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- ==Problem== ...h> \{R_{n}\}_{n\ge 0} </math> obeys the recurrence <math> 7R_{n}= 64-2R_{n-1}+9R_{n-2} </math> for any integers <math>n\ge 2</math>. Additionally, <ma1 KB (178 words) - 13:22, 26 March 2011
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- The '''Mock AIME 2 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 2 Pre 2005 Problems|Entire Exam]]2 KB (181 words) - 09:58, 18 March 2015
- The '''Mock AIME 7 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 7 Pre 2005 Problems|Entire Exam]]1 KB (146 words) - 15:33, 14 October 2022
- The '''Mock AIME 1 2005-2006''' was written by [[Art of Problem Solving]] community member paladin8. * [[Mock AIME 1 2005-2006/Answer Key|Answer Key]]1 KB (135 words) - 16:41, 21 January 2017
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 21:35, 9 January 2016
- ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]7 KB (1,135 words) - 22:53, 24 March 2019
- ==Problem== ...<math>n</math> such that <math>1 \le n \le 1000</math> and <math>n^{12} - 1</math> is divisible by <math>73</math>.714 bytes (105 words) - 22:59, 24 April 2013
- ==Problem== <math>13.</math> Let <math>S</math> denote the value of the sum3 KB (502 words) - 13:53, 19 July 2020
- ==Problem== ...-a)(s-b)(s-c)}</math>, where <math>s=\frac{a+b+c}{2}=\frac{3+6+4}{2}=\frac{13}{2}</math> we have:3 KB (563 words) - 01:05, 25 November 2023
- == Problem == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.2 KB (209 words) - 11:43, 10 August 2019
- ...ath> is divided by <math>2^{101}+2^{51}+1</math>? ([[2020 AMC 10B Problems/Problem 22|2020 AMC 10B, #22]]) ...^4+324)(40^4+324)(52^4+324)}</math>. ([[1987 AIME Problems/Problem 14|1987 AIME, #14]])2 KB (225 words) - 04:42, 30 January 2025
- == Problem == ...> and <math>E</math> are collinear in that order such that <math>AB = BC = 1, CD = 2,</math> and <math>DE = 9</math>. If <math>P</math> can be any point1 KB (217 words) - 05:18, 2 July 2015
- == Problem == ...</math>, we can use a 5-12-13 triangle to determine that <math>AA' = CC' = 13</math>.3 KB (446 words) - 23:18, 9 February 2020
- == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>6 KB (909 words) - 06:27, 12 October 2022
- == Problem == ...56 = 104060657, \end{aligned}</math> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math>517 bytes (55 words) - 19:01, 23 March 2017
- ...ath>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ath>n</math> bins is <math>{n+k-1 \choose n-1}</math> or <math>\dbinom{n+k-1}k</math>.5 KB (795 words) - 16:39, 31 December 2024
- == Problem == ...at <math>\sin \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta</math>. Now, we use law of sines, which gives us the foll2 KB (376 words) - 21:41, 26 December 2016
- == Problem == Note: The problem is meant to be interpreted so that if you cannot produce one arrangement fr692 bytes (107 words) - 12:01, 24 November 2021
- == Problem == <math>CosB=\frac{68^2+100^2-112^2}{2.68.100}=\frac{13}{85}</math>2 KB (282 words) - 09:06, 9 August 2022
- ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</m7 KB (1,094 words) - 14:39, 24 March 2019
- == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,052 words) - 12:52, 9 June 2020