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  • size(200); size(200);
    5 KB (892 words) - 21:52, 1 May 2021
  • == Problem == size(200);
    13 KB (2,080 words) - 13:14, 23 July 2024
  • {{AIME Problems|year=2004|n=I}} == Problem 1 ==
    9 KB (1,434 words) - 13:34, 29 December 2021
  • {{AIME Problems|year=2004|n=II}} == Problem 1 ==
    9 KB (1,410 words) - 05:05, 20 February 2019
  • {{AIME Problems|year=2002|n=II}} == Problem 1 ==
    7 KB (1,177 words) - 15:42, 11 August 2023
  • == Problem == ...f votes they received is <math>\frac{100v_i}s</math>. The condition in the problem statement says that <math>\forall i: \frac{100v_i}s + 1 \leq v_i</math>. (
    4 KB (759 words) - 13:00, 11 December 2022
  • == Problem == ...math> such that <math>1^2+2^2+3^2+\ldots+k^2</math> is a multiple of <math>200</math>.
    3 KB (403 words) - 12:10, 9 September 2023
  • == Problem == ...nd <math>\overline{BD}</math> are perpendicular. Given that <math>AB=\sqrt{11}</math> and <math>AD=\sqrt{1001}</math>, find <math>BC^2</math>.
    4 KB (589 words) - 15:33, 20 January 2025
  • {{AIME Problems|year=2007|n=II}} == Problem 1 ==
    9 KB (1,435 words) - 01:45, 6 December 2021
  • == Problem == ...s 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}</math>
    4 KB (562 words) - 18:37, 30 October 2020
  • {{AIME Problems|year=2008|n=I}} == Problem 1 ==
    9 KB (1,536 words) - 00:46, 26 August 2023
  • {{AIME Problems|year=2011|n=II}} == Problem 1 ==
    8 KB (1,301 words) - 08:43, 11 October 2020
  • {{AIME Problems|year=2015|n=I}} ==Problem 1==
    10 KB (1,615 words) - 17:03, 9 October 2024
  • {{AIME Problems|year=2013|n=I}} == Problem 1 ==
    9 KB (1,580 words) - 13:07, 24 February 2024
  • ==Problem 13== size(200);
    13 KB (2,145 words) - 19:20, 11 August 2024
  • {{AIME Problems|year=2014|n=I}} ==Problem 1==
    9 KB (1,472 words) - 13:59, 30 November 2021
  • {{AIME Problems|year=2017|n=I}} ==Problem 1==
    7 KB (1,163 words) - 16:43, 2 June 2022
  • ==Problem== ...iangle ABC</math> is <math>100\sqrt 3</math> and the circumradius is <math>200 \sqrt 3</math>. Now, consider the line perpendicular to plane <math>ABC</ma
    17 KB (2,861 words) - 19:39, 25 November 2024
  • ...into anything. Using that fact, you can use the Games theorem to solve any problem. 11. Gmaas is known to barf entire universes at will. EDIT: Every once in a whi
    69 KB (11,805 words) - 20:49, 18 December 2019
  • ==Problem== <cmath>HZ+ZI=\frac{IH}{BI}\cdot \left(ID+BD\right)=2\cdot \left(45+55\right)=200</cmath>
    7 KB (1,053 words) - 14:58, 14 January 2024

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