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• 2005 AIME II Problems/Problem 1
  == Problem == ...given by the [[binomial coefficient]] <math>{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}<
  1 KB (239 words) - 11:54, 31 July 2023
• 2005 AIME I Problems/Problem 1
  == Problem == [[Image:2005 AIME I Problem 1.png]]
  1 KB (213 words) - 13:17, 22 July 2017
• Mock AIME 3 Pre 2005 Problems/Problem 1
  ==Problem== {{Mock AIME box|year=Pre 2005|n=3|before=First Question|num-a=2}}
  795 bytes (129 words) - 10:22, 4 April 2012
• Mock AIME 5 2005-2006 Problems/Problem 1
  == Problem == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.
  2 KB (209 words) - 12:43, 10 August 2019
• Mock AIME 1 Pre 2005 Problems/Problem 1
  == 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) - 13:44, 5 September 2012
• Mock AIME 1 Pre 2005 Problems/Problem 3
  == Problem == ...> and <math>E</math> are collinear in that order such that <math>AB = BC = 1, CD = 2,</math> and <math>DE = 9</math>. If <math>P</math> can be any point
  1 KB (217 words) - 06:18, 2 July 2015
• Mock AIME 1 Pre 2005 Problems/Problem 4
  == Problem == When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a rem
  685 bytes (81 words) - 10:51, 11 June 2013
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == 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) - 17:27, 23 February 2013
• Mock AIME 1 Pre 2005 Problems/Problem 9
  == Problem == <cmath> \frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{360}{pqr} = 1</cmath>
  864 bytes (127 words) - 21:47, 21 February 2010
• Mock AIME 1 Pre 2005 Problems/Problem 8
  == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=7|num-a=9|source=14769}}
  3 KB (446 words) - 00:18, 10 February 2020
• Mock AIME 1 Pre 2005 Problems/Problem 10
  == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=9|num-a=11|source=14769}}
  540 bytes (96 words) - 00:28, 23 December 2023
• Mock AIME 1 Pre 2005 Problems/Problem 11
  == Problem == <cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath>
  2 KB (272 words) - 10:51, 2 July 2015
• Mock AIME 1 Pre 2005 Problems/Problem 13
  ==Problem== ...h> \{R_{n}\}_{n\ge 0} </math> obeys the recurrence <math> 7R_{n}= 64-2R_{n-1}+9R_{n-2} </math> for any integers <math>n\ge 2</math>. Additionally, <ma
  1 KB (178 words) - 14:22, 26 March 2011
• Mock AIME 1 Pre 2005 Problems/Problem 12
  == 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 foll
  2 KB (376 words) - 22:41, 26 December 2016
• Mock AIME 1 Pre 2005 Problems/Problem 14
  == Problem == Note: The problem is meant to be interpreted so that if you cannot produce one arrangement fr
  692 bytes (107 words) - 13:01, 24 November 2021
• Mock AIME 1 Pre 2005 Problems/Problem 15
  == Problem == Using the identity <math>\cos A + \cos B + \cos C = 1+\frac{r}{R}</math>, we have that <math>\cos A + \cos B + \cos C = \frac{21}
  2 KB (340 words) - 01:44, 3 March 2020
• Mock AIME 2 Pre 2005 Problems/Problem 1
  == Problem == {{Mock AIME box|year=Pre 2005|n=2|before=First question|num-a=2|source=14769}}
  643 bytes (89 words) - 16:01, 4 August 2019

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
• Combination
  ...as a great introductory video to combinations, permutations, and counting problems in general! [https://bit.ly/CombinationsAndPermutations Permutations & Comb * <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math>
  4 KB (638 words) - 21:55, 5 January 2025
• Brahmagupta's Formula
  Substituting <math>\sin^2B=1-\cos^2B</math> results in <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>
  3 KB (543 words) - 19:35, 29 October 2024
• Constructive counting
  === Example 1 === This is a problem where constructive counting is not the simplest way to proceed. This next e
  13 KB (2,018 words) - 15:31, 10 January 2025
• Ellipse
  MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E); 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
  ...numbers''' arise when we try to solve [[equation]]s such as <math> x^2 = -1 </math>. ...this addition, we are not only able to find the solutions of <math> x^2 = -1 </math> but we can now find ''all'' solutions to ''every'' polynomial. (Se
  5 KB (860 words) - 15:36, 10 December 2023
• Without loss of generality
  ...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
  ...ratio <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 cons ...ogression if and only if <math>a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}</math>. A similar definition holds for infinite geometric sequences. It ap
  4 KB (649 words) - 21:09, 19 July 2024
• Arithmetic sequence
  ...h>1, 2, 3, 4</math> is an arithmetic sequence with common difference <math>1</math> and <math>99, 91, 83, 75</math> is an arithmetic sequence with commo ...ogression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. A similar definition holds for infinite arithmetic sequences. It a
  4 KB (736 words) - 02:00, 7 March 2024
• Diophantine equation
  ...there will always be an infinite number of solutions when <math>\gcd(a,b)=1</math>. If <math>\gcd(a,b)\nmid c</math>, then there are no solutions to t .../math> is an [[odd]] number, then <math>m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}</math> is a Pythagorean triple.
  9 KB (1,434 words) - 01:15, 4 July 2024
• Power of a Point Theorem
  ''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (a circle), so it only remains to prove the theorem for more == Problems ==
  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
  ...set can have multiple 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. ...th>239</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 comp
  11 KB (2,019 words) - 17:20, 7 July 2024
• 2005 AIME II Problems/Problem 1
  == Problem == ...given by the [[binomial coefficient]] <math>{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}<
  1 KB (239 words) - 11:54, 31 July 2023

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