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• 2004 AIME I Problems/Problem 12
  == Problem == {{AIME box|year=2004|n=I|num-b=11|num-a=13}}
  2 KB (303 words) - 18:43, 16 October 2024

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• Overcounting
  An example of a classic problem is as follows: ...ivisible by 3. However, we note that we overcount several numbers, such as 12, which is divisible by both 2 and 3. To correct for this overcounting, we m
  4 KB (635 words) - 12:19, 2 January 2022
• 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
• Fibonacci sequence
  ...iv style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...a triangle. <div style="text-align:right">(Manhattan Mathematical Olympiad 2004)</div>
  7 KB (1,111 words) - 14:57, 24 June 2024
• Complex number
  ...ver, it is possible to define a number, <math> i </math>, such that <math> i = \sqrt{-1} </math>. If we add this new number to the reals, we will have ...rm <math> a + bi </math> where <math> a,b\in \mathbb{R} </math> and <math> i = \sqrt{-1} </math> is the [[imaginary unit]]. The set of complex numbers i
  5 KB (860 words) - 15:36, 10 December 2023
• Arithmetic sequence
  ...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 bet * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]
  4 KB (736 words) - 02:00, 7 March 2024
• 2004 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems]]
  1 KB (135 words) - 18:15, 19 April 2021
• 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
• Mock AMC
  ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...amous theorems and formulas and see if there's any way you can make a good problem out of them.
  51 KB (6,175 words) - 21:41, 27 November 2024
• 2005 AIME I Problems
  {{AIME Problems|year=2005|n=I}} == Problem 1 ==
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME I Problems/Problem 2
  == Problem == ...ible value of <math>k</math>. Thus the requested number of values is <math>12</math>, and the answer is <math>\boxed{012}</math>.
  2 KB (303 words) - 01:31, 5 December 2022
• 2004 AIME I Problems/Problem 15
  == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 13
  == Problem == ...<math>34</math> complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \
  2 KB (298 words) - 20:02, 4 July 2013
• 2004 AIME I Problems/Problem 11
  == Problem == ...has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>,
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 10
  == Problem == ...he same slope, and the equation of <math>A'C'</math> is <math>y = \frac{5}{12}x + c</math>. Manipulating, <math>5x - 12y + 12c = 0</math>. We need to fin
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME I Problems/Problem 3
  == Problem == ...d <math>36</math> faces, <math>24</math> of which are triangular and <math>12</math> of which are quadrilaterals. A space diagonal is a line segment conn
  1 KB (156 words) - 17:56, 1 January 2016
• 2004 AIME I Problems/Problem 7
  == Problem == ...um of the product of each pair of distinct terms, or <math>C = \sum_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that
  7 KB (1,099 words) - 13:41, 30 December 2024
• 2004 AIME I Problems/Problem 8
  == Problem == Let <math>S = \{1,2,3,\ldots, 1000\}</math>, and <math>A_{i}= \{i \in S \mid i\, \textrm{ divides }\,1000\}</math>. The number of <math>m</math>'s that ar
  4 KB (620 words) - 21:26, 5 June 2021
• 2004 AIME I Problems
  {{AIME Problems|year=2004|n=I}} == Problem 1 ==
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 14
  == Problem == ...b,c): n \equiv b \pmod{12}, 0 \le b \le 11</math> and <math>c = \frac{n-b}{12} \le 7</math> (in other words, we take the greatest possible value of <math
  11 KB (1,857 words) - 12:57, 18 July 2024

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