Create the page "Logic Problems" on this wiki! See also the search results found.

• 1992 AIME Problems/Problem 15
  With this logic, we realize that the desired quantity is simply <math>\left \lfloor \frac{7 [[Category:Intermediate Number Theory Problems]]
  2 KB (358 words) - 01:54, 2 October 2020
• 1994 AIME Problems/Problem 10
  ...dot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + [[Category:Intermediate Geometry Problems]]
  3 KB (534 words) - 16:23, 26 August 2018
• 1995 AIME Problems/Problem 15
  Applying similar logic, we get the equations: [[Category:Intermediate Combinatorics Problems]]
  6 KB (979 words) - 13:20, 11 April 2022
• 2001 AIME I Problems/Problem 15
  ..., we are once again left with <math>2</math> choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our [[Category:Intermediate Combinatorics Problems]]
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 14
  Now using the same logic as above we can find <math>M_n = M_{n-2} + M_{n-3}</math> ( the cases are 0 ...an replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candida
  13 KB (2,298 words) - 19:46, 9 July 2020
• 2001 AIME I Problems/Problem 6
  The logic is that, if <math>n</math> is rolled, then the number of valid cases is the [[Category:Intermediate Combinatorics Problems]]
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2003 AIME II Problems/Problem 6
  ...that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles). [[Category: Intermediate Geometry Problems]]
  5 KB (787 words) - 17:38, 30 July 2022
• 2000 AIME II Problems/Problem 15
  ...quence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as [[Category:Intermediate Trigonometry Problems]]
  3 KB (469 words) - 21:14, 7 July 2022
• 2006 Romanian NMO Problems/Grade 9/Problem 4
  ...s <math>n\leq 3</math>, which is clearly false. This is a contradiction in logic, so the three people in second place could not have won exactly two matches *[[2006 Romanian NMO Problems]]
  2 KB (325 words) - 15:53, 11 December 2011
• 2007 AIME I Problems/Problem 10
  ...te, and this implies that <math>k</math> must be even. We can use the same logic for <math>c</math> and <math>d</math>. Therefore, the coefficient of <math> [[Category:Intermediate Combinatorics Problems]]
  13 KB (2,328 words) - 00:12, 29 November 2023
• ITest
  ...est is varied, ranging from simple arithmetic problems to complex Olympiad problems. Winning teams earn recent technology prizes like a video game console of ...luding [[algebra]], [[geometry]], pre-[[calculus]], [[probability]], and [[logic]]. Graphing calculators are allowed, but computers are not allowed.
  2 KB (264 words) - 20:31, 13 December 2018
• 2005 AMC 12A Problems/Problem 13
  Not too bad with some logic and the awesome guess and check. Let <math>A=6</math>. Then let <math>B=7,E [[Category:Introductory Algebra Problems]]
  3 KB (430 words) - 18:52, 11 July 2020
• Primary Mathematics World Contest
  *Individual Round: The individual round consists of 15 problems, numbered I1-I15. *Team Round: The team round consists of 10 problems, numbered T1-T10.
  1 KB (183 words) - 13:57, 15 October 2018
• 2004 AMC 12A Problems/Problem 17
  {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A #24]]}} ~Azjps (Fundamental Logic)
  2 KB (233 words) - 08:14, 6 September 2021
• 2007 AMC 10A Problems/Problem 20
  ~Azjps (Fundamental Logic) ~Rbhale12 (Fundamental Logic)
  4 KB (663 words) - 07:32, 4 November 2022
• 2008 AMC 12B Problems/Problem 25
  ...ude of <math>\triangle{CBY}</math> so <math>BC = BY.</math> Using the same logic on <math>\triangle{BCX}</math> gives <math>BC = BX \iff BYXC</math> is a rh [[Category:Introductory Geometry Problems]]
  12 KB (2,015 words) - 20:54, 9 October 2022
• 2008 AIME I Problems/Problem 3
  == Solution 4 (Logic) == [[Category:Intermediate Number Theory Problems]]
  3 KB (564 words) - 22:15, 28 November 2023
• 2008 AIME I Problems/Problem 11
  Applying this logic to all of the other cases gives us <math>\binom{7}{3}</math>, <math>\binom{ [[Category:Intermediate Combinatorics Problems]]
  7 KB (1,173 words) - 22:39, 28 November 2023
• 2008 AIME II Problems/Problem 14
  ...<math>k=\frac{2}{\sqrt{3}}</math>. This inequality must hold by the above logic, and in fact, the inequality reaches equality when <math>x=0</math>. Thus, [[Category:Intermediate Algebra Problems]]
  8 KB (1,387 words) - 11:56, 29 January 2024
• How should I prepare?
  ...your current abilities, you will want to start out with different practice problems, different books, and in different areas of the forums. You should also try .../102-Combinatorial-Problems-Titu-Andreescu/dp/0817643176 102 Combinatorial Problems] by Titu Andreescu & Zuming Feng
  13 KB (1,926 words) - 11:22, 30 November 2023

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