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• 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
• 1996 AIME Problems/Problem 12
  <cmath>|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.</cmath> ...obtained from these paired sequences are also obtained in another <math>2^5-1</math> ways by permuting the adjacent terms <math>\{a_1,a_2\},\{a_3,a_4\},
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 7
  ...l be <br /> three other equivalent boards.</font></td><td><font style="font-size:85%">For those symmetric about the center, <br /> there is only one oth ...re rotationally symmetric about the center square; there are <math>\frac{49-1}{2}=24</math> such pairs. There are then <math>{49 \choose 2}-24</math> pa
  4 KB (551 words) - 11:44, 26 June 2020
• 1996 AIME Problems/Problem 3
  ...itive [[integer]] <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. ...3)^n</math>. Both [[binomial expansion]]s will contain <math>n+1</math> non-like terms; their product will contain <math>(n+1)^2</math> terms, as each t
  3 KB (515 words) - 04:29, 27 November 2023
• 1998 AIME Problems/Problem 15
  ...s a path that retraces no segment. Each time that such a path reaches a non-terminal vertex, it must leave it. ...it can be arranged that <math>n-2</math> segments will emanate from <math>n-2</math> of the vertices and that an odd number of segments will emanate fro
  9 KB (1,671 words) - 22:10, 15 March 2024
• 1999 AIME Problems/Problem 7
  ...math> take on the values <math>0, 1, \ldots, 9</math>. At step i of a 1000-step process, the <math>i</math>-th switch is advanced one step, and so are ...}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}</math>. In general, the divisor-count of <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a s
  3 KB (475 words) - 13:33, 4 July 2016
• 2001 AIME I Problems/Problem 15
  .../math>, which can be done in <math>4! = 24</math> ways. Then choose a three-edge path along tetrahedron <math>DBEG</math> which, because it must start a ...faces, and one face adjacent to the three B-faces, which we will call the C-face.
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 12
  ...ween a plane and a point <math>I</math> can be calculated as <math>\frac{(I-G) \cdot P}{|P|}</math>, where G is any point on the plane, and P is a vecto ...rpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math>
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2003 AIME I Problems/Problem 3
  ...1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subset of <math> \mathcal{S}, </math> she writes on her list the gre Each [[element]] of the [[set]] will appear in <math>7</math> two-element [[subset]]s, once with each other number.
  2 KB (317 words) - 00:09, 9 January 2024
• 2003 AIME II Problems/Problem 15
  ...that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>. The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \s
  4 KB (675 words) - 17:23, 30 July 2022
• 2003 AIME II Problems/Problem 13
  ...ing the starting vertex in the next move. Thus <math>P_n=\frac{1}{2}(1-P_{n-1})</math>. ...-1</math> steps plus the number of ways to get to <math>C</math> in <math>n-1</math> steps.
  15 KB (2,406 words) - 23:56, 23 November 2023

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