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• Mathematics 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]
  23 KB (3,036 words) - 16:38, 1 March 2025
• AMC 8
  ...Multiple Choice|difficulty=1 - 2|breakdown=<u>Problems 1 - 12</u>: 1<br><u>Problems 13 - 25</u>: 2}} 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. Students are not pe
  4 KB (584 words) - 19:59, 19 February 2025
• Constructive counting
  == Introductory Examples == The exact same logic applies to the third digit; it can be any of the <math>8</math> digits exce
  13 KB (2,018 words) - 15:31, 10 January 2025
• 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== ===Introductory===
  11 KB (2,019 words) - 17:20, 7 July 2024
• 2005 AMC 10A Problems/Problem 24
  ...</math> won't be equal to the <math>\sqrt{n}</math>. We can apply the same logic to <math>\sqrt{n+48}</math>, and since we also the difference between two c [[Category:Introductory Number Theory Problems]]
  8 KB (1,245 words) - 10:53, 4 February 2025
• 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 (434 words) - 11:01, 22 August 2024
• 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
• 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]]
  13 KB (2,248 words) - 14:44, 5 January 2025
• How should I prepare?
  ...your current abilities, you may want to start out with different practice problems, different books, and different areas of the forums. You should also try to ...Alcumus]] tool constantly to practice all of these topics. When you answer problems incorrectly or take a long time to answer one, reread the part in the book
  14 KB (2,062 words) - 11:59, 15 January 2025
• 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
• 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
• 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
• 1991 AJHSME Problems/Problem 20
  Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)< [[Category:Introductory Algebra Problems]]
  1 KB (203 words) - 13:06, 16 August 2020
• 2010 AMC 12A Problems/Problem 3
  ==Video Solution 2 (Logic and Word Analysis)== [[Category:Introductory Geometry Problems]]
  2 KB (322 words) - 10:55, 18 September 2024
• 2010 AMC 12A Problems/Problem 13
  ...there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the gra [[Category:Introductory Geometry Problems]]
  9 KB (1,646 words) - 23:56, 1 November 2024
• 1992 AHSME Problems/Problem 17
  ...is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that <math>N\equiv 19+20+21+22+...+91+92\pmod{3}</math>, and adding [[Category: Introductory Number Theory Problems]]
  2 KB (342 words) - 13:19, 4 January 2021
• 1966 AHSME Problems/Problem 20
  [[Category:Introductory Logic Problems]]
  589 bytes (97 words) - 02:30, 15 September 2014

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