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• 2004 AIME I Problems/Problem 7
  where the left-hand sum can be computed from: <center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center>
  5 KB (833 words) - 19:43, 1 October 2023
• 2004 AIME II Problems/Problem 10
  ...math> is clearly never divisible by 9, so we can just focus on <math>1+2^{y-x}</math>. ...1+2^{y-x} \equiv 0 \pmod 9 \implies 2^{y-x} \equiv -1 \pmod 9 \implies 2^{y-x} \equiv 8 \pmod 9</cmath>
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2019 AIME I Problems/Problem 2
  ...\ne 0</math> because <math>B \ne J</math>, so the probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry. The probability that <math>B-J = 1</math> is <math>\frac{19}{20 \times 19} = \frac{1}{20}</math> because
  5 KB (830 words) - 22:15, 28 December 2023
• 1989 AIME Problems
  <center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center> is true for a unique choice of non-negative integer <math>m^{}_{}</math> and digits <math>a_0,a_1^{},\ldots,a_m
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1990 AIME Problems
  Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>. <center><math>\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0</math></center>
  6 KB (870 words) - 10:14, 19 June 2021
• 1998 AIME Problems
  7 KB (1,084 words) - 02:01, 28 November 2023
• 2000 AIME II Problems
  ...ht distinguishable rings, let <math>n</math> be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order The equation <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n
  6 KB (947 words) - 21:11, 19 February 2019
• 2001 AIME II Problems
  ...the largest number of students who could study both languages. Find <math>M-m</math>. x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2003 AIME II Problems
  ...th>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there? ...ne{EF,}</math> <math>\overline{CD}\parallel \overline{FA},</math> and the y-coordinates of its vertices are distinct elements of the set <math>\{0,2,4,6
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1985 AIME Problems/Problem 10
  ...changes when <math>\{x\}=\frac{m}{n}</math>, where <math>m\in\{1,2,3,...,n-1\}</math>, <math>n\in\{2,4,6,8\}</math>. Using [[Euler's Totient Function]] ...6x\rfloor+\lfloor 8x\rfloor</math>, then <math>k=20a+b</math> for some non-negative integers <math>a</math>, <math>b</math>, where there are <math>12</
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 5
  {{AIME box|year=1985|num-b=4|num-a=6}}
  2 KB (410 words) - 13:37, 1 May 2022
• 1986 AIME Problems/Problem 11
  The [[polynomial]] <math>1-x+x^2-x^3+\cdots+x^{16}-x^{17}</math> may be written in the form <math>a_0+a_1y+a_ ...<math>1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}</math>. Since <math>x = y - 1</math>, this becomes <math>\frac {1
  6 KB (872 words) - 16:51, 9 June 2023
• 1987 AIME Problems/Problem 7
  {{AIME box|year=1987|num-b=6|num-a=8}}
  3 KB (547 words) - 22:54, 4 April 2016
• 1989 AIME Problems/Problem 5
  ...rac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{24 {{AIME box|year=1989|num-b=4|num-a=6}}
  2 KB (258 words) - 00:07, 25 June 2023
• 1990 AIME Problems/Problem 9
  ...ing this pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} ...th>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.
  3 KB (425 words) - 12:36, 12 May 2024
• 1991 AIME Problems/Problem 3
  ...>. Taking logarithms in both sides of this last equation and using the well-known fact <math>\log(a_{}^{}b)=\log a + \log b</math> (valid if <math>a_{}^ ...t[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .
  5 KB (865 words) - 12:13, 21 May 2020
• 2006 AMC 10B Problems/Problem 22
  {{AMC10 box|year=2006|ab=B|num-b=21|num-a=23}}
  2 KB (394 words) - 00:51, 25 November 2023
• 1993 AIME Problems/Problem 13
  ...h>, <math>CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100</math>. ...f this line with <math>AB</math> as <math>S</math>. Then <math>SB=AB-AS=3t-t=2t</math>.
  8 KB (1,231 words) - 20:06, 26 November 2023
• 1994 AIME Problems/Problem 4
  ...th> such that <math>\lfloor\log_2{a}\rfloor=j</math>, and there are <math>n-2^k+1</math> such integers such that <math>\lfloor\log_2{a}\rfloor=k</math>. ...3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994</math>.
  2 KB (264 words) - 13:33, 11 August 2018
• 1995 AIME Problems/Problem 12
  ...>AP = 1.</math> It follows that <math>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <math>OB = OA = \ Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that <math>A = (1,0,0),</math> <math>B =
  8 KB (1,172 words) - 21:57, 22 September 2022

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