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• 2010 AIME I Problems/Problem 10
  ...math>. There are only 2 possible cases where <math>2010 - 10a_1 - a_0 \geq 2000</math>, namely <math>(a_1, a_0) = (1,0), (10,0)</math>. Thus, there are <ma *If <math>1000 \leq 2010 - 10a_1 - a_0 < 2000</math>, then there are 2 valid choices for <math>a_3</math>. Since there ar
  7 KB (1,200 words) - 12:38, 19 June 2024
• 2012 AIME II Problems/Problem 1
  ...<math>1, 6, ...</math> down into <math>0, 5, 10, ...</math>. So <math>20m=2000-12k</math>, where <math>k</math> is a member of the set <math>{0, 5, 10, 15 ...6</math>. Each of this is even (notice that when <math>k=25</math>, <math>2000-12k=1700</math>, it cycles again).
  3 KB (519 words) - 14:56, 2 June 2023
• Mock AIME 6 2006-2007 Problems
  [[Mock AIME 6 2006-2007 Problems/Problem 1|Solution]] [[Mock AIME 6 2006-2007 Problems/Problem 2|Solution]]
  7 KB (1,173 words) - 21:04, 7 December 2018
• 2014 AIME I Problems/Problem 8
  <math>2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}</math> {{AIME box|year=2014|n=I|num-b=7|num-a=9}}
  12 KB (1,962 words) - 08:40, 4 November 2022
• Olympiad Archive
  ...ww.oma.org.ar/enunciados/index.htm ''Full Archive of Mathematical Olympiad Problems''] ** Problems w/o Solutions
  16 KB (1,987 words) - 11:39, 17 February 2024
• 2019 AMC 10C Problems
  Here are the problems from the 2019 AMC 10C, a mock contest created by the AoPS user fidgetboss_4 [[2019 AMC 10C Problems/Problem 1|Solution]]
  12 KB (1,917 words) - 12:14, 29 November 2021
• 2019 AIME II Problems/Problem 9
  ...<math>d(k) = 1</math>, and there is one solution <math>n = 2^4 \cdot 5^3 = 2000</math>. Then we have <math>\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}</math>.
  4 KB (609 words) - 14:46, 4 January 2024
• 2020 AIME II Problems
  {{AIME Problems|year=2020|n=II}} [[2020 AIME II Problems/Problem 1 | Solution]]
  7 KB (1,199 words) - 16:17, 12 March 2021
• 2020 AIME II Problems/Problem 3
  ...equation in, we get <math>8^n\cdot3^{20}=3^{2020}</math>, so <math>8^n=3^{2000}</math>. Taking the 100th root, we get <math>8^{\frac{n}{100}}=3^{20}</math {{AIME box|year=2020|n=II|num-b=2|num-a=4}}
  4 KB (502 words) - 04:23, 23 January 2023
• Know the Facts: Gmaas
  - 2000 - 2005: Reconstruction of Newton's Principia Mathematica Gmaasis. (In Engli ...roblem. Therefore, I proved that you cannot use the Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Gmaas can turn things
  88 KB (14,927 words) - 11:07, 1 July 2024
• Mathboy282
  [[2019 AIME II Problems/Problem 15]] Solution 5 [[2023 USAJMO Problems/Problem 6]] Solution 1
  10 KB (1,116 words) - 12:37, 11 June 2024

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