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- *[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]] *[[Mock AIME 1 2006-2007 Problems/Problem 8 | Next Problem]]3 KB (518 words) - 15:54, 25 November 2015
- {{Mock AIME box|year=2006-2007|n=2|num-b=6|num-a=8}} [[Category:Intermediate Geometry Problems]]1 KB (231 words) - 17:10, 10 July 2014
- ...ion again, we have <math>\lambda(100)=20</math>, so <math>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we <cmath>\begin{align*}3^{87}=(3^{20})^4\cdot 3^7&\equiv 401^4\cdot 187\pmod{1000} \1 KB (127 words) - 23:15, 4 January 2010
- {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=6|num-a=8}} [[Category:Intermediate Number Theory Problems]]853 bytes (134 words) - 20:18, 8 October 2014
- 3 KB (442 words) - 19:51, 26 November 2023
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- ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. ...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) - 20:41, 27 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 ...xOGY2Y2QwOTc3NWZiYjY0LnBkZg==&rn=TWlsZG9yZiBNb2NrIEFJTUUucGRm Mildorf Mock AIME 1]8 KB (901 words) - 19:45, 13 January 2025
- The '''Mock AIME 1 2005-2006''' was written by [[Art of Problem Solving]] community member p * [[Mock AIME 1 2005-2006/Answer Key|Answer Key]]1 KB (135 words) - 16:41, 21 January 2017
- The '''Mock AIME 1 2006-2007''' was written by [[Art of Problem Solving]] community member Altheman. * [[Mock AIME 1 2006-2007/Problems|Entire Exam]]1 KB (155 words) - 15:06, 3 April 2012
- The '''Mock AIME 2 2006-2007''' was written by [[Art of Problem Solving]] community member 4everwise. * [[Mock AIME 2 2006-2007 Problems|Entire Exam]]1 KB (145 words) - 09:55, 4 April 2012
- ...al{S}</math>, we have that <math>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</ ...box and 1 into each of two others. Thus, this gives us <math>m = 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026<1 KB (188 words) - 14:53, 3 April 2012
- Let <math>\triangle ABC</math> have <math>BC=\sqrt{7}</math>, <math>CA=1</math>, and <math>AB=3</math>. If <math>\angle A=\frac{ ...o <math>3^{2000}\equiv 1 \pmod{1000}</math> and so <math>3^{2007} \equiv 3^7 \equiv 2187 \equiv 187 \pmod{1000}</math>963 bytes (135 words) - 14:53, 3 April 2012
- ...> and <math>a = (a + b) - b = 12</math>, so our three vertices are <math>(-7, 49), (-2, 4)</math> and <math>(12, 144)</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]]1 KB (244 words) - 14:21, 5 November 2012
- *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]] *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Next Problem]]3 KB (460 words) - 14:52, 3 April 2012
- *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Previous Problem]] *[[Mock AIME 1 2006-2007 Problems/Problem 9 | Next Problem]]1 KB (244 words) - 13:54, 21 August 2020
- Radius <math>a=\frac{3}{7}</math>, radius <math>b=\frac{6}{11}</math>, radius <math>c=\frac{2}{5}</ma *[[Mock AIME 1 2006-2007 Problems/Problem 9 | Previous Problem]]1 KB (236 words) - 22:58, 24 April 2013
- ...initial values <math>\mathcal{S}_1 = 2, \mathcal{S}_2 = 4, \mathcal{S}_3 = 7</math> we can easily compute that <math>\mathcal{S}_{11} = 927</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 10 | Previous Problem]]2 KB (424 words) - 14:51, 3 April 2012
- ...7</math>, <math>d_{5}=13</math>, and <math>d_{6}=-16</math>, find <math>d_{7}</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 12 | Previous Problem]]3 KB (568 words) - 14:50, 3 April 2012
- [[Mock AIME 1 2006-2007 Problems/Problem 1|Solution]] ...al{S}</math>, we have that <math>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</8 KB (1,355 words) - 13:54, 21 August 2020
- ...tegers. Then <math>x_1 = 8 - x_2 = 7</math> and our [[sequence]] is <math>7, 1, 1, 8, 16, 144, 144(16 + 8) = 3456</math>. {{Mock AIME box|year=2006-2007|n=2|num-b=7|num-a=9}}3 KB (470 words) - 23:33, 9 August 2019
- ....com/wiki/index.php?title=MATHCOUNTS MATHCOUNTS] competition. A number of Mock MATHCOUNTS competitions have been hosted on the Art of Problem Solving mess ...COUNTS competition. There is no guarantee that community members will make Mock MATHCOUNTS in any given year, but it's usually a good bet that someone will26 KB (3,260 words) - 18:28, 15 August 2024
- ...ion again, we have <math>\lambda(100)=20</math>, so <math>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we <cmath>\begin{align*}3^{87}=(3^{20})^4\cdot 3^7&\equiv 401^4\cdot 187\pmod{1000} \1 KB (127 words) - 23:15, 4 January 2010
- .../math>, by [[Euler's Totient Theorem]] <math>2^{20 \cdot 100 + 7} \equiv 2^7 \equiv 3 \pmod{125}</math>. Combining, we have <math>2^{2007} \equiv 128 \p <math>2^{2007} \equiv 2^7\;(mod\;1000)\equiv 128\;(mod\;1000)</math>4 KB (595 words) - 11:14, 25 November 2023
- ...math>, we see that <math>f(1)=0, f(2)=0, f(3)=0, f(4)=1, f(5)=1, f(6)=2, f(7)=2, *[[Mock AIME 4 2006-2007 Problems/Problem 13| Next Problem]]992 bytes (156 words) - 19:34, 27 September 2019
- So all of the prime numbers less than <math>50</math> are <math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,</math> and <math>47</math>. So we <math>\lfloor 99/7\rfloor =14</math>2 KB (209 words) - 11:43, 10 August 2019