Create the page "2005 AIME Problems/Problem 4" on this wiki! See also the search results found.

Page title matches

• 2005 AIME II Problems/Problem 4
  == Problem == {{AIME box|year=2005|n=II|num-b=3|num-a=5}}
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME I Problems/Problem 4
  == Problem == ...erfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the numbe
  8 KB (1,249 words) - 21:25, 20 November 2024
• Mock AIME 3 Pre 2005 Problems/Problem 4
  ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}}
  2 KB (221 words) - 02:49, 19 March 2015
• Mock AIME 1 Pre 2005 Problems/Problem 4
  == Problem == ...<math>1 + 7 + 7^2 + \cdots + 7^{2004} = \frac{7^{2005}-1}{7-1} = \frac{7^{2005}-1}{6}</math>. Since <math>\varphi(1000) = 400</math>, by [[Fermat's Little
  685 bytes (81 words) - 10:51, 11 June 2013
• 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 2 Pre 2005 Problems/Problem 4
  == Problem == {{Mock AIME box|year=Pre 2005|n=2|num-b=3|num-a=5|source=14769}}
  817 bytes (114 words) - 17:16, 4 August 2019
• Mock AIME 4 Pre 2005/Problems/Problem 9
  ==Problem== ...h>S = \sum^{\infty}_{n=1} \frac{(7n + 32) \cdot 3^n}{n \cdot (n + 2) \cdot 4^n}</cmath>
  2 KB (319 words) - 01:21, 30 January 2024
• Mock AIME 5 Pre 2005 Problems/Problem 4
  372 bytes (62 words) - 00:05, 15 February 2024

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
• Brahmagupta's Formula
  <cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath> <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
  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
  ...ts are perpendicular. Drawing all four semi-axes divides the ellipse into 4 [[congruent (geometry)|congruent]] quarters. pair P=(3,12/5), F1=(-4,0), F2=(4,0);
  5 KB (892 words) - 21:52, 1 May 2021
• Complex number
  ...aginary part of a complex number is real: for example, <math>\Im(3 + 4i) = 4</math>. So, if <math>z\in \mathbb C</math>, we can write <math>z=\mathrm{R == Problems ==
  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
• Geometric sequence
  For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <mat == Problems ==
  4 KB (649 words) - 21:09, 19 July 2024
• Arithmetic sequence
  ...on difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference == Problems ==
  4 KB (736 words) - 02:00, 7 March 2024
• Diophantine equation
  An equation of form <math>x^4+y^4=z^2</math> has no [[integer]] solutions, as follows: If <math>\gcd(x_0,y_0)=1</math>, we then proceed with casework, in <math>\mod 4</math>.
  9 KB (1,434 words) - 01:15, 4 July 2024
• Power of a Point Theorem
  == Problems == ...th> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</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
  ...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. ...describe sets should be used with extreme caution. One way to avoid this problem is as follows: given a property <math>P</math>, choose a known set <math>T<
  11 KB (2,019 words) - 17:20, 7 July 2024
• 2005 AIME II Problems/Problem 1
  == Problem == ...cient]] <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}</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

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)