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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 ...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== === Solution 1 (recursive) ===5 KB (795 words) - 15:03, 17 October 2021
- == 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 == 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 == ...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) - 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 == ...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) - 19:01, 23 March 2017
- == Problem == <cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath>2 KB (272 words) - 09:51, 2 July 2015
- .../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 == ...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) - 21:41, 26 December 2016
- == Problem == ...}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have that <math>b = 6\sqrt{23}</math>. By the Law of Cosines, we have that: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
- == Problem == Let <math>S</math> be the set of integers <math>n > 1</math> for which <math>\tfrac1n = 0.d_1d_2d_3d_4\ldots</math>, an infinite1 KB (171 words) - 16:38, 4 August 2019
- ...one by using a form of <math>\sum_{k=1}^{n} f(k)=\sum_{k=1}^{n}(a_{k}-a_{k-1})=a_{k}-a_0</math> for some expression <math>a_{k}</math>. ==Example 1==3 KB (558 words) - 15:37, 21 July 2024