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• 2005 AIME II Problems/Problem 5
  == Problem == ...og_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
  3 KB (547 words) - 19:15, 4 April 2024
• 2005 AIME I Problems/Problem 5
  == Problem == There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
  5 KB (830 words) - 01:51, 1 March 2023
• Mock AIME 3 Pre 2005 Problems/Problem 5
  ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 5 2005-2006 Problems/Problem 1
  == Problem == ...sors of <math>n</math> less than <math>50</math> (e.g. <math>f(12) = 2+3 = 5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)
  2 KB (209 words) - 12:43, 10 August 2019
• Mock AIME 5 Pre 2005 Problems/Problem 12
  == 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) - 20:01, 23 March 2017
• Mock AIME 5 2005-2006 Problems/Problem 2
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=1|num-a=3}}
  752 bytes (117 words) - 21:16, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 3
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=2|num-a=4}}
  795 bytes (133 words) - 08:14, 19 July 2016
• Mock AIME 5 2005-2006 Problems/Problem 4
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=3|num-a=5}}
  645 bytes (109 words) - 20:41, 22 March 2016
• Mock AIME 5 2005-2006 Problems/Problem 6
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=5|num-a=7}}
  522 bytes (77 words) - 21:17, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 7
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=6|num-a=8}}
  853 bytes (134 words) - 21:18, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 8
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=7|num-a=9}}
  511 bytes (79 words) - 21:18, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 9
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=8|num-a=10}}
  461 bytes (62 words) - 21:18, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 10
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=9|num-a=11}}
  328 bytes (45 words) - 14:15, 11 April 2018
• Mock AIME 5 2005-2006 Problems/Problem 11
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=10|num-a=12}}
  786 bytes (131 words) - 21:19, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 12
  == Problem == By pythagoras theorem on <math>\triangle ABD</math> , We get <math>BD=5=CE</math>
  2 KB (294 words) - 16:24, 24 August 2022
• Mock AIME 5 2005-2006 Problems/Problem 13
  == Problem == ...<math>x \in S</math> with <math>n</math> digits must be divisible by <math>5^n</math>. Let <math>A</math> be the sum of the <math>20</math> smallest ele
  539 bytes (83 words) - 21:20, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 14
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=13|num-a=15}}
  2 KB (282 words) - 10:06, 9 August 2022
• Mock AIME 5 2005-2006 Problems/Problem 15
  == Problem == ...labeled <math>a_0</math>, <math>a_1</math>, <math>\ldots</math>, <math>a_{2005}</math> around the circle in order. Two beads <math>a_i</math> and <math>a_
  955 bytes (157 words) - 21:20, 8 October 2014
• Mock AIME 2 Pre 2005 Problems/Problem 5
  == Problem == {{Mock AIME box|year=Pre 2005|n=2|num-b=4|num-a=6|source=14769}}
  1 KB (171 words) - 17:38, 4 August 2019
• Mock AIME 5 Pre 2005 Problems/Problem 2
  ==Problem== Two 5-digit numbers are called "responsible" if they are:
  2 KB (428 words) - 17:27, 23 February 2025

Page text matches

• Simon's Favorite Factoring Trick
  ...nd make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. == Fun Practice Problems ==
  4 KB (682 words) - 10:25, 18 February 2025
• Principle of Inclusion-Exclusion
  ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====
  9 KB (1,703 words) - 01:20, 7 December 2024
• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.
  13 KB (2,018 words) - 15:31, 10 January 2025
• Ellipse
  D((-5,0)--(0,0)--(0,-3)); MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);
  5 KB (892 words) - 21:52, 1 May 2021
• Complex number
  == Problems == *[[2007 AMC 12A Problems/Problem 18]]
  5 KB (860 words) - 15:36, 10 December 2023
• Mathew Crawford
  ...Cameron Matthews. In 2003, Crawford became the first employee of [[Art of Problem Solving]] where he helped to write and teach most of the online classes dur * Perfect score on the [[AIME]] as a freshman.
  2 KB (362 words) - 11:20, 27 September 2024
• Without loss of generality
  ...0, AMC 12, and AIME competitions it is very common to use WLOG in geometry problems by assuming one undefined side length to be equal to 1. == Problems using WLOG ==
  2 KB (318 words) - 23:35, 13 September 2024
• Geometric sequence
  ...tio <math>-1/2</math>; however, <math>1, 3, 9, -27</math> and <math>-3, 1, 5, 9, \ldots</math> are not geometric sequences, as the ratio between consecu == Problems ==
  4 KB (649 words) - 21:09, 19 July 2024
• Arithmetic sequence
  == Problems == Here are some problems with solutions that utilize arithmetic sequences and series.
  4 KB (736 words) - 02:00, 7 March 2024
• Diophantine equation
  ...</math>. Thus, all fourth powers are either <math>0</math> or <math>1 \mod 5</math>. ...\equiv 0\mod 5, y_0^4 \equiv 1\mod 5</math>, so <math>z_0^2 \equiv 1\mod 5</math>.
  9 KB (1,434 words) - 01:15, 4 July 2024
• Power of a Point Theorem
  == Problems == ...h>E</math>. Given that <math>BE=16</math>, <math>DE=4</math>, and <math>AD=5</math>, find <math>CE</math>.
  5 KB (948 words) - 17:04, 21 February 2025
• 2004 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]
  1 KB (154 words) - 12:30, 22 March 2011
• 2006 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]
  1 KB (135 words) - 12:30, 22 March 2011
• Mock AMC
  ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.
  51 KB (6,175 words) - 21:41, 27 November 2024
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} == Problem 1 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• Set
  ...ultiple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset. ...39</math>, for example. (The standard notation for this set is <math>\{1,3,5,239\}</math>. Note that the order in which the terms are listed is complete
  11 KB (2,019 words) - 17:20, 7 July 2024
• 2005 AIME II Problems/Problem 1
  == Problem == ...ath>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>.
  1 KB (239 words) - 11:54, 31 July 2023
• 2005 AIME II Problems/Problem 2
  == Problem == *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>
  4 KB (628 words) - 11:28, 14 April 2024

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