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• 2005 AIME I Problems/Problem 11
  [[Image:2005 AIME I Problem 11.png]] ...+ r\sqrt{2} = 8\sqrt{2}</math> so <math>r = \frac{8\sqrt{2}}{1+\sqrt{2}} = 16 - 8\sqrt{2}</math>. Then the diameter is <math>32 - \sqrt{512}</math> givin
  4 KB (707 words) - 11:11, 16 September 2021
• 2005 AIME II Problems/Problem 15
  ...ns as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*} </cmath>
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2004 AIME I Problems/Problem 15
  6&0&0&0&0&0&1&1&2&4&8&16\\ 7&0&0&0&0&1&1&2&4&8&16&32\\
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 12
  ...\cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots</cmath> {{AIME box|year=2004|n=I|num-b=11|num-a=13}}
  2 KB (303 words) - 22:28, 11 September 2020
• 2004 AIME II Problems/Problem 14
  ...big\rfloor + 1 = 112</math> positive integers that satisfy both conditions i.e. <math>\{1, 10, 19, 28, 37, 46, . . . , 1000\}.</math> ...th>n = 24 + (\cdots)</math> then just put the values of <math>a,b,c</math> i am sure you will get it :) )
  11 KB (1,857 words) - 21:55, 19 June 2023
• 2004 AIME II Problems/Problem 6
  <math>x = \frac{1}{4}b_1 + \frac{1}{8}b_2 + \frac{11}{72}b_3 = \frac{1}{16}b_1 + \frac{1}{8}b_2 + \frac{11}{48}b_3 = \frac{1}{8}b_1 + \frac{3}{8}b_2 + ...d monkey take <math>8y</math>, and the third monkey take <math>24z</math>. I chose these numbers to make it so, when each monkey splits his bananas, the
  6 KB (950 words) - 14:18, 15 January 2024
• 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) - 02:27, 30 June 2024
• 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

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