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- == 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) - 12:43, 10 August 2019
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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== <cmath>\zeta_1+\zeta_2+\zeta_3=1</cmath>2 KB (221 words) - 02:49, 19 March 2015
- ==Problem== === Solution 1 (recursive) ===5 KB (795 words) - 16:03, 17 October 2021
- ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>3 KB (501 words) - 14:48, 29 November 2019
- ==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>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) - 23:59, 24 April 2013
- ==Problem== <math>A_1=\frac{\sqrt{\left(13\right)\left(7\right)\left(1\right)\left(5\right)}}{4}=\frac{\sqrt{455}}{4}</math>3 KB (563 words) - 02:05, 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) - 12:43, 10 August 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 == ...> 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) - 06:18, 2 July 2015
- == 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
