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• Math books
  ...combinatorics, and number theory, along with sets of accompanying practice problems at the end of every section. ...88606&sprefix=after+school+maths+%2Caps%2C268&sr=8-2 100 Challenging Maths Problems]
  24 KB (3,177 words) - 12:53, 20 February 2024
• Logic books
  These '''Logic books''' are recommended by [[Art of Problem Solving]] administrators and m === Digital Logic===
  1 KB (151 words) - 14:14, 19 December 2008
• AMC 8
  ...ltiple Choice|difficulty=1 - 1.5|breakdown=<u>Problems 1 - 12</u>: 1<br><u>Problems 13 - 25</u>: 1.5}} The AMC 8 exam is a 25 problem exam. There are 40 minutes given in the exam. Problems increase in difficulty as the problem number increases.
  4 KB (558 words) - 16:52, 19 February 2024
• Principle of Inclusion-Exclusion
  ...math>|A\cap B|</math>, that's just putting four guys in order. By the same logic as above, this is <math>2!\binom{6}{4}=30</math>. Again, <math>|A\cap C|</m [[2011 AMC 8 Problems/Problem 6]]
  9 KB (1,703 words) - 07:25, 24 March 2024
• Constructive counting
  The exact same logic applies to the third digit; it can be any of the <math>8</math> digits exce ''[[2001 AMC 12 Problems/Problem 16 | 2001 AMC 12 Problem 16]]: A spider has one sock and one shoe f
  12 KB (1,896 words) - 23:55, 27 December 2023
• Casework
  ...emonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an While there are problems where casework produces the most elegant solution, in those where a shorter
  5 KB (709 words) - 10:28, 19 February 2024
• Complementary counting
  ...don't want, then subtracts that from the total number of possibilities. In problems that involve complex or tedious [[casework]], complementary counting is oft ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive intege
  8 KB (1,192 words) - 17:20, 16 June 2023
• Mathematical Kangaroo
  ...e with 5 possible answer choices. The remaining levels have tests with 30 problems, each multiple choice with 5 possible answer choices. ...thin the second one-third of the test is worth 4 points, and the remaining problems are worth 5 points. No penalty or partial credit is given to unanswered or
  6 KB (936 words) - 15:38, 22 February 2024
• 2006 AIME I Problems/Problem 13
  We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form: [[Category:Intermediate Number Theory Problems]]
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AMC 12A Problems/Problem 17
  By logic <math>s=3</math> and <math>sr=5 \implies r=5/3.</math> [[Category:Introductory Geometry Problems]]
  6 KB (958 words) - 23:29, 28 September 2023
• Set
  ==Problems== ([[2005 AMC 12A Problems/Problem 13|Source]])
  11 KB (2,021 words) - 00:00, 17 July 2011
• 1985 AIME Problems/Problem 14
  ...{4} = 10(p-10)</math>. This equation is the same as above, and by the same logic, the answer is <math>n=\boxed{25}</math>. * [[AIME Problems and Solutions]]
  5 KB (772 words) - 22:14, 18 June 2020
• 1985 AIME Problems/Problem 12
  ~Azjps (Fundamental Logic) Here is a similar problem from another AIME test: [[2003 AIME II Problems/Problem 13|2003 AIME II Problem 13]], in which we have an equilateral trian
  17 KB (2,837 words) - 13:34, 4 April 2024
• 1986 AIME Problems/Problem 9
  ...AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. [[Category:Intermediate Geometry Problems]]
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 7
  ...h term is <math>729</math> plus the <math>36</math>th term. Using the same logic, the <math>36</math>th term is <math>243</math> plus the <math>4</math>th t * [[AIME Problems and Solutions]]
  5 KB (866 words) - 00:00, 22 December 2022
• 1988 AIME Problems/Problem 13
  ...x^2-x-1)</math> into <math>1</math>. We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process: [[Category:Intermediate Algebra Problems]]
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1989 AIME Problems/Problem 10
  By identical logic, we can find similar expressions for the sums of the other two cotangents: Identical logic works for the other two angles in the triangle. So, the cotangent of any a
  8 KB (1,401 words) - 21:41, 20 January 2024
• 1989 AIME Problems/Problem 1
  ~Novus677 (Fundamental Logic) ~Vrjmath (Fundamental Logic)
  4 KB (523 words) - 00:12, 8 October 2021
• 1991 AIME Problems/Problem 12
  By similar logic, we have <math>APOS</math> is a cyclic quadrilateral. Let <math>AP = x</mat [[Category:Intermediate Geometry Problems]]
  8 KB (1,270 words) - 23:36, 27 August 2023
• 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

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