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• Mock AIME 1 Pre 2005 Problems/Problem 3
  == Problem == ...>=r, <math>BP = |r - 1|</math>, <math>CP = |r - 2|</math>, <math>DP = |r - 4|</math>, and <math>EP = |r - 13|.</math> Squaring each of these gives:
  1 KB (217 words) - 06:18, 2 July 2015
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == Problem == ...numbers in the middle (those mentioned in condition [2]). There are <math>4-k</math> <tt>A</tt>s amongst the last six numbers then. Also, there are <ma
  1 KB (221 words) - 17:27, 23 February 2013
• Mock AIME 5 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (909 words) - 07:27, 12 October 2022
• Mock AIME 5 Pre 2005 Problems/Problem 12
  == Problem == Let <math>m = 101^4 + 256</math>. Find the sum of the digits of <math>m</math>.
  517 bytes (55 words) - 20:01, 23 March 2017
• Ball-and-urn
  ...h> objects in <math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then
  5 KB (775 words) - 23:53, 13 April 2024
• Mock AIME 1 Pre 2005 Problems/Problem 12
  == Problem == ...{6}</math>. Thus, the area of <math>ABCD</math> is <math>(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 +
  2 KB (376 words) - 22:41, 26 December 2016
• Mock AIME 1 Pre 2005 Problems/Problem 15
  == Problem == ...cdot 23 + b^2+c^2-a^2}{bc} + \frac{36 \cdot 23+a^2+b^2-c^2}{ab} = \frac{7}{4} \implies</math> <math>\frac{36 \cdot 23 + c^2-a^2}{c} + \frac{36 \cdot 23
  2 KB (340 words) - 01:44, 3 March 2020
• 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 6
  == Problem == : <math>P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}</math>
  522 bytes (77 words) - 21:17, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 9
  == Problem == ...assigned to these houses such that there is at least one house with <math>4</math> residents?
  461 bytes (62 words) - 21:18, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 12
  == Problem == So, <math>DE=4</math>.
  2 KB (294 words) - 16:24, 24 August 2022
• Mock AIME 4 Pre 2005/Problems
  ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]
  7 KB (1,094 words) - 15:39, 24 March 2019
• Mock AIME 2 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,052 words) - 13:52, 9 June 2020
• Mock AIME 2 Pre 2005 Problems/Problem 2
  == Problem == ...\dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \left(x + \dfrac{1}{x}\right) = 18 \cdot 47 - 3 = \boxed{843}.</c
  883 bytes (128 words) - 16:14, 4 August 2019
• Mock AIME 2 Pre 2005 Problems/Problem 3
  == Problem == In a box, there are <math>4</math> green balls, <math>4</math> blue balls, <math>2</math> red balls, a brown ball, a white ball, an
  1 KB (170 words) - 17:15, 4 August 2019
• Mock AIME 2 Pre 2005 Problems/Problem 5
  == Problem == ...7 \cdot 3^3 \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{25
  1 KB (171 words) - 17:38, 4 August 2019
• Telescoping series
  ...are positive integers. What is <math>a+b</math>? ([[2022 AMC 10B Problems/Problem 9|Source]]) *Which of the following is equivalent to <math>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</math> (Hint: diffe
  3 KB (500 words) - 14:11, 21 November 2023
• Mock AIME 2 Pre 2005 Problems/Problem 9
  == Problem == k_{4} = {3^3}...</cmath>
  2 KB (232 words) - 00:22, 1 January 2021

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