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- == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=9|num-a=11|source=14769}}540 bytes (96 words) - 00:28, 23 December 2023
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- ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.51 KB (6,175 words) - 21:41, 27 November 2024
- 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) - 10:58, 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) - 16:33, 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) - 17:41, 21 January 2017
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 22:35, 9 January 2016
- ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]7 KB (1,135 words) - 23:53, 24 March 2019
- ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>1 KB (191 words) - 10:22, 4 April 2012
- ==Problem== ...ear at least two consonants. Let <math>N</math> denote the number of <math>10</math>-letter Zuminglish words. Determine the remainder obtained when <math5 KB (795 words) - 16:03, 17 October 2021
- ==Problem== ...<math>(a_1, a_2, \dots, a_8)</math> of real numbers such that <math>a_1 = 10</math> and3 KB (520 words) - 12:55, 11 January 2019
- ==Problem== ==Solution 1==2 KB (278 words) - 16:32, 27 December 2019
- ...ath> is divided by <math>2^{101}+2^{51}+1</math>? ([[2020 AMC 10B Problems/Problem 22|Source]]) *Prove that if <math>n>1</math> then <math>n^4 + 4^n</math> is [[composite]]. (1978 Kurschak Competi2 KB (210 words) - 17:05, 30 January 2025
- == Problem == ...r the tens and units digits. Thus the sum of the hundreds places is <math>(1+2+3+\cdots+9)(72) \times 100 = 45 \cdot 72 \cdot 100 = 324000</math>.1 KB (194 words) - 13:44, 5 September 2012
- == Problem == ...m over triangle <math>ABC</math> to be <math>20</math>. Thus, <math>A'C' = 10</math>. Since the height of this trapezoid is <math>12</math>, and <math>AC3 KB (446 words) - 00:18, 10 February 2020
- == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>6 KB (909 words) - 07:27, 12 October 2022
- == Problem == <cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath>2 KB (272 words) - 10:51, 2 July 2015
- ...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) - 17:39, 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) - 22:41, 26 December 2016
- == Problem == ...even and <math>d_m > d_{m-1}</math> if <math>m</math> is odd for <math>m = 1,2,\ldots,k</math> (and <math>d_0 = 0</math>). Let <math>a</math> be the num795 bytes (133 words) - 08:14, 19 July 2016
- == Problem == ...Define <math>\mu(A) = \sum_{k \in A} \tau(k)</math>, where <math>\tau(k) = 1</math> if <math>k</math> has an odd number of divisors and <math>\tau(k) =786 bytes (131 words) - 21:19, 8 October 2014
- ==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) - 15:39, 24 March 2019
