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- ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:5 KB (795 words) - 15:03, 17 October 2021
- == 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
- == Problem == {{Mock AIME box|year=Pre 2005|n=2|num-b=4|num-a=6|source=14769}}1 KB (171 words) - 16:38, 4 August 2019
- If <math>a=2</math> and <math>b=5</math>, then <math>g=2=b</math>, a contradiction. ...math>c, d, e, h, i, j</math> are distinct digits from the list <math>0, 4, 5, 7, 8, 9</math>.2 KB (335 words) - 05:41, 5 September 2023
- ...ac{x}{y}=\frac{4}{6}=\frac{2}{3}</math>, so the answer is <math>2+3=\boxed{5}</math>.454 bytes (80 words) - 23:02, 14 February 2024
- 372 bytes (62 words) - 23:05, 14 February 2024
- 280 bytes (34 words) - 23:11, 14 February 2024
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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(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4</cmath>1 KB (191 words) - 09:22, 4 April 2012
- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}}2 KB (221 words) - 01:49, 19 March 2015
- ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:5 KB (795 words) - 15:03, 17 October 2021
- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}}3 KB (501 words) - 13:48, 29 November 2019
- == Problem == ...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>9 \c1 KB (194 words) - 12:44, 5 September 2012
- == Problem == This reaches its minimum at <math>r = \frac {40}{2\cdot 5} = 4</math>, at which point the sum of the squares of the distances is <mat1 KB (217 words) - 05:18, 2 July 2015
- == Problem == ...uivalent to finding <math>\frac{7^{400 \cdot 5 + 5} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{801} \pmod{1000}</math>.685 bytes (81 words) - 09:51, 11 June 2013
- == Problem == ...th> <tt>C</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 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]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
- It is used to solve problems of the form: how many ways can one distribute <math>k</math> indistinguisha ...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 (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