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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== === Solution 1 (recursive) ===5 KB (795 words) - 16:03, 17 October 2021
- ==Problem== ...diagonals of <math>ABCD</math> intersect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the insc2 KB (330 words) - 10:23, 4 April 2012
- ==Problem== Here are some thoughts on the problem:3 KB (520 words) - 12:55, 11 January 2019
- ==Problem== <math>\{A_n\}_{n \ge 1}</math> is a sequence of positive integers such that2 KB (306 words) - 10:36, 4 April 2012
- ==Problem== ...> at <math>C</math> and <math>D</math> respectively. If <math>AD = 3, AP = 6, DP = 4,</math> and <math>PQ = 32</math>, then the area of triangle <math>P3 KB (563 words) - 02:05, 25 November 2023
- ...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 == When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a rem685 bytes (81 words) - 10:51, 11 June 2013
- == Problem == ...</tt>s amongst the middle five numbers, and so there are <math>6-(5-k) = k+1</math> <tt>C</tt>s amongst the first four numbers.1 KB (221 words) - 17:27, 23 February 2013
- == Problem == <cmath> \frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{360}{pqr} = 1</cmath>864 bytes (127 words) - 21:47, 21 February 2010
- == Problem == ...top of the frustum is a rectangle <math>A'B'C'D'</math> with <math>A'B' = 6</math> and <math>B'C' = 8</math>.3 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 == ...04060657, \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) - 20:01, 23 March 2017
- == Problem == <cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath>2 KB (272 words) - 10:51, 2 July 2015