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- {{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}} [[Category:Intermediate Algebra Problems]]3 KB (501 words) - 13:48, 29 November 2019
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- ...9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}} ...on the USAMO have considerable experience both solving highly challenging problems and [[writing proofs]].6 KB (874 words) - 22:02, 10 November 2024
- A '''Mock AIME''' is a contest that is intended to mimic the [[AIME]] competition. (In more recent years, recurring competitions will be listed * Pre 20068 KB (901 words) - 19:45, 13 January 2025
- 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
- [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]] [[Mock AIME 1 Pre 2005 Problems/Problem 2|Solution]]6 KB (1,100 words) - 21:35, 9 January 2016
- 6 & 208 & 80 & 64 \ ...th>C</math>, there are <math>2</math> choices. There is a total of <math>2^6=64</math> possible words.5 KB (795 words) - 15:03, 17 October 2021
- ...igits, 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 \cdo Every one of <math>\{0,1,2,3,4,5,6,7,8,9\}</math> may appear as the tens digit; however, since <math>0</math>1 KB (194 words) - 12:44, 5 September 2012
- ...t to finding <math>\frac{7^{400 \cdot 5 + 5} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{801} \pmod{1000}</math>. {{Mock AIME box|year=Pre 2005|n=1|num-b=3|num-a=5|source=14769}}685 bytes (81 words) - 09:51, 11 June 2013
- .../math> <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. ...ath> ways to arrange the middle five numbers, and <math>{6 \choose 4-k} = {6\choose k+2}</math> ways to arrange the last six numbers. Notice that <math>1 KB (221 words) - 16:27, 23 February 2013
- ...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>. {{Mock AIME box|year=Pre 2005|n=1|num-b=7|num-a=9|source=14769}}3 KB (446 words) - 23:18, 9 February 2020
- [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]] [[Mock AIME 5 Pre 2005 Problems/Problem 2|Solution]]6 KB (909 words) - 06:27, 12 October 2022
- ...04060657, \end{aligned}</math> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math> {{Mock AIME box|year=Pre 2005|n=5|num-b=11|num-a=13|source=28368}}517 bytes (55 words) - 19:01, 23 March 2017
- ...1001}\equiv 3 \pmod{500}</math>. Then <math>2\cdot3^{1001}=2(500k+3)=1000k+6</math>. Thus, our answer is <math>\boxed{006}</math>. {{Mock AIME box|year=Pre 2005|n=1|num-b=10|num-a=12|source=14769}}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>. == Problems ==5 KB (795 words) - 16:39, 31 December 2024
- ...rac15, GE = \frac{115}{\sqrt{24}}</math>, and <math>AG = AE - GE = 10\sqrt{6} - \frac{115}{\sqrt{24}} = \frac{5}{\sqrt{24}}</math>. Note that <math>\tri ...)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 + 240 = \boxed{338}</math>.2 KB (376 words) - 21:41, 26 December 2016
- ...}</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: {{Mock AIME box|year=Pre 2005|n=1|num-b=14|num-a=15|source=14769}}2 KB (340 words) - 00:44, 3 March 2020
- 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? [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]7 KB (1,094 words) - 14:39, 24 March 2019
- [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]] [[Mock AIME 2 Pre 2005 Problems/Problem 2|Solution]]6 KB (1,052 words) - 12:52, 9 June 2020
- ...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{255}.</math> {{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
- ==Problems== ...frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math> ([[1959 AHSME Problems/Problem 37|Source]])3 KB (558 words) - 15:37, 21 July 2024