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• 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
• 2009 AIME II Problems/Problem 7
  ...>3</math> on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator o [[Category: Intermediate Number Theory Problems]]
  8 KB (1,312 words) - 16:23, 30 March 2024
• Championnat international des jeux mathématiques et logiques
  ...s jeux mathématiques et logiques'') is a contest that consists of several problems that tests the players' logical capabilities, run jointly by national feder Puzzle] -like problems. Finally, the championship in several countries provides a way to select, v
  3 KB (498 words) - 08:52, 1 September 2022
• 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 1 (Logic and Word Analysis)== [[Category:Introductory Geometry Problems]]
  2 KB (310 words) - 17:12, 1 April 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 (1,622 words) - 20:53, 11 September 2023
• 2010 AIME I Problems/Problem 10
  ...that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. Fo [[Category:Intermediate Number Theory Problems]]
  7 KB (1,147 words) - 21:58, 23 January 2024
• 2010 USAJMO Problems/Problem 1
  ...ways of sorting them. The same goes for <math>m=2, 3, etc.</math> by this logic. Note that the <math>P(n)</math> as stated by the problem requires a <math> [[Category:Olympiad Number Theory Problems]]
  12 KB (2,338 words) - 20:30, 13 February 2024
• 2011 AMC 10A Problems/Problem 22
  ...t of ways to arrange colors among the vertices of a square. Using the same logic as above, there are <math>5^3=125</math> ways, except we must subtract the [[Category:Intermediate Combinatorics Problems]]
  14 KB (2,425 words) - 09:13, 5 November 2023
• 1999 AMC 8 Problems/Problem 18
  ...th>108</math> students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonn ...ts. These students eat <math>81 \cdot 2 = 162</math> cookies. Follow the logic of the second paragraph above to find that there needs to be <math>11</math
  2 KB (296 words) - 02:00, 28 February 2022
• 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

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