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- ...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 (874 words) - 22:02, 10 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) - 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
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 21:35, 9 January 2016
- ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>1 KB (191 words) - 09:22, 4 April 2012
- ==Problem== <cmath>\zeta_1+\zeta_2+\zeta_3=1</cmath>2 KB (221 words) - 01:49, 19 March 2015
- ==Problem== === Solution 1 (recursive) ===5 KB (795 words) - 15:03, 17 October 2021
- ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>3 KB (501 words) - 13:48, 29 November 2019
- == 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) - 12:44, 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) - 05:18, 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) - 09:51, 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) - 16:27, 23 February 2013
- == Problem == ...e of <math>5</math> on either direction of <math>A'C'</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 == ...060657, \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 == ...h> such that <math>\angle{AB'C'} \cong \angle{B'EA}</math>. If <math>AB' = 5</math> and <math>BE = 23</math>, then the area of <math>ABCD</math> can be2 KB (376 words) - 21:41, 26 December 2016
- == Problem == Triangle <math>ABC</math> has an inradius of <math>5</math> and a circumradius of <math>16</math>. If <math>2\cos{B} = \cos{A} +2 KB (340 words) - 00:44, 3 March 2020
- ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</math> distinct positive integers?7 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