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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) - 22:25, 28 April 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 13
  ...additional T's among the first <math>4</math> spaces. We can use identical logic to show why we can divide the two H's among the last four spaces to get exa * [[AIME Problems and Solutions]]
  4 KB (772 words) - 21:09, 7 May 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
• 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
• 1997 PMWC Problems/Problem T5
  [[Category:Logic Problems]]
  1 KB (263 words) - 22:58, 13 April 2013
• 2002 AMC 10A Problems/Problem 20
  ...{AD}{CD}=3=\frac{GA}{HC}</math> which means <math>HC=5</math>. By the same logic, <math>JE=3</math>, so <math>\frac{HC}{JE}=\frac{5}{3}</math>. [[Category:Introductory Geometry Problems]]
  3 KB (439 words) - 22:15, 9 June 2023
• 1997 PMWC Problems/Problem I2
  [[Category:Logic Problems]]
  1 KB (183 words) - 19:32, 10 March 2015
• 1985 AJHSME Problems/Problem 25
  [[Category:Logic Problems]]
  2 KB (300 words) - 09:07, 22 January 2023
• 1986 AJHSME Problems/Problem 17
  We can solve this problem using logic. [[Category:Introductory Number Theory Problems]]
  3 KB (457 words) - 15:02, 4 April 2021
• 1986 AJHSME Problems/Problem 22
  [[Category:Logic Problems]]
  1 KB (257 words) - 21:30, 3 July 2013
• 1987 AJHSME Problems/Problem 20
  [[Category:Introductory Number Theory Problems]] [[Category:Logic Problems]]
  1 KB (170 words) - 16:44, 24 May 2023
• 2009 AMC 10B Problems/Problem 22
  [[Category: Introductory Geometry Problems]] ...gorean theorem, <cmath>MR = \sqrt{1^2+2^2} = \sqrt{5}.</cmath> By the same logic, <math>MQ=\sqrt{5}</math>. The area of <math>\triangle QMR</math> is the ar
  12 KB (1,868 words) - 03:36, 30 September 2023
• 1987 AJHSME Problems/Problem 17
  [[Category:Logic Problems]]
  934 bytes (156 words) - 23:53, 4 July 2013
• 2009 AIME I Problems/Problem 1
  ...>, we get <math>999</math>, but the integers must be distinct. By the same logic, the smallest geometric number is <math>124</math>. The largest geometric n [[Category:Intermediate Number Theory Problems]]
  2 KB (360 words) - 03:32, 16 January 2023

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