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• 2004 AIME II Problems/Problem 5
  ...rd quarter takes is <math>\frac{1000}{800}\cdot\frac{1}{4}\cdot t=\frac{5}{16}t</math>. ...st quarter takes is <math>t\left[1-\left(\frac{1}{4}+\frac{5}{18}+\frac{5}{16}\right)\right]=\frac{23}{144}t</math>.
  4 KB (592 words) - 19:02, 26 September 2020
• 1988 AIME Problems
  Suppose that <math>|x_i| < 1</math> for <math>i = 1, 2, \dots, n</math>. Suppose further that ...gers such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>.
  6 KB (902 words) - 08:57, 19 June 2021
• 1990 AIME Problems
  ...e the probability that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. ax^3 + by^3 &= 16, \\
  6 KB (870 words) - 10:14, 19 June 2021
• 1999 AIME Problems
  ...th>|S(x+2)-S(x)|.</math> For example, <math>T(199)=|S(201)-S(199)|=|3-19|=16.</math> How many values of <math>T(x)</math> do not exceed 1999? ...d so are all the other switches whose labels divide the label on the <math>i</math>-th switch. After step 1000 has been completed, how many switches wi
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME II Problems
  Given that <center><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11! ...h>(f_1,f_2,f_3,\ldots,f_j)</math> is the factorial base expansion of <math>16!-32!+48!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f
  6 KB (947 words) - 21:11, 19 February 2019
• 1983 AIME Problems/Problem 15
  &= \frac{16 - 9}{25} = \frac{7}{25}. \end{align*} </cmath> <cmath> p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0</cmath>
  20 KB (3,497 words) - 15:37, 27 May 2024
• 1984 AIME Problems/Problem 10
  ...differently and now we are adding the score. If this is already confusing, I suggest not looking further.) ...math>4</math> or adding <math>5</math>, we will see we get <math>4, 8, 12, 16, 20,</math> etc. if we add only <math>4</math>s and if we add <math>5</math
  7 KB (1,163 words) - 23:53, 28 March 2022
• 1984 AIME Problems/Problem 15
  ...6, 64\}</math>. After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the equation ...of the two polynomials by a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so
  6 KB (1,051 words) - 04:52, 8 May 2024
• 1984 AIME Problems/Problem 13
  ...ter><p><math>\tan((a+b)+(c+d)) = \frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{16}} = \frac{2}{3}</math>.</p></center> <cmath>(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)</cmath>
  3 KB (473 words) - 12:06, 18 December 2018
• 1985 AIME Problems/Problem 10
  *<math>2</math>: Also simple, for example using <math>\frac 16</math>. <math>\frac{16}{24},\frac{17}{24} \to 12</math>
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1986 AIME Problems/Problem 11
  ...}</math> may be written in the form <math>a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}</math>, where <math>y=x+1</math> and the <math>a_i</math>'s a ...he sum of the first 16 triangular numbers, which evaluates to <math>\frac{(16)(17)(18)}{6} = \boxed{816}</math>.
  6 KB (872 words) - 16:51, 9 June 2023
• 1987 AIME Problems/Problem 14
  <cmath>\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.</cmath> ...8)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}.</cmath>
  7 KB (965 words) - 10:42, 12 April 2024
• 1988 AIME Problems/Problem 13
  ...gers such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>. ...6}\cdot x + F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1989 AIME Problems/Problem 8
  for (real i=1; i<=10; ++i) { label("\boldmath{$"+string(i^2)+"$}",(i-1,0));
  8 KB (1,146 words) - 04:15, 20 November 2023
• 1990 AIME Problems/Problem 10
  ...</math>. We can write the numbers of set <math>A</math> as <math>\{n^8, n^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \l ...right)</math> respectively, where <math>\text{cis}\,\theta = \cos \theta + i \sin \theta</math> and <math>k_1</math> and <math>k_2</math> are integers f
  3 KB (564 words) - 04:47, 4 August 2023
• 1994 AIME Problems/Problem 8
  ...left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.</cmath> ...Equating the two vectors, we get <math>a+b=26\sqrt{3}</math> and <math>a-b=16\sqrt{3}</math>. Therefore, <math>a=21\sqrt{3}</math> and <math>b=5\sqrt{3}<
  5 KB (788 words) - 13:53, 8 July 2023
• 1995 AIME Problems/Problem 15
  ...s the sum of all numbers of the form <math>\frac 32\left( \left(\frac {15}{16}\right)^n \cdot \left(\frac 12\right)^n \right) \cdot \left(\frac 1{32}\rig ...ll <math>4</math> consecutive <tt>H</tt>'s, and there is a <math>\frac{15}{16}</math> probability we roll a <tt>T</tt>. Thus,
  6 KB (979 words) - 13:20, 11 April 2022
• 1995 AIME Problems/Problem 9
  ...ath> such that <math>\text{Arg}((11+xi)^3)=\text{Arg}(1331-33x^2+(363x-x^3)i)=\text{Arg}(1+xi)</math>. This will happen when <math>\frac{363x-x^3}{1331- <cmath>20\cos ^3 y = 16 \cos y</cmath>
  7 KB (1,181 words) - 13:47, 3 February 2023
• 1995 AIME Problems/Problem 5
  ...math> and the sum of the other two roots is <math>3+4i,</math> where <math>i=\sqrt{-1}.</math> Find <math>b.</math> <cmath>m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.</cmath>
  3 KB (451 words) - 15:02, 6 September 2021
• 1995 AIME Problems/Problem 1
  ...> are the perpendicular bisectors of two adjacent sides of square <math>S_{i+2}.</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, ...+ \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2</math>
  2 KB (302 words) - 19:29, 4 July 2013

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