https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathpro12345&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-05T02:56:47Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149281 2021 AIME I Problems 2021-03-12T21:06:50Z <p>Mathpro12345: /* Problem 10 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> WTF<br /> <br /> WTF<br /> <br /> WTF<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149280 2021 AIME I Problems 2021-03-12T21:06:41Z <p>Mathpro12345: /* Problem 9 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> WTF<br /> <br /> WTF<br /> <br /> WTF<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_9&diff=149279 2021 AIME I Problems/Problem 9 2021-03-12T21:06:26Z <p>Mathpro12345: /* Problem */</p> <hr /> <div>==Solution 1==<br /> Construct your isosceles trapezoid. Let, for simplicity, &lt;math&gt;AB = a&lt;/math&gt;, &lt;math&gt;AD = BC = b&lt;/math&gt;, and &lt;math&gt;CD = c&lt;/math&gt;. Extend the sides &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; mark the intersection as &lt;math&gt;P&lt;/math&gt;. Following what the question states, drop a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; labeling the foot as &lt;math&gt;G&lt;/math&gt;. Drop another perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;CD&lt;/math&gt;, calling the foot &lt;math&gt;E&lt;/math&gt;. Lastly, drop a perpendicular from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;BD&lt;/math&gt;, labeling it &lt;math&gt;F&lt;/math&gt;. In addition, drop a perpendicular from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt; calling its foot &lt;math&gt;F'&lt;/math&gt;.<br /> <br /> --DIAGRAM COMING SOON--<br /> <br /> Start out by constructing a triangle &lt;math&gt;ADH&lt;/math&gt; congruent to &lt;math&gt;\triangle ABC&lt;/math&gt; with its side of length &lt;math&gt;a&lt;/math&gt; on line &lt;math&gt;DE&lt;/math&gt;. This works because all isosceles triangles are cyclic and as a result, &lt;math&gt;\angle ADC + \angle ABC = 180^\circ&lt;/math&gt;. <br /> <br /> Notice that &lt;math&gt;\triangle AGC \sim \triangle BF'C&lt;/math&gt; by AA similarity. We are given that &lt;math&gt;AG = 15&lt;/math&gt; and by symmetry we can deduce that &lt;math&gt;F'B = 10&lt;/math&gt;. As a result, &lt;math&gt;\frac{BF}{AG} = \frac{BC}{AC} = \frac{3}{2}&lt;/math&gt;. This gives us that &lt;math&gt;AC = BD = \frac{3}{2} b&lt;/math&gt;.<br /> <br /> The question asks us along the lines of finding the area, &lt;math&gt;K&lt;/math&gt;, of the trapezoid &lt;math&gt;ABCD&lt;/math&gt;. We look at the area of &lt;math&gt;ABC&lt;/math&gt; and notice that it can be represented as &lt;math&gt;\frac{1}{2} \cdot AC \cdot 10 = \frac{1}{2} \cdot a \cdot 18&lt;/math&gt;. Substituting &lt;math&gt;AC = \frac{3}{2} b&lt;/math&gt;, we solve for &lt;math&gt;a&lt;/math&gt;, getting &lt;math&gt;a = \frac{5}{6} b&lt;/math&gt;. <br /> <br /> Now let us focus on isosceles triangle &lt;math&gt;ACH&lt;/math&gt;, where &lt;math&gt;AH = AC = \frac{3}{2} b&lt;/math&gt;. Since, &lt;math&gt;AE&lt;/math&gt; is an altitude from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;CH&lt;/math&gt; of an isosceles triangle, &lt;math&gt;HE&lt;/math&gt; must be equal to &lt;math&gt;EC&lt;/math&gt;. Since &lt;math&gt;DH = a&lt;/math&gt; and &lt;math&gt;DC = c&lt;/math&gt;, we can solve to get that &lt;math&gt;DE = \frac{c-a}{2}&lt;/math&gt; and &lt;math&gt;EC = \frac{a+c}{2}&lt;/math&gt;.<br /> <br /> We must then set up equations using the Pythagorean Theorem, writing everything in terms of &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;. Looking at right triangle &lt;math&gt;AEC&lt;/math&gt; we get &lt;cmath&gt;324 + \frac{(a + c)^2}{4} = \frac{9}{4} b^2&lt;/cmath&gt; Looking at right triangle &lt;math&gt;AED&lt;/math&gt; we get &lt;cmath&gt;b^2 - 324 = \frac{(c-a)^2}{4}&lt;/cmath&gt; Now rearranging and solving, we get two equation &lt;cmath&gt;a+c = 3\sqrt{b^2 - 144}&lt;/cmath&gt; &lt;cmath&gt;c - a = 2\sqrt{b^2 - 324}&lt;/cmath&gt; Those are convenient equations as &lt;math&gt;c+a - (c-a) = 2a = \frac{5}{3} b&lt;/math&gt; which gives us &lt;cmath&gt;3\sqrt{b^2 - 324} - 2\sqrt{b^2 - 324} = \frac{5}{3} b&lt;/cmath&gt; After some &quot;smart&quot; calculation, we get that &lt;math&gt;b = \frac{27}{\sqrt{2}}&lt;/math&gt;.<br /> <br /> Notice that the question asks for &lt;math&gt;K\sqrt{2}&lt;/math&gt;, and &lt;math&gt;K = \frac{1}{2} \cdot 18 \cdot (a+c)&lt;/math&gt; by applying the trapezoid area formula. Fortunately, this is just &lt;math&gt;27\sqrt{b^2 - 144}&lt;/math&gt;, and plugging in the value of &lt;math&gt;b = \frac{27}{\sqrt{2}}&lt;/math&gt;, we get that &lt;math&gt;K\sqrt{2} = \boxed{567}&lt;/math&gt;.<br /> <br /> ~Math_Genius_164<br /> ==Solution 2(LOC and Trig)==<br /> Call AD and BC &lt;math&gt;a&lt;/math&gt;. Draw diagonal AC and call the foot of the perpendicular from B to AC &lt;math&gt;G&lt;/math&gt;. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that &lt;math&gt;\frac{10}{15}&lt;/math&gt;=&lt;math&gt;\frac{a}{AC}&lt;/math&gt; Therefore, &lt;math&gt;AC=1.5a&lt;/math&gt;. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that &lt;math&gt;AB=1.2a&lt;/math&gt;. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is &lt;math&gt;-\frac{1}{3}&lt;/math&gt;. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that &lt;math&gt;a=\frac{27\sqrt{2}}{2}&lt;/math&gt;. Then it can be found that the area &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\frac{567\sqrt{2}}{2}&lt;/math&gt;. Multiplying this by &lt;math&gt;\sqrt{2}&lt;/math&gt;, the answer is &lt;math&gt;\boxed{567}&lt;/math&gt;.<br /> -happykeeper<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149278 2021 AIME I Problems 2021-03-12T21:06:12Z <p>Mathpro12345: /* Problem 8 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> WTF<br /> <br /> WTF<br /> <br /> WTF<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149277 2021 AIME I Problems 2021-03-12T21:06:00Z <p>Mathpro12345: /* Problem 7 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> WTF<br /> <br /> WTF<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149276 2021 AIME I Problems 2021-03-12T21:05:46Z <p>Mathpro12345: /* Problem 6 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> WTF<br /> <br /> ==Problem 7==<br /> Find the number of pairs &lt;math&gt;(m,n)&lt;/math&gt; of positive integers with &lt;math&gt;1\le m&lt;n\le 30&lt;/math&gt; such that there exists a real number &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\sin(mx)+\sin(nx)=2.&lt;/cmath&gt;<br /> <br /> [[2021 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149275 2021 AIME I Problems 2021-03-12T21:05:33Z <p>Mathpro12345: /* Problem 5 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> ==Problem 6==<br /> Segments &lt;math&gt;\overline{AB}, \overline{AC},&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; are edges of a cube and &lt;math&gt;\overline{AG}&lt;/math&gt; is a diagonal through the center of the cube. Point &lt;math&gt;P&lt;/math&gt; satisfies &lt;math&gt;PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},&lt;/math&gt; and &lt;math&gt;PG=36\sqrt{7}&lt;/math&gt;. What is &lt;math&gt;PA&lt;/math&gt;?<br /> <br /> <br /> [[2021 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Find the number of pairs &lt;math&gt;(m,n)&lt;/math&gt; of positive integers with &lt;math&gt;1\le m&lt;n\le 30&lt;/math&gt; such that there exists a real number &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\sin(mx)+\sin(nx)=2.&lt;/cmath&gt;<br /> <br /> [[2021 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149274 2021 AIME I Problems 2021-03-12T21:05:22Z <p>Mathpro12345: /* Problem 4 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> ==Problem 5==<br /> Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br /> <br /> [[2021 AIME I Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Segments &lt;math&gt;\overline{AB}, \overline{AC},&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; are edges of a cube and &lt;math&gt;\overline{AG}&lt;/math&gt; is a diagonal through the center of the cube. Point &lt;math&gt;P&lt;/math&gt; satisfies &lt;math&gt;PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},&lt;/math&gt; and &lt;math&gt;PG=36\sqrt{7}&lt;/math&gt;. What is &lt;math&gt;PA&lt;/math&gt;?<br /> <br /> <br /> [[2021 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Find the number of pairs &lt;math&gt;(m,n)&lt;/math&gt; of positive integers with &lt;math&gt;1\le m&lt;n\le 30&lt;/math&gt; such that there exists a real number &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\sin(mx)+\sin(nx)=2.&lt;/cmath&gt;<br /> <br /> [[2021 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149273 2021 AIME I Problems 2021-03-12T21:05:03Z <p>Mathpro12345: /* Problem 3 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> ==Problem 4==<br /> Find the number of ways &lt;math&gt;66&lt;/math&gt; identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br /> <br /> [[2021 AIME I Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br /> <br /> [[2021 AIME I Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Segments &lt;math&gt;\overline{AB}, \overline{AC},&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; are edges of a cube and &lt;math&gt;\overline{AG}&lt;/math&gt; is a diagonal through the center of the cube. Point &lt;math&gt;P&lt;/math&gt; satisfies &lt;math&gt;PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},&lt;/math&gt; and &lt;math&gt;PG=36\sqrt{7}&lt;/math&gt;. What is &lt;math&gt;PA&lt;/math&gt;?<br /> <br /> <br /> [[2021 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Find the number of pairs &lt;math&gt;(m,n)&lt;/math&gt; of positive integers with &lt;math&gt;1\le m&lt;n\le 30&lt;/math&gt; such that there exists a real number &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\sin(mx)+\sin(nx)=2.&lt;/cmath&gt;<br /> <br /> [[2021 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149272 2021 AIME I Problems 2021-03-12T21:04:53Z <p>Mathpro12345: /* Problem 2 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> ==Problem 3==<br /> Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that can be expressed as the difference of two integral powers of &lt;math&gt;2.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Find the number of ways &lt;math&gt;66&lt;/math&gt; identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br /> <br /> [[2021 AIME I Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br /> <br /> [[2021 AIME I Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Segments &lt;math&gt;\overline{AB}, \overline{AC},&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; are edges of a cube and &lt;math&gt;\overline{AG}&lt;/math&gt; is a diagonal through the center of the cube. Point &lt;math&gt;P&lt;/math&gt; satisfies &lt;math&gt;PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},&lt;/math&gt; and &lt;math&gt;PG=36\sqrt{7}&lt;/math&gt;. What is &lt;math&gt;PA&lt;/math&gt;?<br /> <br /> <br /> [[2021 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Find the number of pairs &lt;math&gt;(m,n)&lt;/math&gt; of positive integers with &lt;math&gt;1\le m&lt;n\le 30&lt;/math&gt; such that there exists a real number &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\sin(mx)+\sin(nx)=2.&lt;/cmath&gt;<br /> <br /> [[2021 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems&diff=149271 2021 AIME I Problems 2021-03-12T21:04:42Z <p>Mathpro12345: /* Problem 1 */</p> <hr /> <div>{{AIME Problems|year=2021|n=I}}<br /> ==Problem 2==<br /> In the diagram below, &lt;math&gt;ABCD&lt;/math&gt; is a rectangle with side lengths &lt;math&gt;AB=3&lt;/math&gt; and &lt;math&gt;BC=11&lt;/math&gt;, and &lt;math&gt;AECF&lt;/math&gt; is a rectangle with side lengths &lt;math&gt;AF=7&lt;/math&gt; and &lt;math&gt;FC=9,&lt;/math&gt; as shown. The area of the shaded region common to the interiors of both rectangles is &lt;math&gt;\frac mn&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 /> &lt;asy&gt;<br /> pair A, B, C, D, E, F;<br /> A = (0,3);<br /> B=(0,0);<br /> C=(11,0);<br /> D=(11,3);<br /> E=foot(C, A, (9/4,0));<br /> F=foot(A, C, (35/4,3));<br /> draw(A--B--C--D--cycle);<br /> draw(A--E--C--F--cycle);<br /> filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;, A, W);<br /> label(&quot;$B$&quot;, B, W);<br /> label(&quot;$C$&quot;, C, (1,0));<br /> label(&quot;$D$&quot;, D, (1,0));<br /> label(&quot;$F$&quot;, F, N);<br /> label(&quot;$E$&quot;, E, S);<br /> &lt;/asy&gt;<br /> <br /> [[2021 AIME I Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Find the number of positive integers less than &lt;math&gt;1000&lt;/math&gt; that can be expressed as the difference of two integral powers of &lt;math&gt;2.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Find the number of ways &lt;math&gt;66&lt;/math&gt; identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.<br /> <br /> [[2021 AIME I Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.<br /> <br /> [[2021 AIME I Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Segments &lt;math&gt;\overline{AB}, \overline{AC},&lt;/math&gt; and &lt;math&gt;\overline{AD}&lt;/math&gt; are edges of a cube and &lt;math&gt;\overline{AG}&lt;/math&gt; is a diagonal through the center of the cube. Point &lt;math&gt;P&lt;/math&gt; satisfies &lt;math&gt;PB=60\sqrt{10}, PC=60\sqrt{5}, PD=120\sqrt{2},&lt;/math&gt; and &lt;math&gt;PG=36\sqrt{7}&lt;/math&gt;. What is &lt;math&gt;PA&lt;/math&gt;?<br /> <br /> <br /> [[2021 AIME I Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Find the number of pairs &lt;math&gt;(m,n)&lt;/math&gt; of positive integers with &lt;math&gt;1\le m&lt;n\le 30&lt;/math&gt; such that there exists a real number &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\sin(mx)+\sin(nx)=2.&lt;/cmath&gt;<br /> <br /> [[2021 AIME I Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Find the number of integers &lt;math&gt;c&lt;/math&gt; such that the equation&lt;cmath&gt;\left||20|x|-x^2|-c\right|=21&lt;/cmath&gt;has &lt;math&gt;12&lt;/math&gt; distinct real solutions.<br /> <br /> [[2021 AIME I Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be an isosceles trapezoid with &lt;math&gt;AD=BC&lt;/math&gt; and &lt;math&gt;AB&lt;CD.&lt;/math&gt; Suppose that the distances from &lt;math&gt;A&lt;/math&gt; to the lines &lt;math&gt;BC,CD,&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; are &lt;math&gt;15,18,&lt;/math&gt; and &lt;math&gt;10,&lt;/math&gt; respectively. Let &lt;math&gt;K&lt;/math&gt; be the area of &lt;math&gt;ABCD.&lt;/math&gt; Find &lt;math&gt;\sqrt2 \cdot K.&lt;/math&gt;<br /> <br /> [[2021 AIME I Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Consider the sequence &lt;math&gt;(a_k)_{k\ge 1}&lt;/math&gt; of positive rational numbers defined by &lt;math&gt;a_1 = \frac{2020}{2021}&lt;/math&gt; and for &lt;math&gt;k\ge 1&lt;/math&gt;, if &lt;math&gt;a_k = \frac{m}{n}&lt;/math&gt; for relatively prime positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, then<br /> <br /> &lt;cmath&gt;a_{k+1} = \frac{m + 18}{n+19}.&lt;/cmath&gt;Determine the sum of all positive integers &lt;math&gt;j&lt;/math&gt; such that the rational number &lt;math&gt;a_j&lt;/math&gt; can be written in the form &lt;math&gt;\frac{t}{t+1}&lt;/math&gt; for some positive integer &lt;math&gt;t&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Let &lt;math&gt;ABCD&lt;/math&gt; be a cyclic quadrilateral with &lt;math&gt;AB=4,BC=5,CD=6,&lt;/math&gt; and &lt;math&gt;DA=7&lt;/math&gt;. Let &lt;math&gt;A_1&lt;/math&gt; and &lt;math&gt;C_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, to line &lt;math&gt;BD,&lt;/math&gt; and let &lt;math&gt;B_1&lt;/math&gt; and &lt;math&gt;D_1&lt;/math&gt; be the feet of the perpendiculars from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D,&lt;/math&gt; respectively, to line &lt;math&gt;AC&lt;/math&gt;. The perimeter of &lt;math&gt;A_1B_1C_1D_1&lt;/math&gt; is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;A_1A_2A_3...A_{12}&lt;/math&gt; be a dodecagon (12-gon). Three frogs initially sit at &lt;math&gt;A_4,A_8,&lt;/math&gt; and &lt;math&gt;A_{12}&lt;/math&gt;. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is &lt;math&gt;\frac mn&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 /> [[2021 AIME I Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Circles &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt; with radii &lt;math&gt;961&lt;/math&gt; and &lt;math&gt;625&lt;/math&gt;, respectively, intersect at distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. A third circle &lt;math&gt;\omega&lt;/math&gt; is externally tangent to both &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;. Suppose line &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\omega&lt;/math&gt; at two points &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;Q&lt;/math&gt; such that the measure of minor arc &lt;math&gt;\widehat{PQ}&lt;/math&gt; is &lt;math&gt;120^{\circ}&lt;/math&gt;. Find the distance between the centers of &lt;math&gt;\omega_1&lt;/math&gt; and &lt;math&gt;\omega_2&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> For any positive integer &lt;math&gt;a,&lt;/math&gt; &lt;math&gt;\sigma(a)&lt;/math&gt; denotes the sum of the positive integer divisors of &lt;math&gt;a&lt;/math&gt;. Let &lt;math&gt;n&lt;/math&gt; be the least positive integer such that &lt;math&gt;\sigma(a^n)-1&lt;/math&gt; is divisible by &lt;math&gt;2021&lt;/math&gt; for all positive integers &lt;math&gt;a&lt;/math&gt;. Find the sum of the prime factors in the prime factorization of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Let &lt;math&gt;S&lt;/math&gt; be the set of positive integers &lt;math&gt;k&lt;/math&gt; such that the two parabolas&lt;cmath&gt;y=x^2-k~~\text{and}~~x=2(y-20)^2-k&lt;/cmath&gt;intersect in four distinct points, and these four points lie on a circle with radius at most &lt;math&gt;21&lt;/math&gt;. Find the sum of the least element of &lt;math&gt;S&lt;/math&gt; and the greatest element of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> [[2021 AIME I Problems/Problem 15|Solution]]<br /> <br /> ==See also==<br /> {{AIME box|year=2021|n=I|before=[[2020 AIME II Problems|2020 AIME II]]|after=[[2021 AIME II Problems|2021 AIME II]]}}<br /> * [[American Invitational Mathematics Examination]]<br /> * [[AIME Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=144797 2017 AMC 10A Problems/Problem 4 2021-02-03T16:46:20Z <p>Mathpro12345: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Mia is &quot;helping&quot; her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds, &lt;math&gt;3-2=1&lt;/math&gt; toys are put in the box, so after &lt;math&gt;27\cdot30&lt;/math&gt; seconds, there will be &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into the box, and we have our total amount of time to be &lt;math&gt;27\cdot30+30=840&lt;/math&gt; seconds, which equals &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> How did you choose to do &lt;math&gt;27\times30?&lt;/math&gt;<br /> ~A question by greenstar<br /> <br /> ==Solution 2==<br /> Though Mia's mom places &lt;math&gt;3&lt;/math&gt; toys every &lt;math&gt;30&lt;/math&gt; seconds, Mia takes out &lt;math&gt;2&lt;/math&gt; toys right after. Therefore, after &lt;math&gt;30&lt;/math&gt; seconds, the two have collectively placed &lt;math&gt;1&lt;/math&gt; toy into the box. Therefore by &lt;math&gt;13.5&lt;/math&gt; minutes, the two would have placed &lt;math&gt;27&lt;/math&gt; toys into the box. Therefore, at &lt;math&gt;14&lt;/math&gt; minutes, the two would have placed &lt;math&gt;30&lt;/math&gt; toys into the box. Though Mia may take &lt;math&gt;2&lt;/math&gt; toys out right after, the number of toys in the box first reaches &lt;math&gt;30&lt;/math&gt; by &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/str7kmcRMY8<br /> <br /> https://youtu.be/1F0IB0y6578<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_4&diff=144795 2017 AMC 10A Problems/Problem 4 2021-02-03T16:45:38Z <p>Mathpro12345: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Mia is &quot;helping&quot; her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Every &lt;math&gt;30&lt;/math&gt; seconds, &lt;math&gt;3-2=1&lt;/math&gt; toys are put in the box, so after &lt;math&gt;27\cdot30&lt;/math&gt; seconds, there will be &lt;math&gt;27&lt;/math&gt; toys in the box. Mia's mom will then put &lt;math&gt;3&lt;/math&gt; toys into the box, and we have our total amount of time to be &lt;math&gt;27\cdot30+30=840&lt;/math&gt; seconds, which equals &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> How did you choose to do &lt;math&gt;27\times30?&lt;/math&gt;<br /> ~A question by greenstarwins<br /> <br /> ==Solution 2==<br /> Though Mia's mom places &lt;math&gt;3&lt;/math&gt; toys every &lt;math&gt;30&lt;/math&gt; seconds, Mia takes out &lt;math&gt;2&lt;/math&gt; toys right after. Therefore, after &lt;math&gt;30&lt;/math&gt; seconds, the two have collectively placed &lt;math&gt;1&lt;/math&gt; toy into the box. Therefore by &lt;math&gt;13.5&lt;/math&gt; minutes, the two would have placed &lt;math&gt;27&lt;/math&gt; toys into the box. Therefore, at &lt;math&gt;14&lt;/math&gt; minutes, the two would have placed &lt;math&gt;30&lt;/math&gt; toys into the box. Though Mia may take &lt;math&gt;2&lt;/math&gt; toys out right after, the number of toys in the box first reaches &lt;math&gt;30&lt;/math&gt; by &lt;math&gt;14&lt;/math&gt; minutes. &lt;math&gt;\boxed{(\textbf{B})\ 14}&lt;/math&gt;<br /> <br /> ==Video Solution==<br /> https://youtu.be/str7kmcRMY8<br /> <br /> https://youtu.be/1F0IB0y6578<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> <br /> {{AMC10 box|year=2017|ab=A|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=IMO_Longlist_Problems&diff=144676 IMO Longlist Problems 2021-02-02T18:56:12Z <p>Mathpro12345: Created page with &quot;Warning: You are recreating a page that was previously deleted.&quot;</p> <hr /> <div>Warning: You are recreating a page that was previously deleted.</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=1971_AHSME_Problems/Problem_12&diff=143742 1971 AHSME Problems/Problem 12 2021-01-29T16:31:39Z <p>Mathpro12345: Created page with &quot;wtf&quot;</p> <hr /> <div>wtf</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_17&diff=141497 1992 AHSME Problems/Problem 17 2021-01-04T17:19:38Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The 2-digit integers from 19 to 92 are written consecutively to form the integer &lt;math&gt;N=192021\cdots9192&lt;/math&gt;. Suppose that &lt;math&gt;3^k&lt;/math&gt; is the highest power of 3 that is a factor of &lt;math&gt;N&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 1\quad<br /> \text{(C) } 2\quad<br /> \text{(D) } 3\quad<br /> \text{(E) more than } 3&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1===<br /> <br /> We can determine if our number is divisible by &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; by summing the digits. Looking at the one's place, we can start out with &lt;math&gt;0, 1, 2, 3, 4, 5, 6, 7, 8, 9&lt;/math&gt; and continue cycling though the numbers from &lt;math&gt;0&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;. For each one of these cycles, we add &lt;math&gt;0 + 1 + ... + 9 = 45&lt;/math&gt;. This is divisible by &lt;math&gt;9&lt;/math&gt;, thus we can ignore the sum. However, this excludes &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;90&lt;/math&gt;, &lt;math&gt;91&lt;/math&gt; and &lt;math&gt;92&lt;/math&gt;. These remaining units digits sum up to &lt;math&gt;9 + 1 + 2 = 12&lt;/math&gt;, which means our units sum is &lt;math&gt;3 \pmod 9&lt;/math&gt;. As for the tens digits, for &lt;math&gt;2, 3, 4, \cdots , 8&lt;/math&gt; we have &lt;math&gt;10&lt;/math&gt; sets of those: &lt;cmath&gt;\frac{8 \cdot 9}{2} - 1 = 35,&lt;/cmath&gt; which is congruent to &lt;math&gt;8 \pmod 9&lt;/math&gt;. We again have &lt;math&gt;19, 90, 91&lt;/math&gt; and &lt;math&gt;92&lt;/math&gt;, so we must add &lt;cmath&gt;1 + 9 \cdot 3 = 28&lt;/cmath&gt; to our total. &lt;math&gt;28&lt;/math&gt; is congruent to &lt;math&gt;1 \pmod 9&lt;/math&gt;. Thus our sum is congruent to &lt;math&gt;3 \pmod 9&lt;/math&gt;, and &lt;math&gt;k = 1 <br /> \implies \boxed{B}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> As our first trial with 3, We can say that &lt;math&gt;N\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\pmod{3}&lt;/math&gt;. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that &lt;math&gt;N\equiv 19+20+21+22+...+91+92\pmod{3}&lt;/math&gt;, and adding that up using the rainbow strategy, we get &lt;math&gt;N\equiv (111\times37)\pmod{3}&lt;/math&gt;. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> <br /> ~mathpro12345<br /> <br /> papa I wrote this second one<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=16|num-a=18}} <br /> <br /> [[Category: Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_17&diff=141496 1992 AHSME Problems/Problem 17 2021-01-04T17:14:16Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The 2-digit integers from 19 to 92 are written consecutively to form the integer &lt;math&gt;N=192021\cdots9192&lt;/math&gt;. Suppose that &lt;math&gt;3^k&lt;/math&gt; is the highest power of 3 that is a factor of &lt;math&gt;N&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 1\quad<br /> \text{(C) } 2\quad<br /> \text{(D) } 3\quad<br /> \text{(E) more than } 3&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1===<br /> <br /> We can determine if our number is divisible by &lt;math&gt;3&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; by summing the digits. Looking at the one's place, we can start out with &lt;math&gt;0, 1, 2, 3, 4, 5, 6, 7, 8, 9&lt;/math&gt; and continue cycling though the numbers from &lt;math&gt;0&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;. For each one of these cycles, we add &lt;math&gt;0 + 1 + ... + 9 = 45&lt;/math&gt;. This is divisible by &lt;math&gt;9&lt;/math&gt;, thus we can ignore the sum. However, this excludes &lt;math&gt;19&lt;/math&gt;, &lt;math&gt;90&lt;/math&gt;, &lt;math&gt;91&lt;/math&gt; and &lt;math&gt;92&lt;/math&gt;. These remaining units digits sum up to &lt;math&gt;9 + 1 + 2 = 12&lt;/math&gt;, which means our units sum is &lt;math&gt;3 \pmod 9&lt;/math&gt;. As for the tens digits, for &lt;math&gt;2, 3, 4, \cdots , 8&lt;/math&gt; we have &lt;math&gt;10&lt;/math&gt; sets of those: &lt;cmath&gt;\frac{8 \cdot 9}{2} - 1 = 35,&lt;/cmath&gt; which is congruent to &lt;math&gt;8 \pmod 9&lt;/math&gt;. We again have &lt;math&gt;19, 90, 91&lt;/math&gt; and &lt;math&gt;92&lt;/math&gt;, so we must add &lt;cmath&gt;1 + 9 \cdot 3 = 28&lt;/cmath&gt; to our total. &lt;math&gt;28&lt;/math&gt; is congruent to &lt;math&gt;1 \pmod 9&lt;/math&gt;. Thus our sum is congruent to &lt;math&gt;3 \pmod 9&lt;/math&gt;, and &lt;math&gt;k = 1 <br /> \implies \boxed{B}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> As our first trial with 3, We can say that &lt;math&gt;N\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\pmod{3}&lt;/math&gt;. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that &lt;math&gt;N\equiv 19+20+21+22+...+91+92\pmod{3}&lt;/math&gt;, and adding that up using the rainbow strategy, we get &lt;math&gt;N\equiv (111\times37)\pmod{3}&lt;/math&gt;. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> <br /> ~mathpro12345<br /> <br /> == See also ==<br /> {{AHSME box|year=1992|num-b=16|num-a=18}} <br /> <br /> [[Category: Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_15&diff=139954 2018 AMC 12B Problems/Problem 15 2020-12-19T01:24:37Z <p>Mathpro12345: /* Solution 4 (easy) */</p> <hr /> <div>== Problem ==<br /> How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?<br /> <br /> &lt;math&gt;\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> Analyze that the three-digit integers divisible by &lt;math&gt;3&lt;/math&gt; start from &lt;math&gt;102&lt;/math&gt;. In the &lt;math&gt;200&lt;/math&gt;'s, it starts from &lt;math&gt;201&lt;/math&gt;. In the &lt;math&gt;300&lt;/math&gt;'s, it starts from &lt;math&gt;300&lt;/math&gt;. We see that the units digits is &lt;math&gt;0, 1, &lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Write out the 1- and 2-digit multiples of &lt;math&gt;3&lt;/math&gt; starting from &lt;math&gt;0, 1,&lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt; Count up the ones that meet the conditions. Then, add up and multiply by &lt;math&gt;3&lt;/math&gt;, since there are three sets of three from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9.&lt;/math&gt; Then, subtract the amount that started from &lt;math&gt;0&lt;/math&gt;, since the &lt;math&gt;300&lt;/math&gt;'s ll contain the digit &lt;math&gt;3&lt;/math&gt;.<br /> <br /> We get: &lt;cmath&gt;3(12+12+12)-12.&lt;/cmath&gt;<br /> <br /> This gives us: &lt;cmath&gt;\boxed{\textbf{(A) } 96}.&lt;/cmath&gt;<br /> <br /> == Solution 2==<br /> <br /> There are &lt;math&gt;4&lt;/math&gt; choices for the last digit (&lt;math&gt;1, 5, 7, 9&lt;/math&gt;), and &lt;math&gt;8&lt;/math&gt; choices for the first digit (exclude &lt;math&gt;0&lt;/math&gt;). We know what the second digit mod &lt;math&gt;3&lt;/math&gt; is, so there are &lt;math&gt;3&lt;/math&gt; choices for it (pick from one of the sets &lt;math&gt;\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}&lt;/math&gt;). The answer is &lt;math&gt;4\cdot 8 \cdot 3 = \boxed{96}&lt;/math&gt; (Plasma_Vortex)<br /> <br /> == Solution 3==<br /> <br /> Consider the number of &lt;math&gt;2&lt;/math&gt;-digit numbers that do not contain the digit &lt;math&gt;3&lt;/math&gt;, which is &lt;math&gt;90-18=72&lt;/math&gt;. For any of these &lt;math&gt;2&lt;/math&gt;-digit numbers, we can append &lt;math&gt;1,5,7,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; to reach a desirable &lt;math&gt;3&lt;/math&gt;-digit number. However, &lt;math&gt;1 \equiv 7 \equiv 1&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt;, and thus we need to count any &lt;math&gt;2&lt;/math&gt;-digit number &lt;math&gt;\equiv 2&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt; twice. There are &lt;math&gt;(98-11)/3+1=30&lt;/math&gt; total such numbers that have remainder &lt;math&gt;2&lt;/math&gt;, but &lt;math&gt;6&lt;/math&gt; of them &lt;math&gt;(23,32,35,38,53,83)&lt;/math&gt; contain &lt;math&gt;3&lt;/math&gt;, so the number we want is &lt;math&gt;30-6=24&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;72+24= \boxed{96}&lt;/math&gt;.<br /> <br /> ==Solution 4 (easy)==<br /> We need to take care of all restrictions. Ranging from &lt;math&gt;101&lt;/math&gt; to &lt;math&gt;999&lt;/math&gt;, there are &lt;math&gt;450&lt;/math&gt; odd 3-digit numbers. Exactly &lt;math&gt;\frac{1}{3}&lt;/math&gt; of these numbers are divisible by 3, which is &lt;math&gt;450\times\frac{1}{3}=150&lt;/math&gt;. Of these 150 numbers, &lt;math&gt;\frac{4}{5}&lt;/math&gt; &lt;math&gt;\textbf{do not}&lt;/math&gt; have 3 in their ones (units) digit, &lt;math&gt;\frac{9}{10}&lt;/math&gt; &lt;math&gt;\textbf{do not}&lt;/math&gt; have 3 in their tens digit, and &lt;math&gt;\frac{8}{9}&lt;/math&gt; &lt;math&gt;\textbf{do not}&lt;/math&gt; have 3 in their hundreds digit. Thus, the total number of 3 digit integers are &lt;math&gt;900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96&lt;/math&gt;, or &lt;math&gt;\boxed{\text{A}}&lt;/math&gt;<br /> <br /> ~mathpro12345<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> [[Category:Introductory Number Theory Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_15&diff=139953 2018 AMC 12B Problems/Problem 15 2020-12-19T01:23:59Z <p>Mathpro12345: /* Solution 4 * (easy)/</p> <hr /> <div>== Problem ==<br /> How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?<br /> <br /> &lt;math&gt;\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 &lt;/math&gt;<br /> <br /> == Solution 1==<br /> Analyze that the three-digit integers divisible by &lt;math&gt;3&lt;/math&gt; start from &lt;math&gt;102&lt;/math&gt;. In the &lt;math&gt;200&lt;/math&gt;'s, it starts from &lt;math&gt;201&lt;/math&gt;. In the &lt;math&gt;300&lt;/math&gt;'s, it starts from &lt;math&gt;300&lt;/math&gt;. We see that the units digits is &lt;math&gt;0, 1, &lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Write out the 1- and 2-digit multiples of &lt;math&gt;3&lt;/math&gt; starting from &lt;math&gt;0, 1,&lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt; Count up the ones that meet the conditions. Then, add up and multiply by &lt;math&gt;3&lt;/math&gt;, since there are three sets of three from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9.&lt;/math&gt; Then, subtract the amount that started from &lt;math&gt;0&lt;/math&gt;, since the &lt;math&gt;300&lt;/math&gt;'s ll contain the digit &lt;math&gt;3&lt;/math&gt;.<br /> <br /> We get: &lt;cmath&gt;3(12+12+12)-12.&lt;/cmath&gt;<br /> <br /> This gives us: &lt;cmath&gt;\boxed{\textbf{(A) } 96}.&lt;/cmath&gt;<br /> <br /> == Solution 2==<br /> <br /> There are &lt;math&gt;4&lt;/math&gt; choices for the last digit (&lt;math&gt;1, 5, 7, 9&lt;/math&gt;), and &lt;math&gt;8&lt;/math&gt; choices for the first digit (exclude &lt;math&gt;0&lt;/math&gt;). We know what the second digit mod &lt;math&gt;3&lt;/math&gt; is, so there are &lt;math&gt;3&lt;/math&gt; choices for it (pick from one of the sets &lt;math&gt;\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}&lt;/math&gt;). The answer is &lt;math&gt;4\cdot 8 \cdot 3 = \boxed{96}&lt;/math&gt; (Plasma_Vortex)<br /> <br /> == Solution 3==<br /> <br /> Consider the number of &lt;math&gt;2&lt;/math&gt;-digit numbers that do not contain the digit &lt;math&gt;3&lt;/math&gt;, which is &lt;math&gt;90-18=72&lt;/math&gt;. For any of these &lt;math&gt;2&lt;/math&gt;-digit numbers, we can append &lt;math&gt;1,5,7,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; to reach a desirable &lt;math&gt;3&lt;/math&gt;-digit number. However, &lt;math&gt;1 \equiv 7 \equiv 1&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt;, and thus we need to count any &lt;math&gt;2&lt;/math&gt;-digit number &lt;math&gt;\equiv 2&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt; twice. There are &lt;math&gt;(98-11)/3+1=30&lt;/math&gt; total such numbers that have remainder &lt;math&gt;2&lt;/math&gt;, but &lt;math&gt;6&lt;/math&gt; of them &lt;math&gt;(23,32,35,38,53,83)&lt;/math&gt; contain &lt;math&gt;3&lt;/math&gt;, so the number we want is &lt;math&gt;30-6=24&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;72+24= \boxed{96}&lt;/math&gt;.<br /> <br /> ==Solution 4 (easy)==<br /> We need to take care of all restrictions. Ranging from &lt;math&gt;101&lt;/math&gt; to &lt;math&gt;999&lt;/math&gt;, there are &lt;math&gt;450&lt;/math&gt; odd 3-digit numbers. Exactly &lt;math&gt;\frac{1}{3}&lt;/math&gt; of these numbers are divisible by 3, which is &lt;math&gt;450\times\frac{1}{3}=150&lt;/math&gt;. Of these 150 numbers, &lt;math&gt;\frac{4}{5}&lt;/math&gt; &lt;math&gt;\textbf{do not}&lt;/math&gt; have 3 in their ones (units) digit, &lt;math&gt;\frac{9}{10}&lt;/math&gt; &lt;math&gt;\textbf{do not}&lt;/math&gt; have 3 in their tens digit, and &lt;math&gt;\frac{8}{9}&lt;/math&gt; &lt;math&gt;\textbf{do not}&lt;/math&gt; have 3 in their hundreds digit. Thus, the total number of 3 digit integers are &lt;math&gt;900\times\frac{1}{2}\times\frac{1}{3}\times\frac{4}{5}\times\frac{9}{10}\times\frac{8}{9}=96&lt;/math&gt;, or &lt;math&gt;\boxed{\text{A}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> [[Category:Introductory Number Theory Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=139888 2003 AMC 10A Problems/Problem 23 2020-12-17T23:42:12Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> &lt;asy&gt;<br /> unitsize(15mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60);<br /> pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp;<br /> pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp;<br /> pair Jp=shift(Gp)*Hp;<br /> pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp};<br /> draw(Ap--Dp--Jp--cycle);<br /> draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle);<br /> for(pair p : points)<br /> {<br /> fill(circle(p, 0.07),white);<br /> }<br /> pair[] Cn=new pair;<br /> Cn=centroid(Ap,Bp,Gp);<br /> Cn=centroid(Gp,Bp,Fp);<br /> Cn=centroid(Bp,Fp,Cp);<br /> Cn=centroid(Cp,Fp,Ep);<br /> Cn=centroid(Cp,Ep,Dp);<br /> label(&quot;$1$&quot;,Cn);<br /> label(&quot;$2$&quot;,Cn);<br /> label(&quot;$3$&quot;,Cn);<br /> label(&quot;$4$&quot;,Cn);<br /> label(&quot;$5$&quot;,Cn);<br /> for (pair p : Cn)<br /> {<br /> draw(circle(p,0.1));<br /> }&lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> <br /> == Solution 1==<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt; <br /> ~dolphin7<br /> <br /> ==Solution 2==<br /> We just need to count upward facing triangles because if we exclude the downward-facing triangles, we won't be overcounting any toothpicks. The first row of triangles has &lt;math&gt;1&lt;/math&gt; upward-facing triangle, the second row has &lt;math&gt;2&lt;/math&gt; upward-facing triangles, the third row has &lt;math&gt;3&lt;/math&gt; upward-facing triangles, and so on having &lt;math&gt;n&lt;/math&gt; upward-facing triangles in the &lt;math&gt;n^\text{th}&lt;/math&gt; row. The last row with &lt;math&gt;2003&lt;/math&gt; small triangles has &lt;math&gt;1002&lt;/math&gt; upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now &lt;math&gt;\frac{1002\times1003}{2}&lt;/math&gt;, meaning that the number of toothpicks are &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, or &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ~mathpro12345<br /> <br /> ===Note===<br /> You don't have to calculate the value of &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, and you can use units digits to find the answer easily. The units digit of &lt;math&gt;1002\times1003&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, and has a unit digit of &lt;math&gt;3&lt;/math&gt; after being divided by &lt;math&gt;2&lt;/math&gt;. Then this is multiplied by &lt;math&gt;3&lt;/math&gt;, now the final number ending with a &lt;math&gt;9&lt;/math&gt;. This leaves only one answer choice possible, which is &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_22&diff=139844 2015 AMC 10A Problems/Problem 22 2020-12-17T02:36:10Z <p>Mathpro12345: /* Solution 5 */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #17]] and [[2015 AMC 10A Problems|2015 AMC 10A #22]]}}<br /> ==Problem==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 2===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. We casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. This yields &lt;math&gt;\dbinom{8}{2} - 8 = 20&lt;/math&gt; arrangements, because the two people cannot be next to each other.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding &lt;math&gt;2&lt;/math&gt; possible arrangements.<br /> <br /> More difficult is:<br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. First, choose the location of the first person standing (&lt;math&gt;8&lt;/math&gt; choices). Next, choose &lt;math&gt;2&lt;/math&gt; of the remaining people in the remaining &lt;math&gt;5&lt;/math&gt; legal seats to stand, amounting to &lt;math&gt;6&lt;/math&gt; arrangements considering that these two people cannot stand next to each other. However, we have to divide by &lt;math&gt;3,&lt;/math&gt; because there are &lt;math&gt;3&lt;/math&gt; ways to choose the first person given any three. This yields &lt;math&gt;\dfrac{8 \cdot 6}{3} = 16&lt;/math&gt; arrangements for Case &lt;math&gt;5.&lt;/math&gt;<br /> <br /> Alternate Case &lt;math&gt;5:&lt;/math&gt; Use complementary counting. Total number of ways to choose 3 people from 8 which is &lt;math&gt;\dbinom{8}{3}&lt;/math&gt;. Sub-case &lt;math&gt;1:&lt;/math&gt; three people are next to each other which is &lt;math&gt;\dbinom{8}{1}&lt;/math&gt;. Sub-case &lt;math&gt;2:&lt;/math&gt; two people are next to each other and the third person is not &lt;math&gt;\dbinom{8}{1}&lt;/math&gt; &lt;math&gt;\dbinom{4}{1}&lt;/math&gt;. This yields &lt;math&gt;\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16&lt;/math&gt; <br /> <br /> Summing gives &lt;math&gt;1 + 8 + 20 + 2 + 16 = 47,&lt;/math&gt; and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We will count how many valid standing arrangements there are counting rotations as distinct and divide by &lt;math&gt;256&lt;/math&gt; at the end.<br /> Line up all &lt;math&gt;8&lt;/math&gt; people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires &lt;math&gt;2&lt;/math&gt; spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.<br /> <br /> If there are &lt;math&gt;4&lt;/math&gt; standing, there are &lt;math&gt;{4 \choose 4}=1&lt;/math&gt; ways to place them.<br /> For &lt;math&gt;3,&lt;/math&gt; there are &lt;math&gt;{3+2 \choose 3}=10&lt;/math&gt; ways.<br /> etc.<br /> Summing, we get &lt;math&gt;{4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34&lt;/math&gt; ways.<br /> <br /> Now we consider that the far right person can be standing as well, so we have<br /> &lt;math&gt;{3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13&lt;/math&gt; ways<br /> <br /> Together we have &lt;math&gt;34+13=47&lt;/math&gt;, and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://www.youtube.com/watch?v=krlnSWWp0I0<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2015|ab=A|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_22&diff=139843 2015 AMC 10A Problems/Problem 22 2020-12-17T02:35:25Z <p>Mathpro12345: /* Solution 4 */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #17]] and [[2015 AMC 10A Problems|2015 AMC 10A #22]]}}<br /> ==Problem==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 2===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. We casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. This yields &lt;math&gt;\dbinom{8}{2} - 8 = 20&lt;/math&gt; arrangements, because the two people cannot be next to each other.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding &lt;math&gt;2&lt;/math&gt; possible arrangements.<br /> <br /> More difficult is:<br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. First, choose the location of the first person standing (&lt;math&gt;8&lt;/math&gt; choices). Next, choose &lt;math&gt;2&lt;/math&gt; of the remaining people in the remaining &lt;math&gt;5&lt;/math&gt; legal seats to stand, amounting to &lt;math&gt;6&lt;/math&gt; arrangements considering that these two people cannot stand next to each other. However, we have to divide by &lt;math&gt;3,&lt;/math&gt; because there are &lt;math&gt;3&lt;/math&gt; ways to choose the first person given any three. This yields &lt;math&gt;\dfrac{8 \cdot 6}{3} = 16&lt;/math&gt; arrangements for Case &lt;math&gt;5.&lt;/math&gt;<br /> <br /> Alternate Case &lt;math&gt;5:&lt;/math&gt; Use complementary counting. Total number of ways to choose 3 people from 8 which is &lt;math&gt;\dbinom{8}{3}&lt;/math&gt;. Sub-case &lt;math&gt;1:&lt;/math&gt; three people are next to each other which is &lt;math&gt;\dbinom{8}{1}&lt;/math&gt;. Sub-case &lt;math&gt;2:&lt;/math&gt; two people are next to each other and the third person is not &lt;math&gt;\dbinom{8}{1}&lt;/math&gt; &lt;math&gt;\dbinom{4}{1}&lt;/math&gt;. This yields &lt;math&gt;\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16&lt;/math&gt; <br /> <br /> Summing gives &lt;math&gt;1 + 8 + 20 + 2 + 16 = 47,&lt;/math&gt; and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We will count how many valid standing arrangements there are counting rotations as distinct and divide by &lt;math&gt;256&lt;/math&gt; at the end.<br /> Line up all &lt;math&gt;8&lt;/math&gt; people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires &lt;math&gt;2&lt;/math&gt; spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.<br /> <br /> If there are &lt;math&gt;4&lt;/math&gt; standing, there are &lt;math&gt;{4 \choose 4}=1&lt;/math&gt; ways to place them.<br /> For &lt;math&gt;3,&lt;/math&gt; there are &lt;math&gt;{3+2 \choose 3}=10&lt;/math&gt; ways.<br /> etc.<br /> Summing, we get &lt;math&gt;{4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34&lt;/math&gt; ways.<br /> <br /> Now we consider that the far right person can be standing as well, so we have<br /> &lt;math&gt;{3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13&lt;/math&gt; ways<br /> <br /> Together we have &lt;math&gt;34+13=47&lt;/math&gt;, and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 5===<br /> We will count the number of valid arrangements and then divide by &lt;math&gt;2^8&lt;/math&gt; at the end. We proceed with casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. To do this, we imagine having 6 people with tails in a line first. Notate &quot;tails&quot; with &lt;math&gt;T&lt;/math&gt;. Thus, we have &lt;math&gt;TTTTTT&lt;/math&gt;. Now, we look to distribute the 2 &lt;math&gt;H&lt;/math&gt;'s into the 7 gaps made by the &lt;math&gt;T&lt;/math&gt;'s. We can do this in &lt;math&gt;{7 \choose 2}&lt;/math&gt; ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have &lt;math&gt;{7 \choose 2}-1=20&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. Similarly, we imagine 5 &lt;math&gt;T&lt;/math&gt;'s. Thus, we have &lt;math&gt;TTTTT&lt;/math&gt;. We distribute 3 &lt;math&gt;H&lt;/math&gt;'s into the gaps, which can be done &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: &lt;math&gt;{4 \choose 1}=4&lt;/math&gt;) Thus, we have &lt;math&gt;{6 \choose 3}-4=16&lt;/math&gt; arrangements. <br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. This can clearly be done in 2 ways: &lt;math&gt;HTHTHTHT&lt;/math&gt; or &lt;math&gt;THTHTHTH&lt;/math&gt;. This yields &lt;math&gt;2&lt;/math&gt; arrangements.<br /> <br /> Summing the cases, we get &lt;math&gt;1+8+20+16+2=47&lt;/math&gt; arrangements.<br /> Thus, the probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;<br /> <br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://www.youtube.com/watch?v=krlnSWWp0I0<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2015|ab=A|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_22&diff=139842 2015 AMC 10A Problems/Problem 22 2020-12-17T02:33:57Z <p>Mathpro12345: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #17]] and [[2015 AMC 10A Problems|2015 AMC 10A #22]]}}<br /> ==Problem==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt;\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256} &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 2===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. We casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. This yields &lt;math&gt;\dbinom{8}{2} - 8 = 20&lt;/math&gt; arrangements, because the two people cannot be next to each other.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. Then the people must be arranged in stand-sit-stand-sit-stand-sit-stand-sit fashion, yielding &lt;math&gt;2&lt;/math&gt; possible arrangements.<br /> <br /> More difficult is:<br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. First, choose the location of the first person standing (&lt;math&gt;8&lt;/math&gt; choices). Next, choose &lt;math&gt;2&lt;/math&gt; of the remaining people in the remaining &lt;math&gt;5&lt;/math&gt; legal seats to stand, amounting to &lt;math&gt;6&lt;/math&gt; arrangements considering that these two people cannot stand next to each other. However, we have to divide by &lt;math&gt;3,&lt;/math&gt; because there are &lt;math&gt;3&lt;/math&gt; ways to choose the first person given any three. This yields &lt;math&gt;\dfrac{8 \cdot 6}{3} = 16&lt;/math&gt; arrangements for Case &lt;math&gt;5.&lt;/math&gt;<br /> <br /> Alternate Case &lt;math&gt;5:&lt;/math&gt; Use complementary counting. Total number of ways to choose 3 people from 8 which is &lt;math&gt;\dbinom{8}{3}&lt;/math&gt;. Sub-case &lt;math&gt;1:&lt;/math&gt; three people are next to each other which is &lt;math&gt;\dbinom{8}{1}&lt;/math&gt;. Sub-case &lt;math&gt;2:&lt;/math&gt; two people are next to each other and the third person is not &lt;math&gt;\dbinom{8}{1}&lt;/math&gt; &lt;math&gt;\dbinom{4}{1}&lt;/math&gt;. This yields &lt;math&gt;\dbinom{8}{3} - \dbinom{8}{1} - \dbinom{8}{1} \dbinom{4}{1} = 16&lt;/math&gt; <br /> <br /> Summing gives &lt;math&gt;1 + 8 + 20 + 2 + 16 = 47,&lt;/math&gt; and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> We will count how many valid standing arrangements there are counting rotations as distinct and divide by &lt;math&gt;256&lt;/math&gt; at the end.<br /> Line up all &lt;math&gt;8&lt;/math&gt; people linearly. In order for no two people standing to be adjacent, we will place a sitting person to the right of each standing person. In effect, each standing person requires &lt;math&gt;2&lt;/math&gt; spaces and the standing people are separated by sitting people. We just need to determine the number of combinations of pairs and singles and the problem becomes very similar to pirates and gold aka stars and bars aka sticks and stones aka balls and urns.<br /> <br /> If there are &lt;math&gt;4&lt;/math&gt; standing, there are &lt;math&gt;{4 \choose 4}=1&lt;/math&gt; ways to place them.<br /> For &lt;math&gt;3,&lt;/math&gt; there are &lt;math&gt;{3+2 \choose 3}=10&lt;/math&gt; ways.<br /> etc.<br /> Summing, we get &lt;math&gt;{4 \choose 4}+{5 \choose 3}+{6 \choose 2}+{7 \choose 1}+{8 \choose 0}=1+10+15+7+1=34&lt;/math&gt; ways.<br /> <br /> Now we consider that the far right person can be standing as well, so we have<br /> &lt;math&gt;{3 \choose 3}+{4 \choose 2}+{5 \choose 1}+{6 \choose 0}=1+6+5+1=13&lt;/math&gt; ways<br /> <br /> Together we have &lt;math&gt;34+13=47&lt;/math&gt;, and so our probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> We will count how many valid standing arrangements there are (counting rotations as distinct), and divide by &lt;math&gt;2^8 = 256&lt;/math&gt; at the end. If we suppose for the moment that the people are in a line, and decide from left to right whether they sit or stand. If the leftmost person sits, we have the same number of arrangements as if there were only &lt;math&gt;7&lt;/math&gt; people. If they stand, we count the arrangements with &lt;math&gt;6&lt;/math&gt; instead because the person second from the left must sit. We notice that this is the Fibonacci sequence, where with &lt;math&gt;1&lt;/math&gt; person there are two ways and with &lt;math&gt;2&lt;/math&gt; people there are three ways. Carrying out the Fibonacci recursion until we get to &lt;math&gt;8&lt;/math&gt; people, we find there are &lt;math&gt;55&lt;/math&gt; standing arrangements. Some of these were illegal however, since both the first and last people stood. In these cases, both the leftmost and rightmost two people are fixed, leaving us to subtract the number of ways for &lt;math&gt;4&lt;/math&gt; people to stand in a line, which is &lt;math&gt;8&lt;/math&gt; from our sequence. Therefore our probability is &lt;math&gt;\frac{55 - 8}{256} = \boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;<br /> <br /> ===Solution 5===<br /> We will count the number of valid arrangements and then divide by &lt;math&gt;2^8&lt;/math&gt; at the end. We proceed with casework on how many people are standing.<br /> <br /> Case &lt;math&gt;1:&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; people are standing. This yields &lt;math&gt;1&lt;/math&gt; arrangement.<br /> <br /> Case &lt;math&gt;2:&lt;/math&gt; &lt;math&gt;1&lt;/math&gt; person is standing. This yields &lt;math&gt;8&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;3:&lt;/math&gt; &lt;math&gt;2&lt;/math&gt; people are standing. To do this, we imagine having 6 people with tails in a line first. Notate &quot;tails&quot; with &lt;math&gt;T&lt;/math&gt;. Thus, we have &lt;math&gt;TTTTTT&lt;/math&gt;. Now, we look to distribute the 2 &lt;math&gt;H&lt;/math&gt;'s into the 7 gaps made by the &lt;math&gt;T&lt;/math&gt;'s. We can do this in &lt;math&gt;{7 \choose 2}&lt;/math&gt; ways. However, note one way does not work, because we have two H's at the end, and the problem states we have a table, not a line. So, we have &lt;math&gt;{7 \choose 2}-1=20&lt;/math&gt; arrangements.<br /> <br /> Case &lt;math&gt;4:&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; people are standing. Similarly, we imagine 5 &lt;math&gt;T&lt;/math&gt;'s. Thus, we have &lt;math&gt;TTTTT&lt;/math&gt;. We distribute 3 &lt;math&gt;H&lt;/math&gt;'s into the gaps, which can be done &lt;math&gt;{6 \choose 3}&lt;/math&gt; ways. However, 4 arrangements will not work. (See this by putting the H's at the ends, and then choosing one of the remaining 4 gaps: &lt;math&gt;{4 \choose 1}=4&lt;/math&gt;) Thus, we have &lt;math&gt;{6 \choose 3}-4=16&lt;/math&gt; arrangements. <br /> <br /> Case &lt;math&gt;5:&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; people are standing. This can clearly be done in 2 ways: &lt;math&gt;HTHTHTHT&lt;/math&gt; or &lt;math&gt;THTHTHTH&lt;/math&gt;. This yields &lt;math&gt;2&lt;/math&gt; arrangements.<br /> <br /> Summing the cases, we get &lt;math&gt;1+8+20+16+2=47&lt;/math&gt; arrangements.<br /> Thus, the probability is &lt;math&gt;\boxed{\textbf{(A) } \dfrac{47}{256}}&lt;/math&gt;<br /> <br /> <br /> === Video Solution by Richard Rusczyk ===<br /> <br /> https://www.youtube.com/watch?v=krlnSWWp0I0<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2015|ab=A|num-b=21|num-a=23}}<br /> {{AMC12 box|year=2015|ab=A|num-b=16|num-a=18}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_I_Problems/Problem_2&diff=139829 2015 AIME I Problems/Problem 2 2020-12-16T23:13:33Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> The nine delegates to the Economic Cooperation Conference include &lt;math&gt;2&lt;/math&gt; officials from Mexico, &lt;math&gt;3&lt;/math&gt; officials from Canada, and &lt;math&gt;4&lt;/math&gt; officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is &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 /> ==Solution==<br /> <br /> The total number of ways to pick &lt;math&gt;3&lt;/math&gt; officials from &lt;math&gt;9&lt;/math&gt; total is &lt;math&gt;\binom{9}{3} = 84&lt;/math&gt;, which will be our denominator. Now we can consider the number of ways for exactly two sleepers to be from the same country for each country individually and add them to find our numerator:<br /> * There are &lt;math&gt;7&lt;/math&gt; different ways to pick &lt;math&gt;3&lt;/math&gt; delegates such that &lt;math&gt;2&lt;/math&gt; are from Mexico, simply because there are &lt;math&gt;9-2=7&lt;/math&gt; &quot;extra&quot; delegates to choose to be the third sleeper once both from Mexico are sleeping.<br /> * There are &lt;math&gt;3\times6=18&lt;/math&gt; ways to pick from Canada, as each Canadian can be left out once and each time one is left out there are &lt;math&gt;9-3=6&lt;/math&gt; &quot;extra&quot; people to select one more sleeper from.<br /> * Lastly, there are &lt;math&gt;6\times5=30&lt;/math&gt; ways to choose for the United States. It is easy to count &lt;math&gt;6&lt;/math&gt; different ways to pick &lt;math&gt;2&lt;/math&gt; of the &lt;math&gt;4&lt;/math&gt; Americans, and each time you do there are &lt;math&gt;9-4=5&lt;/math&gt; officials left over to choose from.<br /> <br /> Thus, the fraction is &lt;math&gt;\frac{7+18+30}{84} = \frac{55}{84}&lt;/math&gt;. Since this does not reduce, the answer is &lt;math&gt;55+84=\boxed{139}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2015|n=I|num-b=1|num-a=3}}<br /> {{MAA Notice}}<br /> [[Category:Introductory Combinatorics Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_17&diff=139797 2014 AMC 10A Problems/Problem 17 2020-12-16T19:15:13Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> First, we note that there are &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt; ways to get sums of &lt;math&gt;2, 3, 4, 5, 6&lt;/math&gt; respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is &lt;cmath&gt;\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.&lt;/cmath&gt; Since there are &lt;math&gt;\dbinom31&lt;/math&gt; ways to choose which die will be the one with the sum of the other two, our answer is &lt;math&gt;3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2014|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_24&diff=139794 2013 AMC 10A Problems/Problem 24 2020-12-16T18:18:43Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let us label the players of the first team &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, and those of the second team, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt;. <br /> <br /> &lt;math&gt;\textbf{1}&lt;/math&gt;. One way of scheduling all six distinct rounds could be:<br /> <br /> Round 1----&gt;&lt;math&gt;AX&lt;/math&gt; &lt;math&gt;BY&lt;/math&gt; &lt;math&gt;CZ&lt;/math&gt;<br /> Round 2----&gt;&lt;math&gt;AX&lt;/math&gt; &lt;math&gt;BZ&lt;/math&gt; &lt;math&gt;CY&lt;/math&gt;<br /> Round 3----&gt;&lt;math&gt;AY&lt;/math&gt; &lt;math&gt;BX&lt;/math&gt; &lt;math&gt;CZ&lt;/math&gt;<br /> Round 4----&gt;&lt;math&gt;AY&lt;/math&gt; &lt;math&gt;BZ&lt;/math&gt; &lt;math&gt;CX&lt;/math&gt;<br /> Round 5----&gt;&lt;math&gt;AZ&lt;/math&gt; &lt;math&gt;BX&lt;/math&gt; &lt;math&gt;CY&lt;/math&gt;<br /> Round 6----&gt;&lt;math&gt;AZ&lt;/math&gt; &lt;math&gt;BY&lt;/math&gt; &lt;math&gt;CX&lt;/math&gt;<br /> <br /> The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permutating those &lt;math&gt;6&lt;/math&gt; rounds and that can be done in &lt;math&gt;6!=720&lt;/math&gt; ways.<br /> <br /> &lt;math&gt;\textbf{2}&lt;/math&gt;. One can also make the schedule in such a way that two rounds are repeated. <br /> <br /> (a)<br /> Round 1----&gt;&lt;math&gt;AX&lt;/math&gt; &lt;math&gt;BZ&lt;/math&gt; &lt;math&gt;CY&lt;/math&gt;<br /> Round 2----&gt;&lt;math&gt;AX&lt;/math&gt; &lt;math&gt;BZ&lt;/math&gt; &lt;math&gt;CY&lt;/math&gt;<br /> Round 3----&gt;&lt;math&gt;AY&lt;/math&gt; &lt;math&gt;BX&lt;/math&gt; &lt;math&gt;CZ&lt;/math&gt;<br /> Round 4----&gt;&lt;math&gt;AY&lt;/math&gt; &lt;math&gt;BX&lt;/math&gt; &lt;math&gt;CZ&lt;/math&gt;<br /> Round 5----&gt;&lt;math&gt;AZ&lt;/math&gt; &lt;math&gt;BY&lt;/math&gt; &lt;math&gt;CX&lt;/math&gt;<br /> Round 6----&gt;&lt;math&gt;AZ&lt;/math&gt; &lt;math&gt;BY&lt;/math&gt; &lt;math&gt;CX&lt;/math&gt;<br /> <br /> (b)<br /> Round 1----&gt;&lt;math&gt;AX&lt;/math&gt; &lt;math&gt;BY&lt;/math&gt; &lt;math&gt;CZ&lt;/math&gt;<br /> Round 2----&gt;&lt;math&gt;AX&lt;/math&gt; &lt;math&gt;BY&lt;/math&gt; &lt;math&gt;CZ&lt;/math&gt;<br /> Round 3----&gt;&lt;math&gt;AY&lt;/math&gt; &lt;math&gt;BZ&lt;/math&gt; &lt;math&gt;CX&lt;/math&gt;<br /> Round 4----&gt;&lt;math&gt;AY&lt;/math&gt; &lt;math&gt;BZ&lt;/math&gt; &lt;math&gt;CX&lt;/math&gt;<br /> Round 5----&gt;&lt;math&gt;AZ&lt;/math&gt; &lt;math&gt;BX&lt;/math&gt; &lt;math&gt;CY&lt;/math&gt;<br /> Round 6----&gt;&lt;math&gt;AZ&lt;/math&gt; &lt;math&gt;BX&lt;/math&gt; &lt;math&gt;CY&lt;/math&gt;<br /> <br /> As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are<br /> &lt;math&gt;\frac{6!}{2!2!2!}&lt;/math&gt; = &lt;math&gt;90&lt;/math&gt;<br /> <br /> <br /> So the total number of schedules is &lt;math&gt;720+90+90&lt;/math&gt; =&lt;math&gt;\boxed{\textbf{(E)} 900}&lt;/math&gt;.<br /> <br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2013amc10a/358<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2013|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=139712 2012 AMC 12B Problems/Problem 16 2020-12-15T16:20:57Z <p>Mathpro12345: /* Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''': How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''': To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> <br /> <br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> === Solution 3===<br /> <br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3}\cdot3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Video Solutions:==<br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/272<br /> <br /> ~dolphin7<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=139711 2012 AMC 12B Problems/Problem 16 2020-12-15T16:20:21Z <p>Mathpro12345: /* Solutions */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''': How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''': To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 2===<br /> <br /> <br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> === Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 ===<br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3}\cdot3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Video Solutions:==<br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/272<br /> <br /> ~dolphin7<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=139710 2012 AMC 12B Problems/Problem 16 2020-12-15T16:19:56Z <p>Mathpro12345: /* Solution 3: A Different Way of Looking at Solution 1 */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''': How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''': To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 2<br /> <br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> === Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 ===<br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3}\cdot3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Video Solutions:==<br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/272<br /> <br /> ~dolphin7<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_16&diff=139705 2012 AMC 12B Problems/Problem 16 2020-12-15T13:17:52Z <p>Mathpro12345: /* Solution 2 (Answer Choices) */</p> <hr /> <div>{{duplicate|[[2012 AMC 12B Problems|2012 AMC 12B #16]] and [[2012 AMC 10B Problems|2012 AMC 10B #24]]}}<br /> <br /> == Problem==<br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> Let the ordered triple &lt;math&gt;(a,b,c)&lt;/math&gt; denote that &lt;math&gt;a&lt;/math&gt; songs are liked by Amy and Beth, &lt;math&gt;b&lt;/math&gt; songs by Beth and Jo, and &lt;math&gt;c&lt;/math&gt; songs by Jo and Amy. We claim that the only possible triples are &lt;math&gt;(1,1,1), (2,1,1), (1,2,1)(1,1,2)&lt;/math&gt;. <br /> <br /> To show this, observe these are all valid conditions. Second, note that none of &lt;math&gt;a,b,c&lt;/math&gt; can be bigger than 3. Suppose otherwise, that &lt;math&gt;a = 3&lt;/math&gt;. Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to be at least 1. In fact, we require either &lt;math&gt;b&lt;/math&gt; or &lt;math&gt;c&lt;/math&gt; to equal 1, otherwise there will be a song liked by all three. Suppose &lt;math&gt;b = 1&lt;/math&gt;. Then we must have &lt;math&gt;c=0&lt;/math&gt; since no song is liked by all three girls, a contradiction.<br /> <br /> '''Case 1''': How many ways are there for &lt;math&gt;(a,b,c)&lt;/math&gt; to equal &lt;math&gt;(1,1,1)&lt;/math&gt;? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So &lt;math&gt;(a,b,c)=(1,1,1)&lt;/math&gt; in &lt;math&gt;4\cdot3\cdot2\cdot4 = 96&lt;/math&gt; ways.<br /> <br /> '''Case 2''': To find the number of ways for &lt;math&gt;(a,b,c) = (2,1,1)&lt;/math&gt;, observe there are &lt;math&gt;\binom{4}{2} = 6&lt;/math&gt; choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are &lt;math&gt;6\cdot2\cdot3=36&lt;/math&gt; ways for the girls to like the songs.<br /> <br /> That gives a total of &lt;math&gt;96 + 36 = \boxed{132}&lt;/math&gt; ways for the girls to like the songs, so the answer is &lt;math&gt;(\textrm{\textbf{B}})&lt;/math&gt;.<br /> <br /> === Solution 3: A Different Way of Looking at Solution 1===<br /> Let &lt;math&gt;AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let &lt;math&gt;A, B, J,&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least &lt;math&gt;1\: AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt;, they must be &lt;math&gt;3&lt;/math&gt; songs out of the &lt;math&gt;4&lt;/math&gt; that Amy, Beth, and Jo listened to. The fourth song can be of any type &lt;math&gt;N, A, B, J, AB, BJ&lt;/math&gt;, and &lt;math&gt;AJ&lt;/math&gt; (there is no &lt;math&gt;ABJ&lt;/math&gt; because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange &lt;math&gt;AB, BJ, AJ&lt;/math&gt;, and a song from the set &lt;math&gt;\{N, A, B, J, AB, BJ, AJ\}&lt;/math&gt;.<br /> <br /> Case 1: Fourth song = &lt;math&gt;N, A, B, J &lt;/math&gt;<br /> <br /> Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 4&lt;/math&gt; choices = &lt;math&gt;96&lt;/math&gt;.<br /> <br /> Case 2: Fourth song = &lt;math&gt;AB, BJ, AJ&lt;/math&gt;<br /> <br /> Note that in Case &lt;math&gt;2&lt;/math&gt;, all three of the choices for the fourth song repeat somewhere in the first three songs.<br /> <br /> Number of ways to rearrange = &lt;math&gt;(4!/2!)&lt;/math&gt; rearrangements for each choice &lt;math&gt;*\: 3&lt;/math&gt; choices = &lt;math&gt;36&lt;/math&gt;.<br /> <br /> &lt;math&gt;96 + 36 = \boxed{\textbf{(B)} \: 132}&lt;/math&gt;.<br /> <br /> <br /> === Solution 4: Another Different Way of Looking at Solution 1 &amp; 3 ===<br /> <br /> There are &lt;math&gt;\binom{4}{3}&lt;/math&gt; ways to choose the three songs that are liked by the three pairs of girls.<br /> <br /> There are &lt;math&gt;3!&lt;/math&gt; ways to determine how the three songs are liked, or which song is liked by which pair of girls.<br /> <br /> In total, there are &lt;math&gt;\binom{4}{3}\cdot3!&lt;/math&gt; possibilities for the first &lt;math&gt;3&lt;/math&gt; songs.<br /> <br /> There are &lt;math&gt;3&lt;/math&gt; cases for the 4th song, call it song D.<br /> <br /> Case &lt;math&gt;1&lt;/math&gt;: D is disliked by all &lt;math&gt;3&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there is only &lt;math&gt;1&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;2&lt;/math&gt;: D is liked by exactly &lt;math&gt;1&lt;/math&gt; girl &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; possibility.<br /> <br /> Case &lt;math&gt;3&lt;/math&gt;: D is liked by exactly &lt;math&gt;2&lt;/math&gt; girls &lt;math&gt;\implies&lt;/math&gt; there are &lt;math&gt;3&lt;/math&gt; pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first &lt;math&gt;3&lt;/math&gt; songs liked by the same girls.<br /> <br /> Counting the overlaps, there are &lt;math&gt;3&lt;/math&gt; ways to choose the pair with overlaps and &lt;math&gt;4\cdot3=12&lt;/math&gt; ways to choose what the other &lt;math&gt;2&lt;/math&gt; pairs like independently. In total, there are &lt;math&gt;3\cdot12=36&lt;/math&gt; overlapped possibilities.<br /> <br /> Finally, there are &lt;math&gt;\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132&lt;/math&gt; ways for the songs to be likely by the girls. &lt;math&gt;\boxed{\mathrm{(B)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Video Solutions:==<br /> ==Video Solution by Richard Rusczyk==<br /> https://artofproblemsolving.com/videos/amc/2012amc10b/272<br /> <br /> ~dolphin7<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> <br /> {{AMC10 box|year=2012|ab=B|num-b=23|num-a=25}}<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=15|num-a=17}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_16&diff=139546 2011 AMC 12A Problems/Problem 16 2020-12-13T18:40:28Z <p>Mathpro12345: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Each vertex of convex pentagon &lt;math&gt;ABCDE&lt;/math&gt; is to be assigned a color. There are &lt;math&gt;6&lt;/math&gt; colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?<br /> <br /> &lt;math&gt;<br /> \textbf{(A)}\ 2520 \qquad<br /> \textbf{(B)}\ 2880 \qquad<br /> \textbf{(C)}\ 3120 \qquad<br /> \textbf{(D)}\ 3250 \qquad<br /> \textbf{(E)}\ 3750 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> We can do some casework when working our way around the pentagon from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;E&lt;/math&gt;. At each stage, there will be a makeshift diagram.<br /> <br /> 1.) For &lt;math&gt;A&lt;/math&gt;, we can choose any of the 6 colors.<br /> <br /> A : 6<br /> <br /> 2.) For &lt;math&gt;B&lt;/math&gt;, we can either have the same color as &lt;math&gt;A&lt;/math&gt;, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and &lt;math&gt;D&lt;/math&gt; will be affected by both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;.<br /> <br /> A : 6<br /> B:1 B:5<br /> <br /> 3.) For &lt;math&gt;C&lt;/math&gt;, we cannot have the same color as &lt;math&gt;A&lt;/math&gt;. Also, we can have the same color as &lt;math&gt;B&lt;/math&gt; (&lt;math&gt;E&lt;/math&gt; will be affected), or any of the other 4 colors. Because &lt;math&gt;C&lt;/math&gt; can't be the same as &lt;math&gt;A&lt;/math&gt;, it can't be the same as &lt;math&gt;B&lt;/math&gt; if &lt;math&gt;B&lt;/math&gt; is the same as &lt;math&gt;A&lt;/math&gt;, so it can be any of the 5 other colors.<br /> <br /> A : 6<br /> B:1 B:5<br /> C:5 C:4 C:1<br /> <br /> 4.) &lt;math&gt;D&lt;/math&gt; is affected by &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. If they are the same, then &lt;math&gt;D&lt;/math&gt; can be any of the other 5 colors. If they are different, then &lt;math&gt;D&lt;/math&gt; can be any of the (6-2)=4 colors.<br /> <br /> A : 6<br /> B:1 B:5<br /> C:5 C:4 C:1<br /> D:5 D:4 D:4<br /> <br /> 5.) &lt;math&gt;E&lt;/math&gt; is affected by &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. If they are the same, then &lt;math&gt;E&lt;/math&gt; can be any of the other 5 colors. If they are different, then &lt;math&gt;E&lt;/math&gt; can be any of the (6-2)=4 colors.<br /> <br /> A : 6<br /> B:1 B:5<br /> C:5 C:4 C:1<br /> D:5 D:4 D:4<br /> E:4 E:4 E:5<br /> <br /> 6.) Now, we can multiply these three paths and add them:<br /> &lt;math&gt;(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)<br /> =600+1920+600=3120&lt;/math&gt;<br /> <br /> 7.) Our answer is &lt;math&gt;C&lt;/math&gt;!<br /> <br /> ==Solution 2==<br /> <br /> Right off the bat, we can analyze three things: <br /> <br /> <br /> 1.) There can only be two of the same color on the pentagon.<br /> <br /> 2.) Any pair of the same color can only be next to each other on the pentagon.<br /> <br /> 3.) There can only be two different pairs of same colors on the pentagon at once.<br /> <br /> <br /> Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.<br /> <br /> <br /> 1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count &lt;math&gt;6!=720&lt;/math&gt; cases. No rotation is necessary because all permutations are accounted for.<br /> <br /> <br /> 2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex. <br /> <br /> We get &lt;math&gt;6\times5\times4\times3=360&lt;/math&gt;.<br /> <br /> However, there are 5 different locations the pair could be at. Therefore we get &lt;math&gt;360\times5=1800&lt;/math&gt; possibilities for one pair.<br /> <br /> <br /> 3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.<br /> <br /> We get &lt;math&gt;6\times5\times4=120&lt;/math&gt;.<br /> <br /> Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get &lt;math&gt;120\times5=600&lt;/math&gt; possibilities for two pairs. <br /> <br /> <br /> 5.) If we add all of three cases together, we get &lt;math&gt;720+1800+600=3120&lt;/math&gt;. The answer is &lt;math&gt;C&lt;/math&gt;.<br /> <br /> Solution by gsaelite<br /> <br /> ==Video Solution==<br /> https://youtu.be/FThly7dRBIE<br /> <br /> ~IceMatrix<br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_17&diff=139542 2010 AMC 12B Problems/Problem 17 2020-12-13T17:53:42Z <p>Mathpro12345: /* Solution 1 */</p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #17]] and [[2010 AMC 10B Problems|2010 AMC 10B #23]]}}<br /> <br /> == Problem ==<br /> The entries in a &lt;math&gt;3 \times 3&lt;/math&gt; array include all the digits from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right. <br /> <br /> <br /> *'''Case 1: Center 4'''<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;4&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;4&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular}&lt;/cmath&gt;<br /> <br /> 3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. &lt;math&gt;2*6=12&lt;/math&gt;<br /> <br /> *'''Case 2: Center 5'''<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;3\\<br /> \hline 4&amp;5&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;5&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;5&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;3\\<br /> \hline 4&amp;5&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular}&lt;/cmath&gt;<br /> <br /> Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that &lt;math&gt;4&lt;5&lt;/math&gt;, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, &lt;math&gt;2*9=18&lt;/math&gt;<br /> <br /> *'''Case 3: Center 6'''<br /> By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured.<br /> &lt;math&gt;2*6=12&lt;/math&gt;<br /> <br /> &lt;cmath&gt;12+18+12=\boxed{\textbf{D)}42}&lt;/cmath&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> == Solution 3==<br /> This solution is trivial by the hook length theorem. The hooks look like this:<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|} \hline 5 &amp; 4 &amp; 3 \\<br /> \hline 4 &amp; 3 &amp; 2\\<br /> \hline 3 &amp; 2 &amp; 1\\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> So, the answer is &lt;math&gt;\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}&lt;/math&gt; = &lt;math&gt;\boxed{\text{(D) }42}&lt;/math&gt;<br /> <br /> P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZfnxbpdFKjU?t=422<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_17&diff=139541 2010 AMC 12B Problems/Problem 17 2020-12-13T17:53:05Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #17]] and [[2010 AMC 10B Problems|2010 AMC 10B #23]]}}<br /> <br /> == Problem ==<br /> The entries in a &lt;math&gt;3 \times 3&lt;/math&gt; array include all the digits from &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;9&lt;/math&gt;, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right. <br /> <br /> <br /> *'''Case 1: Center 4'''<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;4&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;4&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular}&lt;/cmath&gt;<br /> <br /> 3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. &lt;math&gt;2*6=12&lt;/math&gt;<br /> <br /> *'''Case 2: Center 5'''<br /> &lt;cmath&gt;\begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;3\\<br /> \hline 4&amp;5&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;5&amp;\\<br /> \hline &amp;8&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;\\<br /> \hline 3&amp;5&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&amp;2&amp;3\\<br /> \hline 4&amp;5&amp;8\\<br /> \hline &amp;&amp;9\\<br /> \hline \end{tabular}&lt;/cmath&gt;<br /> <br /> Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that &lt;math&gt;4&lt;5&lt;/math&gt;, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, &lt;math&gt;2*9=18&lt;/math&gt;<br /> <br /> *'''Case 3: Center 6'''<br /> By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured.<br /> &lt;math&gt;2*6=12&lt;/math&gt;<br /> <br /> &lt;cmath&gt;12+18+12=\boxed{\textbf{D)}42}&lt;/cmath&gt;<br /> <br /> <br /> ~BJHHar<br /> <br /> <br /> ''P.S.: I like the tetris approach used in Solution 2 but found it a bit arbitrary. Solution 3 is the best, but not many would know hook length theorem. If the initial observations are unclear, make a tableau with a range of possible numbers in each square''.<br /> <br /> == Solution 3==<br /> This solution is trivial by the hook length theorem. The hooks look like this:<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|} \hline 5 &amp; 4 &amp; 3 \\<br /> \hline 4 &amp; 3 &amp; 2\\<br /> \hline 3 &amp; 2 &amp; 1\\<br /> \hline \end{tabular}&lt;/math&gt;<br /> <br /> So, the answer is &lt;math&gt;\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}&lt;/math&gt; = &lt;math&gt;\boxed{\text{(D) }42}&lt;/math&gt;<br /> <br /> P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.<br /> <br /> ==Video Solution==<br /> https://youtu.be/ZfnxbpdFKjU?t=422<br /> <br /> ~IceMatrix<br /> <br /> == See also ==<br /> {{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_21&diff=139489 2009 AMC 12B Problems/Problem 21 2020-12-12T19:28:57Z <p>Mathpro12345: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Ten women sit in &lt;math&gt;10&lt;/math&gt; seats in a line. All of the &lt;math&gt;10&lt;/math&gt; get up and then reseat themselves using all &lt;math&gt;10&lt;/math&gt; seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is<br /> &lt;cmath&gt;\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.&lt;/cmath&gt;<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;S_n&lt;/math&gt; be the number of possible seating arrangements with &lt;math&gt;n&lt;/math&gt; women. Consider &lt;math&gt;n \ge 3,&lt;/math&gt; and focus on the rightmost woman. If she returns back to her seat, then there are &lt;math&gt;S_{n-1}&lt;/math&gt; ways to seat the remaining &lt;math&gt;n-1&lt;/math&gt; women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us &lt;math&gt;S_{n-2}&lt;/math&gt; ways to seat the other &lt;math&gt;n-2&lt;/math&gt; women, so we obtain the recursion<br /> &lt;cmath&gt;S_n = S_{n-1}+S_{n-2}.&lt;/cmath&gt;<br /> <br /> Starting with &lt;math&gt;S_1=1&lt;/math&gt; and &lt;math&gt;S_2=2,&lt;/math&gt; we can calculate &lt;math&gt;S_{10}=\boxed{89}.&lt;/math&gt;<br /> <br /> ==Clarification of Solution 2==<br /> The seating possibilities of woman #&lt;math&gt;10&lt;/math&gt; become the two cases which we work out. &lt;math&gt;S_n&lt;/math&gt; was defined to be the number of different seating arrangements with &lt;math&gt;n&lt;/math&gt; women.<br /> <br /> In the first &quot;case&quot; if woman #&lt;math&gt;10&lt;/math&gt; sits in the seat #&lt;math&gt;10&lt;/math&gt;, this leads to a similar scenario, but with &lt;math&gt;9&lt;/math&gt; women instead. That means that for this case, there are a total of &lt;math&gt;S_{9}&lt;/math&gt; possible arrangements. We don't know how many exactly, but we are able to define it in terms of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> During the second &quot;case&quot;, woman #&lt;math&gt;10&lt;/math&gt; sits in seat #&lt;math&gt;9&lt;/math&gt;. This time, woman #9 must go to seat #&lt;math&gt;10&lt;/math&gt;, as she is the only other person who can go there. This leaves us with &lt;math&gt;8&lt;/math&gt; women, and we again represent this in terms of &lt;math&gt;S \Rightarrow S_8&lt;/math&gt;.<br /> <br /> Therefore, we can write &lt;math&gt;S_{10}&lt;/math&gt; in terms of &lt;math&gt;S_8&lt;/math&gt; and &lt;math&gt;S_9&lt;/math&gt;, like so:<br /> <br /> &lt;cmath&gt;S_{10} = S_8 + S_9.&lt;/cmath&gt;<br /> <br /> We can then generalize this to say<br /> <br /> &lt;cmath&gt;S_n = S_{n-1}+S_{n-2}.&lt;/cmath&gt;<br /> <br /> Calculating &lt;math&gt;S_1 = 1&lt;/math&gt; and &lt;math&gt;S_2 =2,&lt;/math&gt; then following the recursive rule from above, we get &lt;math&gt;S_{10} = 89 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=B|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_21&diff=139488 2009 AMC 12B Problems/Problem 21 2020-12-12T19:28:24Z <p>Mathpro12345: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;Ten women sit in &lt;math&gt;10&lt;/math&gt; seats in a line. All of the &lt;math&gt;10&lt;/math&gt; get up and then reseat themselves using all &lt;math&gt;10&lt;/math&gt; seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is<br /> &lt;cmath&gt;\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.&lt;/cmath&gt;<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;S_n&lt;/math&gt; be the number of possible seating arrangements with &lt;math&gt;n&lt;/math&gt; women. Consider &lt;math&gt;n \ge 3,&lt;/math&gt; and focus on the rightmost woman. If she returns back to her seat, then there are &lt;math&gt;S_{n-1}&lt;/math&gt; ways to seat the remaining &lt;math&gt;n-1&lt;/math&gt; women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us &lt;math&gt;S_{n-2}&lt;/math&gt; ways to seat the other &lt;math&gt;n-2&lt;/math&gt; women, so we obtain the recursion<br /> &lt;cmath&gt;S_n = S_{n-1}+S_{n-2}.&lt;/cmath&gt;<br /> <br /> Starting with &lt;math&gt;S_1=1&lt;/math&gt; and &lt;math&gt;S_2=2,&lt;/math&gt; we can calculate &lt;math&gt;S_{10}=\boxed{89}.&lt;/math&gt;<br /> <br /> ==Clarification of Solution 2==<br /> The seating possibilities of woman #&lt;math&gt;10&lt;/math&gt; become the two cases which we work out. &lt;math&gt;S_n&lt;/math&gt; was defined to be the number of different seating arrangements with &lt;math&gt;n&lt;/math&gt; women.<br /> <br /> In the first &quot;case&quot; if woman #&lt;math&gt;10&lt;/math&gt; sits in the seat #&lt;math&gt;10&lt;/math&gt;, this leads to a similar scenario, but with &lt;math&gt;9&lt;/math&gt; women instead. That means that for this case, there are a total of &lt;math&gt;S_{9}&lt;/math&gt; possible arrangements. We don't know how many exactly, but we are able to define it in terms of &lt;math&gt;S&lt;/math&gt;.<br /> <br /> During the second &quot;case&quot;, woman #&lt;math&gt;10&lt;/math&gt; sits in seat #&lt;math&gt;9&lt;/math&gt;. This time, woman #9 must go to seat #&lt;math&gt;10&lt;/math&gt;, as she is the only other person who can go there. This leaves us with &lt;math&gt;8&lt;/math&gt; women, and we again represent this in terms of &lt;math&gt;S \Rightarrow S_8&lt;/math&gt;.<br /> <br /> Therefore, we can write &lt;math&gt;S_{10}&lt;/math&gt; in terms of &lt;math&gt;S_8&lt;/math&gt; and &lt;math&gt;S_9&lt;/math&gt;, like so:<br /> <br /> &lt;cmath&gt;S_{10} = S_8 + S_9.&lt;/cmath&gt;<br /> <br /> We can then generalize this to say<br /> <br /> &lt;cmath&gt;S_n = S_{n-1}+S_{n-2}.&lt;/cmath&gt;<br /> <br /> Calculating &lt;math&gt;S_1 = 1&lt;/math&gt; and &lt;math&gt;S_2 =2,&lt;/math&gt; then following the recursive rule from above, we get &lt;math&gt;S_{10} = 89 \Rightarrow \boxed{\text{A}}&lt;/math&gt;.<br /> ==Solution 3==<br /> let &lt;math&gt;a_n=&lt;/math&gt; the number of ways to order n women as per the constraints of the problem. We wish to find &lt;math&gt;a_{10}&lt;/math&gt;. Lets take a look at 3 cases regarding the 2 women on the outside:<br /> <br /> Case 1: Both women stay in the same spot<br /> Then, we wish to order the inner people. This gives us &lt;math&gt;a_8&lt;/math&gt; ways to do so<br /> <br /> Case 2: One woman stays in her spot, and the other moves<br /> In this case, the woman that moves has to swap spots with the person right next to her, because that is the only way to use all 10 seats. If this doesn't make sense, look at the diagram below (each star is a woman, and the a represents the woman who will move:<br /> <br /> a * * * * * * * * b<br /> <br /> After &quot;b&quot; moves:<br /> <br /> a * * * * * * * b ?<br /> <br /> Now, the * has to go somewhere. If the star doesn't fill the spot that &quot;b&quot; was in, then only 9 seats would be used. Thus, the star has to go to the spot where &quot;b&quot; was, giving:<br /> <br /> a * * * * * * * b *<br /> <br /> This leaves us to order the 7 people who have not yet decided where they want to go. Since either of the women can move, this gives us &lt;math&gt;2a_7&lt;/math&gt;.<br /> <br /> Finally, Case 3: Both women switch<br /> In this case, we only have to order the inner 6 people, because the two women will swap spots with the people right next to them (see logic above). Thus, this yeilds &lt;math&gt;a_6&lt;/math&gt;<br /> <br /> So: &lt;math&gt;a_10=a_8+a_6+2a_7&lt;/math&gt; We can generalize this to &lt;math&gt;a_{n+2}=a_n+a_{n-2}+2a_{n-1}&lt;/math&gt;<br /> <br /> Now, we calculate by hand that:<br /> &lt;cmath&gt;a_1=1&lt;/cmath&gt;<br /> &lt;cmath&gt;a_2=2&lt;/cmath&gt;<br /> &lt;cmath&gt;a_3=3&lt;/cmath&gt;<br /> &lt;cmath&gt;a_4=5&lt;/cmath&gt;<br /> Finally, we use the recurrence relation we developed to build our way to &lt;math&gt;a_{10}&lt;/math&gt;<br /> &lt;cmath&gt;a_5=8&lt;/cmath&gt;<br /> &lt;cmath&gt;a_6=13&lt;/cmath&gt;<br /> &lt;cmath&gt;a_7=21&lt;/cmath&gt;<br /> &lt;cmath&gt;a_8=34&lt;/cmath&gt;<br /> &lt;cmath&gt;a_9=55&lt;/cmath&gt;<br /> &lt;cmath&gt;a_{10}=89&lt;/cmath&gt;<br /> So, we get &lt;math&gt;\boxed{10 \left(\text{A}\right)}&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=B|num-b=20|num-a=22}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_16&diff=139353 2007 AMC 12A Problems/Problem 16 2020-12-09T21:41:40Z <p>Mathpro12345: /* STOP IT */</p> <hr /> <div>== Problems ==<br /> How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256&lt;/math&gt;<br /> <br /> == Solution 1==<br /> We can find the number of increasing [[arithmetic sequence]]s of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.<br /> <br /> {| class = &quot;wikitable&quot; border=&quot;1px solid&quot;<br /> |-<br /> | Common difference || Sequences possible || Number of sequences<br /> |-<br /> | 1 || &lt;math&gt;012, \ldots, 789&lt;/math&gt; || 8<br /> |-<br /> | 2 || &lt;math&gt;024, \ldots, 579&lt;/math&gt; || 6<br /> |-<br /> | 3 || &lt;math&gt;036, \ldots, 369&lt;/math&gt; || 4<br /> |-<br /> | 4 || &lt;math&gt;048, \ldots, 159&lt;/math&gt; || 2<br /> |}<br /> <br /> This gives us a total of &lt;math&gt;2 + 4 + 6 + 8 = 20&lt;/math&gt; sequences. There are &lt;math&gt;3! = 6&lt;/math&gt; to permute these, for a total of &lt;math&gt;120&lt;/math&gt;.<br /> <br /> However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are &lt;math&gt;2! \cdot 4 = 8&lt;/math&gt; numbers which start with zero, so our answer is &lt;math&gt;120 - 8 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives &lt;math&gt;3! = 6&lt;/math&gt; possible 3-digit numbers; otherwise, &lt;math&gt;4&lt;/math&gt; possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.<br /> <br /> Case 1: 0 is not in the number. Then there are &lt;math&gt;\binom{5}{2} + \binom{4}{2} = 16&lt;/math&gt; ways to choose two nonzero digits of the same parity, and each choice generates &lt;math&gt;3! = 6&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;16 \times 6 = 96&lt;/math&gt; numbers.<br /> <br /> Case 2: 0 is in the number. Then there are &lt;math&gt;4&lt;/math&gt; ways to choose the largest digit (2, 4, 6, or 8), and each choice generates &lt;math&gt;4&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;4 \times 4 = 16&lt;/math&gt; numbers.<br /> <br /> Thus the total is &lt;math&gt;96 + 16 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;. (by scrabbler94)</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_16&diff=139352 2007 AMC 12A Problems/Problem 16 2020-12-09T21:41:04Z <p>Mathpro12345: /* See also */</p> <hr /> <div>== Problems ==<br /> How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256&lt;/math&gt;<br /> <br /> == Solution 1==<br /> We can find the number of increasing [[arithmetic sequence]]s of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.<br /> <br /> {| class = &quot;wikitable&quot; border=&quot;1px solid&quot;<br /> |-<br /> | Common difference || Sequences possible || Number of sequences<br /> |-<br /> | 1 || &lt;math&gt;012, \ldots, 789&lt;/math&gt; || 8<br /> |-<br /> | 2 || &lt;math&gt;024, \ldots, 579&lt;/math&gt; || 6<br /> |-<br /> | 3 || &lt;math&gt;036, \ldots, 369&lt;/math&gt; || 4<br /> |-<br /> | 4 || &lt;math&gt;048, \ldots, 159&lt;/math&gt; || 2<br /> |}<br /> <br /> This gives us a total of &lt;math&gt;2 + 4 + 6 + 8 = 20&lt;/math&gt; sequences. There are &lt;math&gt;3! = 6&lt;/math&gt; to permute these, for a total of &lt;math&gt;120&lt;/math&gt;.<br /> <br /> However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are &lt;math&gt;2! \cdot 4 = 8&lt;/math&gt; numbers which start with zero, so our answer is &lt;math&gt;120 - 8 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives &lt;math&gt;3! = 6&lt;/math&gt; possible 3-digit numbers; otherwise, &lt;math&gt;4&lt;/math&gt; possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.<br /> <br /> Case 1: 0 is not in the number. Then there are &lt;math&gt;\binom{5}{2} + \binom{4}{2} = 16&lt;/math&gt; ways to choose two nonzero digits of the same parity, and each choice generates &lt;math&gt;3! = 6&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;16 \times 6 = 96&lt;/math&gt; numbers.<br /> <br /> Case 2: 0 is in the number. Then there are &lt;math&gt;4&lt;/math&gt; ways to choose the largest digit (2, 4, 6, or 8), and each choice generates &lt;math&gt;4&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;4 \times 4 = 16&lt;/math&gt; numbers.<br /> <br /> Thus the total is &lt;math&gt;96 + 16 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;. (by scrabbler94)<br /> <br /> =STOP IT=</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_16&diff=139351 2007 AMC 12A Problems/Problem 16 2020-12-09T21:40:54Z <p>Mathpro12345: /* == */</p> <hr /> <div>== Problems ==<br /> How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256&lt;/math&gt;<br /> <br /> == Solution 1==<br /> We can find the number of increasing [[arithmetic sequence]]s of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.<br /> <br /> {| class = &quot;wikitable&quot; border=&quot;1px solid&quot;<br /> |-<br /> | Common difference || Sequences possible || Number of sequences<br /> |-<br /> | 1 || &lt;math&gt;012, \ldots, 789&lt;/math&gt; || 8<br /> |-<br /> | 2 || &lt;math&gt;024, \ldots, 579&lt;/math&gt; || 6<br /> |-<br /> | 3 || &lt;math&gt;036, \ldots, 369&lt;/math&gt; || 4<br /> |-<br /> | 4 || &lt;math&gt;048, \ldots, 159&lt;/math&gt; || 2<br /> |}<br /> <br /> This gives us a total of &lt;math&gt;2 + 4 + 6 + 8 = 20&lt;/math&gt; sequences. There are &lt;math&gt;3! = 6&lt;/math&gt; to permute these, for a total of &lt;math&gt;120&lt;/math&gt;.<br /> <br /> However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are &lt;math&gt;2! \cdot 4 = 8&lt;/math&gt; numbers which start with zero, so our answer is &lt;math&gt;120 - 8 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives &lt;math&gt;3! = 6&lt;/math&gt; possible 3-digit numbers; otherwise, &lt;math&gt;4&lt;/math&gt; possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.<br /> <br /> Case 1: 0 is not in the number. Then there are &lt;math&gt;\binom{5}{2} + \binom{4}{2} = 16&lt;/math&gt; ways to choose two nonzero digits of the same parity, and each choice generates &lt;math&gt;3! = 6&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;16 \times 6 = 96&lt;/math&gt; numbers.<br /> <br /> Case 2: 0 is in the number. Then there are &lt;math&gt;4&lt;/math&gt; ways to choose the largest digit (2, 4, 6, or 8), and each choice generates &lt;math&gt;4&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;4 \times 4 = 16&lt;/math&gt; numbers.<br /> <br /> Thus the total is &lt;math&gt;96 + 16 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;. (by scrabbler94)<br /> <br /> =STOP IT=<br /> <br /> =See also=<br /> [[2005 AMC 12A Problems/Problem 11|similar problem]]<br /> {{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_16&diff=139350 2007 AMC 12A Problems/Problem 16 2020-12-09T21:40:35Z <p>Mathpro12345: /* == */</p> <hr /> <div>== Problems ==<br /> How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256&lt;/math&gt;<br /> <br /> == Solution 1==<br /> We can find the number of increasing [[arithmetic sequence]]s of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.<br /> <br /> {| class = &quot;wikitable&quot; border=&quot;1px solid&quot;<br /> |-<br /> | Common difference || Sequences possible || Number of sequences<br /> |-<br /> | 1 || &lt;math&gt;012, \ldots, 789&lt;/math&gt; || 8<br /> |-<br /> | 2 || &lt;math&gt;024, \ldots, 579&lt;/math&gt; || 6<br /> |-<br /> | 3 || &lt;math&gt;036, \ldots, 369&lt;/math&gt; || 4<br /> |-<br /> | 4 || &lt;math&gt;048, \ldots, 159&lt;/math&gt; || 2<br /> |}<br /> <br /> This gives us a total of &lt;math&gt;2 + 4 + 6 + 8 = 20&lt;/math&gt; sequences. There are &lt;math&gt;3! = 6&lt;/math&gt; to permute these, for a total of &lt;math&gt;120&lt;/math&gt;.<br /> <br /> However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are &lt;math&gt;2! \cdot 4 = 8&lt;/math&gt; numbers which start with zero, so our answer is &lt;math&gt;120 - 8 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives &lt;math&gt;3! = 6&lt;/math&gt; possible 3-digit numbers; otherwise, &lt;math&gt;4&lt;/math&gt; possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.<br /> <br /> Case 1: 0 is not in the number. Then there are &lt;math&gt;\binom{5}{2} + \binom{4}{2} = 16&lt;/math&gt; ways to choose two nonzero digits of the same parity, and each choice generates &lt;math&gt;3! = 6&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;16 \times 6 = 96&lt;/math&gt; numbers.<br /> <br /> Case 2: 0 is in the number. Then there are &lt;math&gt;4&lt;/math&gt; ways to choose the largest digit (2, 4, 6, or 8), and each choice generates &lt;math&gt;4&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;4 \times 4 = 16&lt;/math&gt; numbers.<br /> <br /> Thus the total is &lt;math&gt;96 + 16 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;. (by scrabbler94)<br /> <br /> ====<br /> <br /> =See also=<br /> [[2005 AMC 12A Problems/Problem 11|similar problem]]<br /> {{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_16&diff=139347 2007 AMC 12A Problems/Problem 16 2020-12-09T21:25:58Z <p>Mathpro12345: /* Solution 3 */</p> <hr /> <div>== Problems ==<br /> How many three-digit numbers are composed of three distinct digits such that one digit is the [[average]] of the other two? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256&lt;/math&gt;<br /> <br /> == Solution 1==<br /> We can find the number of increasing [[arithmetic sequence]]s of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.<br /> <br /> {| class = &quot;wikitable&quot; border=&quot;1px solid&quot;<br /> |-<br /> | Common difference || Sequences possible || Number of sequences<br /> |-<br /> | 1 || &lt;math&gt;012, \ldots, 789&lt;/math&gt; || 8<br /> |-<br /> | 2 || &lt;math&gt;024, \ldots, 579&lt;/math&gt; || 6<br /> |-<br /> | 3 || &lt;math&gt;036, \ldots, 369&lt;/math&gt; || 4<br /> |-<br /> | 4 || &lt;math&gt;048, \ldots, 159&lt;/math&gt; || 2<br /> |}<br /> <br /> This gives us a total of &lt;math&gt;2 + 4 + 6 + 8 = 20&lt;/math&gt; sequences. There are &lt;math&gt;3! = 6&lt;/math&gt; to permute these, for a total of &lt;math&gt;120&lt;/math&gt;.<br /> <br /> However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are &lt;math&gt;2! \cdot 4 = 8&lt;/math&gt; numbers which start with zero, so our answer is &lt;math&gt;120 - 8 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives &lt;math&gt;3! = 6&lt;/math&gt; possible 3-digit numbers; otherwise, &lt;math&gt;4&lt;/math&gt; possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.<br /> <br /> Case 1: 0 is not in the number. Then there are &lt;math&gt;\binom{5}{2} + \binom{4}{2} = 16&lt;/math&gt; ways to choose two nonzero digits of the same parity, and each choice generates &lt;math&gt;3! = 6&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;16 \times 6 = 96&lt;/math&gt; numbers.<br /> <br /> Case 2: 0 is in the number. Then there are &lt;math&gt;4&lt;/math&gt; ways to choose the largest digit (2, 4, 6, or 8), and each choice generates &lt;math&gt;4&lt;/math&gt; 3-digit numbers, giving &lt;math&gt;4 \times 4 = 16&lt;/math&gt; numbers.<br /> <br /> Thus the total is &lt;math&gt;96 + 16 = 112 \Longrightarrow \mathrm{(C)}&lt;/math&gt;. (by scrabbler94)<br /> <br /> ====<br /> <br /> == See also ==<br /> [[2005 AMC 12A Problems/Problem 11|similar problem]]<br /> {{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}}<br /> <br /> [[Category: Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=139131 2003 AMC 10A Problems/Problem 23 2020-12-06T17:49:44Z <p>Mathpro12345: /* Solution */</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> &lt;asy&gt;<br /> unitsize(15mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60);<br /> pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp;<br /> pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp;<br /> pair Jp=shift(Gp)*Hp;<br /> pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp};<br /> draw(Ap--Dp--Jp--cycle);<br /> draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle);<br /> for(pair p : points)<br /> {<br /> fill(circle(p, 0.07),white);<br /> }<br /> pair[] Cn=new pair;<br /> Cn=centroid(Ap,Bp,Gp);<br /> Cn=centroid(Gp,Bp,Fp);<br /> Cn=centroid(Bp,Fp,Cp);<br /> Cn=centroid(Cp,Fp,Ep);<br /> Cn=centroid(Cp,Ep,Dp);<br /> label(&quot;$1$&quot;,Cn);<br /> label(&quot;$2$&quot;,Cn);<br /> label(&quot;$3$&quot;,Cn);<br /> label(&quot;$4$&quot;,Cn);<br /> label(&quot;$5$&quot;,Cn);<br /> for (pair p : Cn)<br /> {<br /> draw(circle(p,0.1));<br /> }&lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> <br /> == Solution 1==<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt; <br /> ~dolphin7<br /> <br /> ==Solution 2==<br /> The first row of triangles has &lt;math&gt;1&lt;/math&gt; upward-facing triangle, the second row has &lt;math&gt;2&lt;/math&gt; upward-facing triangles, the third row has &lt;math&gt;3&lt;/math&gt; upward-facing triangles, and so on having &lt;math&gt;n&lt;/math&gt; upward-facing triangles in the &lt;math&gt;n^\text{th}&lt;/math&gt; row. The last row with &lt;math&gt;2003&lt;/math&gt; small triangles has &lt;math&gt;1002^\text{th}&lt;/math&gt; upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now &lt;math&gt;\frac{1002\times1003}{2}&lt;/math&gt;, meaning that the number of toothpicks are &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, or &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ~mathpro12345<br /> <br /> ===Note===<br /> You don't have to calculate the value of &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, and you can use units digits to find the answer easily. The units digit of &lt;math&gt;1002\times1003&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, and has a unit digit of &lt;math&gt;3&lt;/math&gt; after being divided by &lt;math&gt;2&lt;/math&gt;. Then this is multiplied by &lt;math&gt;3&lt;/math&gt;, now the final number ending with a &lt;math&gt;9&lt;/math&gt;. This leaves only one answer choice possible, which is &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=139130 2003 AMC 10A Problems/Problem 23 2020-12-06T17:49:26Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> &lt;asy&gt;<br /> unitsize(15mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60);<br /> pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp;<br /> pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp;<br /> pair Jp=shift(Gp)*Hp;<br /> pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp};<br /> draw(Ap--Dp--Jp--cycle);<br /> draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle);<br /> for(pair p : points)<br /> {<br /> fill(circle(p, 0.07),white);<br /> }<br /> pair[] Cn=new pair;<br /> Cn=centroid(Ap,Bp,Gp);<br /> Cn=centroid(Gp,Bp,Fp);<br /> Cn=centroid(Bp,Fp,Cp);<br /> Cn=centroid(Cp,Fp,Ep);<br /> Cn=centroid(Cp,Ep,Dp);<br /> label(&quot;$1$&quot;,Cn);<br /> label(&quot;$2$&quot;,Cn);<br /> label(&quot;$3$&quot;,Cn);<br /> label(&quot;$4$&quot;,Cn);<br /> label(&quot;$5$&quot;,Cn);<br /> for (pair p : Cn)<br /> {<br /> draw(circle(p,0.1));<br /> }&lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1===<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt; <br /> ~dolphin7<br /> <br /> ==Solution 2==<br /> The first row of triangles has &lt;math&gt;1&lt;/math&gt; upward-facing triangle, the second row has &lt;math&gt;2&lt;/math&gt; upward-facing triangles, the third row has &lt;math&gt;3&lt;/math&gt; upward-facing triangles, and so on having &lt;math&gt;n&lt;/math&gt; upward-facing triangles in the &lt;math&gt;n^\text{th}&lt;/math&gt; row. The last row with &lt;math&gt;2003&lt;/math&gt; small triangles has &lt;math&gt;1002^\text{th}&lt;/math&gt; upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now &lt;math&gt;\frac{1002\times1003}{2}&lt;/math&gt;, meaning that the number of toothpicks are &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, or &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ~mathpro12345<br /> <br /> ===Note===<br /> You don't have to calculate the value of &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, and you can use units digits to find the answer easily. The units digit of &lt;math&gt;1002\times1003&lt;/math&gt; is &lt;math&gt;6&lt;/math&gt;, and has a unit digit of &lt;math&gt;3&lt;/math&gt; after being divided by &lt;math&gt;2&lt;/math&gt;. Then this is multiplied by &lt;math&gt;3&lt;/math&gt;, now the final number ending with a &lt;math&gt;9&lt;/math&gt;. This leaves only one answer choice possible, which is &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=139129 2003 AMC 10A Problems/Problem 23 2020-12-06T17:42:49Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> &lt;asy&gt;<br /> unitsize(15mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60);<br /> pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp;<br /> pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp;<br /> pair Jp=shift(Gp)*Hp;<br /> pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp};<br /> draw(Ap--Dp--Jp--cycle);<br /> draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle);<br /> for(pair p : points)<br /> {<br /> fill(circle(p, 0.07),white);<br /> }<br /> pair[] Cn=new pair;<br /> Cn=centroid(Ap,Bp,Gp);<br /> Cn=centroid(Gp,Bp,Fp);<br /> Cn=centroid(Bp,Fp,Cp);<br /> Cn=centroid(Cp,Fp,Ep);<br /> Cn=centroid(Cp,Ep,Dp);<br /> label(&quot;$1$&quot;,Cn);<br /> label(&quot;$2$&quot;,Cn);<br /> label(&quot;$3$&quot;,Cn);<br /> label(&quot;$4$&quot;,Cn);<br /> label(&quot;$5$&quot;,Cn);<br /> for (pair p : Cn)<br /> {<br /> draw(circle(p,0.1));<br /> }&lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1===<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt; <br /> ~dolphin7<br /> <br /> ===Solution 2===<br /> The first row of triangles has &lt;math&gt;1&lt;/math&gt; upward-facing triangle, the second row has &lt;math&gt;2&lt;/math&gt; upward-facing triangles, the third row has &lt;math&gt;3&lt;/math&gt; upward-facing triangles, and so on having &lt;math&gt;n&lt;/math&gt; upward-facing triangles in the &lt;math&gt;n^\text{th}&lt;/math&gt; row. The last row with &lt;math&gt;2003&lt;/math&gt; small triangles has &lt;math&gt;1002^\text{th}&lt;/math&gt; upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now &lt;math&gt;\frac{1002\times1003}{2}&lt;/math&gt;, meaning that the number of toothpicks are &lt;math&gt;\frac{1002\times1003}{2}\times3&lt;/math&gt;, or &lt;math&gt;\boxed{\text{C}}&lt;/math&gt;.<br /> <br /> ~mathpro12345<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=139128 2003 AMC 10A Problems/Problem 23 2020-12-06T17:42:37Z <p>Mathpro12345: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> &lt;asy&gt;<br /> unitsize(15mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60);<br /> pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp;<br /> pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp;<br /> pair Jp=shift(Gp)*Hp;<br /> pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp};<br /> draw(Ap--Dp--Jp--cycle);<br /> draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle);<br /> for(pair p : points)<br /> {<br /> fill(circle(p, 0.07),white);<br /> }<br /> pair[] Cn=new pair;<br /> Cn=centroid(Ap,Bp,Gp);<br /> Cn=centroid(Gp,Bp,Fp);<br /> Cn=centroid(Bp,Fp,Cp);<br /> Cn=centroid(Cp,Fp,Ep);<br /> Cn=centroid(Cp,Ep,Dp);<br /> label(&quot;$1$&quot;,Cn);<br /> label(&quot;$2$&quot;,Cn);<br /> label(&quot;$3$&quot;,Cn);<br /> label(&quot;$4$&quot;,Cn);<br /> label(&quot;$5$&quot;,Cn);<br /> for (pair p : Cn)<br /> {<br /> draw(circle(p,0.1));<br /> }&lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1===<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt; <br /> ~dolphin7<br /> <br /> ===Solution 2===<br /> The first row of triangles has &lt;math&gt;1&lt;/math&gt; upward-facing triangle, the second row has &lt;math&gt;2&lt;/math&gt; upward-facing triangles, the third row has &lt;math&gt;3&lt;/math&gt; upward-facing triangles, and so on having &lt;math&gt;n&lt;/math&gt; upward-facing triangles in the &lt;math&gt;n^\text{th}&lt;/math&gt; row. The last row with &lt;math&gt;2003&lt;/math&gt; small triangles has &lt;math&gt;1002^\text{th} upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now &lt;/math&gt;\frac{1002\times1003}{2}&lt;math&gt;, meaning that the number of toothpicks are &lt;/math&gt;\frac{1002\times1003}{2}\times3&lt;math&gt;, or &lt;/math&gt;\boxed{\text{C}}$.<br /> <br /> ~mathpro12345<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_23&diff=139127 2003 AMC 10A Problems/Problem 23 2020-12-06T17:38:54Z <p>Mathpro12345: /* Solution */</p> <hr /> <div>== Problem ==<br /> A large [[equilateral triangle]] is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have &lt;math&gt;3&lt;/math&gt; rows of small congruent equilateral triangles, with &lt;math&gt;5&lt;/math&gt; small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of &lt;math&gt;2003&lt;/math&gt; small equilateral triangles? <br /> <br /> &lt;asy&gt;<br /> unitsize(15mm);<br /> defaultpen(linewidth(.8pt)+fontsize(8pt));<br /> pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60);<br /> pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp;<br /> pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp;<br /> pair Jp=shift(Gp)*Hp;<br /> pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp};<br /> draw(Ap--Dp--Jp--cycle);<br /> draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle);<br /> for(pair p : points)<br /> {<br /> fill(circle(p, 0.07),white);<br /> }<br /> pair[] Cn=new pair;<br /> Cn=centroid(Ap,Bp,Gp);<br /> Cn=centroid(Gp,Bp,Fp);<br /> Cn=centroid(Bp,Fp,Cp);<br /> Cn=centroid(Cp,Fp,Ep);<br /> Cn=centroid(Cp,Ep,Dp);<br /> label(&quot;$1$&quot;,Cn);<br /> label(&quot;$2$&quot;,Cn);<br /> label(&quot;$3$&quot;,Cn);<br /> label(&quot;$4$&quot;,Cn);<br /> label(&quot;$5\$&quot;,Cn);<br /> for (pair p : Cn)<br /> {<br /> draw(circle(p,0.1));<br /> }&lt;/asy&gt;<br /> &lt;math&gt; \mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018 &lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1===<br /> There are &lt;math&gt;1+3+5+...+2003=1002^{2}=1004004&lt;/math&gt; small equilateral triangles. <br /> <br /> Each small equilateral triangle needs &lt;math&gt;3&lt;/math&gt; toothpicks to make it. <br /> <br /> But, each toothpick that isn't one of the &lt;math&gt;1002\cdot3=3006&lt;/math&gt; toothpicks on the outside of the large equilateral triangle is a side for &lt;math&gt;2&lt;/math&gt; small equilateral triangles. <br /> <br /> So, the number of toothpicks on the inside of the large equilateral triangle is &lt;math&gt;\frac{10040004\cdot3-3006}{2}=1504503&lt;/math&gt;<br /> <br /> Therefore the total number of toothpicks is &lt;math&gt;1504503+3006=\boxed{\mathrm{(C)}\ 1,507,509}&lt;/math&gt; <br /> ~dolphin7<br /> <br /> ===Solution 2===<br /> The first row of triangles has &lt;math&gt;1&lt;/math&gt; upward-facing triangle, the second row has &lt;math&gt;2&lt;/math&gt; upward-facing triangles, the third row has &lt;math&gt;3&lt;/math&gt; upward-facing triangles, and so on having &lt;math&gt;n&lt;/math&gt; upward-facing triangles in the &lt;math&gt;n^\text{th}&lt;/math&gt; row. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are &lt;math&gt;\frac{1002\times1003}{2}&lt;/math&gt;<br /> <br /> ~mathpro12345<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2003|ab=A|num-b=22|num-a=24}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_1&diff=139050 2020 AMC 8 Problems/Problem 1 2020-12-05T16:30:26Z <p>Mathpro12345: /* Problem */</p> <hr /> <div>==Solution 2==<br /> We have that &lt;math&gt;\text{lemonade} : \text{water} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1&lt;/math&gt;, so Luka needs &lt;math&gt;3 \cdot 8 = \boxed{\textbf{(E) }24}&lt;/math&gt; cups.<br /> <br /> ==Video Solution==<br /> https://youtu.be/eSxzI8P9_h8<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_1&diff=139049 2020 AMC 8 Problems/Problem 1 2020-12-05T16:30:16Z <p>Mathpro12345: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> We have that &lt;math&gt;\text{lemonade} : \text{water} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1&lt;/math&gt;, so Luka needs &lt;math&gt;3 \cdot 8 = \boxed{\textbf{(E) }24}&lt;/math&gt; cups.<br /> <br /> ==Video Solution==<br /> https://youtu.be/eSxzI8P9_h8<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|before=First Problem|num-a=2}}<br /> {{MAA Notice}}</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=1997_PMWC_Problems/Problem_T7g&diff=139048 1997 PMWC Problems/Problem T7g 2020-12-05T16:29:15Z <p>Mathpro12345: Created page with &quot;what do you mean??? you're so stupid.... do ur work&quot;</p> <hr /> <div>what do you mean???<br /> <br /> you're so stupid....<br /> <br /> do ur work</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=Principle_of_Inclusion-Exclusion&diff=135755 Principle of Inclusion-Exclusion 2020-10-24T20:57:52Z <p>Mathpro12345: /* Examples */</p> <hr /> <div>The '''Principle of Inclusion-Exclusion''' (abbreviated PIE) provides an organized method/formula to find the number of [[element]]s in the [[union]] of a given group of [[set]]s, the size of each set, and the size of all possible [[intersection]]s among the sets.<br /> <br /> HELLO!<br /> <br /> == See also ==<br /> * [[Combinatorics]]<br /> * [[Overcounting]]<br /> <br /> [[Category:Combinatorics]]</div> Mathpro12345 https://artofproblemsolving.com/wiki/index.php?title=Principle_of_Inclusion-Exclusion&diff=135754 Principle of Inclusion-Exclusion 2020-10-24T20:57:40Z <p>Mathpro12345: /* Remarks */</p> <hr /> <div>The '''Principle of Inclusion-Exclusion''' (abbreviated PIE) provides an organized method/formula to find the number of [[element]]s in the [[union]] of a given group of [[set]]s, the size of each set, and the size of all possible [[intersection]]s among the sets.<br /> <br /> == Examples ==<br /> 2002 AIME I Problems/Problem 1<br /> http://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_1#Problem<br /> <br /> 2011 AMC 8 Problems/Problem 6<br /> https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems/Problem_6<br /> <br /> 2017 AMC 10B Problems/Problem 13<br /> https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_13<br /> <br /> 2005 AMC 12A Problems/Problem 18<br /> https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_18<br /> <br /> == See also ==<br /> * [[Combinatorics]]<br /> * [[Overcounting]]<br /> <br /> [[Category:Combinatorics]]</div> Mathpro12345