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- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}}2 KB (221 words) - 02:49, 19 March 2015
- == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=3|num-a=5|source=14769}}685 bytes (81 words) - 10:51, 11 June 2013
- == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=3|num-a=5}}645 bytes (109 words) - 20:41, 22 March 2016
- == Problem == {{Mock AIME box|year=Pre 2005|n=2|num-b=3|num-a=5|source=14769}}817 bytes (114 words) - 17:16, 4 August 2019
- ==Problem== ...h>S = \sum^{\infty}_{n=1} \frac{(7n + 32) \cdot 3^n}{n \cdot (n + 2) \cdot 4^n}</cmath>2 KB (319 words) - 01:21, 30 January 2024
- 2 bytes (1 word) - 14:19, 28 January 2024
- 372 bytes (62 words) - 00:05, 15 February 2024
Page text matches
- ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat8 KB (1,057 words) - 12:02, 25 February 2024
- ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...ficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}6 KB (869 words) - 12:52, 20 February 2024
- <cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath> <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>3 KB (465 words) - 18:31, 3 July 2023
- ...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) - 20:58, 6 December 2023
- 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== ..., or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(10+r)</math>. <math>84*4=14*24</math>, so <math>r=14</math>. Thus the area of the circle is <math>\b795 bytes (129 words) - 10:22, 4 April 2012
- ==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== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:5 KB (795 words) - 16:03, 17 October 2021
- ==Problem== ...ls 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 inscribed ci2 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== Therefore we have <math>a_n \equiv 6\cdot 16 - 4^2 - 4\cdot 4 - 6 = \boxed{058} \pmod{1000}</math>.2 KB (306 words) - 10:36, 4 April 2012
- ==Problem== ...th>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>PBC</math3 KB (563 words) - 02:05, 25 November 2023
- ==Problem== <math>\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)</math>2 KB (312 words) - 10:38, 4 April 2012
- == Problem == <math>\lfloor 99/23\rfloor =4</math>2 KB (209 words) - 12:43, 10 August 2019
- <div style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> a^4 + 4b^4 &= a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ &= (a^2 + 2b^2)^2 - (2ab)^2 \\ &= (a^2 + 2b^2 - 2ab) (a^2 + 22 KB (222 words) - 15:04, 30 December 2023
- == Problem == ...ds digits, then tens digits, then units digits. Every one of <math>\{1,2,3,4,5,6,7,8,9\}</math> may appear as the hundreds digit, and there are <math>91 KB (194 words) - 13:44, 5 September 2012
- == Problem == ...>=r, <math>BP = |r - 1|</math>, <math>CP = |r - 2|</math>, <math>DP = |r - 4|</math>, and <math>EP = |r - 13|.</math> Squaring each of these gives:1 KB (217 words) - 06:18, 2 July 2015
- == Problem == ...numbers in the middle (those mentioned in condition [2]). There are <math>4-k</math> <tt>A</tt>s amongst the last six numbers then. Also, there are <ma1 KB (221 words) - 17:27, 23 February 2013
- == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]6 KB (909 words) - 07:27, 12 October 2022
- == Problem == Let <math>m = 101^4 + 256</math>. Find the sum of the digits of <math>m</math>.517 bytes (55 words) - 20:01, 23 March 2017
- ...h> objects in <math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then5 KB (775 words) - 23:53, 13 April 2024
- == Problem == ...{6}</math>. Thus, the area of <math>ABCD</math> is <math>(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 +2 KB (376 words) - 22:41, 26 December 2016
- == Problem == ...cdot 23 + b^2+c^2-a^2}{bc} + \frac{36 \cdot 23+a^2+b^2-c^2}{ab} = \frac{7}{4} \implies</math> <math>\frac{36 \cdot 23 + c^2-a^2}{c} + \frac{36 \cdot 232 KB (340 words) - 01:44, 3 March 2020
- == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=2|num-a=4}}795 bytes (133 words) - 08:14, 19 July 2016
- == Problem == : <math>P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}</math>522 bytes (77 words) - 21:17, 8 October 2014
- == Problem == ...assigned to these houses such that there is at least one house with <math>4</math> residents?461 bytes (62 words) - 21:18, 8 October 2014
- == Problem == So, <math>DE=4</math>.2 KB (294 words) - 16:24, 24 August 2022
- ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]7 KB (1,094 words) - 15:39, 24 March 2019
- == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,052 words) - 13:52, 9 June 2020
- == Problem == ...\dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 18 \cdot 47 - 3 = \boxed{843}.</c883 bytes (128 words) - 16:14, 4 August 2019
- == Problem == In a box, there are <math>4</math> green balls, <math>4</math> blue balls, <math>2</math> red balls, a brown ball, a white ball, an1 KB (170 words) - 17:15, 4 August 2019
- == Problem == ...7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{251 KB (171 words) - 17:38, 4 August 2019
- ...are positive integers. What is <math>a+b</math>? ([[2022 AMC 10B Problems/Problem 9|Source]]) *Which of the following is equivalent to <math>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</math> (Hint: diffe3 KB (500 words) - 14:11, 21 November 2023
- == Problem == k_{4} = {3^3}...</cmath>2 KB (232 words) - 00:22, 1 January 2021
