gt;(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)&lt;/math&gt;, and &lt;math&gt;(96, 97)&lt;/math&gt;.<br /> <br /> The last two obviously do not work, since they are too large to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;94&lt;/math&gt;, and by a little testing, only &lt;math&gt;(24, 25)&lt;/math&gt; and &lt;math&gt;(47, 48)&lt;/math&gt; can form the necessary values, so &lt;math&gt;n \in \{24, 47\}&lt;/math&gt;. &lt;math&gt;24 + 47 = \boxed{071}&lt;/math&gt;.<br /> <br /> ~Revision by [[User:emerald_block|emerald_block]]<br /> <br /> ===Note on finding and testing potential pairs===<br /> <br /> In order to find potential &lt;math&gt;(n,n+1)&lt;/math&gt; pairs, we simply test all combinations of &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; that sum to less than &lt;math&gt;4n&lt;/math&gt; (so that &lt;math&gt;n\ge24&lt;/math&gt;) to see if they produce an integer value of &lt;math&gt;n&lt;/math&gt; when their sum is set to &lt;math&gt;96&lt;/math&gt;. Note that, since &lt;math&gt;96&lt;/math&gt; is divisible by &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;, we must either use only &lt;math&gt;n&lt;/math&gt; or only &lt;math&gt;n+1&lt;/math&gt;, as otherwise, the sum is guaranteed to not be divisible by one of the numbers &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c}<br /> \text{Combination} &amp; \text{Sum} &amp; n\text{-value} \\ \hline<br /> n,n,n,n &amp; 4n &amp; 24 \\<br /> n+1,n+1,n+1 &amp; 3n+3 &amp; 31 \\<br /> n,n,n &amp; 3n &amp; 32 \\<br /> n+1,n+1 &amp; 2n+2 &amp; 47 \\<br /> n,n &amp; 2n &amp; 48 \\<br /> n+1 &amp; n+1 &amp; 95 \\<br /> n &amp; n &amp; 96 \\<br /> \end{array}<br /> &lt;/math&gt;<br /> <br /> To test whether a pair works, we simply check that it cannot form &lt;math&gt;91&lt;/math&gt;, and that it can form &lt;math&gt;92&lt;/math&gt;, &lt;math&gt;93&lt;/math&gt;, and &lt;math&gt;94&lt;/math&gt;. (&lt;math&gt;95&lt;/math&gt; can always be formed using only &lt;math&gt;5&lt;/math&gt;s, and the pair is already able to form &lt;math&gt;96&lt;/math&gt; because that was how it was found.) We simply need to reach the residues &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, and &lt;math&gt;4&lt;/math&gt;&lt;math&gt;\pmod{5}&lt;/math&gt; using only &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; without going over the number we are trying to form, while being unable to do so with the residue &lt;math&gt;1&lt;/math&gt;. As stated in the above solution, the last two pairs are clearly too large to work.<br /> <br /> &lt;math&gt;<br /> \begin{array}{c|c|c|c|c}<br /> \text{Pair} &amp; \text{Not }91 &amp; 92 &amp; 93 &amp; 94 \\ \hline<br /> 24,25 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 31,32 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 32,33 &amp; \times &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 47,48 &amp; \checkmark &amp; \checkmark &amp; \checkmark &amp; \checkmark \\<br /> 48,49 &amp; \checkmark &amp; \times &amp; \checkmark &amp; \checkmark \\<br /> \end{array}<br /> &lt;/math&gt;<br /> (Note that if a pair is unable to fulfill a single requirement, there is no need to check the rest.)<br /> <br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> Notice that once we hit all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we'd be able to get any number greater (since we can continually add &lt;math&gt;5&lt;/math&gt; to each residue). Furthermore, &lt;math&gt;n\not\equiv 0,1\pmod{5}&lt;/math&gt; since otherwise &lt;math&gt;91&lt;/math&gt; is obtainable (by repeatedly adding &lt;math&gt;5&lt;/math&gt; to either &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n+1&lt;/math&gt;) Since the given numbers are &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n+1&lt;/math&gt;, we consider two cases: when &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt; and when &lt;math&gt;n&lt;/math&gt; is not that. <br /> <br /> When &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;, we can only hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; once we get to &lt;math&gt;4n&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; only contribute &lt;math&gt;1&lt;/math&gt; more residue &lt;math&gt;\bmod 5&lt;/math&gt;). Looking at multiples of &lt;math&gt;4&lt;/math&gt; greater than &lt;math&gt;91&lt;/math&gt; with &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, we get &lt;math&gt;n=24&lt;/math&gt;. It's easy to check that this works. Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem). <br /> <br /> Now, if &lt;math&gt;n\equiv 2,3\pmod{5}&lt;/math&gt;, then we'd need to go up to &lt;math&gt;2(n+1)=2n+2&lt;/math&gt; until we can hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; create &lt;math&gt;2&lt;/math&gt; distinct residues &lt;math&gt;\bmod{5}&lt;/math&gt;. Checking for such &lt;math&gt;n&lt;/math&gt; gives &lt;math&gt;n=47&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. It's easy to check that &lt;math&gt;n=47&lt;/math&gt; works, but &lt;math&gt;n=48&lt;/math&gt; does not (since &lt;math&gt;92&lt;/math&gt; is unobtainable). Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the &lt;math&gt;3 \pmod{5}&lt;/math&gt; case, the residue &lt;math&gt;2 \pmod{5}&lt;/math&gt; has will not be produced until &lt;math&gt;3(n+1)&lt;/math&gt; while the &lt;math&gt;1\pmod5&lt;/math&gt; case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to &lt;math&gt;1 \pmod5&lt;/math&gt;)<br /> <br /> Since we've checked all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we can be sure that these are all the possible values of &lt;math&gt;n&lt;/math&gt;. Hence, the answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;. - ktong<br /> <br /> ==Solution 3==<br /> Obviously &lt;math&gt;n\le 90&lt;/math&gt;. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If &lt;math&gt;n\equiv 0\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n+1&lt;/math&gt; and some 5's, so there are no solutions for this case. If &lt;math&gt;n\equiv 1\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n&lt;/math&gt; and some 5's, so there are no solutions for this case either.<br /> <br /> For &lt;math&gt;n\equiv 2\pmod{5}&lt;/math&gt;, &lt;math&gt;2n+2&lt;/math&gt; is the smallest value that can be formed which is 1 mod 5, so &lt;math&gt;2n+2=96&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;. We see that &lt;math&gt;92=45+47&lt;/math&gt;, &lt;math&gt;93=48+45&lt;/math&gt;, and &lt;math&gt;94=47+47&lt;/math&gt;, so &lt;math&gt;n=47&lt;/math&gt; does work. If &lt;math&gt;n\equiv 3\pmod{5}&lt;/math&gt;, then the smallest value that can be formed which is 1 mod 5 is &lt;math&gt;2n&lt;/math&gt;, so &lt;math&gt;2n=96&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. We see that &lt;math&gt;94=49+45&lt;/math&gt; and &lt;math&gt;93=48+45&lt;/math&gt;, but 92 cannot be formed, so there are no solutions for this case. If &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, then we can just ignore &lt;math&gt;n+1&lt;/math&gt; since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that &lt;math&gt;5n-n-5=91&lt;/math&gt; meaning &lt;math&gt;4n=96&lt;/math&gt; and &lt;math&gt;n=24&lt;/math&gt;.<br /> Hence, the only two &lt;math&gt;n&lt;/math&gt; that work are &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;, so our answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;.<br /> -Stormersyle<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_14&diff=123938 2019 AIME II Problems/Problem 14 2020-06-05T21:42:10Z <p>Emerald block: revised solution</p> <hr /> <div>==Problem==<br /> Find the sum of all positive integers &lt;math&gt;n&lt;/math&gt; such that, given an unlimited supply of stamps of denominations &lt;math&gt;5,n,&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; cents, &lt;math&gt;91&lt;/math&gt; cents is the greatest postage that cannot be formed.<br /> <br /> ==Solution 1==<br /> By the Chicken McNugget theorem, the least possible value of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;91&lt;/math&gt; cents cannot be formed satisfies &lt;math&gt;5n - (5 + n) = 91 \implies n = 24&lt;/math&gt;, so &lt;math&gt;n&lt;/math&gt; must be at least &lt;math&gt;24&lt;/math&gt;.<br /> <br /> For a value of &lt;math&gt;n&lt;/math&gt; to work, we must not only be unable to form the value &lt;math&gt;91&lt;/math&gt;, but we must also be able to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;96&lt;/math&gt;, as with these five values, we can form any value greater than &lt;math&gt;96&lt;/math&gt; by using additional &lt;math&gt;5&lt;/math&gt; cent stamps.<br /> <br /> Notice that we must form the value &lt;math&gt;96&lt;/math&gt; without forming the value &lt;math&gt;91&lt;/math&gt;. If we use any &lt;math&gt;5&lt;/math&gt; cent stamps when forming &lt;math&gt;96&lt;/math&gt;, we could simply remove one to get &lt;math&gt;91&lt;/math&gt;. This means that we must obtain the value &lt;math&gt;96&lt;/math&gt; using only stamps of denominations &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Recalling that &lt;math&gt;n \geq 24&lt;/math&gt;, we can easily figure out the working &lt;math&gt;(n,n+1)&lt;/math&gt; pairs that can used to obtain &lt;math&gt;96&lt;/math&gt;, as we can use at most &lt;math&gt;\frac{96}{24}=4&lt;/math&gt; stamps without going over. The potential sets are &lt;math&gt;(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)&lt;/math&gt;, and &lt;math&gt;(96, 97)&lt;/math&gt;.<br /> <br /> The last two obviously do not work, since they are too large to form the values &lt;math&gt;92&lt;/math&gt; through &lt;math&gt;94&lt;/math&gt;, and by a little testing, only &lt;math&gt;(24, 25)&lt;/math&gt; and &lt;math&gt;(47, 48)&lt;/math&gt; can form the necessary values, so &lt;math&gt;n \in \{24, 47\}&lt;/math&gt;. &lt;math&gt;24 + 47 = \boxed{071}&lt;/math&gt;.<br /> <br /> ~Revision by [[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> Notice that once we hit all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we'd be able to get any number greater (since we can continually add &lt;math&gt;5&lt;/math&gt; to each residue). Furthermore, &lt;math&gt;n\not\equiv 0,1\pmod{5}&lt;/math&gt; since otherwise &lt;math&gt;91&lt;/math&gt; is obtainable (by repeatedly adding &lt;math&gt;5&lt;/math&gt; to either &lt;math&gt;n&lt;/math&gt; or &lt;math&gt;n+1&lt;/math&gt;) Since the given numbers are &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt;, and &lt;math&gt;n+1&lt;/math&gt;, we consider two cases: when &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt; and when &lt;math&gt;n&lt;/math&gt; is not that. <br /> <br /> When &lt;math&gt;n\equiv 4 \pmod{5}&lt;/math&gt;, we can only hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; once we get to &lt;math&gt;4n&lt;/math&gt; (since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; only contribute &lt;math&gt;1&lt;/math&gt; more residue &lt;math&gt;\bmod 5&lt;/math&gt;). Looking at multiples of &lt;math&gt;4&lt;/math&gt; greater than &lt;math&gt;91&lt;/math&gt; with &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, we get &lt;math&gt;n=24&lt;/math&gt;. It's easy to check that this works. Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem). <br /> <br /> Now, if &lt;math&gt;n\equiv 2,3\pmod{5}&lt;/math&gt;, then we'd need to go up to &lt;math&gt;2(n+1)=2n+2&lt;/math&gt; until we can hit all residues &lt;math&gt;\bmod 5&lt;/math&gt; since &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;n+1&lt;/math&gt; create &lt;math&gt;2&lt;/math&gt; distinct residues &lt;math&gt;\bmod{5}&lt;/math&gt;. Checking for such &lt;math&gt;n&lt;/math&gt; gives &lt;math&gt;n=47&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. It's easy to check that &lt;math&gt;n=47&lt;/math&gt; works, but &lt;math&gt;n=48&lt;/math&gt; does not (since &lt;math&gt;92&lt;/math&gt; is unobtainable). Furthermore, any &lt;math&gt;n&lt;/math&gt; greater than this does not work since &lt;math&gt;91&lt;/math&gt; isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the &lt;math&gt;3 \pmod{5}&lt;/math&gt; case, the residue &lt;math&gt;2 \pmod{5}&lt;/math&gt; has will not be produced until &lt;math&gt;3(n+1)&lt;/math&gt; while the &lt;math&gt;1\pmod5&lt;/math&gt; case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to &lt;math&gt;1 \pmod5&lt;/math&gt;)<br /> <br /> Since we've checked all residues &lt;math&gt;\bmod 5&lt;/math&gt;, we can be sure that these are all the possible values of &lt;math&gt;n&lt;/math&gt;. Hence, the answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;. - ktong<br /> <br /> ==Solution 3==<br /> Obviously &lt;math&gt;n\le 90&lt;/math&gt;. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If &lt;math&gt;n\equiv 0\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n+1&lt;/math&gt; and some 5's, so there are no solutions for this case. If &lt;math&gt;n\equiv 1\pmod{5}&lt;/math&gt;, then 91 can be formed by using &lt;math&gt;n&lt;/math&gt; and some 5's, so there are no solutions for this case either.<br /> <br /> For &lt;math&gt;n\equiv 2\pmod{5}&lt;/math&gt;, &lt;math&gt;2n+2&lt;/math&gt; is the smallest value that can be formed which is 1 mod 5, so &lt;math&gt;2n+2=96&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;. We see that &lt;math&gt;92=45+47&lt;/math&gt;, &lt;math&gt;93=48+45&lt;/math&gt;, and &lt;math&gt;94=47+47&lt;/math&gt;, so &lt;math&gt;n=47&lt;/math&gt; does work. If &lt;math&gt;n\equiv 3\pmod{5}&lt;/math&gt;, then the smallest value that can be formed which is 1 mod 5 is &lt;math&gt;2n&lt;/math&gt;, so &lt;math&gt;2n=96&lt;/math&gt; and &lt;math&gt;n=48&lt;/math&gt;. We see that &lt;math&gt;94=49+45&lt;/math&gt; and &lt;math&gt;93=48+45&lt;/math&gt;, but 92 cannot be formed, so there are no solutions for this case. If &lt;math&gt;n\equiv 4\pmod{5}&lt;/math&gt;, then we can just ignore &lt;math&gt;n+1&lt;/math&gt; since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that &lt;math&gt;5n-n-5=91&lt;/math&gt; meaning &lt;math&gt;4n=96&lt;/math&gt; and &lt;math&gt;n=24&lt;/math&gt;.<br /> Hence, the only two &lt;math&gt;n&lt;/math&gt; that work are &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=47&lt;/math&gt;, so our answer is &lt;math&gt;24+47=\boxed{071}&lt;/math&gt;.<br /> -Stormersyle<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_10&diff=123716 2019 AIME II Problems/Problem 10 2020-06-04T23:09:47Z <p>Emerald block: small tweaks</p> <hr /> <div>==Problem 10==<br /> There is a unique angle &lt;math&gt;\theta&lt;/math&gt; between &lt;math&gt;0^{\circ}&lt;/math&gt; and &lt;math&gt;90^{\circ}&lt;/math&gt; such that for nonnegative integers &lt;math&gt;n&lt;/math&gt;, the value of &lt;math&gt;\tan{\left(2^{n}\theta\right)}&lt;/math&gt; is positive when &lt;math&gt;n&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;, and negative otherwise. The degree measure of &lt;math&gt;\theta&lt;/math&gt; is &lt;math&gt;\tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Note that if &lt;math&gt;\tan \theta&lt;/math&gt; is positive, then &lt;math&gt;\theta&lt;/math&gt; is in the first or third quadrant, so &lt;math&gt;0^{\circ} &lt; \theta &lt; 90^{\circ} \pmod{180^{\circ}}&lt;/math&gt;. Also notice that the only way &lt;math&gt;\tan{\left(2^{n}\theta\right)}&lt;/math&gt; can be positive for all &lt;math&gt;n&lt;/math&gt; that are multiples of &lt;math&gt;3&lt;/math&gt; is when &lt;math&gt;2^0\theta, 2^3\theta, 2^6\theta&lt;/math&gt;, etc. are all the same value &lt;math&gt;\pmod{180^{\circ}}&lt;/math&gt;. (This also must be true in order for &lt;math&gt;\theta&lt;/math&gt; to be unique.) This is the case if &lt;math&gt;8\theta = \theta \pmod{180^{\circ}}&lt;/math&gt;, so &lt;math&gt;7\theta = 0^{\circ} \pmod{180^{\circ}}&lt;/math&gt;. Therefore, the only possible values of &lt;math&gt;\theta&lt;/math&gt; between &lt;math&gt;0^{\circ}&lt;/math&gt; and &lt;math&gt;90^{\circ}&lt;/math&gt; are &lt;math&gt;\frac{180}{7}^{\circ}&lt;/math&gt;, &lt;math&gt;\frac{360}{7}^{\circ}&lt;/math&gt;, and &lt;math&gt;\frac{540}{7}^{\circ}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{180}{7}^{\circ}&lt;/math&gt; does not work since &lt;math&gt;\tan{2 \cdot \frac{180}{7}^{\circ}}&lt;/math&gt; is positive, and &lt;math&gt;\frac{360}{7}^{\circ}&lt;/math&gt; does not work because &lt;math&gt;\tan{4 \cdot \frac{360}{7}^{\circ}}&lt;/math&gt; is positive. Thus, &lt;math&gt;\theta = \frac{540}{7}^{\circ}&lt;/math&gt;, and a quick check verifies that it works. &lt;math&gt;540 + 7 = \boxed{547}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> As in the previous solution, we note that &lt;math&gt;\tan \theta&lt;/math&gt; is positive when &lt;math&gt;\theta&lt;/math&gt; is in the first or third quadrant. In order for &lt;math&gt;\tan\left(2^n\theta\right)&lt;/math&gt; to be positive for all &lt;math&gt;n&lt;/math&gt; divisible by &lt;math&gt;3&lt;/math&gt;, we must have &lt;math&gt;\theta&lt;/math&gt;, &lt;math&gt;2^3\theta&lt;/math&gt;, &lt;math&gt;2^6\theta&lt;/math&gt;, etc to lie in the first or third quadrants. We already know that &lt;math&gt;\theta\in(0,90)&lt;/math&gt;. We can keep track of the range of &lt;math&gt;2^n\theta&lt;/math&gt; for each &lt;math&gt;n&lt;/math&gt; by considering the portion in the desired quadrants, which gives <br /> &lt;cmath&gt;n=1 \implies (90,180)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=2\implies (270,360)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=3 \implies (180,270)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=4 \implies (90,180)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=5\implies(270,360)&lt;/cmath&gt;<br /> &lt;cmath&gt;n=6 \implies (180,270)&lt;/cmath&gt;<br /> &lt;cmath&gt;\cdots&lt;/cmath&gt;<br /> at which point we realize a pattern emerging. Specifically, the intervals repeat every &lt;math&gt;3&lt;/math&gt; after &lt;math&gt;n=1&lt;/math&gt;. We can use these repeating intervals to determine the desired value of &lt;math&gt;\theta&lt;/math&gt; since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.<br /> <br /> Initially, the lower bound is &lt;math&gt;0&lt;/math&gt; (at &lt;math&gt;n=0&lt;/math&gt;), then increases to &lt;math&gt;\frac{90}{2}=45&lt;/math&gt; at &lt;math&gt;n=1&lt;/math&gt;. This then becomes &lt;math&gt;45+\frac{45}{2}&lt;/math&gt; at &lt;math&gt;n=2&lt;/math&gt;, &lt;math&gt;45+\frac{45}{2}&lt;/math&gt; at &lt;math&gt;n=3&lt;/math&gt;, &lt;math&gt;45+\frac{45}{2}+\frac{45}{2^3}&lt;/math&gt; at &lt;math&gt;n=4&lt;/math&gt;,&lt;math&gt;45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}&lt;/math&gt; at &lt;math&gt;n=5&lt;/math&gt;. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as &lt;math&gt;n&lt;/math&gt; approaches infinity, the lower bound converges to <br /> &lt;cmath&gt;\sum_{k=0}^{\infty}\left(45+\frac{45}{2}\right)\cdot \left(\frac{1}{8}\right)^k=\frac{45+\frac{45}{2}}{1-\frac{1}{8}}=\frac{\frac{135}{2}}{\frac{7}{8}}=\frac{540}{7}\implies p+q=540+7=\boxed{547}&lt;/cmath&gt;-ktong<br /> <br /> ==Solution 3==<br /> Since &lt;math&gt;tan\left(\theta\right) &gt; 0&lt;/math&gt;, &lt;math&gt;0 &lt; \theta &lt; 90&lt;/math&gt;. Since &lt;math&gt;tan\left(2\theta\right) &lt; 0&lt;/math&gt;, &lt;math&gt;\theta&lt;/math&gt; has to be in the second half of the interval (0, 90) ie (45, 90). Since &lt;math&gt;tan\left(4\theta\right) &lt; 0&lt;/math&gt;, &lt;math&gt;\theta&lt;/math&gt; has to be in the second half of that interval ie (67.5, 90). And since &lt;math&gt;tan\left(8\theta\right) &gt; 0&lt;/math&gt;, &lt;math&gt;\theta&lt;/math&gt; has to be in the first half of (67.5, 90). Inductively, the pattern repeats: &lt;math&gt;\theta&lt;/math&gt; is in the first half of the second half of the second half of the first half of the second half of the second half... of the interval (0, 90). Consider the binary representation of numbers in the interval (0, 1). Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... . Similarly, numbers in the second half of the second half start with 0.11... etc. So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be &lt;math&gt;0.11011011011..._2 = \frac{6}{7}_{10}&lt;/math&gt;. So we want the number which is 6/7 of the way through the interval (0, 90) so &lt;math&gt;\theta = \frac{6}{7}\cdot 90 = \frac{540}{7}&lt;/math&gt; and &lt;math&gt;p+q = 540 + 7 = \boxed{547}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> With some simple arithmetic and guess and check, we can set the lower bound and upper bounds for the &quot;first round of &lt;math&gt;3&lt;/math&gt; powers of two&quot;, which are &lt;math&gt;\frac{540}{8} = 67.5&lt;/math&gt; and &lt;math&gt;\frac{630}{8} = 78.75&lt;/math&gt;. Going on to the &quot;second round of &lt;math&gt;3&lt;/math&gt; powers of two, we set the new lower and upper bounds as &lt;math&gt;\frac{360 \times 12.5}{64} = 70.3125&lt;/math&gt; and &lt;math&gt;\frac{360 \times 13.75}{64} = 77.34375&lt;/math&gt; using some guess and check and bashing. Now, it is obvious that the bounds for the &quot;zeroth round of &lt;math&gt;3&lt;/math&gt; powers of two&quot; are &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;90&lt;/math&gt;, and notice that &lt;math&gt;90 - 78.75 = 11.25&lt;/math&gt; and &lt;math&gt;78.75 - 77.34375 = 1.40625&lt;/math&gt; and &lt;math&gt;\frac{11.25}{1.40625} = 8&lt;/math&gt;.<br /> This is obviously a geometric series, so setting &lt;math&gt;11.25&lt;/math&gt; as &lt;math&gt;u&lt;/math&gt;, we obtain &lt;math&gt;90 - (u + \frac{u}{8} + \frac{u}{64} + ...)&lt;/math&gt; = &lt;math&gt;90 - \frac{u}{1-\frac{1}{8}}&lt;/math&gt; = &lt;math&gt;\frac{u}{\frac{7}{8}}&lt;/math&gt; = &lt;math&gt;\frac{45}{4} \times \frac{8}{7}&lt;/math&gt; which simplifies to &lt;math&gt;\frac{90}{7}&lt;/math&gt;. We can now finally subtract &lt;math&gt;\frac{90}{7}&lt;/math&gt; from &lt;math&gt;\frac{630}{7}&lt;/math&gt; and then we get &lt;math&gt;\frac{540}{7}&lt;/math&gt; as the unique angle, so &lt;math&gt;\boxed{547}&lt;/math&gt; is our answer.<br /> -fidgetboss_4000<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=9|num-a=11}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_9&diff=123711 2019 AIME II Problems/Problem 9 2020-06-04T22:53:19Z <p>Emerald block: grammar fixes and rewordings</p> <hr /> <div>==Problem 9==<br /> Call a positive integer &lt;math&gt;n&lt;/math&gt; &lt;math&gt;k&lt;/math&gt;-&lt;i&gt;pretty&lt;/i&gt; if &lt;math&gt;n&lt;/math&gt; has exactly &lt;math&gt;k&lt;/math&gt; positive divisors and &lt;math&gt;n&lt;/math&gt; is divisible by &lt;math&gt;k&lt;/math&gt;. For example, &lt;math&gt;18&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;-pretty. Let &lt;math&gt;S&lt;/math&gt; be the sum of positive integers less than &lt;math&gt;2019&lt;/math&gt; that are &lt;math&gt;20&lt;/math&gt;-pretty. Find &lt;math&gt;\tfrac{S}{20}&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Every 20-pretty integer can be written in form &lt;math&gt;n = 2^a 5^b k&lt;/math&gt;, where &lt;math&gt;a \ge 2&lt;/math&gt;, &lt;math&gt;b \ge 1&lt;/math&gt;, &lt;math&gt;\gcd(k,10) = 1&lt;/math&gt;, and &lt;math&gt;d(n) = 20&lt;/math&gt;, where &lt;math&gt;d(n)&lt;/math&gt; is the number of divisors of &lt;math&gt;n&lt;/math&gt;. Thus, we have &lt;math&gt;20 = (a+1)(b+1)d(k)&lt;/math&gt;, using the fact that the divisor function is multiplicative. As &lt;math&gt;(a+1)(b+1)&lt;/math&gt; must be a divisor of 20, there are not many cases to check.<br /> <br /> If &lt;math&gt;a+1 = 4&lt;/math&gt;, then &lt;math&gt;b+1 = 5&lt;/math&gt;. But this leads to no solutions, as &lt;math&gt;(a,b) = (3,4)&lt;/math&gt; gives &lt;math&gt;2^3 5^4 &gt; 2019&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a+1 = 5&lt;/math&gt;, then &lt;math&gt;b+1 = 2&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt;. The first case gives &lt;math&gt;n = 2^4 \cdot 5^1 \cdot p&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; is a prime other than 2 or 5. Thus we have &lt;math&gt;80p &lt; 2019 \implies p = 3, 7, 11, 13, 17, 19, 23&lt;/math&gt;. The sum of all such &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;80(3+7+11+13+17+19+23) = 7440&lt;/math&gt;. In the second case &lt;math&gt;b+1 = 4&lt;/math&gt; and &lt;math&gt;d(k) = 1&lt;/math&gt;, and there is one solution &lt;math&gt;n = 2^4 \cdot 5^3 = 2000&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a+1 = 10&lt;/math&gt;, then &lt;math&gt;b+1 = 2&lt;/math&gt;, but this gives &lt;math&gt;2^9 \cdot 5^1 &gt; 2019&lt;/math&gt;. No other values for &lt;math&gt;a+1&lt;/math&gt; work.<br /> <br /> Then we have &lt;math&gt;\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}&lt;/math&gt;.<br /> <br /> -scrabbler94<br /> <br /> ==Solution 2==<br /> For &lt;math&gt;n&lt;/math&gt; to have exactly &lt;math&gt;20&lt;/math&gt; positive divisors, &lt;math&gt;n&lt;/math&gt; can only take on certain prime factorization forms: namely, &lt;math&gt;p^{19}, p^9q, p^4q^3, p^4qr&lt;/math&gt;. No number that is a multiple of &lt;math&gt;20&lt;/math&gt; can be expressed in the first form, and the only integer divisible by &lt;math&gt;20&lt;/math&gt; that has the second form is &lt;math&gt;2^{9}5&lt;/math&gt;, which is greater than &lt;math&gt;2019&lt;/math&gt;. <br /> <br /> For the third form, the only &lt;math&gt;20&lt;/math&gt;-pretty numbers are &lt;math&gt;2^45^3=2000&lt;/math&gt; and &lt;math&gt;2^35^4=5000&lt;/math&gt;, and only &lt;math&gt;2000&lt;/math&gt; is small enough. <br /> <br /> For the fourth form, any number of the form &lt;math&gt;2^45r&lt;/math&gt; where &lt;math&gt;r&lt;/math&gt; is a prime other than &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; will satisfy the &lt;math&gt;20&lt;/math&gt;-pretty requirement. Since &lt;math&gt;n=80r&lt;2019&lt;/math&gt;, &lt;math&gt;r\le 25&lt;/math&gt;. Therefore, &lt;math&gt;r&lt;/math&gt; can take on &lt;math&gt;3, 7, 11, 13, 17, 19,&lt;/math&gt; or &lt;math&gt;23&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_8&diff=123703 2019 AIME II Problems/Problem 8 2020-06-04T22:28:35Z <p>Emerald block: new solution</p> <hr /> <div>==Problem 8==<br /> The polynomial &lt;math&gt;f(z)=az^{2018}+bz^{2017}+cz^{2016}&lt;/math&gt; has real coefficients not exceeding &lt;math&gt;2019&lt;/math&gt;, and &lt;math&gt;f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i&lt;/math&gt;. Find the remainder when &lt;math&gt;f(1)&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> We have &lt;math&gt;\frac{1+\sqrt{3}i}{2} = \omega&lt;/math&gt; where &lt;math&gt;\omega = e^{\frac{i\pi}{3}}&lt;/math&gt; is a primitive 6th root of unity. Then we have<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> f(\omega) &amp;= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\<br /> &amp;= a\omega^2 + b\omega + c<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> We wish to find &lt;math&gt;f(1) = a+b+c&lt;/math&gt;. We first look at the real parts. As &lt;math&gt;\text{Re}(\omega^2) = -\frac{1}{2}&lt;/math&gt; and &lt;math&gt;\text{Re}(\omega) = \frac{1}{2}&lt;/math&gt;, we have &lt;math&gt;-\frac{1}{2}a + \frac{1}{2}b + c = 2015 \implies -a+b + 2c = 4030&lt;/math&gt;. Looking at imaginary parts, we have &lt;math&gt;\text{Im}(\omega^2) = \text{Im}(\omega) = \frac{\sqrt{3}}{2}&lt;/math&gt;, so &lt;math&gt;\frac{\sqrt{3}}{2}(a+b) = 2019\sqrt{3} \implies a+b = 4038&lt;/math&gt;. As &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; do not exceed 2019, we must have &lt;math&gt;a = 2019&lt;/math&gt; and &lt;math&gt;b = 2019&lt;/math&gt;. Then &lt;math&gt;c = \frac{4030}{2} = 2015&lt;/math&gt;, so &lt;math&gt;f(1) = 4038 + 2015 = 6053 \implies f(1) \pmod{1000} = \boxed{053}&lt;/math&gt;.<br /> <br /> -scrabbler94<br /> <br /> ==Solution 2==<br /> <br /> Denote &lt;math&gt;\frac{1+\sqrt{3}i}{2}&lt;/math&gt; with &lt;math&gt;\omega&lt;/math&gt;.<br /> <br /> By using the quadratic formula (&lt;math&gt;\frac{-b\pm\sqrt{b^2-4ac}}{2a}&lt;/math&gt;) in reverse, we can find that &lt;math&gt;\omega&lt;/math&gt; is the solution to a quadratic equation of the form &lt;math&gt;ax^2+bx+c=0&lt;/math&gt; such that &lt;math&gt;2a=2&lt;/math&gt;, &lt;math&gt;-b=1&lt;/math&gt;, and &lt;math&gt;b^2-4ac=-3&lt;/math&gt;. This clearly solves to &lt;math&gt;a=1&lt;/math&gt;, &lt;math&gt;b=-1&lt;/math&gt;, and &lt;math&gt;c=1&lt;/math&gt;, so &lt;math&gt;\omega&lt;/math&gt; solves &lt;math&gt;x^2-x+1=0&lt;/math&gt;.<br /> <br /> Multiplying &lt;math&gt;x^2-x+1=0&lt;/math&gt; by &lt;math&gt;x+1&lt;/math&gt; on both sides yields &lt;math&gt;x^3+1=0&lt;/math&gt;. Muliplying this by &lt;math&gt;x^3-1&lt;/math&gt; on both sides yields &lt;math&gt;x^6-1=0&lt;/math&gt;, or &lt;math&gt;x^6=1&lt;/math&gt;. This means that &lt;math&gt;\omega^6=1&lt;/math&gt;.<br /> <br /> We can use this to simplify the equation &lt;math&gt;a\omega^{2018}+b\omega^{2017}+c\omega^{2016}=f(\omega)=2015+2019\sqrt{3}i&lt;/math&gt; to &lt;math&gt;a\omega^2+b\omega+c=2015+2019\sqrt{3}i.&lt;/math&gt;<br /> <br /> As in Solution 1, we use the values &lt;math&gt;\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i&lt;/math&gt; and &lt;math&gt;\omega^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i&lt;/math&gt; to find that &lt;math&gt;-\frac{1}{2}a+\frac{1}{2}b+c=2015&lt;/math&gt; and &lt;math&gt;\frac{\sqrt{3}}{2}a+\frac{\sqrt{3}}{2}b=2019\sqrt{3} \implies a+b=4038.&lt;/math&gt; Since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; can exceed &lt;math&gt;2019&lt;/math&gt;, they must both be equal to &lt;math&gt;2019&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are equal, they cancel out in the first equation, resulting in &lt;math&gt;c=2015&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;f(1)=a+b+c=2019+2019+2015=6053&lt;/math&gt;, and &lt;math&gt;6053\bmod1000=\boxed{053}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_5&diff=123620 2019 AIME II Problems/Problem 5 2020-06-04T05:03:21Z <p>Emerald block: reworked the solution due to poor wording and lack of explanation in original</p> <hr /> <div>==Problem==<br /> Four ambassadors and one advisor for each of them are to be seated at a round table with &lt;math&gt;12&lt;/math&gt; chairs numbered in order &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt;. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are &lt;math&gt;N&lt;/math&gt; ways for the &lt;math&gt;8&lt;/math&gt; people to be seated at the table under these conditions. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> ==Solution 1==<br /> Let us first consider the &lt;math&gt;4&lt;/math&gt; ambassadors and the &lt;math&gt;6&lt;/math&gt; even-numbered chairs. If we consider only their relative positions, they can sit in one of &lt;math&gt;3&lt;/math&gt; distinct ways: Such that the &lt;math&gt;2&lt;/math&gt; empty even-numbered chairs are adjacent, such that they are separated by one occupied even-numbered chair, and such that they are opposite each other. For each way, there are &lt;math&gt;4!=24&lt;/math&gt; ways to assign the &lt;math&gt;4&lt;/math&gt; ambassadors to the &lt;math&gt;4&lt;/math&gt; selected seats.<br /> <br /> In the first way, there are &lt;math&gt;6&lt;/math&gt; distinct orientations. The &lt;math&gt;4&lt;/math&gt; advisors can be placed in any of the &lt;math&gt;5&lt;/math&gt; odd-numbered chairs adjacent to the ambassadors, and for each placement, there is exactly one way to assign them to the ambassadors. This means that there are &lt;math&gt;24\cdot6\cdot\binom{5}{4}=720&lt;/math&gt; total seating arrangements for this way.<br /> <br /> In the second way, there are &lt;math&gt;6&lt;/math&gt; distinct orientations. &lt;math&gt;3&lt;/math&gt; advisors can be placed in any of the &lt;math&gt;4&lt;/math&gt; chairs adjacent to the &quot;chunk&quot; of &lt;math&gt;3&lt;/math&gt; ambassadors, and &lt;math&gt;1&lt;/math&gt; advisor can be placed in either of the &lt;math&gt;2&lt;/math&gt; chairs adjacent to the &quot;lonely&quot; ambassador. Once again, for each placement, there is exactly one way to assign the advisors to the ambassadors. This means that there are &lt;math&gt;24\cdot6\cdot\binom{4}{3}\cdot\binom{2}{1}=1152&lt;/math&gt; total seating arrangements for this way.<br /> <br /> In the third way, there are &lt;math&gt;3&lt;/math&gt; distinct orientations. For both &quot;chunks&quot; of &lt;math&gt;2&lt;/math&gt; ambassadors, &lt;math&gt;2&lt;/math&gt; advisors can be placed in any of the &lt;math&gt;3&lt;/math&gt; chairs adjacent to them, and once again, there is exactly one way to assign them for each placement. This means that there are &lt;math&gt;24\cdot3\cdot\binom{3}{2}\cdot\binom{3}{2}=648&lt;/math&gt; total seating arrangements for this way.<br /> <br /> Totalling up the arrangements, there are &lt;math&gt;720+1152+648=2520&lt;/math&gt; total ways to seat the people, and the remainer when &lt;math&gt;2520&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;\boxed{520}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 2==<br /> [[File:Ambassador_Table.png|200px]]<br /> <br /> In the diagram, the seats are numbered 1...12. Rather than picking seats for each person, however, each ambassador/assistant team picks a gap between the seats (A...L) and the ambassador sits in the even seat while the assistant sits in the odd seat. For example, if team 1 picks gap C then Ambassador 1 will sit in seat 2 while assistant 1 will sit in seat 3. No two teams can pick adjacent gaps. For example, if team 1 chooses gap C then team 2 cannot pick gaps B or D. In the diagram, the teams have picked gaps C, F, H and J. Note that the gap-gaps - distances between the chosen gaps - (in the diagram, 2, 1, 1, 4) must sum to 8. So, to get the number of seatings, we:<br /> #Choose a gap for team 1 (&lt;math&gt;12&lt;/math&gt; options)<br /> #Choose 3 other gaps around the table with positive gap-gaps. The number of ways to do this is the number of ways to partition 8 with 4 positive integers. This is the same as partitioning 4 with 4 non-negative integers, and using stars-and-bars, this is &lt;math&gt;\dbinom{4+4-1}{4-1}=\dbinom{7}{3}=35&lt;/math&gt;<br /> #Place the other 3 teams in the chosen gaps (&lt;math&gt;6&lt;/math&gt; permutations)<br /> So the total is &lt;math&gt;12\cdot35\cdot6=2520&lt;/math&gt;<br /> And the remainder is &lt;math&gt;\boxed{520}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> There are &lt;math&gt;360\cdot16&lt;/math&gt; total ways to let everyone sit. However this may lead to advisors sitting in the same chair, leading to awkward situations. So we find how many ways this happens. There are 6 ways to choose which advisors end up sitting together, times 12 ways to find neighboring even seats and sitting down, 12 ways for the rest of the ambassador to sit, and 4 ways for their advisors to sit to get 3456 ways for this to happen. However we overcounted the case when two pairs of advisors run out of room to sit, where there are &lt;math&gt;6\cdot9\cdot4=216&lt;/math&gt; ways to happen. So 5760-3456+216=&lt;math&gt;2\boxed{520}&lt;/math&gt;. <br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_4&diff=123619 2019 AIME II Problems/Problem 4 2020-06-04T03:25:15Z <p>Emerald block: fixed small errors in headers</p> <hr /> <div>==Problem 4==<br /> A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are &lt;math&gt;6^4 = 1296&lt;/math&gt; outcomes).<br /> <br /> Case 1 (easy): Four 5's are rolled. This has probability &lt;math&gt;\frac{1}{6^4}&lt;/math&gt; of occurring.<br /> <br /> Case 2: Two 5's are rolled.<br /> <br /> Case 3: No 5's are rolled.<br /> <br /> To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For &lt;math&gt;n \ge 1&lt;/math&gt;, let &lt;math&gt;a_n&lt;/math&gt; equal the number of outcomes after rolling the die &lt;math&gt;n&lt;/math&gt; times, with the property that the product is a square. Thus, &lt;math&gt;a_1 = 2&lt;/math&gt; as 1 and 4 are the only possibilities.<br /> <br /> To find &lt;math&gt;a_{n+1}&lt;/math&gt; given &lt;math&gt;a_n&lt;/math&gt; (where &lt;math&gt;n \ge 1&lt;/math&gt;), we observe that if the first &lt;math&gt;n&lt;/math&gt; rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives &lt;math&gt;2a_n&lt;/math&gt; outcomes. Otherwise, the first &lt;math&gt;n&lt;/math&gt; rolls do not multiply to a perfect square (&lt;math&gt;5^n - a_n&lt;/math&gt; outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first &lt;math&gt;n&lt;/math&gt; rolls is &lt;math&gt;2^x 3^y&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are not both even, then we observe that if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both odd, then the last roll must be 6; if only &lt;math&gt;x&lt;/math&gt; is odd, the last roll must be 2, and if only &lt;math&gt;y&lt;/math&gt; is odd, the last roll must be 3. Thus, we have &lt;math&gt;5^n - a_n&lt;/math&gt; outcomes in this case, and &lt;math&gt;a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n&lt;/math&gt;.<br /> <br /> Computing &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, &lt;math&gt;a_4&lt;/math&gt; gives &lt;math&gt;a_2 = 7&lt;/math&gt;, &lt;math&gt;a_3 = 32&lt;/math&gt;, and &lt;math&gt;a_4 = 157&lt;/math&gt;. Thus for Case 3, there are 157 outcomes. For case 2, we multiply by &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; to distribute the two 5's among four rolls. Thus the probability is<br /> <br /> &lt;cmath&gt; \frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187} &lt;/cmath&gt;<br /> <br /> -scrabbler94<br /> <br /> ==Solution 2==<br /> <br /> We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square). <br /> <br /> Probability of rolling 4 duds: &lt;math&gt; \left(\frac{1}{3}\right)^4 &lt;/math&gt;<br /> <br /> Probability of rolling 3 duds: &lt;math&gt; 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} &lt;/math&gt;<br /> <br /> Probability of rolling 2 duds: &lt;math&gt; 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 &lt;/math&gt;<br /> <br /> Probability of rolling 1 dud: &lt;math&gt; 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 &lt;/math&gt;<br /> <br /> Probability of rolling 0 duds: &lt;math&gt; \left(\frac{2}{3}\right)^4 &lt;/math&gt;<br /> <br /> Now we will find the probability of a square product given we have rolled each amount of duds<br /> <br /> Probability of getting a square product given 4 duds: 1<br /> <br /> Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)<br /> <br /> Probability of getting a square product given 2 duds: &lt;math&gt; \frac{1}{4}&lt;/math&gt; (as long as our two non-duds are the same, our product will be square)<br /> <br /> Probability of getting a square product given 1 dud: &lt;math&gt;\frac{3!}{4^3}&lt;/math&gt; = &lt;math&gt;\frac{3}{32}&lt;/math&gt; (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of &lt;math&gt;4^3&lt;/math&gt; ways to roll 3 non-duds).<br /> <br /> Probability of getting a square product given 0 duds: &lt;math&gt; \frac{40}{4^4} &lt;/math&gt;= &lt;math&gt;\frac{5}{32} &lt;/math&gt; (We can have any two non-duds twice. For example, 2,2,5,5. There are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; ways of choosing which two non-duds to use and &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).<br /> <br /> We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.<br /> &lt;cmath&gt; \left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}. &lt;/cmath&gt;<br /> <br /> &lt;math&gt;25+162&lt;/math&gt; = &lt;math&gt;\boxed{187}&lt;/math&gt; <br /> <br /> -dnaidu (silverlizard)<br /> <br /> ==Solution 3==<br /> Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:<br /> <br /> If there are four 1/4's, then there are &lt;math&gt;2^4=16&lt;/math&gt; combinations. <br /> If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. <br /> If there are two 1/4's, there are &lt;math&gt;2^2=4&lt;/math&gt; ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and &lt;math&gt;\frac{4!}{2!2!}=6&lt;/math&gt; ways to arrange, so there are &lt;math&gt;4\cdot 4\cdot 6=96&lt;/math&gt; combinations for this case. <br /> If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are &lt;math&gt;4!&lt;/math&gt; ways to order, meaning there are &lt;math&gt;2\cdot 4!=48&lt;/math&gt; combinations for this case. <br /> Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there &lt;math&gt;\binom{4}{2}&lt;/math&gt; to choose the numbers and &lt;math&gt;\frac{4!}{2!2!}=6&lt;/math&gt; ways to arrange them, so &lt;math&gt;6\cdot 6=36&lt;/math&gt;. If all four numbers are the same there are &lt;math&gt;4&lt;/math&gt; combinations, so there are &lt;math&gt;4+36=40&lt;/math&gt; combinations for this case.<br /> <br /> Hence there are &lt;math&gt;16+0+96+48+40=200&lt;/math&gt; combinations where the product of the dice is a perfect square, and there are &lt;math&gt;6^4=1296&lt;/math&gt; total combinations, so the desired probability is &lt;math&gt;\frac{200}{1296}=\frac{25}{162}&lt;/math&gt;, yielding an answer of &lt;math&gt;25+162=\boxed{187}&lt;/math&gt;.<br /> <br /> -Stormersyle<br /> <br /> ==Solution 4 (Casework)==<br /> Another way to solve this problem is to do casework on all the perfect squares from &lt;math&gt;1^2&lt;/math&gt; to &lt;math&gt;36^2&lt;/math&gt;, and how many ways they can be ordered<br /> &lt;math&gt;1^2&lt;/math&gt;- &lt;math&gt;1,1,1,1&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;2^2&lt;/math&gt;- &lt;math&gt;4,1,1,1&lt;/math&gt; or &lt;math&gt;2,2,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+4=10&lt;/math&gt; ways.<br /> &lt;math&gt;3^2&lt;/math&gt;- &lt;math&gt;3,3,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;4^2&lt;/math&gt;- &lt;math&gt;4,4,1,1&lt;/math&gt;, &lt;math&gt;2,2,2,2&lt;/math&gt;, or &lt;math&gt;2,2,4,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+1+12=19&lt;/math&gt; ways.<br /> &lt;math&gt;5^2&lt;/math&gt;- &lt;math&gt;5,5,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;6^2&lt;/math&gt;- &lt;math&gt;6,6,1,1&lt;/math&gt;, &lt;math&gt;1,2,3,6&lt;/math&gt;, &lt;math&gt;2,3,2,3&lt;/math&gt;, &lt;math&gt;3,3,4,1&lt;/math&gt;- &lt;math&gt;2*\binom{4}{2}+4!+12=48&lt;/math&gt; ways.<br /> &lt;math&gt;7^2&lt;/math&gt;- Since there is a prime greater than 6 in its prime factorization there are &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;8^2&lt;/math&gt;- &lt;math&gt;4,4,4,1&lt;/math&gt; or &lt;math&gt;2,4,2,4&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+4=10&lt;/math&gt; ways.<br /> &lt;math&gt;9^2&lt;/math&gt;- &lt;math&gt;3,3,3,3&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;10^2&lt;/math&gt;- &lt;math&gt;2,2,5,5&lt;/math&gt; or &lt;math&gt;1,4,5,5&lt;/math&gt;- &lt;math&gt;6+12=18&lt;/math&gt; ways.<br /> &lt;math&gt;11^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways for the same reason as &lt;math&gt;7^2&lt;/math&gt;.<br /> &lt;math&gt;12^2&lt;/math&gt;- &lt;math&gt;6,6,2,2&lt;/math&gt;, &lt;math&gt;4,4,3,3&lt;/math&gt;, &lt;math&gt;2,3,4,6&lt;/math&gt;, or &lt;math&gt;1,4,6,6&lt;/math&gt;- &lt;math&gt;2*\binom{4}{2}+4!+12=48&lt;/math&gt; ways.<br /> &lt;math&gt;13^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;14^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;15^2&lt;/math&gt;- &lt;math&gt;3,3,5,5&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;16^2&lt;/math&gt;- &lt;math&gt;4,4,4,4&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;17^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;18^2&lt;/math&gt;- &lt;math&gt;3,3,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;19^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;20^2&lt;/math&gt;- &lt;math&gt;4,4,5,5&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;21^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;22^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;23^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;24^2&lt;/math&gt;-&lt;math&gt;4,4,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;25^2&lt;/math&gt;- &lt;math&gt;5,5,5,5&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;26^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;27^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;28^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;29^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;30^2&lt;/math&gt;- &lt;math&gt;5,5,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways.<br /> &lt;math&gt;31^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;32^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;33^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;34^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;35^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;36^2&lt;/math&gt;- &lt;math&gt;6,6,6,6&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> <br /> There are &lt;math&gt;6^4=1296&lt;/math&gt; ways that the dice can land. Summing up the ways, it is easy to see that there are &lt;math&gt;200&lt;/math&gt; ways. <br /> This results in a probability of &lt;math&gt;\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}&lt;/math&gt;<br /> -superninja2000<br /> <br /> ==Solution 5 (Recursion)==<br /> We can do recursion on the number of rolls to find the number of ways we can get &lt;math&gt;4&lt;/math&gt; rolls to multiply to a square.<br /> <br /> After &lt;math&gt;n&lt;/math&gt; rolls, let us say that the product is &lt;math&gt;p = 2^a3^b5^c&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Define the following:}&lt;/math&gt;<br /> <br /> &lt;math&gt;A_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a&lt;/math&gt; is odd, and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are even<br /> <br /> &lt;math&gt;B_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;b&lt;/math&gt; is odd, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are even<br /> <br /> &lt;math&gt;C_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;c&lt;/math&gt; is odd, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are even<br /> <br /> &lt;math&gt;D_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;c&lt;/math&gt; is even, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;E_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;b&lt;/math&gt; is even, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;F_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a&lt;/math&gt; is even, and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;G_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt; c&lt;/math&gt; are all odd<br /> <br /> &lt;math&gt;S_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt; c&lt;/math&gt; are all even (square!)<br /> <br /> &lt;math&gt;\text{We have the following equations after considering the possible values of the nth roll:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{We have the following values after considering the possible values of the 1st roll:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{After applying recursion twice, we get:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38&lt;/cmath&gt;<br /> <br /> Finally, we have&lt;math&gt;S_4 = 200&lt;/math&gt;, &lt;math&gt;\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}&lt;/math&gt;meaning our answer is &lt;math&gt;\boxed{187}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Consider all the distinct &quot;fundamental&quot; groups of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt; whose product is a perfect square. A &quot;fundamental&quot; group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, &lt;math&gt;\{2,2\}&lt;/math&gt; is a fundamental group, while &lt;math&gt;\{3,3,4\}&lt;/math&gt; is not, because it can be broken up into &lt;math&gt;\{3,3\}&lt;/math&gt; and &lt;math&gt;\{4\}&lt;/math&gt;.<br /> <br /> &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; are already perfect squares, so they each form a &quot;fundamental&quot; group and cannot belong in any other group. Pairs of the other &lt;math&gt;4&lt;/math&gt; numbers (&lt;math&gt;\{2,2\}&lt;/math&gt;,&lt;math&gt;\{3,3\}&lt;/math&gt;, etc. ) form &quot;fundamental&quot; groups as well. The last &quot;fundamental&quot; group is &lt;math&gt;\{2,3,6\}&lt;/math&gt;. It can be easily seen that no more groups exist.<br /> <br /> Thus, we have the &quot;fundamental&quot; groups &lt;math&gt;\{1\}&lt;/math&gt;,&lt;math&gt;\{4\}&lt;/math&gt;,&lt;math&gt;\{2,2\}&lt;/math&gt;,&lt;math&gt;\{3,3\}&lt;/math&gt;,&lt;math&gt;\{5,5\}&lt;/math&gt;,&lt;math&gt;\{6,6\}&lt;/math&gt;, and &lt;math&gt;\{2,3,6\}&lt;/math&gt;.<br /> <br /> We now consider the ways to use these groups to form a sequence of &lt;math&gt;4&lt;/math&gt; numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; and fill in any remaining spots with &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s. We can do this in one of &lt;math&gt;5&lt;/math&gt; ways: Using only &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, using one group of size &lt;math&gt;2&lt;/math&gt;, using one group of size &lt;math&gt;3&lt;/math&gt;, using two different groups of size &lt;math&gt;2&lt;/math&gt;, and using the same group of size &lt;math&gt;2&lt;/math&gt; twice.<br /> <br /> If we only use &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, each of the &lt;math&gt;4&lt;/math&gt; slots can be filled with one of the &lt;math&gt;2&lt;/math&gt; numbers, so there are &lt;math&gt;2^4=16&lt;/math&gt; possibilities.<br /> <br /> If we use one group of size &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;4&lt;/math&gt; options for the group to use, &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways to place the two numbers (since they are identical), and &lt;math&gt;2^2&lt;/math&gt; ways to fill in the remaining slots with &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, so there are &lt;math&gt;4\cdot\binom{4}{2}\cdot2^2=96&lt;/math&gt; possibilities.<br /> <br /> If we use one group of size &lt;math&gt;3&lt;/math&gt;, there is only &lt;math&gt;1&lt;/math&gt; option for the group to use, &lt;math&gt;4\cdot3\cdot2&lt;/math&gt; ways to place the three numbers (since they are distinct), and &lt;math&gt;2&lt;/math&gt; ways to fill in the remaining slot, so there are &lt;math&gt;4\cdot3\cdot2\cdot2=48&lt;/math&gt; possibilities.<br /> <br /> If we use two different groups of size &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;\binom{4}{2}&lt;/math&gt; options for the groups to use and &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways to place the four numbers (since there are &lt;math&gt;2&lt;/math&gt; groups of identical numbers, and one group's placement uniquely determines the other's), so there are &lt;math&gt;\binom{4}{2}\cdot\binom{4}{2}=36&lt;/math&gt; possibilities.<br /> <br /> If we use the same group of size &lt;math&gt;2&lt;/math&gt; twice, there are &lt;math&gt;4&lt;/math&gt; options for the group to use and &lt;math&gt;1&lt;/math&gt; way to place the four numbers (since they are all identical), so there are &lt;math&gt;4=4&lt;/math&gt; possibilities.<br /> <br /> This gives us a total of &lt;math&gt;16+96+48+36+4=200&lt;/math&gt; possibilities, and since there are &lt;math&gt;6^4=1296&lt;/math&gt; total sequences that can be rolled, the probability is equal to &lt;math&gt;\frac{200}{1296}=\frac{25}{162}&lt;/math&gt;, so the answer is &lt;math&gt;25+162=\boxed{187}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_4&diff=123618 2019 AIME II Problems/Problem 4 2020-06-04T03:23:23Z <p>Emerald block: added credits to solution</p> <hr /> <div>==Problem 4==<br /> A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are &lt;math&gt;6^4 = 1296&lt;/math&gt; outcomes).<br /> <br /> Case 1 (easy): Four 5's are rolled. This has probability &lt;math&gt;\frac{1}{6^4}&lt;/math&gt; of occurring.<br /> <br /> Case 2: Two 5's are rolled.<br /> <br /> Case 3: No 5's are rolled.<br /> <br /> To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For &lt;math&gt;n \ge 1&lt;/math&gt;, let &lt;math&gt;a_n&lt;/math&gt; equal the number of outcomes after rolling the die &lt;math&gt;n&lt;/math&gt; times, with the property that the product is a square. Thus, &lt;math&gt;a_1 = 2&lt;/math&gt; as 1 and 4 are the only possibilities.<br /> <br /> To find &lt;math&gt;a_{n+1}&lt;/math&gt; given &lt;math&gt;a_n&lt;/math&gt; (where &lt;math&gt;n \ge 1&lt;/math&gt;), we observe that if the first &lt;math&gt;n&lt;/math&gt; rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives &lt;math&gt;2a_n&lt;/math&gt; outcomes. Otherwise, the first &lt;math&gt;n&lt;/math&gt; rolls do not multiply to a perfect square (&lt;math&gt;5^n - a_n&lt;/math&gt; outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first &lt;math&gt;n&lt;/math&gt; rolls is &lt;math&gt;2^x 3^y&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are not both even, then we observe that if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both odd, then the last roll must be 6; if only &lt;math&gt;x&lt;/math&gt; is odd, the last roll must be 2, and if only &lt;math&gt;y&lt;/math&gt; is odd, the last roll must be 3. Thus, we have &lt;math&gt;5^n - a_n&lt;/math&gt; outcomes in this case, and &lt;math&gt;a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n&lt;/math&gt;.<br /> <br /> Computing &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, &lt;math&gt;a_4&lt;/math&gt; gives &lt;math&gt;a_2 = 7&lt;/math&gt;, &lt;math&gt;a_3 = 32&lt;/math&gt;, and &lt;math&gt;a_4 = 157&lt;/math&gt;. Thus for Case 3, there are 157 outcomes. For case 2, we multiply by &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; to distribute the two 5's among four rolls. Thus the probability is<br /> <br /> &lt;cmath&gt; \frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187} &lt;/cmath&gt;<br /> <br /> -scrabbler94<br /> <br /> ==Solution 2==<br /> <br /> We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square). <br /> <br /> Probability of rolling 4 duds: &lt;math&gt; \left(\frac{1}{3}\right)^4 &lt;/math&gt;<br /> <br /> Probability of rolling 3 duds: &lt;math&gt; 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} &lt;/math&gt;<br /> <br /> Probability of rolling 2 duds: &lt;math&gt; 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 &lt;/math&gt;<br /> <br /> Probability of rolling 1 dud: &lt;math&gt; 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 &lt;/math&gt;<br /> <br /> Probability of rolling 0 duds: &lt;math&gt; \left(\frac{2}{3}\right)^4 &lt;/math&gt;<br /> <br /> Now we will find the probability of a square product given we have rolled each amount of duds<br /> <br /> Probability of getting a square product given 4 duds: 1<br /> <br /> Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)<br /> <br /> Probability of getting a square product given 2 duds: &lt;math&gt; \frac{1}{4}&lt;/math&gt; (as long as our two non-duds are the same, our product will be square)<br /> <br /> Probability of getting a square product given 1 dud: &lt;math&gt;\frac{3!}{4^3}&lt;/math&gt; = &lt;math&gt;\frac{3}{32}&lt;/math&gt; (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of &lt;math&gt;4^3&lt;/math&gt; ways to roll 3 non-duds).<br /> <br /> Probability of getting a square product given 0 duds: &lt;math&gt; \frac{40}{4^4} &lt;/math&gt;= &lt;math&gt;\frac{5}{32} &lt;/math&gt; (We can have any two non-duds twice. For example, 2,2,5,5. There are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; ways of choosing which two non-duds to use and &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).<br /> <br /> We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.<br /> &lt;cmath&gt; \left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}. &lt;/cmath&gt;<br /> <br /> &lt;math&gt;25+162&lt;/math&gt; = &lt;math&gt;\boxed{187}&lt;/math&gt; <br /> <br /> -dnaidu (silverlizard)<br /> <br /> ==Solution 3==<br /> Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:<br /> <br /> If there are four 1/4's, then there are &lt;math&gt;2^4=16&lt;/math&gt; combinations. <br /> If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. <br /> If there are two 1/4's, there are &lt;math&gt;2^2=4&lt;/math&gt; ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and &lt;math&gt;\frac{4!}{2!2!}=6&lt;/math&gt; ways to arrange, so there are &lt;math&gt;4\cdot 4\cdot 6=96&lt;/math&gt; combinations for this case. <br /> If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are &lt;math&gt;4!&lt;/math&gt; ways to order, meaning there are &lt;math&gt;2\cdot 4!=48&lt;/math&gt; combinations for this case. <br /> Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there &lt;math&gt;\binom{4}{2}&lt;/math&gt; to choose the numbers and &lt;math&gt;\frac{4!}{2!2!}=6&lt;/math&gt; ways to arrange them, so &lt;math&gt;6\cdot 6=36&lt;/math&gt;. If all four numbers are the same there are &lt;math&gt;4&lt;/math&gt; combinations, so there are &lt;math&gt;4+36=40&lt;/math&gt; combinations for this case.<br /> <br /> Hence there are &lt;math&gt;16+0+96+48+40=200&lt;/math&gt; combinations where the product of the dice is a perfect square, and there are &lt;math&gt;6^4=1296&lt;/math&gt; total combinations, so the desired probability is &lt;math&gt;\frac{200}{1296}=\frac{25}{162}&lt;/math&gt;, yielding an answer of &lt;math&gt;25+162=\boxed{187}&lt;/math&gt;.<br /> <br /> -Stormersyle<br /> <br /> ==Solution 4(Casework)==<br /> Another way to solve this problem is to do casework on all the perfect squares from &lt;math&gt;1^2&lt;/math&gt; to &lt;math&gt;36^2&lt;/math&gt;, and how many ways they can be ordered<br /> &lt;math&gt;1^2&lt;/math&gt;- &lt;math&gt;1,1,1,1&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;2^2&lt;/math&gt;- &lt;math&gt;4,1,1,1&lt;/math&gt; or &lt;math&gt;2,2,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+4=10&lt;/math&gt; ways.<br /> &lt;math&gt;3^2&lt;/math&gt;- &lt;math&gt;3,3,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;4^2&lt;/math&gt;- &lt;math&gt;4,4,1,1&lt;/math&gt;, &lt;math&gt;2,2,2,2&lt;/math&gt;, or &lt;math&gt;2,2,4,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+1+12=19&lt;/math&gt; ways.<br /> &lt;math&gt;5^2&lt;/math&gt;- &lt;math&gt;5,5,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;6^2&lt;/math&gt;- &lt;math&gt;6,6,1,1&lt;/math&gt;, &lt;math&gt;1,2,3,6&lt;/math&gt;, &lt;math&gt;2,3,2,3&lt;/math&gt;, &lt;math&gt;3,3,4,1&lt;/math&gt;- &lt;math&gt;2*\binom{4}{2}+4!+12=48&lt;/math&gt; ways.<br /> &lt;math&gt;7^2&lt;/math&gt;- Since there is a prime greater than 6 in its prime factorization there are &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;8^2&lt;/math&gt;- &lt;math&gt;4,4,4,1&lt;/math&gt; or &lt;math&gt;2,4,2,4&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+4=10&lt;/math&gt; ways.<br /> &lt;math&gt;9^2&lt;/math&gt;- &lt;math&gt;3,3,3,3&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;10^2&lt;/math&gt;- &lt;math&gt;2,2,5,5&lt;/math&gt; or &lt;math&gt;1,4,5,5&lt;/math&gt;- &lt;math&gt;6+12=18&lt;/math&gt; ways.<br /> &lt;math&gt;11^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways for the same reason as &lt;math&gt;7^2&lt;/math&gt;.<br /> &lt;math&gt;12^2&lt;/math&gt;- &lt;math&gt;6,6,2,2&lt;/math&gt;, &lt;math&gt;4,4,3,3&lt;/math&gt;, &lt;math&gt;2,3,4,6&lt;/math&gt;, or &lt;math&gt;1,4,6,6&lt;/math&gt;- &lt;math&gt;2*\binom{4}{2}+4!+12=48&lt;/math&gt; ways.<br /> &lt;math&gt;13^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;14^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;15^2&lt;/math&gt;- &lt;math&gt;3,3,5,5&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;16^2&lt;/math&gt;- &lt;math&gt;4,4,4,4&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;17^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;18^2&lt;/math&gt;- &lt;math&gt;3,3,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;19^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;20^2&lt;/math&gt;- &lt;math&gt;4,4,5,5&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;21^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;22^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;23^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;24^2&lt;/math&gt;-&lt;math&gt;4,4,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;25^2&lt;/math&gt;- &lt;math&gt;5,5,5,5&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;26^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;27^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;28^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;29^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;30^2&lt;/math&gt;- &lt;math&gt;5,5,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways.<br /> &lt;math&gt;31^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;32^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;33^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;34^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;35^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;36^2&lt;/math&gt;- &lt;math&gt;6,6,6,6&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> <br /> There are &lt;math&gt;6^4=1296&lt;/math&gt; ways that the dice can land. Summing up the ways, it is easy to see that there are &lt;math&gt;200&lt;/math&gt; ways. <br /> This results in a probability of &lt;math&gt;\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}&lt;/math&gt;<br /> -superninja2000<br /> <br /> ==Solution 5 (Recursion)==<br /> We can do recursion on the number of rolls to find the number of ways we can get &lt;math&gt;4&lt;/math&gt; rolls to multiply to a square.<br /> <br /> After &lt;math&gt;n&lt;/math&gt; rolls, let us say that the product is &lt;math&gt;p = 2^a3^b5^c&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Define the following:}&lt;/math&gt;<br /> <br /> &lt;math&gt;A_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a&lt;/math&gt; is odd, and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are even<br /> <br /> &lt;math&gt;B_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;b&lt;/math&gt; is odd, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are even<br /> <br /> &lt;math&gt;C_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;c&lt;/math&gt; is odd, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are even<br /> <br /> &lt;math&gt;D_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;c&lt;/math&gt; is even, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;E_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;b&lt;/math&gt; is even, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;F_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a&lt;/math&gt; is even, and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;G_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt; c&lt;/math&gt; are all odd<br /> <br /> &lt;math&gt;S_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt; c&lt;/math&gt; are all even (square!)<br /> <br /> &lt;math&gt;\text{We have the following equations after considering the possible values of the nth roll:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{We have the following values after considering the possible values of the 1st roll:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{After applying recursion twice, we get:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38&lt;/cmath&gt;<br /> <br /> Finally, we have&lt;math&gt;S_4 = 200&lt;/math&gt;, &lt;math&gt;\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}&lt;/math&gt;meaning our answer is &lt;math&gt;\boxed{187}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Consider all the distinct &quot;fundamental&quot; groups of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt; whose product is a perfect square. A &quot;fundamental&quot; group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, &lt;math&gt;\{2,2\}&lt;/math&gt; is a fundamental group, while &lt;math&gt;\{3,3,4\}&lt;/math&gt; is not, because it can be broken up into &lt;math&gt;\{3,3\}&lt;/math&gt; and &lt;math&gt;\{4\}&lt;/math&gt;.<br /> <br /> &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; are already perfect squares, so they each form a &quot;fundamental&quot; group and cannot belong in any other group. Pairs of the other &lt;math&gt;4&lt;/math&gt; numbers (&lt;math&gt;\{2,2\}&lt;/math&gt;,&lt;math&gt;\{3,3\}&lt;/math&gt;, etc. ) form &quot;fundamental&quot; groups as well. The last &quot;fundamental&quot; group is &lt;math&gt;\{2,3,6\}&lt;/math&gt;. It can be easily seen that no more groups exist.<br /> <br /> Thus, we have the &quot;fundamental&quot; groups &lt;math&gt;\{1\}&lt;/math&gt;,&lt;math&gt;\{4\}&lt;/math&gt;,&lt;math&gt;\{2,2\}&lt;/math&gt;,&lt;math&gt;\{3,3\}&lt;/math&gt;,&lt;math&gt;\{5,5\}&lt;/math&gt;,&lt;math&gt;\{6,6\}&lt;/math&gt;, and &lt;math&gt;\{2,3,6\}&lt;/math&gt;.<br /> <br /> We now consider the ways to use these groups to form a sequence of &lt;math&gt;4&lt;/math&gt; numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; and fill in any remaining spots with &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s. We can do this in one of &lt;math&gt;5&lt;/math&gt; ways: Using only &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, using one group of size &lt;math&gt;2&lt;/math&gt;, using one group of size &lt;math&gt;3&lt;/math&gt;, using two different groups of size &lt;math&gt;2&lt;/math&gt;, and using the same group of size &lt;math&gt;2&lt;/math&gt; twice.<br /> <br /> If we only use &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, each of the &lt;math&gt;4&lt;/math&gt; slots can be filled with one of the &lt;math&gt;2&lt;/math&gt; numbers, so there are &lt;math&gt;2^4=16&lt;/math&gt; possibilities.<br /> <br /> If we use one group of size &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;4&lt;/math&gt; options for the group to use, &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways to place the two numbers (since they are identical), and &lt;math&gt;2^2&lt;/math&gt; ways to fill in the remaining slots with &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, so there are &lt;math&gt;4\cdot\binom{4}{2}\cdot2^2=96&lt;/math&gt; possibilities.<br /> <br /> If we use one group of size &lt;math&gt;3&lt;/math&gt;, there is only &lt;math&gt;1&lt;/math&gt; option for the group to use, &lt;math&gt;4\cdot3\cdot2&lt;/math&gt; ways to place the three numbers (since they are distinct), and &lt;math&gt;2&lt;/math&gt; ways to fill in the remaining slot, so there are &lt;math&gt;4\cdot3\cdot2\cdot2=48&lt;/math&gt; possibilities.<br /> <br /> If we use two different groups of size &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;\binom{4}{2}&lt;/math&gt; options for the groups to use and &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways to place the four numbers (since there are &lt;math&gt;2&lt;/math&gt; groups of identical numbers, and one group's placement uniquely determines the other's), so there are &lt;math&gt;\binom{4}{2}\cdot\binom{4}{2}=36&lt;/math&gt; possibilities.<br /> <br /> If we use the same group of size &lt;math&gt;2&lt;/math&gt; twice, there are &lt;math&gt;4&lt;/math&gt; options for the group to use and &lt;math&gt;1&lt;/math&gt; way to place the four numbers (since they are all identical), so there are &lt;math&gt;4=4&lt;/math&gt; possibilities.<br /> <br /> This gives us a total of &lt;math&gt;16+96+48+36+4=200&lt;/math&gt; possibilities, and since there are &lt;math&gt;6^4=1296&lt;/math&gt; total sequences that can be rolled, the probability is equal to &lt;math&gt;\frac{200}{1296}=\frac{25}{162}&lt;/math&gt;, so the answer is &lt;math&gt;25+162=\boxed{187}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_4&diff=123617 2019 AIME II Problems/Problem 4 2020-06-04T03:21:30Z <p>Emerald block: new solution</p> <hr /> <div>==Problem 4==<br /> A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is &lt;math&gt;\tfrac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution==<br /> Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are &lt;math&gt;6^4 = 1296&lt;/math&gt; outcomes).<br /> <br /> Case 1 (easy): Four 5's are rolled. This has probability &lt;math&gt;\frac{1}{6^4}&lt;/math&gt; of occurring.<br /> <br /> Case 2: Two 5's are rolled.<br /> <br /> Case 3: No 5's are rolled.<br /> <br /> To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For &lt;math&gt;n \ge 1&lt;/math&gt;, let &lt;math&gt;a_n&lt;/math&gt; equal the number of outcomes after rolling the die &lt;math&gt;n&lt;/math&gt; times, with the property that the product is a square. Thus, &lt;math&gt;a_1 = 2&lt;/math&gt; as 1 and 4 are the only possibilities.<br /> <br /> To find &lt;math&gt;a_{n+1}&lt;/math&gt; given &lt;math&gt;a_n&lt;/math&gt; (where &lt;math&gt;n \ge 1&lt;/math&gt;), we observe that if the first &lt;math&gt;n&lt;/math&gt; rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives &lt;math&gt;2a_n&lt;/math&gt; outcomes. Otherwise, the first &lt;math&gt;n&lt;/math&gt; rolls do not multiply to a perfect square (&lt;math&gt;5^n - a_n&lt;/math&gt; outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first &lt;math&gt;n&lt;/math&gt; rolls is &lt;math&gt;2^x 3^y&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are not both even, then we observe that if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both odd, then the last roll must be 6; if only &lt;math&gt;x&lt;/math&gt; is odd, the last roll must be 2, and if only &lt;math&gt;y&lt;/math&gt; is odd, the last roll must be 3. Thus, we have &lt;math&gt;5^n - a_n&lt;/math&gt; outcomes in this case, and &lt;math&gt;a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n&lt;/math&gt;.<br /> <br /> Computing &lt;math&gt;a_2&lt;/math&gt;, &lt;math&gt;a_3&lt;/math&gt;, &lt;math&gt;a_4&lt;/math&gt; gives &lt;math&gt;a_2 = 7&lt;/math&gt;, &lt;math&gt;a_3 = 32&lt;/math&gt;, and &lt;math&gt;a_4 = 157&lt;/math&gt;. Thus for Case 3, there are 157 outcomes. For case 2, we multiply by &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; to distribute the two 5's among four rolls. Thus the probability is<br /> <br /> &lt;cmath&gt; \frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187} &lt;/cmath&gt;<br /> <br /> -scrabbler94<br /> <br /> ==Solution 2==<br /> <br /> We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square). <br /> <br /> Probability of rolling 4 duds: &lt;math&gt; \left(\frac{1}{3}\right)^4 &lt;/math&gt;<br /> <br /> Probability of rolling 3 duds: &lt;math&gt; 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} &lt;/math&gt;<br /> <br /> Probability of rolling 2 duds: &lt;math&gt; 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 &lt;/math&gt;<br /> <br /> Probability of rolling 1 dud: &lt;math&gt; 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 &lt;/math&gt;<br /> <br /> Probability of rolling 0 duds: &lt;math&gt; \left(\frac{2}{3}\right)^4 &lt;/math&gt;<br /> <br /> Now we will find the probability of a square product given we have rolled each amount of duds<br /> <br /> Probability of getting a square product given 4 duds: 1<br /> <br /> Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)<br /> <br /> Probability of getting a square product given 2 duds: &lt;math&gt; \frac{1}{4}&lt;/math&gt; (as long as our two non-duds are the same, our product will be square)<br /> <br /> Probability of getting a square product given 1 dud: &lt;math&gt;\frac{3!}{4^3}&lt;/math&gt; = &lt;math&gt;\frac{3}{32}&lt;/math&gt; (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of &lt;math&gt;4^3&lt;/math&gt; ways to roll 3 non-duds).<br /> <br /> Probability of getting a square product given 0 duds: &lt;math&gt; \frac{40}{4^4} &lt;/math&gt;= &lt;math&gt;\frac{5}{32} &lt;/math&gt; (We can have any two non-duds twice. For example, 2,2,5,5. There are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; ways of choosing which two non-duds to use and &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).<br /> <br /> We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values.<br /> &lt;cmath&gt; \left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}. &lt;/cmath&gt;<br /> <br /> &lt;math&gt;25+162&lt;/math&gt; = &lt;math&gt;\boxed{187}&lt;/math&gt; <br /> <br /> -dnaidu (silverlizard)<br /> <br /> ==Solution 3==<br /> Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:<br /> <br /> If there are four 1/4's, then there are &lt;math&gt;2^4=16&lt;/math&gt; combinations. <br /> If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. <br /> If there are two 1/4's, there are &lt;math&gt;2^2=4&lt;/math&gt; ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and &lt;math&gt;\frac{4!}{2!2!}=6&lt;/math&gt; ways to arrange, so there are &lt;math&gt;4\cdot 4\cdot 6=96&lt;/math&gt; combinations for this case. <br /> If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are &lt;math&gt;4!&lt;/math&gt; ways to order, meaning there are &lt;math&gt;2\cdot 4!=48&lt;/math&gt; combinations for this case. <br /> Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there &lt;math&gt;\binom{4}{2}&lt;/math&gt; to choose the numbers and &lt;math&gt;\frac{4!}{2!2!}=6&lt;/math&gt; ways to arrange them, so &lt;math&gt;6\cdot 6=36&lt;/math&gt;. If all four numbers are the same there are &lt;math&gt;4&lt;/math&gt; combinations, so there are &lt;math&gt;4+36=40&lt;/math&gt; combinations for this case.<br /> <br /> Hence there are &lt;math&gt;16+0+96+48+40=200&lt;/math&gt; combinations where the product of the dice is a perfect square, and there are &lt;math&gt;6^4=1296&lt;/math&gt; total combinations, so the desired probability is &lt;math&gt;\frac{200}{1296}=\frac{25}{162}&lt;/math&gt;, yielding an answer of &lt;math&gt;25+162=\boxed{187}&lt;/math&gt;.<br /> <br /> -Stormersyle<br /> <br /> ==Solution 4(Casework)==<br /> Another way to solve this problem is to do casework on all the perfect squares from &lt;math&gt;1^2&lt;/math&gt; to &lt;math&gt;36^2&lt;/math&gt;, and how many ways they can be ordered<br /> &lt;math&gt;1^2&lt;/math&gt;- &lt;math&gt;1,1,1,1&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;2^2&lt;/math&gt;- &lt;math&gt;4,1,1,1&lt;/math&gt; or &lt;math&gt;2,2,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+4=10&lt;/math&gt; ways.<br /> &lt;math&gt;3^2&lt;/math&gt;- &lt;math&gt;3,3,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;4^2&lt;/math&gt;- &lt;math&gt;4,4,1,1&lt;/math&gt;, &lt;math&gt;2,2,2,2&lt;/math&gt;, or &lt;math&gt;2,2,4,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+1+12=19&lt;/math&gt; ways.<br /> &lt;math&gt;5^2&lt;/math&gt;- &lt;math&gt;5,5,1,1&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;6^2&lt;/math&gt;- &lt;math&gt;6,6,1,1&lt;/math&gt;, &lt;math&gt;1,2,3,6&lt;/math&gt;, &lt;math&gt;2,3,2,3&lt;/math&gt;, &lt;math&gt;3,3,4,1&lt;/math&gt;- &lt;math&gt;2*\binom{4}{2}+4!+12=48&lt;/math&gt; ways.<br /> &lt;math&gt;7^2&lt;/math&gt;- Since there is a prime greater than 6 in its prime factorization there are &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;8^2&lt;/math&gt;- &lt;math&gt;4,4,4,1&lt;/math&gt; or &lt;math&gt;2,4,2,4&lt;/math&gt;- &lt;math&gt;\binom{4}{2}+4=10&lt;/math&gt; ways.<br /> &lt;math&gt;9^2&lt;/math&gt;- &lt;math&gt;3,3,3,3&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;10^2&lt;/math&gt;- &lt;math&gt;2,2,5,5&lt;/math&gt; or &lt;math&gt;1,4,5,5&lt;/math&gt;- &lt;math&gt;6+12=18&lt;/math&gt; ways.<br /> &lt;math&gt;11^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways for the same reason as &lt;math&gt;7^2&lt;/math&gt;.<br /> &lt;math&gt;12^2&lt;/math&gt;- &lt;math&gt;6,6,2,2&lt;/math&gt;, &lt;math&gt;4,4,3,3&lt;/math&gt;, &lt;math&gt;2,3,4,6&lt;/math&gt;, or &lt;math&gt;1,4,6,6&lt;/math&gt;- &lt;math&gt;2*\binom{4}{2}+4!+12=48&lt;/math&gt; ways.<br /> &lt;math&gt;13^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;14^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;15^2&lt;/math&gt;- &lt;math&gt;3,3,5,5&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;16^2&lt;/math&gt;- &lt;math&gt;4,4,4,4&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;17^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;18^2&lt;/math&gt;- &lt;math&gt;3,3,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;19^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;20^2&lt;/math&gt;- &lt;math&gt;4,4,5,5&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;21^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;22^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;23^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;24^2&lt;/math&gt;-&lt;math&gt;4,4,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}=6&lt;/math&gt; ways.<br /> &lt;math&gt;25^2&lt;/math&gt;- &lt;math&gt;5,5,5,5&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> &lt;math&gt;26^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;27^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;28^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;29^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;30^2&lt;/math&gt;- &lt;math&gt;5,5,6,6&lt;/math&gt;- &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways.<br /> &lt;math&gt;31^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;32^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;33^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;34^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;35^2&lt;/math&gt;- &lt;math&gt;0&lt;/math&gt; ways.<br /> &lt;math&gt;36^2&lt;/math&gt;- &lt;math&gt;6,6,6,6&lt;/math&gt;- &lt;math&gt;1&lt;/math&gt; way.<br /> <br /> There are &lt;math&gt;6^4=1296&lt;/math&gt; ways that the dice can land. Summing up the ways, it is easy to see that there are &lt;math&gt;200&lt;/math&gt; ways. <br /> This results in a probability of &lt;math&gt;\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}&lt;/math&gt;<br /> -superninja2000<br /> <br /> ==Solution 5 (Recursion)==<br /> We can do recursion on the number of rolls to find the number of ways we can get &lt;math&gt;4&lt;/math&gt; rolls to multiply to a square.<br /> <br /> After &lt;math&gt;n&lt;/math&gt; rolls, let us say that the product is &lt;math&gt;p = 2^a3^b5^c&lt;/math&gt;.<br /> <br /> &lt;math&gt;\text{Define the following:}&lt;/math&gt;<br /> <br /> &lt;math&gt;A_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a&lt;/math&gt; is odd, and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are even<br /> <br /> &lt;math&gt;B_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;b&lt;/math&gt; is odd, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are even<br /> <br /> &lt;math&gt;C_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;c&lt;/math&gt; is odd, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are even<br /> <br /> &lt;math&gt;D_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;c&lt;/math&gt; is even, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;E_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;b&lt;/math&gt; is even, and &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;F_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a&lt;/math&gt; is even, and &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt; are odd<br /> <br /> &lt;math&gt;G_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt; c&lt;/math&gt; are all odd<br /> <br /> &lt;math&gt;S_{n} = &lt;/math&gt; the number of ways to have a product after &lt;math&gt;n&lt;/math&gt; rolls where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt; c&lt;/math&gt; are all even (square!)<br /> <br /> &lt;math&gt;\text{We have the following equations after considering the possible values of the nth roll:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{We have the following values after considering the possible values of the 1st roll:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\text{After applying recursion twice, we get:}&lt;/math&gt;<br /> <br /> &lt;cmath&gt;A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38&lt;/cmath&gt;<br /> <br /> Finally, we have&lt;math&gt;S_4 = 200&lt;/math&gt;, &lt;math&gt;\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}&lt;/math&gt;meaning our answer is &lt;math&gt;\boxed{187}&lt;/math&gt;.<br /> <br /> ==Solution 6==<br /> Consider all the distinct &quot;fundamental&quot; groups of integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6&lt;/math&gt; whose product is a perfect square. A &quot;fundamental&quot; group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, &lt;math&gt;\{2,2\}&lt;/math&gt; is a fundamental group, while &lt;math&gt;\{3,3,4\}&lt;/math&gt; is not, because it can be broken up into &lt;math&gt;\{3,3\}&lt;/math&gt; and &lt;math&gt;\{4\}&lt;/math&gt;.<br /> <br /> &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; are already perfect squares, so they each form a &quot;fundamental&quot; group and cannot belong in any other group. Pairs of the other &lt;math&gt;4&lt;/math&gt; numbers (&lt;math&gt;\{2,2\}&lt;/math&gt;,&lt;math&gt;\{3,3\}&lt;/math&gt;, etc. ) form &quot;fundamental&quot; groups as well. The last &quot;fundamental&quot; group is &lt;math&gt;\{2,3,6\}&lt;/math&gt;. It can be easily seen that no more groups exist.<br /> <br /> Thus, we have the &quot;fundamental&quot; groups &lt;math&gt;\{1\}&lt;/math&gt;,&lt;math&gt;\{4\}&lt;/math&gt;,&lt;math&gt;\{2,2\}&lt;/math&gt;,&lt;math&gt;\{3,3\}&lt;/math&gt;,&lt;math&gt;\{5,5\}&lt;/math&gt;,&lt;math&gt;\{6,6\}&lt;/math&gt;, and &lt;math&gt;\{2,3,6\}&lt;/math&gt;.<br /> <br /> We now consider the ways to use these groups to form a sequence of &lt;math&gt;4&lt;/math&gt; numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size &lt;math&gt;2&lt;/math&gt; or &lt;math&gt;3&lt;/math&gt; and fill in any remaining spots with &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s. We can do this in one of &lt;math&gt;5&lt;/math&gt; ways: Using only &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, using one group of size &lt;math&gt;2&lt;/math&gt;, using one group of size &lt;math&gt;3&lt;/math&gt;, using two different groups of size &lt;math&gt;2&lt;/math&gt;, and using the same group of size &lt;math&gt;2&lt;/math&gt; twice.<br /> <br /> If we only use &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, each of the &lt;math&gt;4&lt;/math&gt; slots can be filled with one of the &lt;math&gt;2&lt;/math&gt; numbers, so there are &lt;math&gt;2^4=16&lt;/math&gt; possibilities.<br /> <br /> If we use one group of size &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;4&lt;/math&gt; options for the group to use, &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways to place the two numbers (since they are identical), and &lt;math&gt;2^2&lt;/math&gt; ways to fill in the remaining slots with &lt;math&gt;1&lt;/math&gt;s and &lt;math&gt;4&lt;/math&gt;s, so there are &lt;math&gt;4\cdot\binom{4}{2}\cdot2^2=96&lt;/math&gt; possibilities.<br /> <br /> If we use one group of size &lt;math&gt;3&lt;/math&gt;, there is only &lt;math&gt;1&lt;/math&gt; option for the group to use, &lt;math&gt;4\cdot3\cdot2&lt;/math&gt; ways to place the three numbers (since they are distinct), and &lt;math&gt;2&lt;/math&gt; ways to fill in the remaining slot, so there are &lt;math&gt;4\cdot3\cdot2\cdot2=48&lt;/math&gt; possibilities.<br /> <br /> If we use two different groups of size &lt;math&gt;2&lt;/math&gt;, there are &lt;math&gt;\binom{4}{2}&lt;/math&gt; options for the groups to use and &lt;math&gt;\binom{4}{2}&lt;/math&gt; ways to place the four numbers (since there are &lt;math&gt;2&lt;/math&gt; groups of identical numbers, and one group's placement uniquely determines the other's), so there are &lt;math&gt;\binom{4}{2}\cdot\binom{4}{2}=36&lt;/math&gt; possibilities.<br /> <br /> If we use the same group of size &lt;math&gt;2&lt;/math&gt; twice, there are &lt;math&gt;4&lt;/math&gt; options for the group to use and &lt;math&gt;1&lt;/math&gt; way to place the four numbers (since they are all identical), so there are &lt;math&gt;4=4&lt;/math&gt; possibilities.<br /> <br /> This gives us a total of &lt;math&gt;16+96+48+36+4=200&lt;/math&gt; possibilities, and since there are &lt;math&gt;6^4=1296&lt;/math&gt; total sequences that can be rolled, the probability is equal to &lt;math&gt;\frac{200}{1296}=\frac{25}{162}&lt;/math&gt;, so the answer is &lt;math&gt;25+162=\boxed{187}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_3&diff=123614 2019 AIME II Problems/Problem 3 2020-06-04T02:29:08Z <p>Emerald block: converted number to latex</p> <hr /> <div>==Problem 3==<br /> Find the number of &lt;math&gt;7&lt;/math&gt;-tuples of positive integers &lt;math&gt;(a,b,c,d,e,f,g)&lt;/math&gt; that satisfy the following systems of equations:<br /> &lt;cmath&gt;\begin{align*}<br /> abc&amp;=70,\\<br /> cde&amp;=71,\\<br /> efg&amp;=72.<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> ==Solution 1==<br /> As 71 is prime, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; must be 1, 1, and 71 (in some order). However, since &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;e&lt;/math&gt; are divisors of 70 and 72 respectively, the only possibility is &lt;math&gt;(c,d,e) = (1,71,1)&lt;/math&gt;. Now we are left with finding the number of solutions &lt;math&gt;(a,b,f,g)&lt;/math&gt; satisfying &lt;math&gt;ab = 70&lt;/math&gt; and &lt;math&gt;fg = 72&lt;/math&gt;, which separates easily into two subproblems. The number of positive integer solutions to &lt;math&gt;ab = 70&lt;/math&gt; simply equals the number of divisors of 70 (as we can choose a divisor for &lt;math&gt;a&lt;/math&gt;, which uniquely determines &lt;math&gt;b&lt;/math&gt;). As &lt;math&gt;70 = 2^1 \cdot 5^1 \cdot 7^1&lt;/math&gt;, we have &lt;math&gt;d(70) = (1+1)(1+1)(1+1) = 8&lt;/math&gt; solutions. Similarly, &lt;math&gt;72 = 2^3 \cdot 3^2&lt;/math&gt;, so &lt;math&gt;d(72) = 4 \times 3 = 12&lt;/math&gt;.<br /> <br /> Then the answer is simply &lt;math&gt;8 \times 12 = \boxed{096}&lt;/math&gt;.<br /> <br /> -scrabbler94<br /> <br /> ==Solution 2==<br /> We know that any two consecutive numbers are coprime. Using this, we can figure out that &lt;math&gt;c=1&lt;/math&gt; and &lt;math&gt;e=1&lt;/math&gt;. &lt;math&gt;d&lt;/math&gt; then has to be &lt;math&gt;71&lt;/math&gt;. Now we have two equations left. &lt;math&gt;ab=70&lt;/math&gt; and &lt;math&gt;fg=72&lt;/math&gt;. To solve these we just need to figure out all of the factors. Doing the prime factorization of &lt;math&gt;70&lt;/math&gt; and &lt;math&gt;72&lt;/math&gt;, we find that they have &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;12&lt;/math&gt; factors, respectively. The answer is &lt;math&gt;8 \times 12=\boxed{096}&lt;/math&gt;<br /> <br /> ~Hithere22702<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=II|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123588 2019 AIME I Problems/Problem 11 2020-06-03T18:16:20Z <p>Emerald block: fixed asymptote tags</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2 (Lots of Pythagorean Theorem)==<br /> <br /> &lt;asy&gt;<br /> <br /> unitsize(1cm);<br /> <br /> <br /> var x = 9;<br /> <br /> pair A = (0,sqrt(x^2-1));<br /> pair B = (-1,0);<br /> pair C = (1,0);<br /> <br /> dot(Label(&quot;$A$&quot;,A,NE),A);<br /> dot(Label(&quot;$B$&quot;,B,SW),B);<br /> dot(Label(&quot;$C$&quot;,C,SE),C);<br /> <br /> draw(A--B--C--cycle);<br /> <br /> <br /> var r = sqrt((x-1)/(x+1));<br /> <br /> pair I = (0,r);<br /> dot(Label(&quot;$I$&quot;,I,SE),I);<br /> draw(circle(I,r));<br /> draw(Label(&quot;$r$&quot;),I--I+r*SSW,dashed);<br /> <br /> <br /> pair M = intersectionpoint(A--B,circle(I,r));<br /> pair N = (0,0);<br /> pair O = intersectionpoint(A--C,circle(I,r));<br /> <br /> dot(Label(&quot;$M$&quot;,M,W),M);<br /> dot(Label(&quot;$N$&quot;,N,S),N);<br /> dot(Label(&quot;$O$&quot;,O,E),O);<br /> <br /> var rN = sqrt((x+1)/(x-1));<br /> <br /> pair EN = (0,-rN);<br /> dot(Label(&quot;$E_N$&quot;,EN,SE),EN);<br /> draw(circle(EN,rN));<br /> draw(Label(&quot;$r_N$&quot;),EN--EN+rN*SSW,dashed);<br /> <br /> <br /> pair AB = (-1-2/(x-1),-2rN);<br /> pair AC = (1+2/(x-1),-2rN);<br /> <br /> draw(B--AB,EndArrow);<br /> draw(C--AC,EndArrow);<br /> <br /> pair H = intersectionpoint(B--AB,circle(EN,rN));<br /> dot(Label(&quot;$H$&quot;,H,W),H);<br /> <br /> <br /> var rM = sqrt(x^2-1);<br /> <br /> pair EM = (-x,rM);<br /> dot(Label(&quot;$E_M$&quot;,EM,SW),EM);<br /> draw(Label(&quot;$r_M$&quot;),EM--EM+rM*SSE,dashed);<br /> <br /> <br /> pair CB = (-x-1,0);<br /> pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));<br /> <br /> draw(B--CB,EndArrow);<br /> draw(A--CA,EndArrow);<br /> <br /> <br /> pair J = intersectionpoint(A--B,circle(EM,rM));<br /> pair K = intersectionpoint(B--CB,circle(EM,rM));<br /> <br /> dot(Label(&quot;$J$&quot;,J,W),J);<br /> dot(Label(&quot;$K$&quot;,K,S),K);<br /> <br /> <br /> draw(arc(EM,rM,-100,15),Arrows);<br /> <br /> &lt;/asy&gt;<br /> <br /> First, assume &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;. The triangle can be scaled later if necessary. Let &lt;math&gt;I&lt;/math&gt; be the incenter and let &lt;math&gt;r&lt;/math&gt; be the inradius. Let the points at which the incircle intersects &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, and &lt;math&gt;CA&lt;/math&gt; be denoted &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt;, respectively.<br /> <br /> <br /> Next, we calculate &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;. Note the right triangle formed by &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt;. The length &lt;math&gt;IM&lt;/math&gt; is equal to &lt;math&gt;r&lt;/math&gt;. Using the Pythagorean Theorem, the length &lt;math&gt;AN&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so the length &lt;math&gt;AI&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}-r&lt;/math&gt;. Note that &lt;math&gt;BN&lt;/math&gt; is half of &lt;math&gt;BC=2&lt;/math&gt;, and by symmetry caused by the incircle, &lt;math&gt;BN=BM&lt;/math&gt; and &lt;math&gt;BM=1&lt;/math&gt;, so &lt;math&gt;MA=x-1&lt;/math&gt;. Applying the Pythagorean Theorem to &lt;math&gt;AIM&lt;/math&gt;, we get<br /> &lt;cmath&gt;r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.&lt;/cmath&gt;<br /> Expanding yields<br /> &lt;cmath&gt;r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;2r\sqrt{x^2-1}=2x-2.&lt;/cmath&gt;<br /> Dividing by &lt;math&gt;2&lt;/math&gt; and then squaring results in<br /> &lt;cmath&gt;r^2(x^2-1)=(x-1)^2,&lt;/cmath&gt;<br /> and isolating &lt;math&gt;r^2&lt;/math&gt; gets us<br /> &lt;cmath&gt;r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},&lt;/cmath&gt;<br /> so &lt;math&gt;r=\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;.<br /> <br /> <br /> We then calculate the radius of the excircle tangent to &lt;math&gt;BC&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_N&lt;/math&gt; and the radius &lt;math&gt;r_N&lt;/math&gt;.<br /> <br /> Consider the quadrilateral formed by &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;E_N&lt;/math&gt;, and the point at which the excircle intersects the extension of &lt;math&gt;AB&lt;/math&gt;, which we denote &lt;math&gt;H&lt;/math&gt;. By symmetry caused by the excircle, &lt;math&gt;BN=BH&lt;/math&gt;, so &lt;math&gt;BH=1&lt;/math&gt;.<br /> <br /> Note that triangles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; are congruent, and &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; are also congruent. Denoting the measure of angles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; measure &lt;math&gt;\alpha&lt;/math&gt; and the measure of angles &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; measure &lt;math&gt;\beta&lt;/math&gt;, straight angle &lt;math&gt;MBH=2\alpha+2\beta&lt;/math&gt;, so &lt;math&gt;\alpha + \beta=90^\circ&lt;/math&gt;. This means that angle &lt;math&gt;IBE&lt;/math&gt; is a right angle, so it forms a right triangle.<br /> <br /> Setting the base of the right triangle to &lt;math&gt;IE&lt;/math&gt;, the height is &lt;math&gt;BN=1&lt;/math&gt; and the base consists of &lt;math&gt;IN=r&lt;/math&gt; and &lt;math&gt;EN=r_N&lt;/math&gt;. Triangles &lt;math&gt;INB&lt;/math&gt; and &lt;math&gt;BNE&lt;/math&gt; are similar to &lt;math&gt;IBE&lt;/math&gt;, so &lt;math&gt;\frac{IN}{BN}=\frac{BN}{EN}&lt;/math&gt;, or &lt;math&gt;\frac{r}{1}=\frac{1}{r_N}&lt;/math&gt;. This makes &lt;math&gt;r_N&lt;/math&gt; the reciprocal of &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r_N=\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;.<br /> <br /> <br /> Circle &lt;math&gt;\omega&lt;/math&gt;'s radius can be expressed by the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the bottom of the excircle with center &lt;math&gt;E_N&lt;/math&gt;. This length is equal to &lt;math&gt;r+2r_N&lt;/math&gt;, or &lt;math&gt;\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;. Denote this value &lt;math&gt;r_\omega&lt;/math&gt;.<br /> <br /> <br /> Finally, we calculate the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the closest point on the excircle tangent to &lt;math&gt;AB&lt;/math&gt;, which forms another radius of circle &lt;math&gt;\omega&lt;/math&gt; and is equal to &lt;math&gt;r_\omega&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_M&lt;/math&gt; and the radius &lt;math&gt;r_M&lt;/math&gt;. We also denote the points where the excircle intersects &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt; using &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;, respectively. In order to calculate the distance, we must find the distance between &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E_M&lt;/math&gt; and subtract off the radius &lt;math&gt;r_M&lt;/math&gt;.<br /> <br /> We first must calculate the radius of the excircle. Because the excircle is tangent to both &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;AC&lt;/math&gt;, its center must lie on the angle bisector formed by the two lines, which is parallel to &lt;math&gt;BC&lt;/math&gt;. This means that the distance from &lt;math&gt;E_M&lt;/math&gt; to &lt;math&gt;K&lt;/math&gt; is equal to the length of &lt;math&gt;AN&lt;/math&gt;, so the radius is also &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;.<br /> <br /> Next, we find the length of &lt;math&gt;IE_M&lt;/math&gt;. We can do this by forming the right triangle &lt;math&gt;IAE_M&lt;/math&gt;. The length of leg &lt;math&gt;AI&lt;/math&gt; is equal to &lt;math&gt;AN&lt;/math&gt; minus &lt;math&gt;r&lt;/math&gt;, or &lt;math&gt;\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;. In order to calculate the length of leg &lt;math&gt;AE_M&lt;/math&gt;, note that right triangles &lt;math&gt;AJE_M&lt;/math&gt; and &lt;math&gt;BNA&lt;/math&gt; are congruent, as &lt;math&gt;JE_M&lt;/math&gt; and &lt;math&gt;NA&lt;/math&gt; share a length of &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, and angles &lt;math&gt;E_MAJ&lt;/math&gt; and &lt;math&gt;NAB&lt;/math&gt; add up to the right angle &lt;math&gt;NAE_M&lt;/math&gt;. This means that &lt;math&gt;AE_M=BA=x&lt;/math&gt;.<br /> <br /> Using Pythagorean Theorem, we get<br /> &lt;cmath&gt;IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.&lt;/cmath&gt;<br /> Bringing back<br /> &lt;cmath&gt;r_\omega=IE_M-r_M&lt;/cmath&gt;<br /> and substituting in some values, the equation becomes<br /> &lt;cmath&gt;r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.&lt;/cmath&gt;<br /> Rearranging and squaring both sides gets<br /> &lt;cmath&gt;\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.&lt;/cmath&gt;<br /> Distributing both sides yields<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Canceling terms results in<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),&lt;/cmath&gt;<br /> We can further simplify to<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Substituting out &lt;math&gt;r_\omega&lt;/math&gt; gets<br /> &lt;cmath&gt;\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2&lt;/cmath&gt;<br /> which when distributed yields<br /> &lt;cmath&gt;\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> After some canceling, distributing, and rearranging, we obtain<br /> &lt;cmath&gt;4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x-1&lt;/math&gt; results in<br /> &lt;cmath&gt;4x+4=x^3-x^2-8x^2+8x-4x+4,&lt;/cmath&gt;<br /> which can be rearranged into<br /> &lt;cmath&gt;x^3-9x^2=0&lt;/cmath&gt;<br /> and factored into<br /> &lt;cmath&gt;x^2(x-9)=0.&lt;/cmath&gt;<br /> This means that &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, and since a side length of &lt;math&gt;0&lt;/math&gt; cannot exist, &lt;math&gt;x=9&lt;/math&gt;.<br /> <br /> <br /> As a result, the triangle must have sides in the ratio of &lt;math&gt;9:2:9&lt;/math&gt;. Since the triangle must have integer side lengths, and these values share no common factors greater than &lt;math&gt;1&lt;/math&gt;, the triangle with the smallest possible perimeter under these restrictions has a perimeter of &lt;math&gt;9+2+9=\boxed{020}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3==<br /> <br /> Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123586 2019 AIME I Problems/Problem 11 2020-06-03T18:12:45Z <p>Emerald block: added a diagram</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2 (Lots of Pythagorean Theorem)==<br /> <br /> [asy]<br /> <br /> unitsize(1cm);<br /> <br /> <br /> var x = 9;<br /> <br /> pair A = (0,sqrt(x^2-1));<br /> pair B = (-1,0);<br /> pair C = (1,0);<br /> <br /> dot(Label(&quot;&lt;math&gt;A&lt;/math&gt;&quot;,A,NE),A);<br /> dot(Label(&quot;&lt;math&gt;B&lt;/math&gt;&quot;,B,SW),B);<br /> dot(Label(&quot;&lt;math&gt;C&lt;/math&gt;&quot;,C,SE),C);<br /> <br /> draw(A--B--C--cycle);<br /> <br /> <br /> var r = sqrt((x-1)/(x+1));<br /> <br /> pair I = (0,r);<br /> dot(Label(&quot;&lt;math&gt;I&lt;/math&gt;&quot;,I,SE),I);<br /> draw(circle(I,r));<br /> draw(Label(&quot;&lt;math&gt;r&lt;/math&gt;&quot;),I--I+r*SSW,dashed);<br /> <br /> <br /> pair M = intersectionpoint(A--B,circle(I,r));<br /> pair N = (0,0);<br /> pair O = intersectionpoint(A--C,circle(I,r));<br /> <br /> dot(Label(&quot;&lt;math&gt;M&lt;/math&gt;&quot;,M,W),M);<br /> dot(Label(&quot;&lt;math&gt;N&lt;/math&gt;&quot;,N,S),N);<br /> dot(Label(&quot;&lt;math&gt;O&lt;/math&gt;&quot;,O,E),O);<br /> <br /> var rN = sqrt((x+1)/(x-1));<br /> <br /> pair EN = (0,-rN);<br /> dot(Label(&quot;&lt;math&gt;E_N&lt;/math&gt;&quot;,EN,SE),EN);<br /> draw(circle(EN,rN));<br /> draw(Label(&quot;&lt;math&gt;r_N&lt;/math&gt;&quot;),EN--EN+rN*SSW,dashed);<br /> <br /> <br /> pair AB = (-1-2/(x-1),-2rN);<br /> pair AC = (1+2/(x-1),-2rN);<br /> <br /> draw(B--AB,EndArrow);<br /> draw(C--AC,EndArrow);<br /> <br /> pair H = intersectionpoint(B--AB,circle(EN,rN));<br /> dot(Label(&quot;&lt;math&gt;H&lt;/math&gt;&quot;,H,W),H);<br /> <br /> <br /> var rM = sqrt(x^2-1);<br /> <br /> pair EM = (-x,rM);<br /> dot(Label(&quot;&lt;math&gt;E_M&lt;/math&gt;&quot;,EM,SW),EM);<br /> draw(Label(&quot;&lt;math&gt;r_M&lt;/math&gt;&quot;),EM--EM+rM*SSE,dashed);<br /> <br /> <br /> pair CB = (-x-1,0);<br /> pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x));<br /> <br /> draw(B--CB,EndArrow);<br /> draw(A--CA,EndArrow);<br /> <br /> <br /> pair J = intersectionpoint(A--B,circle(EM,rM));<br /> pair K = intersectionpoint(B--CB,circle(EM,rM));<br /> <br /> dot(Label(&quot;&lt;math&gt;J&lt;/math&gt;&quot;,J,W),J);<br /> dot(Label(&quot;&lt;math&gt;K&lt;/math&gt;&quot;,K,S),K);<br /> <br /> <br /> draw(arc(EM,rM,-100,15),Arrows);<br /> <br /> [/asy]<br /> <br /> First, assume &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;. The triangle can be scaled later if necessary. Let &lt;math&gt;I&lt;/math&gt; be the incenter and let &lt;math&gt;r&lt;/math&gt; be the inradius. Let the points at which the incircle intersects &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, and &lt;math&gt;CA&lt;/math&gt; be denoted &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt;, respectively.<br /> <br /> <br /> Next, we calculate &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;. Note the right triangle formed by &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt;. The length &lt;math&gt;IM&lt;/math&gt; is equal to &lt;math&gt;r&lt;/math&gt;. Using the Pythagorean Theorem, the length &lt;math&gt;AN&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so the length &lt;math&gt;AI&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}-r&lt;/math&gt;. Note that &lt;math&gt;BN&lt;/math&gt; is half of &lt;math&gt;BC=2&lt;/math&gt;, and by symmetry caused by the incircle, &lt;math&gt;BN=BM&lt;/math&gt; and &lt;math&gt;BM=1&lt;/math&gt;, so &lt;math&gt;MA=x-1&lt;/math&gt;. Applying the Pythagorean Theorem to &lt;math&gt;AIM&lt;/math&gt;, we get<br /> &lt;cmath&gt;r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.&lt;/cmath&gt;<br /> Expanding yields<br /> &lt;cmath&gt;r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;2r\sqrt{x^2-1}=2x-2.&lt;/cmath&gt;<br /> Dividing by &lt;math&gt;2&lt;/math&gt; and then squaring results in<br /> &lt;cmath&gt;r^2(x^2-1)=(x-1)^2,&lt;/cmath&gt;<br /> and isolating &lt;math&gt;r^2&lt;/math&gt; gets us<br /> &lt;cmath&gt;r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},&lt;/cmath&gt;<br /> so &lt;math&gt;r=\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;.<br /> <br /> <br /> We then calculate the radius of the excircle tangent to &lt;math&gt;BC&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_N&lt;/math&gt; and the radius &lt;math&gt;r_N&lt;/math&gt;.<br /> <br /> Consider the quadrilateral formed by &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;E_N&lt;/math&gt;, and the point at which the excircle intersects the extension of &lt;math&gt;AB&lt;/math&gt;, which we denote &lt;math&gt;H&lt;/math&gt;. By symmetry caused by the excircle, &lt;math&gt;BN=BH&lt;/math&gt;, so &lt;math&gt;BH=1&lt;/math&gt;.<br /> <br /> Note that triangles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; are congruent, and &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; are also congruent. Denoting the measure of angles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; measure &lt;math&gt;\alpha&lt;/math&gt; and the measure of angles &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; measure &lt;math&gt;\beta&lt;/math&gt;, straight angle &lt;math&gt;MBH=2\alpha+2\beta&lt;/math&gt;, so &lt;math&gt;\alpha + \beta=90^\circ&lt;/math&gt;. This means that angle &lt;math&gt;IBE&lt;/math&gt; is a right angle, so it forms a right triangle.<br /> <br /> Setting the base of the right triangle to &lt;math&gt;IE&lt;/math&gt;, the height is &lt;math&gt;BN=1&lt;/math&gt; and the base consists of &lt;math&gt;IN=r&lt;/math&gt; and &lt;math&gt;EN=r_N&lt;/math&gt;. Triangles &lt;math&gt;INB&lt;/math&gt; and &lt;math&gt;BNE&lt;/math&gt; are similar to &lt;math&gt;IBE&lt;/math&gt;, so &lt;math&gt;\frac{IN}{BN}=\frac{BN}{EN}&lt;/math&gt;, or &lt;math&gt;\frac{r}{1}=\frac{1}{r_N}&lt;/math&gt;. This makes &lt;math&gt;r_N&lt;/math&gt; the reciprocal of &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r_N=\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;.<br /> <br /> <br /> Circle &lt;math&gt;\omega&lt;/math&gt;'s radius can be expressed by the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the bottom of the excircle with center &lt;math&gt;E_N&lt;/math&gt;. This length is equal to &lt;math&gt;r+2r_N&lt;/math&gt;, or &lt;math&gt;\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;. Denote this value &lt;math&gt;r_\omega&lt;/math&gt;.<br /> <br /> <br /> Finally, we calculate the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the closest point on the excircle tangent to &lt;math&gt;AB&lt;/math&gt;, which forms another radius of circle &lt;math&gt;\omega&lt;/math&gt; and is equal to &lt;math&gt;r_\omega&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_M&lt;/math&gt; and the radius &lt;math&gt;r_M&lt;/math&gt;. We also denote the points where the excircle intersects &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt; using &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;, respectively. In order to calculate the distance, we must find the distance between &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E_M&lt;/math&gt; and subtract off the radius &lt;math&gt;r_M&lt;/math&gt;.<br /> <br /> We first must calculate the radius of the excircle. Because the excircle is tangent to both &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;AC&lt;/math&gt;, its center must lie on the angle bisector formed by the two lines, which is parallel to &lt;math&gt;BC&lt;/math&gt;. This means that the distance from &lt;math&gt;E_M&lt;/math&gt; to &lt;math&gt;K&lt;/math&gt; is equal to the length of &lt;math&gt;AN&lt;/math&gt;, so the radius is also &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;.<br /> <br /> Next, we find the length of &lt;math&gt;IE_M&lt;/math&gt;. We can do this by forming the right triangle &lt;math&gt;IAE_M&lt;/math&gt;. The length of leg &lt;math&gt;AI&lt;/math&gt; is equal to &lt;math&gt;AN&lt;/math&gt; minus &lt;math&gt;r&lt;/math&gt;, or &lt;math&gt;\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;. In order to calculate the length of leg &lt;math&gt;AE_M&lt;/math&gt;, note that right triangles &lt;math&gt;AJE_M&lt;/math&gt; and &lt;math&gt;BNA&lt;/math&gt; are congruent, as &lt;math&gt;JE_M&lt;/math&gt; and &lt;math&gt;NA&lt;/math&gt; share a length of &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, and angles &lt;math&gt;E_MAJ&lt;/math&gt; and &lt;math&gt;NAB&lt;/math&gt; add up to the right angle &lt;math&gt;NAE_M&lt;/math&gt;. This means that &lt;math&gt;AE_M=BA=x&lt;/math&gt;.<br /> <br /> Using Pythagorean Theorem, we get<br /> &lt;cmath&gt;IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.&lt;/cmath&gt;<br /> Bringing back<br /> &lt;cmath&gt;r_\omega=IE_M-r_M&lt;/cmath&gt;<br /> and substituting in some values, the equation becomes<br /> &lt;cmath&gt;r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.&lt;/cmath&gt;<br /> Rearranging and squaring both sides gets<br /> &lt;cmath&gt;\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.&lt;/cmath&gt;<br /> Distributing both sides yields<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Canceling terms results in<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),&lt;/cmath&gt;<br /> We can further simplify to<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Substituting out &lt;math&gt;r_\omega&lt;/math&gt; gets<br /> &lt;cmath&gt;\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2&lt;/cmath&gt;<br /> which when distributed yields<br /> &lt;cmath&gt;\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> After some canceling, distributing, and rearranging, we obtain<br /> &lt;cmath&gt;4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x-1&lt;/math&gt; results in<br /> &lt;cmath&gt;4x+4=x^3-x^2-8x^2+8x-4x+4,&lt;/cmath&gt;<br /> which can be rearranged into<br /> &lt;cmath&gt;x^3-9x^2=0&lt;/cmath&gt;<br /> and factored into<br /> &lt;cmath&gt;x^2(x-9)=0.&lt;/cmath&gt;<br /> This means that &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, and since a side length of &lt;math&gt;0&lt;/math&gt; cannot exist, &lt;math&gt;x=9&lt;/math&gt;.<br /> <br /> <br /> As a result, the triangle must have sides in the ratio of &lt;math&gt;9:2:9&lt;/math&gt;. Since the triangle must have integer side lengths, and these values share no common factors greater than &lt;math&gt;1&lt;/math&gt;, the triangle with the smallest possible perimeter under these restrictions has a perimeter of &lt;math&gt;9+2+9=\boxed{020}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3==<br /> <br /> Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123539 2019 AIME I Problems/Problem 11 2020-06-03T03:09:44Z <p>Emerald block: </p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2 (Lots of Pythagorean Theorem)==<br /> <br /> First, assume &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;. The triangle can be scaled later if necessary. Let &lt;math&gt;I&lt;/math&gt; be the incenter and let &lt;math&gt;r&lt;/math&gt; be the inradius. Let the points at which the incircle intersects &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, and &lt;math&gt;CA&lt;/math&gt; be denoted &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt;, respectively.<br /> <br /> <br /> Next, we calculate &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;. Note the right triangle formed by &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt;. The length &lt;math&gt;IM&lt;/math&gt; is equal to &lt;math&gt;r&lt;/math&gt;. Using the Pythagorean Theorem, the length &lt;math&gt;AN&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so the length &lt;math&gt;AI&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}-r&lt;/math&gt;. Note that &lt;math&gt;BN&lt;/math&gt; is half of &lt;math&gt;BC=2&lt;/math&gt;, and by symmetry caused by the incircle, &lt;math&gt;BN=BM&lt;/math&gt; and &lt;math&gt;BM=1&lt;/math&gt;, so &lt;math&gt;MA=x-1&lt;/math&gt;. Applying the Pythagorean Theorem to &lt;math&gt;AIM&lt;/math&gt;, we get<br /> &lt;cmath&gt;r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.&lt;/cmath&gt;<br /> Expanding yields<br /> &lt;cmath&gt;r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;2r\sqrt{x^2-1}=2x-2.&lt;/cmath&gt;<br /> Dividing by &lt;math&gt;2&lt;/math&gt; and then squaring results in<br /> &lt;cmath&gt;r^2(x^2-1)=(x-1)^2,&lt;/cmath&gt;<br /> and isolating &lt;math&gt;r^2&lt;/math&gt; gets us<br /> &lt;cmath&gt;r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},&lt;/cmath&gt;<br /> so &lt;math&gt;r=\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;.<br /> <br /> <br /> We then calculate the radius of the excircle tangent to &lt;math&gt;BC&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_N&lt;/math&gt; and the radius &lt;math&gt;r_N&lt;/math&gt;.<br /> <br /> Consider the quadrilateral formed by &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;E_N&lt;/math&gt;, and the point at which the excircle intersects the extension of &lt;math&gt;AB&lt;/math&gt;, which we denote &lt;math&gt;H&lt;/math&gt;. By symmetry caused by the excircle, &lt;math&gt;BN=BH&lt;/math&gt;, so &lt;math&gt;BH=1&lt;/math&gt;.<br /> <br /> Note that triangles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; are congruent, and &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; are also congruent. Denoting the measure of angles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; measure &lt;math&gt;\alpha&lt;/math&gt; and the measure of angles &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; measure &lt;math&gt;\beta&lt;/math&gt;, straight angle &lt;math&gt;MBH=2\alpha+2\beta&lt;/math&gt;, so &lt;math&gt;\alpha + \beta=90^\circ&lt;/math&gt;. This means that angle &lt;math&gt;IBE&lt;/math&gt; is a right angle, so it forms a right triangle.<br /> <br /> Setting the base of the right triangle to &lt;math&gt;IE&lt;/math&gt;, the height is &lt;math&gt;BN=1&lt;/math&gt; and the base consists of &lt;math&gt;IN=r&lt;/math&gt; and &lt;math&gt;EN=r_N&lt;/math&gt;. Triangles &lt;math&gt;INB&lt;/math&gt; and &lt;math&gt;BNE&lt;/math&gt; are similar to &lt;math&gt;IBE&lt;/math&gt;, so &lt;math&gt;\frac{IN}{BN}=\frac{BN}{EN}&lt;/math&gt;, or &lt;math&gt;\frac{r}{1}=\frac{1}{r_N}&lt;/math&gt;. This makes &lt;math&gt;r_N&lt;/math&gt; the reciprocal of &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r_N=\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;.<br /> <br /> <br /> Circle &lt;math&gt;\omega&lt;/math&gt;'s radius can be expressed by the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the bottom of the excircle with center &lt;math&gt;E_N&lt;/math&gt;. This length is equal to &lt;math&gt;r+2r_N&lt;/math&gt;, or &lt;math&gt;\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;. Denote this value &lt;math&gt;r_\omega&lt;/math&gt;.<br /> <br /> <br /> Finally, we calculate the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the closest point on the excircle tangent to &lt;math&gt;AB&lt;/math&gt;, which forms another radius of circle &lt;math&gt;\omega&lt;/math&gt; and is equal to &lt;math&gt;r_\omega&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_M&lt;/math&gt; and the radius &lt;math&gt;r_M&lt;/math&gt;. We also denote the points where the excircle intersects &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt; using &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;, respectively. In order to calculate the distance, we must find the distance between &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E_M&lt;/math&gt; and subtract off the radius &lt;math&gt;r_M&lt;/math&gt;.<br /> <br /> We first must calculate the radius of the excircle. Because the excircle is tangent to both &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;AC&lt;/math&gt;, its center must lie on the angle bisector formed by the two lines, which is parallel to &lt;math&gt;BC&lt;/math&gt;. This means that the distance from &lt;math&gt;E_M&lt;/math&gt; to &lt;math&gt;K&lt;/math&gt; is equal to the length of &lt;math&gt;AN&lt;/math&gt;, so the radius is also &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;.<br /> <br /> Next, we find the length of &lt;math&gt;IE_M&lt;/math&gt;. We can do this by forming the right triangle &lt;math&gt;IAE_M&lt;/math&gt;. The length of leg &lt;math&gt;AI&lt;/math&gt; is equal to &lt;math&gt;AN&lt;/math&gt; minus &lt;math&gt;r&lt;/math&gt;, or &lt;math&gt;\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;. In order to calculate the length of leg &lt;math&gt;AE_M&lt;/math&gt;, note that right triangles &lt;math&gt;AJE_M&lt;/math&gt; and &lt;math&gt;BNA&lt;/math&gt; are congruent, as &lt;math&gt;JE_M&lt;/math&gt; and &lt;math&gt;NA&lt;/math&gt; share a length of &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, and angles &lt;math&gt;E_MAJ&lt;/math&gt; and &lt;math&gt;NAB&lt;/math&gt; add up to the right angle &lt;math&gt;NAE_M&lt;/math&gt;. This means that &lt;math&gt;AE_M=BA=x&lt;/math&gt;.<br /> <br /> Using Pythagorean Theorem, we get<br /> &lt;cmath&gt;IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.&lt;/cmath&gt;<br /> Bringing back<br /> &lt;cmath&gt;r_\omega=IE_M-r_M&lt;/cmath&gt;<br /> and substituting in some values, the equation becomes<br /> &lt;cmath&gt;r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.&lt;/cmath&gt;<br /> Rearranging and squaring both sides gets<br /> &lt;cmath&gt;\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.&lt;/cmath&gt;<br /> Distributing both sides yields<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Canceling terms results in<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),&lt;/cmath&gt;<br /> We can further simplify to<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Substituting out &lt;math&gt;r_\omega&lt;/math&gt; gets<br /> &lt;cmath&gt;\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2&lt;/cmath&gt;<br /> which when distributed yields<br /> &lt;cmath&gt;\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> After some canceling, distributing, and rearranging, we obtain<br /> &lt;cmath&gt;4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x-1&lt;/math&gt; results in<br /> &lt;cmath&gt;4x+4=x^3-x^2-8x^2+8x-4x+4&lt;/cmath&gt;<br /> which can be rearranged into<br /> &lt;cmath&gt;x^3-9x^2=0&lt;/cmath&gt;<br /> and factored into<br /> &lt;cmath&gt;x^2(x-9)=0.&lt;/cmath&gt;<br /> This means that &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, and since a side length of &lt;math&gt;0&lt;/math&gt; cannot exist, &lt;math&gt;x=9&lt;/math&gt;.<br /> <br /> <br /> As a result, the triangle must have sides in the ratio of &lt;math&gt;9:2:9&lt;/math&gt;. Since the triangle must have integer side lengths, and these values share no common factors greater than 1, the triangle with the smallest possible perimeter under these restrictions has a perimeter of &lt;math&gt;9+2+9=\boxed{020}&lt;/math&gt;.<br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3==<br /> <br /> Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123538 2019 AIME I Problems/Problem 11 2020-06-03T03:09:05Z <p>Emerald block: new solution</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2 (Lots of Pythagorean Theorem)==<br /> <br /> First, assume &lt;math&gt;BC=2&lt;/math&gt; and &lt;math&gt;AB=AC=x&lt;/math&gt;. The triangle can be scaled later if necessary. Let &lt;math&gt;I&lt;/math&gt; be the incenter and let &lt;math&gt;r&lt;/math&gt; be the inradius. Let the points at which the incircle intersects &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt;, and &lt;math&gt;CA&lt;/math&gt; be denoted &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;N&lt;/math&gt;, and &lt;math&gt;O&lt;/math&gt;, respectively.<br /> <br /> <br /> Next, we calculate &lt;math&gt;r&lt;/math&gt; in terms of &lt;math&gt;x&lt;/math&gt;. Note the right triangle formed by &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, and &lt;math&gt;M&lt;/math&gt;. The length &lt;math&gt;IM&lt;/math&gt; is equal to &lt;math&gt;r&lt;/math&gt;. Using the Pythagorean Theorem, the length &lt;math&gt;AN&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, so the length &lt;math&gt;AI&lt;/math&gt; is &lt;math&gt;\sqrt{x^2-1}-r&lt;/math&gt;. Note that &lt;math&gt;BN&lt;/math&gt; is half of &lt;math&gt;BC=2&lt;/math&gt;, and by symmetry caused by the incircle, &lt;math&gt;BN=BM&lt;/math&gt; and &lt;math&gt;BM=1&lt;/math&gt;, so &lt;math&gt;MA=x-1&lt;/math&gt;. Applying the Pythagorean Theorem to &lt;math&gt;AIM&lt;/math&gt;, we get<br /> &lt;cmath&gt;r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.&lt;/cmath&gt;<br /> Expanding yields<br /> &lt;cmath&gt;r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,&lt;/cmath&gt;<br /> which can be simplified to<br /> &lt;cmath&gt;2r\sqrt{x^2-1}=2x-2.&lt;/cmath&gt;<br /> Dividing by &lt;math&gt;2&lt;/math&gt; and then squaring results in<br /> &lt;cmath&gt;r^2(x^2-1)=(x-1)^2,&lt;/cmath&gt;<br /> and isolating &lt;math&gt;r^2&lt;/math&gt; gets us<br /> &lt;cmath&gt;r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},&lt;/cmath&gt;<br /> so &lt;math&gt;r=\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;.<br /> <br /> <br /> We then calculate the radius of the excircle tangent to &lt;math&gt;BC&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_N&lt;/math&gt; and the radius &lt;math&gt;r_N&lt;/math&gt;.<br /> <br /> Consider the quadrilateral formed by &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;I&lt;/math&gt;, &lt;math&gt;E_N&lt;/math&gt;, and the point at which the excircle intersects the extension of &lt;math&gt;AB&lt;/math&gt;, which we denote &lt;math&gt;H&lt;/math&gt;. By symmetry caused by the excircle, &lt;math&gt;BN=BH&lt;/math&gt;, so &lt;math&gt;BH=1&lt;/math&gt;.<br /> <br /> Note that triangles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; are congruent, and &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; are also congruent. Denoting the measure of angles &lt;math&gt;MBI&lt;/math&gt; and &lt;math&gt;NBI&lt;/math&gt; measure &lt;math&gt;\alpha&lt;/math&gt; and the measure of angles &lt;math&gt;HBE&lt;/math&gt; and &lt;math&gt;NBE&lt;/math&gt; measure &lt;math&gt;\beta&lt;/math&gt;, straight angle &lt;math&gt;MBH=2\alpha+2\beta&lt;/math&gt;, so &lt;math&gt;\alpha + \beta=90^\circ&lt;/math&gt;. This means that angle &lt;math&gt;IBE&lt;/math&gt; is a right angle, so it forms a right triangle.<br /> <br /> Setting the base of the right triangle to &lt;math&gt;IE&lt;/math&gt;, the height is &lt;math&gt;BN=1&lt;/math&gt; and the base consists of &lt;math&gt;IN=r&lt;/math&gt; and &lt;math&gt;EN=r_N&lt;/math&gt;. Triangles &lt;math&gt;INB&lt;/math&gt; and &lt;math&gt;BNE&lt;/math&gt; are similar to &lt;math&gt;IBE&lt;/math&gt;, so &lt;math&gt;\frac{IN}{BN}=\frac{BN}{EN}&lt;/math&gt;, or &lt;math&gt;\frac{r}{1}=\frac{1}{r_N}&lt;/math&gt;. This makes &lt;math&gt;r_N&lt;/math&gt; the reciprocal of &lt;math&gt;r&lt;/math&gt;, so &lt;math&gt;r_N=\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;.<br /> <br /> <br /> Circle &lt;math&gt;\omega&lt;/math&gt;'s radius can be expressed by the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the bottom of the excircle with center &lt;math&gt;E_N&lt;/math&gt;. This length is equal to &lt;math&gt;r+2r_N&lt;/math&gt;, or &lt;math&gt;\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}&lt;/math&gt;. Denote this value &lt;math&gt;r_\omega&lt;/math&gt;.<br /> <br /> <br /> Finally, we calculate the distance from the incenter &lt;math&gt;I&lt;/math&gt; to the closest point on the excircle tangent to &lt;math&gt;AB&lt;/math&gt;, which forms another radius of circle &lt;math&gt;\omega&lt;/math&gt; and is equal to &lt;math&gt;r_\omega&lt;/math&gt;. We denote the center of the excircle &lt;math&gt;E_M&lt;/math&gt; and the radius &lt;math&gt;r_M&lt;/math&gt;. We also denote the points where the excircle intersects &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;BC&lt;/math&gt; using &lt;math&gt;J&lt;/math&gt; and &lt;math&gt;K&lt;/math&gt;, respectively. In order to calculate the distance, we must find the distance between &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;E_M&lt;/math&gt; and subtract off the radius &lt;math&gt;r_M&lt;/math&gt;.<br /> <br /> We first must calculate the radius of the excircle. Because the excircle is tangent to both &lt;math&gt;AB&lt;/math&gt; and the extension of &lt;math&gt;AC&lt;/math&gt;, its center must lie on the angle bisector formed by the two lines, which is parallel to &lt;math&gt;BC&lt;/math&gt;. This means that the distance from &lt;math&gt;E_M&lt;/math&gt; to &lt;math&gt;K&lt;/math&gt; is equal to the length of &lt;math&gt;AN&lt;/math&gt;, so the radius is also &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;.<br /> <br /> Next, we find the length of &lt;math&gt;IE_M&lt;/math&gt;. We can do this by forming the right triangle &lt;math&gt;IAE_M&lt;/math&gt;. The length of leg &lt;math&gt;AI&lt;/math&gt; is equal to &lt;math&gt;AN&lt;/math&gt; minus &lt;math&gt;r&lt;/math&gt;, or &lt;math&gt;\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}&lt;/math&gt;. In order to calculate the length of leg &lt;math&gt;AE_M&lt;/math&gt;, note that right triangles &lt;math&gt;AJE_M&lt;/math&gt; and &lt;math&gt;BNA&lt;/math&gt; are congruent, as &lt;math&gt;JE_M&lt;/math&gt; and &lt;math&gt;NA&lt;/math&gt; share a length of &lt;math&gt;\sqrt{x^2-1}&lt;/math&gt;, and angles &lt;math&gt;E_MAJ&lt;/math&gt; and &lt;math&gt;NAB&lt;/math&gt; add up to the right angle &lt;math&gt;NAE_M&lt;/math&gt;. This means that &lt;math&gt;AE_M=BA=x&lt;/math&gt;.<br /> <br /> Using Pythagorean Theorem, we get<br /> &lt;cmath&gt;IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.&lt;/cmath&gt;<br /> Bringing back<br /> &lt;cmath&gt;r_\omega=IE_M-r_M&lt;/cmath&gt;<br /> and substituting in some values, the equation becomes<br /> &lt;cmath&gt;r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.&lt;/cmath&gt;<br /> Rearranging and squaring both sides gets<br /> &lt;cmath&gt;\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.&lt;/cmath&gt;<br /> Distributing both sides yields<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Canceling terms results in<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Since<br /> &lt;cmath&gt;-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),&lt;/cmath&gt;<br /> We can further simplify to<br /> &lt;cmath&gt;r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> Substituting out &lt;math&gt;r_\omega&lt;/math&gt; gets<br /> &lt;cmath&gt;\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2&lt;/cmath&gt;<br /> which when distributed yields<br /> &lt;cmath&gt;\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.&lt;/cmath&gt;<br /> After some canceling, distributing, and rearranging, we obtain<br /> &lt;cmath&gt;4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.&lt;/cmath&gt;<br /> Multiplying both sides by &lt;math&gt;x-1&lt;/math&gt; results in<br /> &lt;cmath&gt;4x+4=x^3-x^2-8x^2+8x-4x+4&lt;/cmath&gt;<br /> which can be rearranged into<br /> &lt;cmath&gt;x^3-9x^2=0&lt;/cmath&gt;<br /> and factored into<br /> &lt;cmath&gt;x^2(x-9)=0.&lt;/cmath&gt;<br /> This means that &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;0&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt;, and since a side length of &lt;math&gt;0&lt;/math&gt; cannot exist, &lt;math&gt;x=9&lt;/math&gt;.<br /> <br /> As a result, the triangle must have sides in the ratio of &lt;math&gt;9:2:9&lt;/math&gt;. Since the triangle must have integer side lengths, and these values share no common factors greater than 1, the triangle with the smallest possible perimeter under these restrictions has a perimeter of &lt;math&gt;9+2+9=\boxed{020}&lt;/math&gt;.<br /> ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Solution 3==<br /> <br /> Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&diff=123537 2019 AIME I Problems/Problem 11 2020-06-03T02:59:00Z <p>Emerald block: fixed typo</p> <hr /> <div>==Problem 11==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, the sides have integer lengths and &lt;math&gt;AB=AC&lt;/math&gt;. Circle &lt;math&gt;\omega&lt;/math&gt; has its center at the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;. An ''excircle'' of &lt;math&gt;\triangle ABC&lt;/math&gt; is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to &lt;math&gt;\overline{BC}&lt;/math&gt; is internally tangent to &lt;math&gt;\omega&lt;/math&gt;, and the other two excircles are both externally tangent to &lt;math&gt;\omega&lt;/math&gt;. Find the minimum possible value of the perimeter of &lt;math&gt;\triangle ABC&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let the tangent circle be &lt;math&gt;\omega&lt;/math&gt;. Some notation first: let &lt;math&gt;BC=a&lt;/math&gt;, &lt;math&gt;AB=b&lt;/math&gt;, &lt;math&gt;s&lt;/math&gt; be the semiperimeter, &lt;math&gt;\theta=\angle ABC&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; be the inradius. Intuition tells us that the radius of &lt;math&gt;\omega&lt;/math&gt; is &lt;math&gt;r+\frac{2rs}{s-a}&lt;/math&gt; (using the exradius formula). However, the sum of the radius of &lt;math&gt;\omega&lt;/math&gt; and &lt;math&gt;\frac{rs}{s-b}&lt;/math&gt; is equivalent to the distance between the incenter and the the &lt;math&gt;B/C&lt;/math&gt; excenter. Denote the B excenter as &lt;math&gt;I_B&lt;/math&gt; and the incenter as &lt;math&gt;I&lt;/math&gt;. <br /> Lemma: &lt;math&gt;I_BI=\frac{2b*IB}{a}&lt;/math&gt;<br /> We draw the circumcircle of &lt;math&gt;\triangle ABC&lt;/math&gt;. Let the angle bisector of &lt;math&gt;\angle ABC&lt;/math&gt; hit the circumcircle at a second point &lt;math&gt;M&lt;/math&gt;. By the incenter-excenter lemma, &lt;math&gt;AM=CM=IM&lt;/math&gt;. Let this distance be &lt;math&gt;\alpha&lt;/math&gt;. Ptolemy's theorem on &lt;math&gt;ABCM&lt;/math&gt; gives us &lt;cmath&gt;a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}&lt;/cmath&gt; Again, by the incenter-excenter lemma, &lt;math&gt;II_B=2IM&lt;/math&gt; so &lt;math&gt;II_b=\frac{2b*IB}{a}&lt;/math&gt; as desired.<br /> Using this gives us the following equation: &lt;cmath&gt;\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}&lt;/cmath&gt; <br /> Motivated by the &lt;math&gt;s-a&lt;/math&gt; and &lt;math&gt;s-b&lt;/math&gt;, we make the following substitution: &lt;math&gt;x=s-a, y=s-b&lt;/math&gt;<br /> This changes things quite a bit. Here's what we can get from it: &lt;cmath&gt;a=2y, b=x+y, s=x+2y&lt;/cmath&gt; It is known (easily proved with Heron's and a=rs) that &lt;cmath&gt;r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}&lt;/cmath&gt; Using this, we can also find &lt;math&gt;IB&lt;/math&gt;: let the midpoint of &lt;math&gt;BC&lt;/math&gt; be &lt;math&gt;N&lt;/math&gt;. Using Pythagorean's Theorem on &lt;math&gt;\triangle INB&lt;/math&gt;, &lt;cmath&gt;IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y} &lt;/cmath&gt; We now look at the RHS of the main equation: &lt;cmath&gt;r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}&lt;/cmath&gt;<br /> Cancelling some terms, we have &lt;cmath&gt;\frac{r(x+4y)}{x}=IB&lt;/cmath&gt; <br /> Squaring, &lt;cmath&gt;\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)&lt;/cmath&gt; Expanding and moving terms around gives &lt;cmath&gt;(x-8y)(x+2y)=0\to x=8y&lt;/cmath&gt; Reverse substituting, &lt;cmath&gt;s-a=8s-8b\to b=\frac{9}{2}a&lt;/cmath&gt; Clearly the smallest solution is &lt;math&gt;a=2&lt;/math&gt; and &lt;math&gt;b=9&lt;/math&gt;, so our answer is &lt;math&gt;2+9+9=\boxed{020}&lt;/math&gt;<br /> -franchester<br /> <br /> ==Solution 2==<br /> <br /> Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_4&diff=123382 2019 AIME I Problems/Problem 4 2020-06-01T02:50:16Z <p>Emerald block: fixed small error in equation</p> <hr /> <div>==Problem 4==<br /> A soccer team has &lt;math&gt;22&lt;/math&gt; available players. A fixed set of &lt;math&gt;11&lt;/math&gt; players starts the game, while the other &lt;math&gt;11&lt;/math&gt; are available as substitutes. During the game, the coach may make as many as &lt;math&gt;3&lt;/math&gt; substitutions, where any one of the &lt;math&gt;11&lt;/math&gt; players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let &lt;math&gt;n&lt;/math&gt; be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when &lt;math&gt;n&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> ==Solution 1 (Recursion)==<br /> There are &lt;math&gt;0-3&lt;/math&gt; substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for &lt;math&gt;0&lt;/math&gt; subs is &lt;math&gt;1&lt;/math&gt;, and the ways to reorganize after &lt;math&gt;n&lt;/math&gt; subs is the product of the number of new subs (&lt;math&gt;12-n&lt;/math&gt;) and the players that can be ejected (&lt;math&gt;11&lt;/math&gt;). The formula for &lt;math&gt;n&lt;/math&gt; subs is then &lt;math&gt;a_n=11(12-n)a_{n-1}&lt;/math&gt; with &lt;math&gt;a_0=1&lt;/math&gt;. <br /> <br /> Summing from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;3&lt;/math&gt; gives &lt;math&gt;1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9&lt;/math&gt;. Notice that &lt;math&gt;10+9\cdot11\cdot10=10+990=1000&lt;/math&gt;. Then, rearrange it into &lt;math&gt;1+11^2+11^3\cdot (10+11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)&lt;/math&gt;. When taking modulo &lt;math&gt;1000&lt;/math&gt;, the last term goes away. What is left is &lt;math&gt;1+11^2=\boxed{122}&lt;/math&gt;.<br /> <br /> ~BJHHar<br /> <br /> ==Solution 2 (Casework) ==<br /> We can perform casework. Call the substitution area the &quot;bench&quot;.<br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;: No substitutions. There is &lt;math&gt;1&lt;/math&gt; way of doing this: leaving everybody on the field.<br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;: One substitution. Choose one player on the field to sub out, and one player on the bench. This corresponds to &lt;math&gt;11\cdot 11 = 121&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 3}&lt;/math&gt;: Two substitutions. Choose one player on the field to sub out, and one player on the bench. Once again, this is &lt;math&gt;11\cdot 11&lt;/math&gt;. Now choose one more player on the field to sub out and one player on the bench that was not the original player subbed out. This gives us a total of &lt;math&gt;11\cdot 11\cdot 11\cdot 10 = 13310\equiv 310 \bmod{1000}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 4}&lt;/math&gt;: Three substitutions. Using similar logic as &lt;math&gt;\textbf{Case 3}&lt;/math&gt;, we get &lt;math&gt;(11\cdot 11)\cdot (11\cdot 10)\cdot (11\cdot 9)&lt;/math&gt;. The resulting number ends in &lt;math&gt;690&lt;/math&gt;.<br /> <br /> Therefore, the answer is &lt;math&gt;1 + 121 + 310 + 690 = \boxed{122}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=MATHCOUNTS&diff=122486 MATHCOUNTS 2020-05-15T19:33:11Z <p>Emerald block: deleted irrelevant information</p> <hr /> <div>Many AoPS Community members and online school students have been participants at National MATHCOUNTS, including many Nationals Countdown Round participants in the past decade. '''MATHCOUNTS''' is a large national [[mathematics competition]] and [[mathematics coaching]] program that has served millions of middle school students since 1984. Sponsored by the [http://cna.com CNA Foundation], [http://www.nspe.org/ National Society of Professional Engineers], the [http://www.nctm.org/ National Council of Teachers of Mathematics], and others including Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete beyond their eighth grade year.<br /> <br /> {{Contest Info|name=MATHCOUNTS|region=USA|type=Free Response|difficulty=0.5 - 2.5|breakdown=&lt;u&gt;Countdown&lt;/u&gt;: 0.5 (School/Chapter), 1 (State/National)&lt;br&gt;&lt;u&gt;Sprint&lt;/u&gt;: 1-1.5 (School/Chapter), 2-2.5 (State/National)&lt;br&gt;&lt;u&gt;Target:&lt;/u&gt; 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}}<br /> <br /> == MATHCOUNTS Resources ==<br /> === MATHCOUNTS Books ===<br /> Art of Problem Solving's [http://artofproblemsolving.com/store/list/aops-curriculum Introductory subject textbooks] are ideal for students preparing for MATHCOUNTS, as are [http://artofproblemsolving.com/store/item/aops-vol1 AoPS Volume 1] and [http://artofproblemsolving.com/store/item/competition-math Competition Math for Middle School]<br /> <br /> === MATHCOUNTS Classes ===<br /> Art of Problem Solving hosts a [http://artofproblemsolving.com/school/course/mathcounts-basics Basic] and an [http://artofproblemsolving.com/school/course/mathcounts-advanced Advanced] MATHCOUNTS course. The AoPS Introduction-level subject courses also include a great deal of MATHCOUNTS preparation. Many AoPS instructors are former National MATHCOUNTS Mathletes.<br /> <br /> === MATHCOUNTS Online ===<br /> * [http://www.mathcounts.org Official MATHCOUNTS Homepage]<br /> * Art of Problem Solving hosts a large [http://artofproblemsolving.com/community/c3_middle_school_math Middle School Math Forum] as well as a private [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23209 MATHCOUNTS Coaches Forum].<br /> * The AoPS MATHCOUNTS Trainer is available on the [http://artofproblemsolving.com/mathcounts_trainer AoPS website], as well as on the [https://itunes.apple.com/us/app/mathcounts-trainer-math-contest/id1023961880?ls=1&amp;mt=8 iPhone and iPad].<br /> * The free [http://www.artofproblemsolving.com/alcumus AoPS Alcumus learning system] includes thousands of MATHCOUNTS problems.<br /> * [http://artofproblemsolving.com/ftw/ftw.php For the Win!] gives students free Countdown Round-like practice against other AoPS students.<br /> * AoPS founder Richard Rusczyk has created dozens of [http://artofproblemsolving.com/videos/mathcounts MATHCOUNTS Mini video lessons].<br /> * [http://mathcounts.saab.org/ Elias Saab's MATHCOUNTS Preparation Homepage]<br /> * [http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html The MATHCOUNTS Bible According to Mr. Diaz]<br /> *[http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php/ Building a Successful MATHCOUNTS Program] by [[Jeff Boyd]], who coached the 2005, 2007, and 2008 National Champion [[Texas MathCounts]] team.<br /> *[http://mathweb.scranton.edu/monks/courses/ProblemSolving/MathCountsPlaybookBW.pdf Coach Monk's MathCounts Playbook]<br /> * MathCounts Minis make hard problems easy<br /> <br /> == MATHCOUNTS Curriculum ==<br /> MATHCOUNTS curriculum includes [[arithmetic]], [[algebra]], [[counting]], [[geometry]], [[number theory]], [[probability]], and [[statistics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.<br /> <br /> Before 1990, MATHCOUNTS chose particular areas of mathematics to highlight each year before changing the focus of the competition more broadly to problem solving.<br /> <br /> == Past State Team Winners ==<br /> * 1984: Virginia<br /> * 1985: Florida<br /> * 1986: California<br /> * 1987: New York<br /> * 1988: New York<br /> * 1989: North Carolina<br /> * 1990: Ohio<br /> * 1991: Alabama<br /> * 1992: California<br /> * 1993: Kansas<br /> * 1994: Pennsylvania<br /> * 1995: Indiana<br /> * 1996: Wisconsin<br /> * 1997: Massachusetts<br /> * 1998: Wisconsin<br /> * 1999: Massachusetts<br /> * 2000: California<br /> * 2001: Virginia<br /> * 2002: California<br /> * 2003: California<br /> * 2004: Illinois<br /> * 2005: Texas<br /> * 2006: Virginia<br /> * 2007: Texas<br /> * 2008: Texas<br /> * 2009: Texas<br /> * 2010: California<br /> * 2011: California<br /> * 2012: Massachusetts<br /> * 2013: Massachusetts<br /> * 2014: California<br /> * 2015: Indiana<br /> * 2016: Texas<br /> * 2017: Texas<br /> * 2018: Texas<br /> * 2019: Massachusetts<br /> <br /> == MATHCOUNTS Competition Structure ==<br /> <br /> === Sprint Round ===<br /> <br /> 30 problems are given all at once. Students have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The earlier problems are usually the easiest problems in the competition, and the later problems can be as hard as some of the Team Round questions. No calculators are allowed during this round.<br /> <br /> === Target Round ===<br /> 8 problems given 2 at a time. Students have 6 minutes to complete each set of two problems. Students may not go back to previous rounds (or forwards to future rounds) even if they finish before time is called. Calculators are allowed for the Target Round. Usually comprised of one &quot;confidence booster&quot; and one hard problem.<br /> <br /> === Team Round ===<br /> <br /> 10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions).<br /> <br /> === Countdown Round ===<br /> High scoring individuals compete head-to-head until a champion is crowned. People compete from off a screen taking 45 seconds or less to finish the problem. The Countdown round is run differently in various different chapter, state, and national competitions. In the national competitions, it is the round that determines the champion. Calculators are not allowed, but scratch paper will be provided.<br /> <br /> <br /> ====Chapter and State Competitions====<br /> <br /> In the chapter and state competitions, the countdown round is not mandatory. However, if it is deemed official by the chapter or state, the following format must be used:<br /> <br /> *The 10th place written finisher competes against the 9th place written finisher. A problem is displayed, and both competitors have 45 seconds to answer the question, and the first competitor to correctly answer the question receives one point. The person who gets the most correct out of three questions (not necessarily two out of three) is the winner.<br /> <br /> *The winner of the first round goes up against the 8th place finisher.<br /> <br /> *The winner of the second round goes up against the 7th place finisher.<br /> <br /> This process is continued until the countdown round reaches the top four written competitors. Starting then, the first person to get three questions correct wins (as opposed to the best-out-of-three rule).<br /> <br /> If the countdown round is unofficial, any format may be used. Single-elimination bracket-style tournaments are common.<br /> <br /> ====National Competition====<br /> <br /> At the national competition, there are some structural changes to the countdown round. The top 12 (not the top 10) written finishers make it to the countdown round, and the format is changed from a ladder competition to a single elimination tournament where the top four written competitors get a bye. This setup makes it far more likely for a 12th place finisher to become champion, and it makes it less likely for a first place written finisher to become champion, equalizing the field. But even then, a 12th place written competitor will have less of a chance to become champion than the top 4, because the top 4 get a bye. Until the semi-finals, the scoring is best out of five advances.<br /> <br /> At the first round and the second round, the person to correctly answer the most out of 5 questions wins. However, at the semifinals, the rules slightly change&amp;mdash;the first person to correctly answer four questions wins.<br /> <br /> === Ciphering Round ===<br /> In some states, (most notably Florida) there is an optional ciphering round. Very similar to countdown (in both difficulty and layout), a team sends up a representative to go against all representatives from the other teams. A problem is shown on a screen and students work fast to answer the problem. The students give their answer and after 45 seconds the answer is shown and the answers are checked to see if they are right. The fastest correct answer gets five points, the next fastest gets 4, etc. There are 4 questions per individual and teams send up 4 people. A perfect score is then 80. Often times the questions take clever reading skills. For example, one question was &quot;How much dirt is in a 3 ft by 3 ft by 4 ft hole?&quot; The answer was 0 because there is no dirt in a hole.<br /> <br /> === Masters Round ===<br /> Top students give in-depth explanations to challenging problems. This round is optional at the state level competition and is mandatory at the national competition (up to 2011). At nationals the top two on the written and countdown participate. <br /> In 2012, it was replaced by the Reel Math Challenge (now called the Math Video Challenge).<br /> <br /> === Scoring and Ranking ===<br /> An individual's score is their total number of correct sprint round answers plus 2 times their total number of correct target round answers. This total is out of a maximum of &lt;math&gt;30 + 2(8) = 46&lt;/math&gt; points.<br /> <br /> A team's score is the individual scores of its members divided by 4 plus 2 points for every correct team round answer, making a team's maximum possible score 66 points. Therefore, it is possible to win with a relatively low team score and a phenomenal individual score, as the team score is only roughly 30% of the total team score. Note that when there are less than four members the score will become less.<br /> <br /> == MATHCOUNTS Competition Levels ==<br /> === School Competition ===<br /> Students vie for the chance to make their school teams. Problems at this level are generally the easiest and most basic in curriculum.<br /> <br /> === Chapter Competition ===<br /> Chapter competitions serve as a selection filter for state competitions. A few states don't need to host chapter competitions due to a small population size.<br /> <br /> === State Competition ===<br /> The top 4 students in each state form the state team for the national competition. The coach of the top school team at the state level is invited to coach the state team at the national competition. Interestingly, the coach of a state team is not necessarily the coach of any of the state's team members.<br /> <br /> === National Competition ===<br /> ==== National Competition Sites ====<br /> For many years, the National MATHCOUNTS competition was held in Washington, D.C. More recently, the competition has changed venues often.<br /> <br /> * The 2020 competition was canceled due to the COVID-19 pandemic.<br /> * The 2019 competition was held in Orlando, Florida.<br /> * The 2018 competition was held in Washington, D.C.<br /> * The 2017 competition was held in Orlando, Florida.<br /> * The 2016 competition was held in Washington, D.C.<br /> * The 2015 competition was held in Boston, Massachusetts.<br /> * The 2014 competition was held in Orlando, Florida.<br /> * The 2013 competition was held in Washington, D.C.<br /> * The 2012 competition was held in Orlando, Florida.<br /> * The 2011 competition was held in Washington, D.C.<br /> * The 2009 and 2010 competitions were held in Orlando, Florida.<br /> * The 2008 competition was held in Denver, Colorado.<br /> * The 2007 competition was held in Fort Worth, Texas.<br /> * The 2006 competition was held in Arlington, Virginia.<br /> * The 2005 competition was held in Detroit, Michigan.<br /> * The 2004 competition was held in Washington, D.C.<br /> * The 2002 and 2003 competitions were held in Chicago, Illinois.<br /> <br /> == What comes after MATHCOUNTS? ==<br /> <br /> Give the following competitions a try and take a look at the [[List of United States high school mathematics competitions]].<br /> * [[American Mathematics Competitions]]<br /> * [[American Regions Math League]]<br /> * [[Mandelbrot Competition]]<br /> * [[Mu Alpha Theta]]<br /> <br /> [[Category:Mathematics competitions]]<br /> <br /> == See also... ==<br /> * [[List of national MATHCOUNTS teams]]<br /> * [[Mathematics competition resources]]<br /> * [[Math contest books]]<br /> * [[Math books]]<br /> * [[List of United States middle school mathematics competitions]]<br /> * [[List of United States high school mathematics competitions]]<br /> * [http://www.mathcounts.org/webarticles/anmviewer.asp?a=921&amp;z=71 2006 MATHCOUNTS Countdown Video]<br /> <br /> [[Category:Introductory mathematics competitions]]</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_23&diff=120339 2002 AMC 10B Problems/Problem 23 2020-03-31T18:38:53Z <p>Emerald block: new solution built off of additional comment</p> <hr /> <div>== Problem 23 ==<br /> <br /> Let &lt;math&gt;\{a_k\}&lt;/math&gt; be a sequence of integers such that &lt;math&gt;a_1=1&lt;/math&gt; and &lt;math&gt;a_{m+n}=a_m+a_n+mn,&lt;/math&gt; for all positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n.&lt;/math&gt; Then &lt;math&gt;a_{12}&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> When &lt;math&gt;m=1&lt;/math&gt;, &lt;math&gt;a_{n+1}=1+a_n+n&lt;/math&gt;. Hence,<br /> &lt;cmath&gt;a_{2}=1+a_1+2&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{3}=1+a_2+3&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{4}=1+a_3+4&lt;/cmath&gt;<br /> &lt;cmath&gt;\dots&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{12}=1+a_{11}+11&lt;/cmath&gt;<br /> Adding these equations up, we have that &lt;math&gt;a_{12}=12+(1+2+3+...+11)=\boxed{78}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> == Solution 2 ==<br /> <br /> Substituting &lt;math&gt;n=1&lt;/math&gt; into &lt;math&gt;a_{m+n}=a_m+a_n+mn&lt;/math&gt;: &lt;math&gt;a_{m+1}=a_m+a_{1}+m&lt;/math&gt;. Since &lt;math&gt;a_1 = 1&lt;/math&gt;, &lt;math&gt;a_{m+1}=a_m+m+1&lt;/math&gt;. Therefore, &lt;math&gt;a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)&lt;/math&gt;, and so on until &lt;math&gt;a_2 = a_1 + 2&lt;/math&gt;. Adding the Left Hand Sides of all of these equations gives &lt;math&gt;a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2&lt;/math&gt;; adding the Right Hand Sides of these equations gives &lt;math&gt;(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)&lt;/math&gt;. These two expressions must be equal; hence &lt;math&gt;a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)&lt;/math&gt; and &lt;math&gt;a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)&lt;/math&gt;. Substituting &lt;math&gt;a_1 = 1&lt;/math&gt;: &lt;math&gt;a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}&lt;/math&gt;. Thus we have a general formula for &lt;math&gt;a_m&lt;/math&gt; and substituting &lt;math&gt;m=12&lt;/math&gt;: &lt;math&gt;a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D) \ } 78}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since &lt;math&gt;a_{m+n} = a_m+a_n +mn&lt;/math&gt;, we know &lt;math&gt;a_2=a_1+a_1+1\cdot1=1+1+1=3&lt;/math&gt;. After this, we can use &lt;math&gt;a_2&lt;/math&gt; to find &lt;math&gt;a_4&lt;/math&gt;. &lt;math&gt;a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10&lt;/math&gt;. Now, we can use &lt;math&gt;a_2&lt;/math&gt; and &lt;math&gt;a_4&lt;/math&gt; to find &lt;math&gt;a_6&lt;/math&gt;, or &lt;math&gt;a_6=a_4+a_2+4\cdot 2 = 10+3+8=21&lt;/math&gt;. Lastly, we can use &lt;math&gt;a_6&lt;/math&gt; to find &lt;math&gt;a_{12}&lt;/math&gt;. &lt;math&gt;a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{(D) 78}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> <br /> We can set &lt;math&gt;n&lt;/math&gt; equal to &lt;math&gt;m&lt;/math&gt;, so we can say that <br /> &lt;cmath&gt;a_{m + m} = a_m + a_m + m*m&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{2m} = 2a_m + m^2&lt;/cmath&gt;<br /> <br /> We set &lt;math&gt;2m = 12&lt;/math&gt;, we get &lt;math&gt;m = 6&lt;/math&gt;.<br /> &lt;cmath&gt;a_{12} = 2a_6 + 36&lt;/cmath&gt;<br /> <br /> We set &lt;math&gt;2m = 6&lt;/math&gt;m, we get &lt;math&gt;m = 3&lt;/math&gt;.<br /> &lt;cmath&gt;a_6 = 2a_3 + 9&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;a_3&lt;/math&gt; is easy, just direct substitution.<br /> &lt;cmath&gt;a_2 = 1 + 1 + 1 = 3&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3 = a_{2 + 1} = 3 + 1 + 2 = 6&lt;/cmath&gt;<br /> <br /> Substituting, we get<br /> &lt;cmath&gt;a_6 = 2(6) + 9 = 21&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{12} = 2(21) + 36 = 78&lt;/cmath&gt;<br /> <br /> Thus, the answer is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ~ euler123<br /> <br /> == Solution 5 ==<br /> <br /> Note that the sequence of triangular numbers &lt;math&gt;T_n=1+2+3+...+n&lt;/math&gt; satisfies these conditions. It is immediately obvious that it satisfies &lt;math&gt;a_1=1&lt;/math&gt;, and &lt;math&gt;a_{m+n}=a_m+a_n+mn&lt;/math&gt; can be visually proven with the diagram below.<br /> <br /> &lt;asy&gt;<br /> for(int i=5; i &gt; 0; --i) {<br /> for(int j=0; j &lt; i; ++j) {<br /> draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2));<br /> };<br /> };<br /> path m1 = brace((2,-.3),(0,-.3),.2);<br /> draw(m1);<br /> label(&quot;$m$&quot;,m1,S);<br /> <br /> path n1 = brace((4,-.3),(3,-.3),.2);<br /> draw(n1);<br /> label(&quot;$n$&quot;,n1,S);<br /> <br /> <br /> draw((-.2*sqrt(3),-.2)--(2+.2*sqrt(3),-.2)--(1,.4+sqrt(3))--cycle);<br /> label(&quot;$T_m$&quot;,(1,1/3*sqrt(3)));<br /> <br /> draw((3-.2*sqrt(3),-.2)--(4+.2*sqrt(3),-.2)--(3.5,.4+.5*sqrt(3))--cycle);<br /> label(&quot;$T_n$&quot;,(3.5,.5/3*sqrt(3)));<br /> <br /> <br /> path m2 = brace((2+.15*sqrt(3),.15+2*sqrt(3)),(3+.15*sqrt(3),.15+sqrt(3)),.2);<br /> draw(m2);<br /> label(&quot;$m$&quot;,m2,(.5*sqrt(3),.5));<br /> <br /> path n2 = brace((1.5-.15*sqrt(3),.15+1.5*sqrt(3)),(2-.15*sqrt(3),.15+2*sqrt(3)),.2);<br /> draw(n2);<br /> label(&quot;$n$&quot;,n2,(-.5*sqrt(3),.5));<br /> <br /> <br /> draw((2.5,-.4+.5*sqrt(3))--(3+.4/3*sqrt(3),sqrt(3))--(2,.4+2*sqrt(3))--(1.5-.4/3*sqrt(3),1.5*sqrt(3))--cycle);<br /> label(&quot;$mn$&quot;,(2.25,1.25*sqrt(3)));<br /> &lt;/asy&gt;<br /> <br /> This means that we can use the triangular number formula &lt;math&gt;T_n = \frac{n(n+1)}{2}&lt;/math&gt;, so the answer is &lt;math&gt;T_{12} = \frac{12(12+1)}{2} = \boxed{\mathrm{(D) \ } 78}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_23&diff=120207 2016 AMC 12A Problems/Problem 23 2020-03-27T21:37:46Z <p>Emerald block: fixed image</p> <hr /> <div>==Problem==<br /> <br /> Three numbers in the interval &lt;math&gt;\left[0,1\right]&lt;/math&gt; are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1: Super WLOG===<br /> <br /> WLOG assume &lt;math&gt;a&lt;/math&gt; is the largest. Scale the triangle to &lt;math&gt;1,{b}/{a},{c}/{a}&lt;/math&gt; or &lt;math&gt;1,x,y&lt;/math&gt; with &lt;math&gt;x,y \in [0,1]&lt;/math&gt; and equally likely to be any pair of numbers within the interval. Then, &lt;math&gt;x+y&gt;1&lt;/math&gt;, meaning the solution is &lt;math&gt;\boxed{\textbf{(C)}\;1/2}&lt;/math&gt;, as shown in the graph below.<br /> &lt;asy&gt;<br /> pair A = (0,0);<br /> pair B = (1,0);<br /> pair C = (1,1);<br /> pair D = (0,1);<br /> pair E = (0,0);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--D,dashed);<br /> fill(B--D--C--cycle,gray);<br /> <br /> label(&quot;$0$&quot;,A,SW);<br /> label(&quot;$1$&quot;,B,S);<br /> label(&quot;$1$&quot;,D,W);<br /> label(&quot;$y$&quot;,(0,.5),W);<br /> label(&quot;$x$&quot;,(.5,0),S);<br /> label(&quot;$x+y&gt;1$&quot;,(5/7,5/7));<br /> &lt;/asy&gt;<br /> <br /> ===Solution 2: Conditional Probability===<br /> <br /> WLOG, let the largest of the three numbers drawn be &lt;math&gt;a&gt;0&lt;/math&gt;. Then the other two numbers are drawn uniformly and independently from the interval &lt;math&gt;[0,a]&lt;/math&gt;. The probability that their sum is greater than &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\;1/2.}&lt;/math&gt;<br /> <br /> ===Solution 3: Calculus===<br /> <br /> When &lt;math&gt;a&gt;b&lt;/math&gt;, consider two cases:<br /> <br /> 1) &lt;math&gt;0&lt;a&lt;\frac{1}{2}&lt;/math&gt;, then <br /> &lt;math&gt;\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}&lt;/math&gt;<br /> <br /> 2)&lt;math&gt;\frac{1}{2}&lt;a&lt;1&lt;/math&gt;, then <br /> &lt;math&gt;\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}&lt;/math&gt;<br /> <br /> &lt;math&gt;a&lt;b&lt;/math&gt; is the same. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> ===Solution 4: Geometry===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. The region where, WLOG, side &lt;math&gt;z&lt;/math&gt; is too long, &lt;math&gt;z\geq x+y&lt;/math&gt;, is a pyramid with a base of area &lt;math&gt;\frac{1}{2}&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so its volume is &lt;math&gt;\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}&lt;/math&gt;. Accounting for the corresponding cases in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; multiplies our answer by &lt;math&gt;3&lt;/math&gt;, so we have excluded a total volume of &lt;math&gt;\frac{1}{2}&lt;/math&gt; from the space of possible probabilities. Subtracting this from &lt;math&gt;1&lt;/math&gt; leaves us with a final answer of &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> === Solution 5: More Calculus ===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when &lt;math&gt;x + y &lt; z&lt;/math&gt;, which has area &lt;math&gt;\frac{z^2}{2}&lt;/math&gt; or when &lt;math&gt;x+z&lt;y&lt;/math&gt; or &lt;math&gt;y+z&lt;x&lt;/math&gt;, which have an area of &lt;math&gt;\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.&lt;/math&gt; Integrating this expression from 0 to 1 in the form<br /> <br /> &lt;math&gt;\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg[\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}&lt;/math&gt;<br /> <br /> === Solution 6: Geometry in 2-D ===<br /> WLOG assume that &lt;math&gt;z&lt;/math&gt; is the largest number and hence the largest side. Then &lt;math&gt;x,y \leq z&lt;/math&gt;. We can set up a square that is &lt;math&gt;z&lt;/math&gt; by &lt;math&gt;z&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt; plane. We are wanting all the points within this square that satisfy &lt;math&gt;x+y &gt; z&lt;/math&gt;. This happens to be a line dividing the square into 2 equal regions. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> <br /> [][] diagram for this problem goes here (z by z square)<br /> <br /> === Solution 7: More WLOG, Complementary Probability ===<br /> The triangle inequality simplifies to considering only one case: &lt;math&gt;\text{the smallest side+ the second smallest side} &gt; \text{the largest side}&lt;/math&gt;. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) &lt;math&gt;a&lt;/math&gt; is the largest, so on average &lt;math&gt;a=1/2&lt;/math&gt; (now equal to becomes a degenerate case with probability &lt;math&gt;0&lt;/math&gt;, so we no longer need to consider it). We now want &lt;math&gt;b+c&lt;1/2&lt;/math&gt;, so imagine choosing &lt;math&gt;b+c&lt;/math&gt; at once rather than independently. But we know that &lt;math&gt;b+c&lt;/math&gt; is between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. The complement is thus: &lt;math&gt;(1/2-0)/2=1/4&lt;/math&gt;. But keep in mind that we choose each &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; randomly and independently, so if there are &lt;math&gt;k&lt;/math&gt; ways to choose &lt;math&gt;b+c&lt;/math&gt; together, there are &lt;math&gt;2k&lt;/math&gt; ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if &lt;math&gt;b+c=3&lt;/math&gt;, then we only count this once, but in reality: we have two cases &lt;math&gt;1+2&lt;/math&gt;, and &lt;math&gt;2+1&lt;/math&gt;; similar reasoning also generalizes to non-integral values). The complement is then actually &lt;math&gt;2(1/4)=1/2&lt;/math&gt;. Therefore, our desired probability is given by &lt;math&gt;1-\text{complement}=1/2, C&lt;/math&gt;<br /> <br /> === Solution 8: 3D geometry ===<br /> <br /> We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines &lt;math&gt;x+y&gt;z, x+z&gt;y,&lt;/math&gt; and &lt;math&gt;y+z&gt;x,&lt;/math&gt; We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length &lt;math&gt;\sqrt(2)&lt;/math&gt; and the other has 3 sides of length &lt;math&gt;\sqrt(2)&lt;/math&gt; and 3 sides of length &lt;math&gt;1.&lt;/math&gt; The volume of this region is &lt;math&gt;\frac 1 2&lt;/math&gt;. Hence our solution is &lt;math&gt;C.&lt;/math&gt;<br /> <br /> === Solution 9: Cheap Solution ===<br /> <br /> Pretend that the values of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are integers ranging from &lt;math&gt;[1,n]&lt;/math&gt;. Test out the probability of the first few values of &lt;math&gt;n&lt;/math&gt; (for example, &lt;math&gt;P(2) = \frac{5}{8}, P(3) = \frac{15}{27}, P(4) = \frac{34}{64}, P(5) = \frac{65}{125}&lt;/math&gt;). Since real numbers contain infinite increments, the answer is the limit as &lt;math&gt;n&lt;/math&gt; approaches infinity of &lt;math&gt;P(n)&lt;/math&gt;, which is easily hypothesized as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;C.&lt;/math&gt;<br /> -solution by fidgetboss_4000 get rect<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_23&diff=120206 2016 AMC 12A Problems/Problem 23 2020-03-27T21:37:09Z <p>Emerald block: /* Solution 1: Super WLOG */</p> <hr /> <div>==Problem==<br /> <br /> Three numbers in the interval &lt;math&gt;\left[0,1\right]&lt;/math&gt; are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1: Super WLOG===<br /> <br /> WLOG assume &lt;math&gt;a&lt;/math&gt; is the largest. Scale the triangle to &lt;math&gt;1,{b}/{a},{c}/{a}&lt;/math&gt; or &lt;math&gt;1,x,y&lt;/math&gt; with &lt;math&gt;x,y \in [0,1]&lt;/math&gt; and equally likely to be any pair of numbers within the interval. Then, &lt;math&gt;x+y&gt;1&lt;/math&gt;, meaning the solution is &lt;math&gt;\boxed{\textbf{(C)}\;1/2}&lt;/math&gt;, as shown in the graph below.<br /> <br /> ===Solution 2: Conditional Probability===<br /> <br /> WLOG, let the largest of the three numbers drawn be &lt;math&gt;a&gt;0&lt;/math&gt;. Then the other two numbers are drawn uniformly and independently from the interval &lt;math&gt;[0,a]&lt;/math&gt;. The probability that their sum is greater than &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\;1/2.}&lt;/math&gt;<br /> <br /> ===Solution 3: Calculus===<br /> <br /> When &lt;math&gt;a&gt;b&lt;/math&gt;, consider two cases:<br /> <br /> 1) &lt;math&gt;0&lt;a&lt;\frac{1}{2}&lt;/math&gt;, then <br /> &lt;math&gt;\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}&lt;/math&gt;<br /> <br /> 2)&lt;math&gt;\frac{1}{2}&lt;a&lt;1&lt;/math&gt;, then <br /> &lt;math&gt;\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}&lt;/math&gt;<br /> <br /> &lt;math&gt;a&lt;b&lt;/math&gt; is the same. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> ===Solution 4: Geometry===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. The region where, WLOG, side &lt;math&gt;z&lt;/math&gt; is too long, &lt;math&gt;z\geq x+y&lt;/math&gt;, is a pyramid with a base of area &lt;math&gt;\frac{1}{2}&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so its volume is &lt;math&gt;\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}&lt;/math&gt;. Accounting for the corresponding cases in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; multiplies our answer by &lt;math&gt;3&lt;/math&gt;, so we have excluded a total volume of &lt;math&gt;\frac{1}{2}&lt;/math&gt; from the space of possible probabilities. Subtracting this from &lt;math&gt;1&lt;/math&gt; leaves us with a final answer of &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> === Solution 5: More Calculus ===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when &lt;math&gt;x + y &lt; z&lt;/math&gt;, which has area &lt;math&gt;\frac{z^2}{2}&lt;/math&gt; or when &lt;math&gt;x+z&lt;y&lt;/math&gt; or &lt;math&gt;y+z&lt;x&lt;/math&gt;, which have an area of &lt;math&gt;\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.&lt;/math&gt; Integrating this expression from 0 to 1 in the form<br /> <br /> &lt;math&gt;\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg[\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}&lt;/math&gt;<br /> <br /> === Solution 6: Geometry in 2-D ===<br /> WLOG assume that &lt;math&gt;z&lt;/math&gt; is the largest number and hence the largest side. Then &lt;math&gt;x,y \leq z&lt;/math&gt;. We can set up a square that is &lt;math&gt;z&lt;/math&gt; by &lt;math&gt;z&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt; plane. We are wanting all the points within this square that satisfy &lt;math&gt;x+y &gt; z&lt;/math&gt;. This happens to be a line dividing the square into 2 equal regions. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> <br /> [][] diagram for this problem goes here (z by z square)<br /> <br /> === Solution 7: More WLOG, Complementary Probability ===<br /> The triangle inequality simplifies to considering only one case: &lt;math&gt;\text{the smallest side+ the second smallest side} &gt; \text{the largest side}&lt;/math&gt;. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) &lt;math&gt;a&lt;/math&gt; is the largest, so on average &lt;math&gt;a=1/2&lt;/math&gt; (now equal to becomes a degenerate case with probability &lt;math&gt;0&lt;/math&gt;, so we no longer need to consider it). We now want &lt;math&gt;b+c&lt;1/2&lt;/math&gt;, so imagine choosing &lt;math&gt;b+c&lt;/math&gt; at once rather than independently. But we know that &lt;math&gt;b+c&lt;/math&gt; is between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. The complement is thus: &lt;math&gt;(1/2-0)/2=1/4&lt;/math&gt;. But keep in mind that we choose each &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; randomly and independently, so if there are &lt;math&gt;k&lt;/math&gt; ways to choose &lt;math&gt;b+c&lt;/math&gt; together, there are &lt;math&gt;2k&lt;/math&gt; ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if &lt;math&gt;b+c=3&lt;/math&gt;, then we only count this once, but in reality: we have two cases &lt;math&gt;1+2&lt;/math&gt;, and &lt;math&gt;2+1&lt;/math&gt;; similar reasoning also generalizes to non-integral values). The complement is then actually &lt;math&gt;2(1/4)=1/2&lt;/math&gt;. Therefore, our desired probability is given by &lt;math&gt;1-\text{complement}=1/2, C&lt;/math&gt;<br /> <br /> === Solution 8: 3D geometry ===<br /> <br /> We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines &lt;math&gt;x+y&gt;z, x+z&gt;y,&lt;/math&gt; and &lt;math&gt;y+z&gt;x,&lt;/math&gt; We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length &lt;math&gt;\sqrt(2)&lt;/math&gt; and the other has 3 sides of length &lt;math&gt;\sqrt(2)&lt;/math&gt; and 3 sides of length &lt;math&gt;1.&lt;/math&gt; The volume of this region is &lt;math&gt;\frac 1 2&lt;/math&gt;. Hence our solution is &lt;math&gt;C.&lt;/math&gt;<br /> <br /> === Solution 9: Cheap Solution ===<br /> <br /> Pretend that the values of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are integers ranging from &lt;math&gt;[1,n]&lt;/math&gt;. Test out the probability of the first few values of &lt;math&gt;n&lt;/math&gt; (for example, &lt;math&gt;P(2) = \frac{5}{8}, P(3) = \frac{15}{27}, P(4) = \frac{34}{64}, P(5) = \frac{65}{125}&lt;/math&gt;). Since real numbers contain infinite increments, the answer is the limit as &lt;math&gt;n&lt;/math&gt; approaches infinity of &lt;math&gt;P(n)&lt;/math&gt;, which is easily hypothesized as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;C.&lt;/math&gt;<br /> -solution by fidgetboss_4000 get rect<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_23&diff=120205 2016 AMC 12A Problems/Problem 23 2020-03-27T21:35:49Z <p>Emerald block: added clarification</p> <hr /> <div>==Problem==<br /> <br /> Three numbers in the interval &lt;math&gt;\left[0,1\right]&lt;/math&gt; are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1: Super WLOG===<br /> <br /> WLOG assume &lt;math&gt;a&lt;/math&gt; is the largest. Scale the triangle to &lt;math&gt;1,{b}/{a},{c}/{a}&lt;/math&gt; or &lt;math&gt;1,x,y&lt;/math&gt; with &lt;math&gt;x,y \in [0,1]&lt;/math&gt; and equally likely to be any pair of numbers within the interval. Then, &lt;math&gt;x+y&gt;1&lt;/math&gt;, meaning the solution is &lt;math&gt;\boxed{\textbf{(C)}\;1/2}&lt;/math&gt;, as shown in the graph below.<br /> &lt;asy&gt;<br /> pair A = (0,0);<br /> pair B = (1,0);<br /> pair C = (1,1);<br /> pair D = (0,1);<br /> pair E = (0,0);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(A--C,dashed);<br /> fill(A--D--C--cycle,gray);<br /> <br /> label(&quot;$0$&quot;,A,SW);<br /> label(&quot;$1$&quot;,B,S);<br /> label(&quot;$1$&quot;,D,W);<br /> label(&quot;$y$&quot;,(0,.5),W);<br /> label(&quot;$x$&quot;,(.5,0),S);<br /> label(&quot;$x+y&gt;1$&quot;,(2/7,5/7));<br /> &lt;/asy&gt;<br /> <br /> ===Solution 2: Conditional Probability===<br /> <br /> WLOG, let the largest of the three numbers drawn be &lt;math&gt;a&gt;0&lt;/math&gt;. Then the other two numbers are drawn uniformly and independently from the interval &lt;math&gt;[0,a]&lt;/math&gt;. The probability that their sum is greater than &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\;1/2.}&lt;/math&gt;<br /> <br /> ===Solution 3: Calculus===<br /> <br /> When &lt;math&gt;a&gt;b&lt;/math&gt;, consider two cases:<br /> <br /> 1) &lt;math&gt;0&lt;a&lt;\frac{1}{2}&lt;/math&gt;, then <br /> &lt;math&gt;\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}&lt;/math&gt;<br /> <br /> 2)&lt;math&gt;\frac{1}{2}&lt;a&lt;1&lt;/math&gt;, then <br /> &lt;math&gt;\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}&lt;/math&gt;<br /> <br /> &lt;math&gt;a&lt;b&lt;/math&gt; is the same. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> ===Solution 4: Geometry===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. The region where, WLOG, side &lt;math&gt;z&lt;/math&gt; is too long, &lt;math&gt;z\geq x+y&lt;/math&gt;, is a pyramid with a base of area &lt;math&gt;\frac{1}{2}&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so its volume is &lt;math&gt;\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}&lt;/math&gt;. Accounting for the corresponding cases in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; multiplies our answer by &lt;math&gt;3&lt;/math&gt;, so we have excluded a total volume of &lt;math&gt;\frac{1}{2}&lt;/math&gt; from the space of possible probabilities. Subtracting this from &lt;math&gt;1&lt;/math&gt; leaves us with a final answer of &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> === Solution 5: More Calculus ===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when &lt;math&gt;x + y &lt; z&lt;/math&gt;, which has area &lt;math&gt;\frac{z^2}{2}&lt;/math&gt; or when &lt;math&gt;x+z&lt;y&lt;/math&gt; or &lt;math&gt;y+z&lt;x&lt;/math&gt;, which have an area of &lt;math&gt;\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.&lt;/math&gt; Integrating this expression from 0 to 1 in the form<br /> <br /> &lt;math&gt;\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg[\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}&lt;/math&gt;<br /> <br /> === Solution 6: Geometry in 2-D ===<br /> WLOG assume that &lt;math&gt;z&lt;/math&gt; is the largest number and hence the largest side. Then &lt;math&gt;x,y \leq z&lt;/math&gt;. We can set up a square that is &lt;math&gt;z&lt;/math&gt; by &lt;math&gt;z&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt; plane. We are wanting all the points within this square that satisfy &lt;math&gt;x+y &gt; z&lt;/math&gt;. This happens to be a line dividing the square into 2 equal regions. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> <br /> [][] diagram for this problem goes here (z by z square)<br /> <br /> === Solution 7: More WLOG, Complementary Probability ===<br /> The triangle inequality simplifies to considering only one case: &lt;math&gt;\text{the smallest side+ the second smallest side} &gt; \text{the largest side}&lt;/math&gt;. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) &lt;math&gt;a&lt;/math&gt; is the largest, so on average &lt;math&gt;a=1/2&lt;/math&gt; (now equal to becomes a degenerate case with probability &lt;math&gt;0&lt;/math&gt;, so we no longer need to consider it). We now want &lt;math&gt;b+c&lt;1/2&lt;/math&gt;, so imagine choosing &lt;math&gt;b+c&lt;/math&gt; at once rather than independently. But we know that &lt;math&gt;b+c&lt;/math&gt; is between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. The complement is thus: &lt;math&gt;(1/2-0)/2=1/4&lt;/math&gt;. But keep in mind that we choose each &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; randomly and independently, so if there are &lt;math&gt;k&lt;/math&gt; ways to choose &lt;math&gt;b+c&lt;/math&gt; together, there are &lt;math&gt;2k&lt;/math&gt; ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if &lt;math&gt;b+c=3&lt;/math&gt;, then we only count this once, but in reality: we have two cases &lt;math&gt;1+2&lt;/math&gt;, and &lt;math&gt;2+1&lt;/math&gt;; similar reasoning also generalizes to non-integral values). The complement is then actually &lt;math&gt;2(1/4)=1/2&lt;/math&gt;. Therefore, our desired probability is given by &lt;math&gt;1-\text{complement}=1/2, C&lt;/math&gt;<br /> <br /> === Solution 8: 3D geometry ===<br /> <br /> We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines &lt;math&gt;x+y&gt;z, x+z&gt;y,&lt;/math&gt; and &lt;math&gt;y+z&gt;x,&lt;/math&gt; We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length &lt;math&gt;\sqrt(2)&lt;/math&gt; and the other has 3 sides of length &lt;math&gt;\sqrt(2)&lt;/math&gt; and 3 sides of length &lt;math&gt;1.&lt;/math&gt; The volume of this region is &lt;math&gt;\frac 1 2&lt;/math&gt;. Hence our solution is &lt;math&gt;C.&lt;/math&gt;<br /> <br /> === Solution 9: Cheap Solution ===<br /> <br /> Pretend that the values of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are integers ranging from &lt;math&gt;[1,n]&lt;/math&gt;. Test out the probability of the first few values of &lt;math&gt;n&lt;/math&gt; (for example, &lt;math&gt;P(2) = \frac{5}{8}, P(3) = \frac{15}{27}, P(4) = \frac{34}{64}, P(5) = \frac{65}{125}&lt;/math&gt;). Since real numbers contain infinite increments, the answer is the limit as &lt;math&gt;n&lt;/math&gt; approaches infinity of &lt;math&gt;P(n)&lt;/math&gt;, which is easily hypothesized as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;C.&lt;/math&gt;<br /> -solution by fidgetboss_4000 get rect<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_23&diff=120204 2016 AMC 12A Problems/Problem 23 2020-03-27T21:34:44Z <p>Emerald block: added clarification and image</p> <hr /> <div>==Problem==<br /> <br /> Three numbers in the interval &lt;math&gt;\left[0,1\right]&lt;/math&gt; are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1: Super WLOG===<br /> <br /> WLOG assume &lt;math&gt;a&lt;/math&gt; is the largest. Scale the triangle to &lt;math&gt;1,{b}/{a},{c}/{a}&lt;/math&gt; or &lt;math&gt;1,x,y&lt;/math&gt; with &lt;math&gt;x,y \in [0,1]&lt;/math&gt; and equally likely to be any pair of numbers. Then, &lt;math&gt;x+y&gt;1&lt;/math&gt;, meaning the solution is &lt;math&gt;\boxed{\textbf{(C)}\;1/2}&lt;/math&gt;, as shown in the graph below.<br /> &lt;asy&gt;<br /> pair A = (0,0);<br /> pair B = (1,0);<br /> pair C = (1,1);<br /> pair D = (0,1);<br /> pair E = (0,0);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(A--C,dashed);<br /> fill(A--D--C--cycle,gray);<br /> <br /> label(&quot;$0$&quot;,A,SW);<br /> label(&quot;$1$&quot;,B,S);<br /> label(&quot;$1$&quot;,D,W);<br /> label(&quot;$y$&quot;,(0,.5),W);<br /> label(&quot;$x$&quot;,(.5,0),S);<br /> label(&quot;$x+y&gt;1$&quot;,(2/7,5/7));<br /> &lt;/asy&gt;<br /> <br /> ===Solution 2: Conditional Probability===<br /> <br /> WLOG, let the largest of the three numbers drawn be &lt;math&gt;a&gt;0&lt;/math&gt;. Then the other two numbers are drawn uniformly and independently from the interval &lt;math&gt;[0,a]&lt;/math&gt;. The probability that their sum is greater than &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\;1/2.}&lt;/math&gt;<br /> <br /> ===Solution 3: Calculus===<br /> <br /> When &lt;math&gt;a&gt;b&lt;/math&gt;, consider two cases:<br /> <br /> 1) &lt;math&gt;0&lt;a&lt;\frac{1}{2}&lt;/math&gt;, then <br /> &lt;math&gt;\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}&lt;/math&gt;<br /> <br /> 2)&lt;math&gt;\frac{1}{2}&lt;a&lt;1&lt;/math&gt;, then <br /> &lt;math&gt;\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}&lt;/math&gt;<br /> <br /> &lt;math&gt;a&lt;b&lt;/math&gt; is the same. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> ===Solution 4: Geometry===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. The region where, WLOG, side &lt;math&gt;z&lt;/math&gt; is too long, &lt;math&gt;z\geq x+y&lt;/math&gt;, is a pyramid with a base of area &lt;math&gt;\frac{1}{2}&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so its volume is &lt;math&gt;\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}&lt;/math&gt;. Accounting for the corresponding cases in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; multiplies our answer by &lt;math&gt;3&lt;/math&gt;, so we have excluded a total volume of &lt;math&gt;\frac{1}{2}&lt;/math&gt; from the space of possible probabilities. Subtracting this from &lt;math&gt;1&lt;/math&gt; leaves us with a final answer of &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> === Solution 5: More Calculus ===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when &lt;math&gt;x + y &lt; z&lt;/math&gt;, which has area &lt;math&gt;\frac{z^2}{2}&lt;/math&gt; or when &lt;math&gt;x+z&lt;y&lt;/math&gt; or &lt;math&gt;y+z&lt;x&lt;/math&gt;, which have an area of &lt;math&gt;\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.&lt;/math&gt; Integrating this expression from 0 to 1 in the form<br /> <br /> &lt;math&gt;\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg[\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}&lt;/math&gt;<br /> <br /> === Solution 6: Geometry in 2-D ===<br /> WLOG assume that &lt;math&gt;z&lt;/math&gt; is the largest number and hence the largest side. Then &lt;math&gt;x,y \leq z&lt;/math&gt;. We can set up a square that is &lt;math&gt;z&lt;/math&gt; by &lt;math&gt;z&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt; plane. We are wanting all the points within this square that satisfy &lt;math&gt;x+y &gt; z&lt;/math&gt;. This happens to be a line dividing the square into 2 equal regions. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> <br /> [][] diagram for this problem goes here (z by z square)<br /> <br /> === Solution 7: More WLOG, Complementary Probability ===<br /> The triangle inequality simplifies to considering only one case: &lt;math&gt;\text{the smallest side+ the second smallest side} &gt; \text{the largest side}&lt;/math&gt;. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) &lt;math&gt;a&lt;/math&gt; is the largest, so on average &lt;math&gt;a=1/2&lt;/math&gt; (now equal to becomes a degenerate case with probability &lt;math&gt;0&lt;/math&gt;, so we no longer need to consider it). We now want &lt;math&gt;b+c&lt;1/2&lt;/math&gt;, so imagine choosing &lt;math&gt;b+c&lt;/math&gt; at once rather than independently. But we know that &lt;math&gt;b+c&lt;/math&gt; is between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. The complement is thus: &lt;math&gt;(1/2-0)/2=1/4&lt;/math&gt;. But keep in mind that we choose each &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; randomly and independently, so if there are &lt;math&gt;k&lt;/math&gt; ways to choose &lt;math&gt;b+c&lt;/math&gt; together, there are &lt;math&gt;2k&lt;/math&gt; ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if &lt;math&gt;b+c=3&lt;/math&gt;, then we only count this once, but in reality: we have two cases &lt;math&gt;1+2&lt;/math&gt;, and &lt;math&gt;2+1&lt;/math&gt;; similar reasoning also generalizes to non-integral values). The complement is then actually &lt;math&gt;2(1/4)=1/2&lt;/math&gt;. Therefore, our desired probability is given by &lt;math&gt;1-\text{complement}=1/2, C&lt;/math&gt;<br /> <br /> === Solution 8: 3D geometry ===<br /> <br /> We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines &lt;math&gt;x+y&gt;z, x+z&gt;y,&lt;/math&gt; and &lt;math&gt;y+z&gt;x,&lt;/math&gt; We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length &lt;math&gt;\sqrt(2)&lt;/math&gt; and the other has 3 sides of length &lt;math&gt;\sqrt(2)&lt;/math&gt; and 3 sides of length &lt;math&gt;1.&lt;/math&gt; The volume of this region is &lt;math&gt;\frac 1 2&lt;/math&gt;. Hence our solution is &lt;math&gt;C.&lt;/math&gt;<br /> <br /> === Solution 9: Cheap Solution ===<br /> <br /> Pretend that the values of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are integers ranging from &lt;math&gt;[1,n]&lt;/math&gt;. Test out the probability of the first few values of &lt;math&gt;n&lt;/math&gt; (for example, &lt;math&gt;P(2) = \frac{5}{8}, P(3) = \frac{15}{27}, P(4) = \frac{34}{64}, P(5) = \frac{65}{125}&lt;/math&gt;). Since real numbers contain infinite increments, the answer is the limit as &lt;math&gt;n&lt;/math&gt; approaches infinity of &lt;math&gt;P(n)&lt;/math&gt;, which is easily hypothesized as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;C.&lt;/math&gt;<br /> -solution by fidgetboss_4000 get rect<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=120086 2005 AMC 12A Problems/Problem 22 2020-03-25T17:44:54Z <p>Emerald block: fixed spacing error</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 112. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Its surface area is &lt;cmath&gt;2lw+2lh+2wl=384,&lt;/cmath&gt;<br /> and the sum of all its edges is &lt;cmath&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}.&lt;/cmath&gt;<br /> Notice that &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,&lt;/cmath&gt; so the diameter is<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.&lt;/cmath&gt; The radius is half of the diameter, so<br /> &lt;cmath&gt;r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> As in the previous solution, we have that &lt;math&gt;2lw+2lh+2wl=384&lt;/math&gt; and &lt;math&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28&lt;/math&gt;, and the diameter of the sphere is the space diagonal of the prism, &lt;math&gt;\sqrt{l^2 + w^2 + h^2}&lt;/math&gt;.<br /> <br /> <br /> Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that &lt;math&gt;h=0&lt;/math&gt;. (This essentially means that we have an infinitesimally thin box.) We now have that &lt;math&gt;2lw = 384&lt;/math&gt; and &lt;math&gt;l + w = 28&lt;/math&gt;, and we are solving for &lt;math&gt;\sqrt{l^2 + w^2}&lt;/math&gt;. Because &lt;cmath&gt;(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,&lt;/cmath&gt; this means that &lt;cmath&gt;l^2 + w^2 = 28^2 - 384 = 400,&lt;/cmath&gt; so the space diagonal is &lt;math&gt;\sqrt{400} = 20&lt;/math&gt;. Since the diameter of the sphere is &lt;math&gt;20&lt;/math&gt;, the radius is &lt;math&gt;\boxed{\textbf{(B) } 10}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=120085 2005 AMC 12A Problems/Problem 22 2020-03-25T17:44:22Z <p>Emerald block: new solution</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 112. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Its surface area is &lt;cmath&gt;2lw+2lh+2wl=384,&lt;/cmath&gt;<br /> and the sum of all its edges is &lt;cmath&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}.&lt;/cmath&gt;<br /> Notice that &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,&lt;/cmath&gt; so the diameter is<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.&lt;/cmath&gt; The radius is half of the diameter, so<br /> &lt;cmath&gt;r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> As in the previous solution, we have that &lt;math&gt;2lw+2lh+2wl=384&lt;/math&gt; and &lt;math&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28&lt;/math&gt;, and the diameter of the sphere is the space diagonal of the prism, &lt;math&gt;\sqrt{l^2 + w^2 + h^2}&lt;/math&gt;.<br /> <br /> <br /> Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that &lt;math&gt;h=0&lt;/math&gt;. (This essentially means that we have an infinitesimally thin box.) We now have that &lt;math&gt;2lw = 384&lt;/math&gt; and &lt;math&gt;l + w = 28&lt;/math&gt;, and we are solving for &lt;math&gt;\sqrt{l^2 + w^2}&lt;/math&gt;. Because &lt;cmath&gt;(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,&lt;/cmath&gt; this means that &lt;cmath&gt;l^2 + w^2 = 28^2 - 384 = 400,&lt;/cmath&gt; so the space diagonal is &lt;math&gt;\sqrt{400} = 20&lt;/math&gt;. Since the diameter of the sphere is &lt;math&gt;20&lt;/math&gt;, the radius is &lt;math&gt;\boxed{\textbf{(B)} 10}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_25&diff=119960 2007 AMC 12A Problems/Problem 25 2020-03-22T19:07:37Z <p>Emerald block: New solution</p> <hr /> <div>== Problem ==<br /> Call a set of integers ''spacy'' if it contains no more than one out of any three consecutive integers. How many [[subset]]s of &lt;math&gt;\{1,2,3,\ldots,12\},&lt;/math&gt; including the [[empty set]], are spacy?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129&lt;/math&gt;<br /> <br /> __TOC__<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> Let &lt;math&gt;S_{n}&lt;/math&gt; denote the number of spacy subsets of &lt;math&gt;\{ 1, 2, ... n \}&lt;/math&gt;. We have &lt;math&gt;S_{0} = 1, S_{1} = 2, S_{2} = 3&lt;/math&gt;. <br /> <br /> The spacy subsets of &lt;math&gt;S_{n + 1}&lt;/math&gt; can be divided into two groups:<br /> *&lt;math&gt;A = &lt;/math&gt; those not containing &lt;math&gt;n + 1&lt;/math&gt;. Clearly &lt;math&gt;|A|=S_{n}&lt;/math&gt;. <br /> *&lt;math&gt;B = &lt;/math&gt; those containing &lt;math&gt;n + 1&lt;/math&gt;. We have &lt;math&gt;|B|=S_{n - 2}&lt;/math&gt;, since removing &lt;math&gt;n + 1&lt;/math&gt; from any set in &lt;math&gt;B&lt;/math&gt; produces a spacy set with all elements at most equal to &lt;math&gt;n - 2,&lt;/math&gt; and each such spacy set can be constructed from exactly one spacy set in &lt;math&gt;B&lt;/math&gt;.<br /> Hence,<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;S_{n + 1} = S_{n} + S_{n - 2}&lt;/math&gt;&lt;/div&gt;<br /> <br /> From this [[recursion]], we find that<br /> <br /> {| class=&quot;wikitable&quot; border=&quot;1px solid&quot;<br /> |-<br /> | &lt;math&gt;S(0)&lt;/math&gt; || &lt;math&gt;S(1)&lt;/math&gt; || &lt;math&gt;S(2)&lt;/math&gt; || &lt;math&gt;S(3)&lt;/math&gt; || &lt;math&gt;S(4)&lt;/math&gt; || &lt;math&gt;S(5)&lt;/math&gt; || &lt;math&gt;S(6)&lt;/math&gt; || &lt;math&gt;S(7)&lt;/math&gt; || &lt;math&gt;S(8)&lt;/math&gt; || &lt;math&gt;S(9)&lt;/math&gt; || &lt;math&gt;S(10)&lt;/math&gt; || &lt;math&gt;S(11)&lt;/math&gt; || &lt;math&gt;S(12)&lt;/math&gt; || <br /> |-<br /> | 1 || 2 || 3 || 4 || 6 || 9 || 13 || 19 || 28 || 41 || 60 || 88 || 129<br /> |}<br /> And so the answer is &lt;math&gt;(E)&lt;/math&gt; &lt;math&gt;129&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.<br /> <br /> From the set &lt;math&gt;\{1,2,3,4,5,6,7,8,9,10,11,12\}&lt;/math&gt; we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents &lt;math&gt;{1,4,7,10}&lt;/math&gt;.<br /> <br /> For subsets of size &lt;math&gt;k&lt;/math&gt; there must be &lt;math&gt;2(k - 1)&lt;/math&gt; dividers between the balls, leaving &lt;math&gt;12 - k - 2(k - 1) = 12 - 3k + 2&lt;/math&gt; dividers to be be placed in &lt;math&gt;k + 1&lt;/math&gt; spots between the balls. The number of way this can be done is &lt;math&gt;\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k&lt;/math&gt;.<br /> <br /> Therefore, the number of spacy subsets is &lt;math&gt;\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = \boxed{129} &lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most &lt;math&gt;4&lt;/math&gt; elements. Given any arrangment, we subract &lt;math&gt;2i-2&lt;/math&gt; from the &lt;math&gt;i-th&lt;/math&gt; element in our subset, when the elements are arranged in increasing order. This creates a [[bijection]] with the number of size &lt;math&gt;k&lt;/math&gt; subsets of the set of the first &lt;math&gt;14-2k&lt;/math&gt; positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for &lt;math&gt;k&lt;/math&gt;: &lt;math&gt;{14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}&lt;/math&gt;<br /> <br /> In general, the number of subsets of a set with &lt;math&gt;n&lt;/math&gt; elements and with no &lt;math&gt;k&lt;/math&gt; consecutive numbers is &lt;math&gt;\sum^{\lfloor{\frac{n}{k}}\rfloor}_{i=0}{{n-(k-1)(i-1) \choose i}}&lt;/math&gt;.<br /> <br /> === Solution 4 (Casework) ===<br /> Let us consider each size of subset individually. Since each integer in the subset must be at least &lt;math&gt;3&lt;/math&gt; away from any other integer in the subset, the largest spacey subset contains &lt;math&gt;4&lt;/math&gt; elements.<br /> <br /> <br /> First, it is clear that there is &lt;math&gt;1&lt;/math&gt; spacey set with &lt;math&gt;0&lt;/math&gt; elements in it, the empty set. Next, there are &lt;math&gt;12&lt;/math&gt; spacey subsets with &lt;math&gt;1&lt;/math&gt; element in them, one for each integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;12&lt;/math&gt;.<br /> <br /> <br /> Now, let us consider the spacey subsets with &lt;math&gt;2&lt;/math&gt; elements in them. If the smaller integer is &lt;math&gt;1&lt;/math&gt;, the larger integer is any of the &lt;math&gt;9&lt;/math&gt; integers from &lt;math&gt;4&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt;. If the smaller integer is &lt;math&gt;2&lt;/math&gt;, the larger integer is any of the &lt;math&gt;8&lt;/math&gt; integers from &lt;math&gt;5&lt;/math&gt; to &lt;math&gt;12&lt;/math&gt;. This continues, up to a smaller integer of &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt; choice for the larger integer, &lt;math&gt;12&lt;/math&gt;. This means that there are &lt;math&gt;9 + 8 + \cdots + 1 = 45&lt;/math&gt; spacey subsets with &lt;math&gt;2&lt;/math&gt; elements.<br /> <br /> <br /> For spacey subsets with &lt;math&gt;3&lt;/math&gt; elements, we first consider the middle integer. The smallest such integer is &lt;math&gt;4&lt;/math&gt;, and it allows for &lt;math&gt;1&lt;/math&gt; possible value for the smaller integer (&lt;math&gt;1&lt;/math&gt;) and possible &lt;math&gt;6&lt;/math&gt; values for the larger integer (&lt;math&gt;7&lt;/math&gt; through &lt;math&gt;12&lt;/math&gt;), for a total of &lt;math&gt;1 \cdot 6 = 6&lt;/math&gt; possible subsets. The next middle integer, &lt;math&gt;5&lt;/math&gt;, allows for &lt;math&gt;2&lt;/math&gt; smaller integers and &lt;math&gt;5&lt;/math&gt; larger integers, and this pattern continues up until the middle integer of &lt;math&gt;9&lt;/math&gt;, which has &lt;math&gt;6&lt;/math&gt; values for the smaller integer and &lt;math&gt;1&lt;/math&gt; value for the larger integer. This means that there are &lt;math&gt;1 \cdot 6 + 2 \cdot 5 + \cdots + 6 \cdot 1 = 56&lt;/math&gt; spacey subsets with &lt;math&gt;3&lt;/math&gt; elements.<br /> <br /> <br /> Lastly, there are &lt;math&gt;3&lt;/math&gt; main categories for spacey subsets with &lt;math&gt;4&lt;/math&gt; elements, defined by the difference between their smallest and largest values. The difference ranges from &lt;math&gt;9&lt;/math&gt; to &lt;math&gt;11&lt;/math&gt;. If it is &lt;math&gt;9&lt;/math&gt;, there is only &lt;math&gt;1&lt;/math&gt; set of places to put the two middle values (&lt;math&gt;n + 3&lt;/math&gt; and &lt;math&gt;n + 6&lt;/math&gt;, where &lt;math&gt;n&lt;/math&gt; is the smallest value). Since there are &lt;math&gt;3&lt;/math&gt; possible sets of smallest and largest values, there are &lt;math&gt;1 \cdot 3 = 3&lt;/math&gt; sets in this category. If the difference is &lt;math&gt;10&lt;/math&gt;, there are now &lt;math&gt;3&lt;/math&gt; sets of places to put the two middle values (&lt;math&gt;n + 3&lt;/math&gt; and &lt;math&gt;n + 6&lt;/math&gt; or &lt;math&gt;7&lt;/math&gt;, and &lt;math&gt;n + 4&lt;/math&gt; and &lt;math&gt;n + 7&lt;/math&gt;). There are &lt;math&gt;2&lt;/math&gt; possible sets of smallest and largest values, so there are &lt;math&gt;3 \cdot 2 = 6&lt;/math&gt; sets in this category. Finally, if the difference is &lt;math&gt;11&lt;/math&gt;, there are &lt;math&gt;6&lt;/math&gt; possible sets of places to put the two middle values (&lt;math&gt;n + 3&lt;/math&gt; and &lt;math&gt;n + 6&lt;/math&gt;, &lt;math&gt;7&lt;/math&gt;, or &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;n + 4&lt;/math&gt; and &lt;math&gt;n + 7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;, and &lt;math&gt;n + 5&lt;/math&gt; and &lt;math&gt;n + 8&lt;/math&gt;) and one possible set of smallest and largest values, meaning that there are &lt;math&gt;6 \cdot 1 = 6&lt;/math&gt; sets in this category. Adding them up, there are &lt;math&gt;3 + 6 + 6 = 15&lt;/math&gt; spacey subsets with &lt;math&gt;4&lt;/math&gt; elements.<br /> <br /> <br /> Adding these all up, we have a total of &lt;math&gt;1 + 12 + 45 + 56 + 15 = \boxed{\mathrm{(E)}\ 129}&lt;/math&gt; spacey subsets. ~[[User:emerald_block|emerald_block]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_2&diff=119725 2010 AIME II Problems/Problem 2 2020-03-18T18:36:44Z <p>Emerald block: fixed grammatical errors and added clarification</p> <hr /> <div>== Problem 2 ==<br /> A point &lt;math&gt;P&lt;/math&gt; is chosen at random in the interior of a unit square &lt;math&gt;S&lt;/math&gt;. Let &lt;math&gt;d(P)&lt;/math&gt; denote the distance from &lt;math&gt;P&lt;/math&gt; to the closest side of &lt;math&gt;S&lt;/math&gt;. The probability that &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is equal to &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> <br /> == Solution ==<br /> <br /> Any point outside the square with side length &lt;math&gt;\frac{1}{3}&lt;/math&gt; that has the same center and orientation as the unit square and inside the square with side length &lt;math&gt;\frac{3}{5}&lt;/math&gt; that has the same center and orientation as the unit square has &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt;.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(1mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> draw((0,0)--(0,30)--(30,30)--(30,0)--cycle);<br /> draw((6,6)--(6,24)--(24,24)--(24,6)--cycle);<br /> draw((10,10)--(10,20)--(20,20)--(20,10)--cycle);<br /> fill((6,6)--(6,24)--(24,24)--(24,6)--cycle,gray);<br /> fill((10,10)--(10,20)--(20,20)--(20,10)--cycle,white);<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> Since the area of the unit square is &lt;math&gt;1&lt;/math&gt;, the probability of a point &lt;math&gt;P&lt;/math&gt; with &lt;math&gt;\frac{1}{5}\le d(P)\le\frac{1}{3}&lt;/math&gt; is the area of the shaded region, which is the difference of the area of two squares.<br /> <br /> &lt;math&gt;\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}&lt;/math&gt;<br /> <br /> Thus, the answer is &lt;math&gt;56 + 225 = \boxed{281}.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2010|num-b=1|num-a=3|n=II}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_1&diff=119724 2010 AIME II Problems/Problem 1 2020-03-18T18:33:35Z <p>Emerald block: Fixed modulo equation</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;N&lt;/math&gt; be the greatest integer multiple of &lt;math&gt;36&lt;/math&gt; all of whose digits are even and no two of whose digits are the same. Find the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> == Solution ==<br /> If an integer is divisible by &lt;math&gt;36&lt;/math&gt;, it must also be divisible by &lt;math&gt;9&lt;/math&gt; since &lt;math&gt;9&lt;/math&gt; is a factor of &lt;math&gt;36&lt;/math&gt;. It is a well-known fact that, if &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is a multiple of &lt;math&gt;9&lt;/math&gt;. Hence, if &lt;math&gt;N&lt;/math&gt; contains all the even digits, the sum of the digits would be &lt;math&gt;0 + 2 + 4 + 6 + 8 = 20&lt;/math&gt;, which is not divisible by &lt;math&gt;9&lt;/math&gt; and thus &lt;math&gt;36&lt;/math&gt;.<br /> The next logical try would be &lt;math&gt;8640&lt;/math&gt;, which happens to be divisible by &lt;math&gt;36&lt;/math&gt;. Thus, &lt;math&gt;N = 8640 \equiv \boxed{640} \pmod {1000}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2010|before=First Problem|num-a=2|n=II}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_1&diff=119722 2006 AMC 8 Problems/Problem 1 2020-03-18T17:14:17Z <p>Emerald block: fixed small grammar mistake and added clarification</p> <hr /> <div>== Problem ==<br /> Mindy made three purchases for &lt;math&gt; \textdollar 1.98&lt;/math&gt; dollars, &lt;math&gt; \textdollar 5.04 &lt;/math&gt; dollars, and &lt;math&gt; \textdollar 9.89&lt;/math&gt; dollars. What was her total, to the nearest dollar? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> == Solution ==<br /> The three prices round to &lt;math&gt; \textdollar 2 &lt;/math&gt;, &lt;math&gt; \textdollar 5 &lt;/math&gt;, and &lt;math&gt; \textdollar 10 &lt;/math&gt;, which have a sum of &lt;math&gt; \boxed{\textbf{(D)}\ 17} &lt;/math&gt;.<br /> <br /> We know that there will not be a rounding error, as the total amount rounded is clearly less than &lt;math&gt; \textdollar 0.50 &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2006|before=First &lt;br /&gt;Question|num-a=2}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_21&diff=119660 2017 AMC 12A Problems/Problem 21 2020-03-18T00:19:30Z <p>Emerald block: added shorter expressions in section for determining elements</p> <hr /> <div>==Problem==<br /> <br /> A set &lt;math&gt;S&lt;/math&gt; is constructed as follows. To begin, &lt;math&gt;S = \{0,10\}&lt;/math&gt;. Repeatedly, as long as possible, if &lt;math&gt;x&lt;/math&gt; is an integer root of some polynomial &lt;math&gt;a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0&lt;/math&gt; for some &lt;math&gt;n\geq{1}&lt;/math&gt;, all of whose coefficients &lt;math&gt;a_i&lt;/math&gt; are elements of &lt;math&gt;S&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; is put into &lt;math&gt;S&lt;/math&gt;. When no more elements can be added to &lt;math&gt;S&lt;/math&gt;, how many elements does &lt;math&gt;S&lt;/math&gt; have?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4<br /> \qquad \textbf{(B)}\ 5<br /> \qquad\textbf{(C)}\ 7<br /> \qquad\textbf{(D)}\ 9<br /> \qquad\textbf{(E)}\ 11&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> At first, &lt;math&gt;S=\{0,10\}&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;\begin{tabular}{r c l c l}<br /> $$10x+10$$ &amp; has root &amp; $$x=-1$$ &amp; so now &amp; $$S=\{-1,0,10\}$$ \\<br /> $$-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10$$ &amp; has root &amp; $$x=1$$ &amp; so now &amp; $$S=\{-1,0,1,10\}$$ \\<br /> $$x+10$$ &amp; has root &amp; $$x=-10$$ &amp; so now &amp; $$S=\{-10,-1,0,1,10\}$$ \\<br /> $$x^3+x-10$$ &amp; has root &amp; $$x=2$$ &amp; so now &amp; $$S=\{-10,-1,0,1,2,10\}$$ \\<br /> $$x+2$$ &amp; has root &amp; $$x=-2$$ &amp; so now &amp; $$S=\{-10,-2,-1,0,1,2,10\}$$ \\<br /> $$2x-10$$ &amp; has root &amp; $$x=5$$ &amp; so now &amp; $$S=\{-10,-2,-1,0,1,2,5,10\}$$ \\<br /> $$x+5$$ &amp; has root &amp; $$x=-5$$ &amp; so now &amp; $$S=\{-10,-5,-2,-1,0,1,2,5,10\}$$<br /> \end{tabular}&lt;/cmath&gt;<br /> <br /> At this point, no more elements can be added to &lt;math&gt;S&lt;/math&gt;. To see this, let<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &amp;= 0 \\<br /> x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &amp;= 0 \\<br /> x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &amp;= -a_0<br /> \end{align*}&lt;/cmath&gt;<br /> <br /> with each &lt;math&gt;a_i&lt;/math&gt; in &lt;math&gt;S&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is a factor of &lt;math&gt;a_0&lt;/math&gt;, and &lt;math&gt;a_0&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt;, so &lt;math&gt;x&lt;/math&gt; has to be a factor of some element in &lt;math&gt;S&lt;/math&gt;. There are no such integers left, so there can be no more additional elements. &lt;math&gt;\{-10,-5,-2,-1,0,1,2,5,10\}&lt;/math&gt; has &lt;math&gt;9&lt;/math&gt; elements &lt;math&gt;\to \boxed{\textbf{(D)}}&lt;/math&gt;<br /> <br /> ==Solution 2 (If you are short on time)==<br /> <br /> By Rational Root Theorem, the only rational roots for this function we're dealing with must be in the form &lt;math&gt; \pm \frac p{q} &lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are co-prime, &lt;math&gt;p&lt;/math&gt; is a factor of &lt;math&gt;a_0&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; is a factor of &lt;math&gt;a_n&lt;/math&gt;. We can easily see &lt;math&gt;-1&lt;/math&gt; is in &lt;math&gt;S&lt;/math&gt; because of &lt;math&gt;10x + 10 = 0&lt;/math&gt; has root &lt;math&gt;-1&lt;/math&gt;. Since we want set &lt;math&gt;S&lt;/math&gt; to be as large as possible, we let &lt;math&gt;p=10&lt;/math&gt; and &lt;math&gt;q=-1&lt;/math&gt;, and quickly see that all possible integer roots are &lt;math&gt;\pm 1&lt;/math&gt;, &lt;math&gt;\pm 2&lt;/math&gt;, &lt;math&gt;\pm 5&lt;/math&gt;, &lt;math&gt;\pm 10&lt;/math&gt;, plus the &lt;math&gt;0&lt;/math&gt; we started with, we get a total of &lt;math&gt;9&lt;/math&gt; elements &lt;math&gt;\to \boxed{\textbf{(D)}}&lt;/math&gt;<br /> <br /> -BochTheNerd<br /> <br /> == Solution 3 (If you are also short on time) ==<br /> <br /> By the Rational Root theorem, notice that we must have &lt;math&gt;x | a_0&lt;/math&gt;. Since &lt;math&gt;a_0 \in S&lt;/math&gt;, this implies that any &lt;math&gt;x&lt;/math&gt; added must be a factor of a certain element in &lt;math&gt;S&lt;/math&gt; before. This therefore implies that any &lt;math&gt;x&lt;/math&gt;'s added must be a factor of &lt;math&gt;10&lt;/math&gt;. Thus, the largest possible set is all the positive and negative factors of &lt;math&gt;10&lt;/math&gt;, hence &lt;math&gt;\boxed{9}&lt;/math&gt;. <br /> <br /> Note: this solution is not a real solution because it does not show that each &lt;math&gt;x&lt;/math&gt; actually works (basically we have found the maximum possible elements but we have not shown that there is a polynomial for each of them to work).<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2017|ab=A|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=1999_AHSME_Problems/Problem_6&diff=119613 1999 AHSME Problems/Problem 6 2020-03-17T03:59:18Z <p>Emerald block: Deleted non-solution</p> <hr /> <div>==Problem==<br /> <br /> What is the sum of the digits of the decimal form of the product &lt;math&gt; 2^{1999}\cdot 5^{2001}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}&lt;/math&gt;, a number with the digits &quot;25&quot; followed by 1999 zeros. The sum of the digits in the decimal form would be &lt;math&gt;2+5=7&lt;/math&gt;, thus making the answer &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1999|num-b=5|num-a=7}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_18&diff=119485 2010 AMC 10A Problems/Problem 18 2020-03-16T03:06:04Z <p>Emerald block: small grammar fix</p> <hr /> <div>== Problem ==<br /> Bernardo randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,4,5,6,7,8,9\}&lt;/math&gt; and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,4,5,6,7,8\}&lt;/math&gt; and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}&lt;/math&gt;<br /> <br /> == Solution ==<br /> We can solve this by breaking the problem down into &lt;math&gt;2&lt;/math&gt; cases and adding up the probabilities.<br /> <br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Bernardo picks &lt;math&gt;9&lt;/math&gt;.<br /> If Bernardo picks a &lt;math&gt;9&lt;/math&gt; then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;\dfrac{3}{9} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Bernardo does not pick &lt;math&gt;9&lt;/math&gt;.<br /> Since the chance of Bernardo picking &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, the probability of not picking &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;\frac{2}{3}&lt;/math&gt;.<br /> <br /> If Bernardo does not pick 9, then he can pick any number from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;8&lt;/math&gt;. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.<br /> <br /> Ignoring the &lt;math&gt;9&lt;/math&gt; for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.<br /> <br /> We get this probability to be &lt;math&gt;\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}&lt;/math&gt;<br /> <br /> (This was confusing for me, so think of it like this: Bernardo chooses his number. Silva has one out of the total number of options to get the same number. The total number of options is 8 choose 3: &lt;math&gt;\frac{8\cdot{7}\cdot{6}}{3!} = 56&lt;/math&gt; . There are 8 numbers and you choose three to make a three-digit number. Setting it up to be descending doesn't affect the number of ways there are.<br /> )<br /> <br /> Probability of Bernardo's number being greater is<br /> &lt;cmath&gt;\frac{1-\frac{1}{56}}{2} = \frac{55}{112}&lt;/cmath&gt;<br /> <br /> Factoring in the fact that Bernardo could've picked a &lt;math&gt;9&lt;/math&gt; but didn't:<br /> <br /> &lt;cmath&gt;\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}&lt;/cmath&gt;<br /> <br /> Adding up the two cases we get &lt;math&gt;\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=17|num-a=19|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_18&diff=119484 2010 AMC 10A Problems/Problem 18 2020-03-16T03:05:11Z <p>Emerald block: fixed error in equation</p> <hr /> <div>== Problem ==<br /> Bernardo randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,4,5,6,7,8,9\}&lt;/math&gt; and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set &lt;math&gt;\{1,2,3,4,5,6,7,8\}&lt;/math&gt; and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}&lt;/math&gt;<br /> <br /> == Solution ==<br /> We can solve this by breaking the problem down into &lt;math&gt;2&lt;/math&gt; cases and adding up the probabilities.<br /> <br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: Bernardo picks &lt;math&gt;9&lt;/math&gt;.<br /> If Bernardo picks a &lt;math&gt;9&lt;/math&gt; then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;\dfrac{3}{9} = \frac{1}{3}&lt;/math&gt;.<br /> <br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: Bernardo does not pick &lt;math&gt;9&lt;/math&gt;.<br /> Since the chance of Bernardo picking &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;\frac{1}{3}&lt;/math&gt;, the probability of not picking &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;\frac{2}{3}&lt;/math&gt;.<br /> <br /> If Bernardo does not pick 9, then he can pick any number from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;8&lt;/math&gt;. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.<br /> <br /> Ignoring the &lt;math&gt;9&lt;/math&gt; for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.<br /> <br /> We get this probability to be &lt;math&gt;\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}&lt;/math&gt;<br /> <br /> (This was confusing for me, so think of it like this: Bernardo chooses his number. Silva has one out of the total number of options to get the same number. The total number of options is 8 choose 3: &lt;math&gt;\frac{8\cdot{7}\cdot{6}}{3!} = 56&lt;/math&gt; . There are 8 numbers and you choose three to make a three-digit number. Setting it up to be descending doesn't affect the number of ways there are.<br /> )<br /> <br /> Probability of Bernardo's number being greater is<br /> &lt;cmath&gt;\frac{1-\frac{1}{56}}{2} = \frac{55}{112}&lt;/cmath&gt;<br /> <br /> Factoring the fact that Bernardo could've picked a &lt;math&gt;9&lt;/math&gt; but didn't:<br /> <br /> &lt;cmath&gt;\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}&lt;/cmath&gt;<br /> <br /> Adding up the two cases we get &lt;math&gt;\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=17|num-a=19|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_19&diff=119483 2010 AMC 10A Problems/Problem 19 2020-03-16T02:59:21Z <p>Emerald block: Rephrased sentence</p> <hr /> <div>== Problem ==<br /> Equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; has side lengths &lt;math&gt;AB=CD=EF=1&lt;/math&gt; and &lt;math&gt;BC=DE=FA=r&lt;/math&gt;. The area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;70\%&lt;/math&gt; of the area of the hexagon. What is the sum of all possible values of &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> == Solutions ==<br /> This diagram only shows one of the possible values of r.<br /> &lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle);<br /> draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle);<br /> <br /> label(&quot;$E$&quot;,(0,0),SW);<br /> label(&quot;$F$&quot;,(-0.5,0.866),W);<br /> label(&quot;$A$&quot;,(-0.414, 1.015),NW);<br /> label(&quot;$B$&quot;,(0.586,1.015),NE);<br /> label(&quot;$C$&quot;,(0.6716,0.866),E);<br /> label(&quot;$D$&quot;,(0.1716, 0),SE);<br /> label(&quot;1&quot;,(-0.25,0.433),SW);<br /> label(&quot;r&quot;,(-0.457,0.9405),NW);<br /> label(&quot;1&quot;,(0.086,1.015),N);<br /> label(&quot;r&quot;,(0.6288,0.9405),NE);<br /> label(&quot;1&quot;,(0.4216,0.433),SE);<br /> label(&quot;r&quot;,(0.0858,0),S);<br /> <br /> &lt;/asy&gt;<br /> ~Brian<br /> ===Solution 1===<br /> It is clear that &lt;math&gt;\triangle ACE&lt;/math&gt; is an equilateral triangle. From the [[Law of Cosines]] on triangle ABC, we get that &lt;math&gt;AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1&lt;/math&gt;. Therefore, the area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;\frac{\sqrt{3}}{4}(r^2+r+1)&lt;/math&gt;.<br /> <br /> If we extend &lt;math&gt;BC&lt;/math&gt;, &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; so that &lt;math&gt;FA&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt; meet at &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;DE&lt;/math&gt; meet at &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;DE&lt;/math&gt; and &lt;math&gt;FA&lt;/math&gt; meet at &lt;math&gt;Z&lt;/math&gt;, we find that hexagon &lt;math&gt;ABCDEF&lt;/math&gt; is formed by taking equilateral triangle &lt;math&gt;XYZ&lt;/math&gt; of side length &lt;math&gt;r+2&lt;/math&gt; and removing three equilateral triangles, &lt;math&gt;ABX&lt;/math&gt;, &lt;math&gt;CDY&lt;/math&gt; and &lt;math&gt;EFZ&lt;/math&gt;, of side length &lt;math&gt;1&lt;/math&gt;. The area of &lt;math&gt;ABCDEF&lt;/math&gt; is therefore<br /> <br /> &lt;math&gt;\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)&lt;/math&gt;.<br /> <br /> <br /> Based on the initial conditions,<br /> <br /> &lt;cmath&gt;\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)&lt;/cmath&gt;<br /> <br /> Simplifying this gives us &lt;math&gt;r^2-6r+1 = 0&lt;/math&gt;. By [[Vieta's Formulas]] (or Girard identities, or Newton-Girard identities) we know that the sum of the possible value of &lt;math&gt;r&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(E)}\ 6}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> As above, we find that the area of &lt;math&gt;\triangle ACE&lt;/math&gt; is &lt;math&gt;\frac{\sqrt3}4(r^2+r+1)&lt;/math&gt;.<br /> <br /> We also find by the sine [[triangle]] area formula that &lt;math&gt;ABC=CDE=EFA=\frac12\cdot1\cdot r\cdot\frac{\sqrt3}2=\frac{r\sqrt3}4&lt;/math&gt;, and thus<br /> &lt;cmath&gt;\frac{\frac{\sqrt3}4(r^2+r+1)}{\frac{\sqrt3}4(r^2+r+1)+3\left(\frac{r\sqrt3}4\right)}=\frac{r^2+r+1}{r^2+4r+1}=\frac7{10}&lt;/cmath&gt;<br /> This simplifies to &lt;math&gt;r^2-6r+1=0\Rightarrow \boxed{\textbf{(E)}\ 6}&lt;/math&gt;.<br /> <br /> ===Solution 3 (no trig)===<br /> &lt;asy&gt;<br /> unitsize(5cm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> <br /> draw((0,0)--(-0.5,0.866)--(-0.414, 1.015)--(0.586,1.015)--(0.6716,0.866)--(0.1716, 0)--cycle);<br /> draw((0,0)--(-0.414, 1.015)--(0.6716,0.866)--cycle);<br /> draw((0.586,1.015)--(0.6716,1.015)--(0.6716,0.866));<br /> <br /> label(&quot;$E$&quot;,(0,0),SW);<br /> label(&quot;$F$&quot;,(-0.5,0.866),W);<br /> label(&quot;$A$&quot;,(-0.414, 1.015),NW);<br /> label(&quot;$B$&quot;,(0.586,1.015),N);<br /> label(&quot;$C$&quot;,(0.6716,0.866),E);<br /> label(&quot;$D$&quot;,(0.1716, 0),SE);<br /> label(&quot;$M$&quot;,(0.6716,1.015),NE);<br /> label(&quot;1&quot;,(-0.25,0.433),SW);<br /> label(&quot;r&quot;,(-0.457,0.9405),NW);<br /> label(&quot;1&quot;,(0.086,1.015),N);<br /> label(&quot;r&quot;,(0.6288,0.9405),W);<br /> label(&quot;1&quot;,(0.4216,0.433),SE);<br /> label(&quot;r&quot;,(0.0858,0),S);<br /> label(&quot;$\frac{r}{2}$&quot;,(0.6289,1.015),N);<br /> label(&quot;$\frac{\sqrt{3}}{2}r$&quot;,(0.6716,0.9405),E);<br /> <br /> &lt;/asy&gt;<br /> Extend &lt;math&gt;AB&lt;/math&gt; to point M so that it creates right triangle &lt;math&gt;\triangle AMC&lt;/math&gt; where &lt;math&gt;\angle M = 90^\circ&lt;/math&gt;. It is given that the hexagon is equiangular, therefore &lt;math&gt;\angle MBC = \frac{360}{6} = 60^\circ&lt;/math&gt;. (exterior angles of a polygon add up to 360 &lt;math&gt;^\circ&lt;/math&gt;)<br /> <br /> <br /> We can use either Pythagorean theorem or the properties of a &lt;math&gt;30-60-90&lt;/math&gt; triangle to find the length of &lt;math&gt;BM={r \over 2}&lt;/math&gt; and &lt;math&gt;CM = {\sqrt 3 \over 2 }r&lt;/math&gt;. The legs of &lt;math&gt;\triangle AMC&lt;/math&gt; are &lt;math&gt;1 + {r \over 2}&lt;/math&gt; and &lt;math&gt;{\sqrt 3 \over 2 }r&lt;/math&gt;. <br /> <br /> <br /> Using Pythagorean theorem, we get &lt;math&gt;AC^{2} = (r^2+r+1)&lt;/math&gt;. We can then follow &lt;math&gt;\textbf {Solution 1}&lt;/math&gt; to solve for &lt;math&gt;r&lt;/math&gt;. &lt;math&gt;\boxed{\textbf{(E)}\ 6}&lt;/math&gt;.<br /> <br /> <br /> Alternatively, we can find the area of &lt;math&gt;\triangle ABC&lt;/math&gt;. We know that the three smaller triangles: &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;\triangle CDE&lt;/math&gt;, and &lt;math&gt;\triangle EFA&lt;/math&gt; are congruent because of &lt;math&gt;S-A-S&lt;/math&gt;. Therefore one of the smaller triangles accounts for &lt;math&gt;10\%&lt;/math&gt; of the total area. The height of the smaller triangle &lt;math&gt;\triangle ABC&lt;/math&gt; is just &lt;math&gt;CM&lt;/math&gt; so the area is &lt;math&gt;{1 \cdot {\sqrt 3 \over 2 }r \over 2}&lt;/math&gt;. We can then find the area of the hexagon using &lt;math&gt;\textbf {Solution 1}&lt;/math&gt;.<br /> <br /> <br /> We can even find the area of &lt;math&gt;\triangle ACE&lt;/math&gt; and &lt;math&gt;\triangle ABC&lt;/math&gt; and solve for &lt;math&gt;r&lt;/math&gt; because the ratio of the areas is &lt;math&gt;7&lt;/math&gt; to &lt;math&gt;1&lt;/math&gt;.<br /> <br /> ~Zeric Hang<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=18|num-a=20|ab=A}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_19&diff=117333 2020 AMC 10B Problems/Problem 19 2020-02-08T05:01:01Z <p>Emerald block: revised solution</p> <hr /> <div>==Problem==<br /> <br /> In a certain card game, a player is dealt a hand of &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt; distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as &lt;math&gt;158A00A4AA0&lt;/math&gt;. What is the digit &lt;math&gt;A&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;math&gt;158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}&lt;/math&gt;<br /> <br /> We're looking for the amount of ways we can get &lt;math&gt;10&lt;/math&gt; cards from a deck of &lt;math&gt;52&lt;/math&gt;, which is represented by &lt;math&gt;\binom{52}{10}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}&lt;/math&gt;<br /> <br /> We need to get rid of the multiples of &lt;math&gt;3&lt;/math&gt;, which will subsequently get rid of the multiples of &lt;math&gt;9&lt;/math&gt; (if we didn't, the zeroes would mess with the equation since you can't divide by 0)<br /> <br /> &lt;math&gt;9\cdot5=45&lt;/math&gt;, &lt;math&gt;8\cdot6=48&lt;/math&gt;, &lt;math&gt;\frac{51}{3}&lt;/math&gt; leaves us with 17.<br /> <br /> &lt;math&gt;\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}&lt;/math&gt;<br /> <br /> Converting these into&lt;math&gt;\pmod{9}&lt;/math&gt;, we have<br /> <br /> &lt;math&gt;\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9} &lt;/math&gt;<br /> <br /> &lt;math&gt;4A\equiv8\pmod{9} \implies A=\boxed{\textbf{(A) }2}&lt;/math&gt; ~quacker88<br /> <br /> ==Solution 2==<br /> &lt;math&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43&lt;/math&gt;<br /> <br /> Since this number is divisible by &lt;math&gt;4&lt;/math&gt; but not &lt;math&gt;8&lt;/math&gt;, the last &lt;math&gt;2&lt;/math&gt; digits must be divisible by &lt;math&gt;4&lt;/math&gt; but the last &lt;math&gt;3&lt;/math&gt; digits cannot be divisible by &lt;math&gt;8&lt;/math&gt;. This narrows the options down to &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;. <br /> <br /> Also, the number cannot be divisible by &lt;math&gt;3&lt;/math&gt;. Adding up the digits, we get &lt;math&gt;18+4A&lt;/math&gt;. If &lt;math&gt;A=6&lt;/math&gt;, then the expression equals &lt;math&gt;42&lt;/math&gt;, a multiple of &lt;math&gt;3&lt;/math&gt;. This would mean that the entire number would be divisible by &lt;math&gt;3&lt;/math&gt;, which is not what we want. Therefore, the only option is &lt;math&gt;\boxed{\textbf{(A) }2}&lt;/math&gt;-PCChess<br /> <br /> ==Solution 3==<br /> It is not hard to check that &lt;math&gt;13&lt;/math&gt; divides the number,<br /> &lt;cmath&gt;\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.&lt;/cmath&gt; As &lt;math&gt;10^3\equiv-1\pmod{13}&lt;/math&gt;, using &lt;math&gt;\pmod{13}&lt;/math&gt; we have &lt;math&gt;13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781&lt;/math&gt;. Thus &lt;math&gt;6A+1\equiv0\pmod{13}&lt;/math&gt;, implying &lt;math&gt;A\equiv2\pmod{13}&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> As mentioned above, &lt;br&gt;<br /> &lt;cmath&gt;\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.&lt;/cmath&gt;<br /> We can divide both sides of &lt;math&gt;10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0&lt;/math&gt; by 10 to obtain<br /> &lt;cmath&gt;17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,&lt;/cmath&gt;<br /> which means &lt;math&gt;A&lt;/math&gt; is simply the units digit of the left-hand side. This value is<br /> &lt;cmath&gt;7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{\textbf{(A) }2} \pmod{10}.&lt;/cmath&gt;<br /> ~[[User:i_equal_tan_90|i_equal_tan_90]], revised by [[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/3BvJeZU3T-M<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_25&diff=117315 2020 AMC 10B Problems/Problem 25 2020-02-08T04:51:38Z <p>Emerald block: New solution</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;D(n)&lt;/math&gt; denote the number of ways of writing the positive integer &lt;math&gt;n&lt;/math&gt; as a product&lt;cmath&gt;n = f_1\cdot f_2\cdots f_k,&lt;/cmath&gt;where &lt;math&gt;k\ge1&lt;/math&gt;, the &lt;math&gt;f_i&lt;/math&gt; are integers strictly greater than &lt;math&gt;1&lt;/math&gt;, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number &lt;math&gt;6&lt;/math&gt; can be written as &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;2\cdot 3&lt;/math&gt;, and &lt;math&gt;3\cdot2&lt;/math&gt;, so &lt;math&gt;D(6) = 3&lt;/math&gt;. What is &lt;math&gt;D(96)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Note that &lt;math&gt;96 = 2^5 \cdot 3&lt;/math&gt;. Since there are at most six not nexxessarily distinct factors &lt;math&gt;&gt;1&lt;/math&gt; multiplying to &lt;math&gt;96&lt;/math&gt;, we have six cases: &lt;math&gt;k=1, 2, ..., 6.&lt;/math&gt; Now we look at each of the six cases.<br /> <br /> <br /> &lt;math&gt;k=1&lt;/math&gt;: We see that there is &lt;math&gt;1&lt;/math&gt; way, merely &lt;math&gt;96&lt;/math&gt;.<br /> <br /> &lt;math&gt;k=2&lt;/math&gt;: This way, we have the &lt;math&gt;3&lt;/math&gt; in one slot and &lt;math&gt;2&lt;/math&gt; in another, and symmetry. The four other &lt;math&gt;2&lt;/math&gt;'s leave us with &lt;math&gt;5&lt;/math&gt; ways and symmetry doubles us so we have &lt;math&gt;10&lt;/math&gt;.<br /> <br /> &lt;math&gt;k=3&lt;/math&gt;: We have &lt;math&gt;3, 2, 2&lt;/math&gt; as our baseline. We need to multiply by &lt;math&gt;2&lt;/math&gt; in &lt;math&gt;3&lt;/math&gt; places, and see that we can split the remaining three powers of 2 in a manner that is 3-0-0, 2-1-0 or 1-1-1. A 3-0-0 split has &lt;math&gt;6 + 3 = 9&lt;/math&gt; ways of happening (24-2-2 and symmetry; 2-3-16 and symmetry), a 2-1-0 split has &lt;math&gt;6 \cdot 3 = 18&lt;/math&gt; ways of happening (due to all being distinct) and a 1-1-1 split has &lt;math&gt;3&lt;/math&gt; ways of happening (6-4-4 and symmetry) so in this case we have &lt;math&gt;9+18+3=30&lt;/math&gt; ways.<br /> <br /> &lt;math&gt;k=4&lt;/math&gt;: We have &lt;math&gt;3, 2, 2, 2&lt;/math&gt; as our baseline, and for the two other &lt;math&gt;2&lt;/math&gt;'s, we have a 2-0-0-0 or 1-1-0-0 split. The former grants us &lt;math&gt;4+12=16&lt;/math&gt; ways (12-2-2-2 and symmetry and 3-8-2-2 and symmetry) and the latter grants us also &lt;math&gt;12+12=24&lt;/math&gt; ways (6-4-2-2 and symmetry and 3-4-4-2 and symmetry) for a total of &lt;math&gt;16+24=40&lt;/math&gt; ways.<br /> <br /> &lt;math&gt;k=5&lt;/math&gt;: We have &lt;math&gt;3, 2, 2, 2, 2&lt;/math&gt; as our baseline and one place to put the last two: on another two or on the three. On the three gives us &lt;math&gt;5&lt;/math&gt; ways due to symmetry and on another two gives us &lt;math&gt;5 \cdot 4 = 20&lt;/math&gt; ways due to symmetry. Thus, we have &lt;math&gt;5+20=25&lt;/math&gt; ways.<br /> <br /> &lt;math&gt;k=6&lt;/math&gt;: We have &lt;math&gt;3, 2, 2, 2, 2, 2&lt;/math&gt; and symmetry and no more twos to multiply, so by symmetry, we have &lt;math&gt;6&lt;/math&gt; ways.<br /> <br /> <br /> Thus, adding, we have &lt;math&gt;1+10+30+40+25+6=\boxed{\textbf{(A) } 112}&lt;/math&gt;.<br /> <br /> ~kevinmathz<br /> <br /> ==Solution 2==<br /> <br /> As before, note that &lt;math&gt;96=2^5\cdot3&lt;/math&gt;, and we need to consider 6 different cases, one for each possible value of &lt;math&gt;k&lt;/math&gt;, the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with &lt;math&gt;k&lt;/math&gt; factors. First, the factorization needs to contain one factor that is itself a multiple of &lt;math&gt;3&lt;/math&gt;, and there are &lt;math&gt;k&lt;/math&gt; to choose from, and the rest must contain at least one factor of &lt;math&gt;2&lt;/math&gt;. Next, consider the remaining &lt;math&gt;6-n&lt;/math&gt; factors of &lt;math&gt;2&lt;/math&gt; left to assign to the &lt;math&gt;k&lt;/math&gt; factors. Using stars and bars, the number of ways to do this is &lt;cmath&gt;{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}&lt;/cmath&gt; This makes &lt;math&gt;k{5\choose{6-k}}&lt;/math&gt; possibilities for each k.<br /> <br /> To obtain the total number of factorizations, add all possible values for k: &lt;cmath&gt;\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{\textbf{(A) } \text{112}}&lt;/cmath&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Begin examining &lt;math&gt;f_1&lt;/math&gt;. &lt;math&gt;f_1&lt;/math&gt; can take on any value that is a factor of &lt;math&gt;96&lt;/math&gt; except &lt;math&gt;1&lt;/math&gt;. For each choice of &lt;math&gt;f_1&lt;/math&gt;, the resulting &lt;math&gt;f_2...f_k&lt;/math&gt; must have a product of &lt;math&gt;96/f_1&lt;/math&gt;. This means the number of ways the rest &lt;math&gt;f_a&lt;/math&gt; &lt;math&gt;1&lt;a&lt;=k&lt;/math&gt; can be written by the scheme stated in the problem for each &lt;math&gt;f_1&lt;/math&gt; is &lt;math&gt;D(96/f_1)&lt;/math&gt;, since the product of &lt;math&gt;f_1 \cdot f_2&lt;/math&gt;...&lt;math&gt; \cdot f_k&lt;/math&gt; is 96 if and only if &lt;math&gt;f_1 \cdot x=96&lt;/math&gt; and the product &lt;math&gt;x&lt;/math&gt; has the properties that factors are greater than &lt;math&gt;1&lt;/math&gt;, and order matters in counting the solutions, which is how &lt;math&gt;D&lt;/math&gt; is defined.<br /> <br /> For example, say the first factor is &lt;math&gt;2&lt;/math&gt;, the remaining numbers must multiply to &lt;math&gt;48&lt;/math&gt;, so the number of ways the product can be written beginning with &lt;math&gt;2&lt;/math&gt; is &lt;math&gt;D(48)&lt;/math&gt;. To add up all possible starting factors, &lt;math&gt;D(96/f_a)&lt;/math&gt; must be calculated and summed for all &lt;math&gt;f_a&lt;/math&gt; except &lt;math&gt;96&lt;/math&gt; and &lt;math&gt;1&lt;/math&gt;, since a single &lt;math&gt;1&lt;/math&gt; is not counted according to the problem statement. The &lt;math&gt;96&lt;/math&gt; however, is counted, but only results in &lt;math&gt;1&lt;/math&gt; possibility, the first and only factor being &lt;math&gt;96&lt;/math&gt;. This means &lt;math&gt;D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(4)+D(2)+1&lt;/math&gt;. Instead of calculating D for the larger factors first, reduce &lt;math&gt;D(48)&lt;/math&gt;, &lt;math&gt;D(32)&lt;/math&gt;, and &lt;math&gt;D(24)&lt;/math&gt; into sums of &lt;math&gt;D(m)&lt;/math&gt; where &lt;math&gt;m&lt;=16&lt;/math&gt; to ease calculation.<br /> <br /> Following the definition &lt;math&gt;D(n)=(&lt;/math&gt;sums of &lt;math&gt;D(c))+1&lt;/math&gt; where c takes on every divisor of n except for 1 and itself, the sum simplifies to &lt;math&gt;(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)+<br /> (D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(4)+D(2)+1=\boxed{\textbf{(A) } 112}&lt;/math&gt;. <br /> <br /> ~monmath a.k.a Fmirza<br /> <br /> ==Solution 4==<br /> Note that &lt;math&gt;96 = 3 \cdot 2^5&lt;/math&gt;, and that &lt;math&gt;D&lt;/math&gt; of a perfect power of a prime is relatively easy to calculate. Also note that you can find &lt;math&gt;D(96)&lt;/math&gt; from &lt;math&gt;D(32)&lt;/math&gt; by simply totaling the number of ways there are to insert a &lt;math&gt;3&lt;/math&gt; into a set of numbers that multiply to &lt;math&gt;32&lt;/math&gt;.<br /> <br /> First, calculate &lt;math&gt;D(32)&lt;/math&gt;. Since &lt;math&gt;32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2&lt;/math&gt;, all you have to do was find the number of ways to divide the &lt;math&gt;2&lt;/math&gt;'s into groups, such that each group has at least one &lt;math&gt;2&lt;/math&gt;. By stars and bars, this results in &lt;math&gt;1&lt;/math&gt; way with five terms, &lt;math&gt;4&lt;/math&gt; ways with four terms, &lt;math&gt;6&lt;/math&gt; ways with three terms, &lt;math&gt;4&lt;/math&gt; ways with two terms, and &lt;math&gt;1&lt;/math&gt; way with one term. (The total, &lt;math&gt;16&lt;/math&gt;, is not needed for the remaining calculations.)<br /> <br /> Then, to get &lt;math&gt;D(96)&lt;/math&gt;, in each possible &lt;math&gt;D(32)&lt;/math&gt; sequence, insert a &lt;math&gt;3&lt;/math&gt; somewhere in it, either by placing it somewhere next to the original numbers (in one of &lt;math&gt;n+1&lt;/math&gt; ways, where &lt;math&gt;n&lt;/math&gt; is the number of terms in the &lt;math&gt;D(32)&lt;/math&gt; sequence), or by multiplying one of the numbers by &lt;math&gt;3&lt;/math&gt; (in one of &lt;math&gt;n&lt;/math&gt; ways). There are &lt;math&gt;2+1=3&lt;/math&gt; ways to do this with one term, &lt;math&gt;3+2=5&lt;/math&gt; with two, &lt;math&gt;7&lt;/math&gt; with three, &lt;math&gt;9&lt;/math&gt; with four, and &lt;math&gt;11&lt;/math&gt; with five.<br /> <br /> The resulting number of possible sequences is &lt;math&gt;3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{\textbf{(A) }112}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_22&diff=116565 2020 AMC 10A Problems/Problem 22 2020-02-02T05:35:53Z <p>Emerald block: fixed mistakes, elaborated more</p> <hr /> <div><br /> ==Problem==<br /> For how many positive integers &lt;math&gt;n \le 1000&lt;/math&gt; is&lt;cmath&gt;\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor&lt;/cmath&gt;not divisible by &lt;math&gt;3&lt;/math&gt;? (Recall that &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt;.)<br /> <br /> &lt;math&gt;\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> <br /> == Solution (Casework) ==<br /> <br /> &lt;b&gt;Expression:&lt;/b&gt;<br /> &lt;cmath&gt;\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor&lt;/cmath&gt;<br /> <br /> &lt;b&gt;Solution:&lt;/b&gt;<br /> <br /> Let &lt;math&gt;a = \left\lfloor \frac{998}n \right\rfloor&lt;/math&gt;<br /> <br /> Since &lt;math&gt;\frac{1000}n - \frac{998}n = \frac{2}n&lt;/math&gt;, for any integer &lt;math&gt;n \geq 2&lt;/math&gt;, the difference between the largest and smallest terms before the &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; function is applied is less than or equal to &lt;math&gt;1&lt;/math&gt;, and thus the terms must have a range of &lt;math&gt;1&lt;/math&gt; or less after the function is applied.<br /> <br /> This means that for every integer &lt;math&gt;n \geq 2&lt;/math&gt;,<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; if &lt;math&gt;\frac{1000}n&lt;/math&gt; is an integer, then the three terms in the expression above must be &lt;math&gt;(a, a, a + 1)&lt;/math&gt;,<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; if &lt;math&gt;\frac{999}n&lt;/math&gt; is an integer, then the three terms in the expression above must be &lt;math&gt;(a, a + 1, a + 1)&lt;/math&gt;,<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; if &lt;math&gt;\frac{998}n&lt;/math&gt; is an integer and &lt;math&gt;n \neq 2&lt;/math&gt; (since if &lt;math&gt;n = 2&lt;/math&gt;, &lt;math&gt;\frac{1000}n&lt;/math&gt; will be an integer, and it will be &lt;math&gt;1&lt;/math&gt; greater than &lt;math&gt;\frac{998}n&lt;/math&gt;), then the three terms in the expression above must be &lt;math&gt;(a, a, a)&lt;/math&gt;, and<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; if none of &lt;math&gt;\{\frac{998}n, \frac{999}n, \frac{1000}n\}&lt;/math&gt; are integral, then the three terms in the expression above must be &lt;math&gt;(a, a, a)&lt;/math&gt;.<br /> <br /> The last statement is true because in order for the terms to be different, there must be some integer in the interval &lt;math&gt;(\frac{998}n, \frac{999}n)&lt;/math&gt; or the interval &lt;math&gt;(\frac{999}n, \frac{1000}n)&lt;/math&gt;. However, this means that multiplying the integer by &lt;math&gt;n&lt;/math&gt; should produce a new integer between &lt;math&gt;998&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt; or &lt;math&gt;999&lt;/math&gt; and &lt;math&gt;1000&lt;/math&gt;, exclusive, but because no such integers exist, the original integer cannot exist, and thus, the terms must be equal.<br /> <br /> <br /> Note that &lt;math&gt;n = 1&lt;/math&gt; does not work; to prove this, we just have to substitute &lt;math&gt;1&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt; in the expression.<br /> This gives us<br /> &lt;math&gt;\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor = 998 + 999 + 1000 = 2997 = 999 \cdot 3&lt;/math&gt; which is divisible by 3.<br /> <br /> <br /> Now, we test the four cases listed above.<br /> <br /> <br /> &lt;b&gt;Case 1:&lt;/b&gt; &lt;math&gt;n&lt;/math&gt; divides &lt;math&gt;998&lt;/math&gt;<br /> <br /> As mentioned above, the three terms in the expression are &lt;math&gt;(a, a, a)&lt;/math&gt;, so the sum is &lt;math&gt;3a&lt;/math&gt;, which is divisible by &lt;math&gt;3&lt;/math&gt;.<br /> Therefore, the first case does not work.<br /> <br /> <br /> &lt;b&gt;Case 2:&lt;/b&gt; &lt;math&gt;n&lt;/math&gt; divides &lt;math&gt;999&lt;/math&gt;<br /> <br /> Because &lt;math&gt;n&lt;/math&gt; divides &lt;math&gt;999&lt;/math&gt;, the number of possibilities for &lt;math&gt;n&lt;/math&gt; is the same as the number of factors of &lt;math&gt;999&lt;/math&gt;.<br /> <br /> &lt;math&gt;999&lt;/math&gt; = &lt;math&gt;3^3 \cdot 37^1&lt;/math&gt;.<br /> So, the total number of factors of &lt;math&gt;999&lt;/math&gt; is &lt;math&gt;4 \cdot 2 = 8&lt;/math&gt;.<br /> <br /> However, we have to subtract &lt;math&gt;1&lt;/math&gt;, because the case &lt;math&gt;n = 1&lt;/math&gt; does not work, as mentioned previously. This leaves &lt;math&gt;8 - 1 = 7&lt;/math&gt; cases.<br /> <br /> <br /> &lt;b&gt;Case 3:&lt;/b&gt; &lt;math&gt;n&lt;/math&gt; divides &lt;math&gt;1000&lt;/math&gt;<br /> <br /> Because &lt;math&gt;n&lt;/math&gt; divides &lt;math&gt;1000&lt;/math&gt;, the number of possibilities for &lt;math&gt;n&lt;/math&gt; is the same as the number of factors of &lt;math&gt;1000&lt;/math&gt;.<br /> <br /> &lt;math&gt;1000&lt;/math&gt; = &lt;math&gt;5^3 \cdot 2^3&lt;/math&gt;.<br /> So, the total number of factors of &lt;math&gt;1000&lt;/math&gt; is &lt;math&gt;4 \cdot 4 = 16&lt;/math&gt;.<br /> <br /> Again, we have to subtract &lt;math&gt;1&lt;/math&gt;, so this leaves &lt;math&gt;16 - 1 = 15&lt;/math&gt; cases.<br /> <br /> <br /> &lt;b&gt;Case 4:&lt;/b&gt; &lt;math&gt;n&lt;/math&gt; divides none of &lt;math&gt;\{998, 999, 1000\}&lt;/math&gt;<br /> <br /> As in Case 1, the value of the terms of the expression are &lt;math&gt;(a, a, a)&lt;/math&gt;. The sum is &lt;math&gt;3a&lt;/math&gt;, which is divisible by 3, so this case does not work.<br /> <br /> <br /> Now that we have counted all of the cases, we add them.<br /> <br /> &lt;math&gt;0 + 7 + 15 + 0 = 22&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(A)}22}&lt;/math&gt;.<br /> <br /> ~dragonchomper, additional edits by [[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/Ozp3k2464u4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_21&diff=116551 2020 AMC 10A Problems/Problem 21 2020-02-02T04:31:12Z <p>Emerald block: signed solution</p> <hr /> <div>There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First, substitute &lt;math&gt;2^{17}&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;. <br /> Then, the given equation becomes &lt;math&gt;\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0&lt;/math&gt;.<br /> Now consider only &lt;math&gt;a^{16}-a^{15}&lt;/math&gt;. This equals &lt;math&gt;a^{15}(a-1)=a^{15}*(2^{17}-1)&lt;/math&gt;.<br /> Note that &lt;math&gt;2^{17}-1&lt;/math&gt; equals &lt;math&gt;2^{16}+2^{15}+...+1&lt;/math&gt;, since the sum of a geometric sequence is &lt;math&gt;\frac{a^n-1}{a-1}&lt;/math&gt;.<br /> Thus, we can see that &lt;math&gt;a^{16}-a^{15}&lt;/math&gt; forms the sum of 17 different powers of 2. <br /> Applying the same method to each of &lt;math&gt;a^{14}-a^{13}&lt;/math&gt;, &lt;math&gt;a^{12}-a^{11}&lt;/math&gt;, ... , &lt;math&gt;a^{2}-a^{1}&lt;/math&gt;, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us &lt;math&gt;17*8=136&lt;/math&gt;.<br /> But we must count also the &lt;math&gt;a^0&lt;/math&gt; term. <br /> Thus, Our answer is &lt;math&gt;136+1=\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~seanyoon777<br /> <br /> == Solution 2 ==<br /> (This is similar to solution 1)<br /> Let &lt;math&gt;x = 2^{17}&lt;/math&gt;. Then, &lt;math&gt;2^{289} = x^{17}&lt;/math&gt;.<br /> The LHS can be rewritten as &lt;math&gt;\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1&lt;/math&gt;. Plugging &lt;math&gt;2^{17}&lt;/math&gt; back in for &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;(2^{17}-1)(2^{15\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1&lt;/math&gt;. When expanded, this will have &lt;math&gt;17\cdot8+1=137&lt;/math&gt; terms. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Note that the expression is equal to something slightly lower than &lt;math&gt;2^{272}&lt;/math&gt;. Clearly, answer choices &lt;math&gt;(D)&lt;/math&gt; and &lt;math&gt;(E)&lt;/math&gt; make no sense because the lowest sum for &lt;math&gt;273&lt;/math&gt; terms is &lt;math&gt;2^{273}-1&lt;/math&gt;. &lt;math&gt;(A)&lt;/math&gt; just makes no sense. &lt;math&gt;(B)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt; are 1 apart, but because the expression is odd, it will have to contain &lt;math&gt;2^0=1&lt;/math&gt;, and because &lt;math&gt;(C)&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; bigger, the answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~Lcz<br /> <br /> == Solution 4 ==<br /> In order to shorten expressions, &lt;math&gt;\#&lt;/math&gt; will represent &lt;math&gt;16&lt;/math&gt; consecutive &lt;math&gt;0&lt;/math&gt;s when expressing numbers. &lt;br&gt;<br /> &lt;br&gt;<br /> Think of the problem in binary. We have &lt;br&gt;<br /> &lt;math&gt;\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}&lt;/math&gt; &lt;br&gt;<br /> Note that &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> and &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Since &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2&lt;/math&gt; &lt;br&gt;<br /> this means that&lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}&lt;/math&gt; &lt;br&gt;<br /> so &lt;br&gt;<br /> &lt;math&gt;\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Expressing each of the pairs of the form &lt;math&gt;2^{n + 17} - 2^n&lt;/math&gt; in binary, we have &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1000000000000000000 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{10000000000000000} 10 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= \phantom{1} 111111111111111110 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> or &lt;br&gt;<br /> &lt;math&gt;2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}&lt;/math&gt; &lt;br&gt;<br /> This means that each pair has &lt;math&gt;17&lt;/math&gt; terms of the form &lt;math&gt;2^n&lt;/math&gt;. &lt;br&gt;<br /> &lt;br&gt;<br /> Since there are &lt;math&gt;8&lt;/math&gt; of these pairs, there are a total of &lt;math&gt;8 \cdot 17 = 136&lt;/math&gt; terms. Accounting for the &lt;math&gt;2^0&lt;/math&gt; term, which was not in the pair, we have a total of &lt;math&gt;136 + 1 = \boxed{\textbf{(C) } 137}&lt;/math&gt; terms. ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/Ozp3k2464u4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_21&diff=116550 2020 AMC 10A Problems/Problem 21 2020-02-02T04:30:15Z <p>Emerald block: New solution</p> <hr /> <div>There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> First, substitute &lt;math&gt;2^{17}&lt;/math&gt; with &lt;math&gt;a&lt;/math&gt;. <br /> Then, the given equation becomes &lt;math&gt;\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}...-a^1+a^0&lt;/math&gt;.<br /> Now consider only &lt;math&gt;a^{16}-a^{15}&lt;/math&gt;. This equals &lt;math&gt;a^{15}(a-1)=a^{15}*(2^{17}-1)&lt;/math&gt;.<br /> Note that &lt;math&gt;2^{17}-1&lt;/math&gt; equals &lt;math&gt;2^{16}+2^{15}+...+1&lt;/math&gt;, since the sum of a geometric sequence is &lt;math&gt;\frac{a^n-1}{a-1}&lt;/math&gt;.<br /> Thus, we can see that &lt;math&gt;a^{16}-a^{15}&lt;/math&gt; forms the sum of 17 different powers of 2. <br /> Applying the same method to each of &lt;math&gt;a^{14}-a^{13}&lt;/math&gt;, &lt;math&gt;a^{12}-a^{11}&lt;/math&gt;, ... , &lt;math&gt;a^{2}-a^{1}&lt;/math&gt;, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us &lt;math&gt;17*8=136&lt;/math&gt;.<br /> But we must count also the &lt;math&gt;a^0&lt;/math&gt; term. <br /> Thus, Our answer is &lt;math&gt;136+1=\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~seanyoon777<br /> <br /> == Solution 2 ==<br /> (This is similar to solution 1)<br /> Let &lt;math&gt;x = 2^{17}&lt;/math&gt;. Then, &lt;math&gt;2^{289} = x^{17}&lt;/math&gt;.<br /> The LHS can be rewritten as &lt;math&gt;\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+\cdots+x^2-x+1=(x-1)(x^{15}+x^{13}+\cdots+x^{1})+1&lt;/math&gt;. Plugging &lt;math&gt;2^{17}&lt;/math&gt; back in for &lt;math&gt;x&lt;/math&gt;, we have &lt;math&gt;(2^{17}-1)(2^{15\cdot{17}}+2^{13\cdot{17}}+\cdots+2^{1\cdot{17}})+1=(2^{16}+2^{15}+\cdots+2^{0})(2^{15\cdot17}+2^{13\cdot17}+\cdots+2^{1\cdot17})+1&lt;/math&gt;. When expanded, this will have &lt;math&gt;17\cdot8+1=137&lt;/math&gt; terms. Therefore, our answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Note that the expression is equal to something slightly lower than &lt;math&gt;2^{272}&lt;/math&gt;. Clearly, answer choices &lt;math&gt;(D)&lt;/math&gt; and &lt;math&gt;(E)&lt;/math&gt; make no sense because the lowest sum for &lt;math&gt;273&lt;/math&gt; terms is &lt;math&gt;2^{273}-1&lt;/math&gt;. &lt;math&gt;(A)&lt;/math&gt; just makes no sense. &lt;math&gt;(B)&lt;/math&gt; and &lt;math&gt;(C)&lt;/math&gt; are 1 apart, but because the expression is odd, it will have to contain &lt;math&gt;2^0=1&lt;/math&gt;, and because &lt;math&gt;(C)&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; bigger, the answer is &lt;math&gt;\boxed{\textbf{(C) } 137}&lt;/math&gt;.<br /> <br /> ~Lcz<br /> <br /> == Solution 4 ==<br /> In order to shorten expressions, &lt;math&gt;\#&lt;/math&gt; will represent &lt;math&gt;16&lt;/math&gt; consecutive &lt;math&gt;0&lt;/math&gt;s when expressing numbers. &lt;br&gt;<br /> &lt;br&gt;<br /> Think of the problem in binary. We have &lt;br&gt;<br /> &lt;math&gt;\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}&lt;/math&gt; &lt;br&gt;<br /> Note that &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> and &lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Since &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2&lt;/math&gt; &lt;br&gt;<br /> this means that&lt;br&gt;<br /> &lt;math&gt;(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}&lt;/math&gt; &lt;br&gt;<br /> so &lt;br&gt;<br /> &lt;math&gt;\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})&lt;/math&gt; &lt;br&gt;<br /> &lt;br&gt;<br /> Expressing each of the pairs of the form &lt;math&gt;2^{n + 17} - 2^n&lt;/math&gt; in binary, we have &lt;br&gt;<br /> &lt;math&gt;\phantom{=\ } 1000000000000000000 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;-\ \phantom{10000000000000000} 10 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> &lt;math&gt;= \phantom{1} 111111111111111110 \cdots 0_2&lt;/math&gt; &lt;br&gt;<br /> or &lt;br&gt;<br /> &lt;math&gt;2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}&lt;/math&gt; &lt;br&gt;<br /> This means that each pair has &lt;math&gt;17&lt;/math&gt; terms of the form &lt;math&gt;2^n&lt;/math&gt;. &lt;br&gt;<br /> &lt;br&gt;<br /> Since there are &lt;math&gt;8&lt;/math&gt; of these pairs, there are a total of &lt;math&gt;8 \cdot 17 = 136&lt;/math&gt; terms. Accounting for the &lt;math&gt;2^0&lt;/math&gt; term, which was not in the pair, we have a total of &lt;math&gt;136 + 1 = \boxed{\textbf{(C) } 137}&lt;/math&gt; terms.<br /> <br /> ==Video Solution==<br /> https://youtu.be/Ozp3k2464u4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=116374 2020 AMC 10A Problems/Problem 19 2020-02-01T22:41:17Z <p>Emerald block: fixed coding error</p> <hr /> <div>== Problem ==<br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;asy&gt;<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> &lt;/asy&gt;<br /> <br /> == Solution ==<br /> Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br /> <br /> We have &lt;math&gt;5&lt;/math&gt; choices for which face we visit first on the top ring. From there, we have &lt;math&gt;9&lt;/math&gt; choices for how far around the top ring we go before moving down: &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around clockwise, &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring. We then have &lt;math&gt;2&lt;/math&gt; choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly &lt;math&gt;2&lt;/math&gt; lower-ring faces) and then once again &lt;math&gt;9&lt;/math&gt; choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br /> <br /> Multiplying together all the numbers of choices we have, we get &lt;math&gt;5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Swap the faces as vertices and the vertices as faces. Then, this problem is the same as 2016 AIME I #3, which had an answer of &lt;math&gt;\boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br /> {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_19&diff=116372 2020 AMC 10A Problems/Problem 19 2020-02-01T22:39:44Z <p>Emerald block: fixed spelling/grammar mistake</p> <hr /> <div>== Problem ==<br /> As shown in the figure below, a regular dodecahedron (the polyhedron consisting of &lt;math&gt;12&lt;/math&gt; congruent regular pentagonal faces) floats in space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?<br /> <br /> &lt;math&gt;\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810&lt;/math&gt;<br /> <br /> == Diagram ==<br /> &lt;math&gt;Asymptote code below<br /> [asy]<br /> import graph;<br /> unitsize(5cm);<br /> pair A = (0.082, 0.378);<br /> pair B = (0.091, 0.649);<br /> pair C = (0.249, 0.899);<br /> pair D = (0.479, 0.939);<br /> pair E = (0.758, 0.893);<br /> pair F = (0.862, 0.658);<br /> pair G = (0.924, 0.403);<br /> pair H = (0.747, 0.194);<br /> pair I = (0.526, 0.075);<br /> pair J = (0.251, 0.170);<br /> pair K = (0.568, 0.234);<br /> pair L = (0.262, 0.449);<br /> pair M = (0.373, 0.813);<br /> pair N = (0.731, 0.813);<br /> pair O = (0.851, 0.461);<br /> path[] f;<br /> f = A--B--C--M--L--cycle;<br /> f = C--D--E--N--M--cycle;<br /> f = E--F--G--O--N--cycle;<br /> f = G--H--I--K--O--cycle;<br /> f = I--J--A--L--K--cycle;<br /> f = K--L--M--N--O--cycle;<br /> draw(f);<br /> axialshade(f, white, M, gray(0.5), (C+2*D)/3);<br /> draw(f);<br /> filldraw(f, gray);<br /> filldraw(f, gray);<br /> axialshade(f, white, L, gray(0.7), J);<br /> draw(f);<br /> draw(f);<br /> [/asy]&lt;/math&gt;<br /> <br /> == Solution ==<br /> Since we start at the top face and end at the bottom face without moving from the lower ring to the upper ring or revisiting a face, our journey must consist of the top face, a series of faces in the upper ring, a series of faces in the lower ring, and the bottom face, in that order.<br /> <br /> We have &lt;math&gt;5&lt;/math&gt; choices for which face we visit first on the top ring. From there, we have &lt;math&gt;9&lt;/math&gt; choices for how far around the top ring we go before moving down: &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around clockwise, &lt;math&gt;1,2,3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt; faces around counterclockwise, or immediately going down to the lower ring without visiting any other faces in the upper ring. We then have &lt;math&gt;2&lt;/math&gt; choices for which lower ring face to visit first (since every upper-ring face is adjacent to exactly &lt;math&gt;2&lt;/math&gt; lower-ring faces) and then once again &lt;math&gt;9&lt;/math&gt; choices for how to travel around the lower ring. We then proceed to the bottom face, completing the trip.<br /> <br /> Multiplying together all the numbers of choices we have, we get &lt;math&gt;5 \cdot 9 \cdot 2 \cdot 9 = \boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Swap the faces as vertices and the vertices as faces. Then, this problem is the same as 2016 AIME I #3, which had an answer of &lt;math&gt;\boxed{\textbf{(E) } 810}&lt;/math&gt;.<br /> &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_3<br /> {{AMC10 box|year=2020|ab=A|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&diff=116369 2020 AMC 10A Problems/Problem 16 2020-02-01T22:37:44Z <p>Emerald block: added alternate solution</p> <hr /> <div>== Problem ==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> We consider an individual one-by-one block.<br /> <br /> If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius &lt;math&gt;d&lt;/math&gt;, the area covered by the circles should be &lt;math&gt;0.5&lt;/math&gt;. Because of this, and the fact that there are four circles, we write<br /> <br /> &lt;cmath&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/cmath&gt;<br /> <br /> Solving for &lt;math&gt;d&lt;/math&gt;, we obtain &lt;math&gt;d = \frac{1}{\sqrt{2\pi}}&lt;/math&gt;, where with &lt;math&gt;\pi \approx 3&lt;/math&gt;, we get &lt;math&gt;d = \frac{1}{\sqrt{6}}&lt;/math&gt;, and from here, we simplify and see that &lt;math&gt;d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}&lt;/math&gt; ~Crypthes<br /> <br /> &lt;math&gt;\textbf{Note:}&lt;/math&gt; To be more rigorous, note that &lt;math&gt;d&lt;0.5&lt;/math&gt; since if &lt;math&gt;d\geq0.5&lt;/math&gt; then clearly the probability is greater than &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This would make sure the above solution works, as if &lt;math&gt;d\geq0.5&lt;/math&gt; there is overlap with the quartercircles. &lt;math&gt;\textbf{- Emathmaster}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> As in the previous solution, we obtain the equation &lt;math&gt;4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}&lt;/math&gt;, which simplifies to &lt;math&gt;\pi d^2 = \frac{1}{2} = 0.5&lt;/math&gt;. Since &lt;math&gt;\pi&lt;/math&gt; is slightly more than &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;d^2&lt;/math&gt; is slightly less than &lt;math&gt;\frac{0.5}{3} = 0.1\bar{6}&lt;/math&gt;. We notice that &lt;math&gt;0.1\bar{6}&lt;/math&gt; is slightly more than &lt;math&gt;0.4^2 = 0.16&lt;/math&gt;, so &lt;math&gt;d&lt;/math&gt; is roughly &lt;math&gt;\boxed{\textbf{(B) } 0.4}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/RKlG6oZq9so<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=116365 2020 AMC 10A Problems/Problem 15 2020-02-01T22:27:36Z <p>Emerald block: changed phrasing of bottom formula</p> <hr /> <div>==Problem==<br /> <br /> A positive integer divisor of &lt;math&gt;12!&lt;/math&gt; is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m+n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23&lt;/math&gt;<br /> <br /> ==Solution==<br /> The prime factorization of &lt;math&gt;12!&lt;/math&gt; is &lt;math&gt;2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11&lt;/math&gt;. <br /> This yields a total of &lt;math&gt;11 \cdot 6 \cdot 3 \cdot 2 \cdot 2&lt;/math&gt; divisors of &lt;math&gt;12!.&lt;/math&gt;<br /> In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that &lt;math&gt;7&lt;/math&gt; and &lt;math&gt;11&lt;/math&gt; can not be in the prime factorization of a perfect square because there is only one of each in &lt;math&gt;12!.&lt;/math&gt; Thus, there are &lt;math&gt;6 \cdot 3 \cdot 2&lt;/math&gt; perfect squares. (For &lt;math&gt;2&lt;/math&gt;, you can have &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, or &lt;math&gt;1&lt;/math&gt;0 &lt;math&gt;2&lt;/math&gt;s, etc.)<br /> The probability that the divisor chosen is a perfect square is &lt;cmath&gt;\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }&lt;/cmath&gt;<br /> <br /> ~mshell214, edited by Rzhpamath<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Emerald block https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_11&diff=116364 2020 AMC 10A Problems/Problem 11 2020-02-01T22:25:00Z <p>Emerald block: signed solution</p> <hr /> <div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #8]] and [[2020 AMC 10A Problems|2020 AMC 10A #11]]}}<br /> <br /> ==Problem 11==<br /> <br /> What is the median of the following list of &lt;math&gt;4040&lt;/math&gt; numbers&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;cmath&gt;1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2&lt;/cmath&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> We can see that &lt;math&gt;44^2&lt;/math&gt; is less than 2020. Therefore, there are &lt;math&gt;1976&lt;/math&gt; of the &lt;math&gt;4040&lt;/math&gt; numbers after &lt;math&gt;2020&lt;/math&gt;. Also, there are &lt;math&gt;2064&lt;/math&gt; numbers that are under and equal to &lt;math&gt;2020&lt;/math&gt;. Since &lt;math&gt;44^2&lt;/math&gt; is &lt;math&gt;1936&lt;/math&gt; it, with the other squares will shift our median's placement up &lt;math&gt;44&lt;/math&gt;. We can find that the median of the whole set is &lt;math&gt;2020.5&lt;/math&gt;, and &lt;math&gt;2020.5-44&lt;/math&gt; gives us &lt;math&gt;1976.5&lt;/math&gt;. Our answer is &lt;math&gt;\boxed{\textbf{(C) } 1976.5}&lt;/math&gt;.<br /> <br /> ~aryam<br /> <br /> == Solution 2 ==<br /> As we are trying to find the median of a &lt;math&gt;4040&lt;/math&gt;-term set, we must find the average of the &lt;math&gt;2020&lt;/math&gt;th and &lt;math&gt;2021&lt;/math&gt;st terms. &lt;br&gt;<br /> &lt;br&gt;<br /> Since &lt;math&gt;45^2 = 2025&lt;/math&gt; is slightly greater than &lt;math&gt;2020&lt;/math&gt;, we know that the &lt;math&gt;44&lt;/math&gt; perfect squares &lt;math&gt;1^2&lt;/math&gt; through &lt;math&gt;44^2&lt;/math&gt; are less than &lt;math&gt;2020&lt;/math&gt;, and the rest are greater. Thus, from the number &lt;math&gt;1&lt;/math&gt; to the number &lt;math&gt;2020&lt;/math&gt;, there are &lt;math&gt;2020 + 44 = 2064&lt;/math&gt; terms. Since &lt;math&gt;44^2&lt;/math&gt; is &lt;math&gt;44 + 45 = 89&lt;/math&gt; less than &lt;math&gt;45^2 = 2025&lt;/math&gt; and &lt;math&gt;84&lt;/math&gt; less than &lt;math&gt;2020&lt;/math&gt;, we will only need to consider the perfect square terms going down from the &lt;math&gt;2064&lt;/math&gt;th term, &lt;math&gt;2020&lt;/math&gt;, after going down &lt;math&gt;84&lt;/math&gt; terms. Since the &lt;math&gt;2020&lt;/math&gt;th and &lt;math&gt;2021&lt;/math&gt;st terms are only &lt;math&gt;44&lt;/math&gt; and &lt;math&gt;43&lt;/math&gt; terms away from the &lt;math&gt;2064&lt;/math&gt;th term, we can simply subtract &lt;math&gt;44&lt;/math&gt; from &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;43&lt;/math&gt; from &lt;math&gt;2020&lt;/math&gt; to get the two terms, which are &lt;math&gt;1976&lt;/math&gt; and &lt;math&gt;1977&lt;/math&gt;. Averaging the two, we get &lt;math&gt;\boxed{\textbf{(C) } 1976.5}.&lt;/math&gt; ~[[User:emerald_block|emerald_block]]<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZGwAasE32Y4<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=A|num-b=10|num-a=12}}<br /> {{AMC12 box|year=2020|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Emerald block