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• Pigeonhole Principle
  # Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to r ...n</math> such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee. ([[Pigeonhole Princi
  11 KB (1,985 words) - 21:03, 5 August 2023
• Relatively prime
  ...n) = 1</math>. Equivalently, <math>m</math> and <math>n</math> must have no [[prime]] divisors in common. The positive integers <math>m</math> and <ma
  2 KB (245 words) - 15:51, 25 February 2020
• 1965 AHSME Problems/Problem 40
  ...>x \neq 10</math> and <math>x < -9</math> or <math>x > 3</math>, there are no solutions for <math>x</math>. ...e force and checking each value individually, we can assert that there are no such solutions for <math>x</math>, leaving us with only <math>1</math> solu
  3 KB (571 words) - 00:42, 22 October 2021
• Principle of Inclusion-Exclusion
  .... Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to b
  9 KB (1,703 words) - 07:25, 24 March 2024
• 2020 CIME II Problems/Problem 10
  ...+b</math>; hence <math>a+b</math> is a multiple of <math>c</math> which is no more than <math>2c+6</math>. It follows that <math>a+b\in\{c,2c,3c,4c,5c\}< Case IV) <math>a+b=4c\Rightarrow (4a-1)(4b-1)=65</math> for which there are no solutions.
  2 KB (332 words) - 09:37, 30 December 2021
• Cauchy-Schwarz Inequality
  ...triangle <math>T </math> whose side lengths are all positive integers with no common divisor and determine those integers.
  13 KB (2,048 words) - 15:28, 22 February 2024
• Fermat's Little Theorem
  ...es of the theorem by <math>a</math>. The restated form is nice because we no longer need to restrict ourselves to integers <math>{a}</math> not divisibl
  16 KB (2,675 words) - 10:57, 7 March 2024
• Number theory/Advanced
  ...written as an expansion in the parameter <math>q=e^{2\pi i z}</math> with no negative exponents:
  5 KB (849 words) - 16:14, 18 May 2021
• 1959 IMO Problems/Problem 2
  This tells there that there is no solution for (b), since we must have <math>A^2 \ge 2</math> ...h>x \in \left[\frac{1}{2}, 1\right]</math> is a solution for (a); there is no solution for (b); there is one solution for (c), which is <math>x=\frac{3}{
  3 KB (466 words) - 12:04, 12 April 2024
• 1970 IMO Problems/Problem 6
  ...onsider all possible triangles having these point as vertices. Prove that no more than <math>70 \%</math> of these triangles are acute-angled.
  1 KB (234 words) - 19:26, 14 July 2017
• Divisibility rules
  ...550</math>. Difference: <math>D = 49</math>. <math>A = 50 \cdot D</math>? No!
  8 KB (1,315 words) - 18:18, 2 March 2024
• Triangle
  A '''scalene''' triangle has no congruent sides, and also no congruent angles (this can be proven by the converse of the Hinge Theorem).
  4 KB (628 words) - 17:17, 17 May 2018
• Constructive counting
  ...that each entry is between <math>2</math> and <math>9</math> inclusive and no two consecutive entries are equal?'' ...luding <math>2</math> and <math>9</math>; the key restriction here is that no two boxes right next to each other can have the same number.
  12 KB (1,896 words) - 23:55, 27 December 2023
• Casework
  ...>. But the question mandates that <math>y</math> is positive, so there are no solutions when <math>x \geq 3</math>. Thus, there are <math>68+53=121</math
  5 KB (709 words) - 10:28, 19 February 2024
• Prime number
  ...6</math>; <math>2^{2^6} + 1 = 274177\cdot 67280421310721</math>. In fact, no other Fermat primes have been found to date.
  6 KB (985 words) - 12:38, 25 February 2024
• Complementary counting
  ...use a complementary approach. The total number of positive integers, with no restrictions, is <math>99</math> integers. What we don't want are the multi ...git positive integers. What we don't want are the four-digit integers with no digit that is a two or three, Using a [[Constructive counting | constructiv
  8 KB (1,192 words) - 17:20, 16 June 2023
• Integer
  ...depending on whether <math>a</math> can be divided by <math>b</math> with no remainder.
  2 KB (296 words) - 15:04, 5 August 2022
• Mathematical Kangaroo
  ...he test is worth 4 points, and the remaining problems are worth 5 points. No penalty or partial credit is given to unanswered or incorrectly answered qu ...nutes is given to solve all problems, each with exactly 5 answer choices. No problems require the use of a calculator, and calculators of all types are
  6 KB (936 words) - 15:38, 22 February 2024
• Pi
  ...lementary Chebyshev's estimate <math>D_n\le n^{\sqrt n}\cdot 4^n</math> is no longer sufficient).
  8 KB (1,469 words) - 21:11, 16 September 2022
• Fibonacci sequence
  # Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to r
  6 KB (957 words) - 23:49, 7 March 2024

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