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• Divisibility rules/Rule for 5 and powers of 5 proof
  ...y <math>5^n</math> if the last <math>n</math> digits are divisible by that power of 5.
  2 KB (292 words) - 09:58, 17 August 2006
• Power set
  The '''power set''' of a given [[set]] <math>S</math> is the set <math>\mathcal{P}(S)</m Similarly, for any [[finite]] set with <math>n</math> elements, the power set has <math>2^n</math> elements.
  4 KB (757 words) - 11:44, 8 March 2018
• North Fulton Varsity Invitational Math Tournament
  ...rty-question written test, ten ciphering questions, and a proof-based team power round. Calculutors will not be permitted for any part of the competition.
  785 bytes (102 words) - 19:40, 27 August 2006
• Partially ordered set
  ...= (\mathcal{P}(S), \leq)</math> where <math>\mathcal{P}(S)</math> is the [[power set]] (that is, the set of [[subset]]s) of <math>S</math> and for <math>x,
  4 KB (717 words) - 20:01, 25 April 2009
• Rockdale Mathematics Competition
  ...h problem after the first relies on answers to the previous problems. The power round, worth sixty points, is a multi-part question for which the team has
  2 KB (267 words) - 21:50, 6 March 2016
• Boolean lattice
  ...dered set]] whose elements are those of <math>\mathcal{P}(S)</math>, the [[power set]] of <math>S</math>, ordered by inclusion (<math>\subset</math>).
  1 KB (168 words) - 22:46, 20 April 2008
• 1981 IMO Problems/Problem 4
  ...equal to 3 must be divisible by a number that is greater than two and is a power of a prime. ...note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.
  3 KB (516 words) - 09:43, 28 March 2012
• 1981 IMO Problems/Problem 6
  ...^a 2</math> represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where <math>a</math> amount of 2s. So therefore the answer is <math>f
  2 KB (306 words) - 18:15, 12 April 2024
• Duke Math Meet
  ...t of [[ARML]]. Teams are of six students each. The test commences with a power round, a multi-part proof round that lasts for one hour. This is followed
  1 KB (174 words) - 09:48, 10 November 2012
• Euler line
  The power of point <math>O</math> with respect to <math>\omega_1, \omega_2,</math> an The power of point <math>H</math> with respect to <math>\omega_1</math> is <math>AH \
  59 KB (10,203 words) - 04:47, 30 August 2023
• Division
  When dividing by decimals, multiply both sides by a power of 10 so the divisor is an integer.
  2 KB (259 words) - 09:52, 23 January 2020
• 2003 AMC 10A Problems/Problem 16
  ...on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the t
  1 KB (218 words) - 15:52, 19 August 2023
• Chinese Remainder Theorem
  ...iquity derives from the fact that many results can be easily proven mod (a power of a prime), and can then be generalized to mod <math>m</math> using the Ch
  6 KB (1,022 words) - 14:57, 6 May 2023
• 1985 IMO Problems
  ...a subset of <math>4</math> elements whose product is the <math>4</math>th power of an integer.
  3 KB (465 words) - 03:00, 29 March 2021
• 1985 IMO Problems/Problem 3
  Let <math>k</math> be the largest power of 2 that is less than or equal to <math>i_n</math>. We proceed by inducti
  2 KB (354 words) - 04:56, 11 March 2023
• Perfect power
  ...math> is a perfect <math>2</math>nd, <math>3</math>rd and <math>6</math>th power. ...h>st power" is a meaningless property: every integer is a <math>1</math>st power of itself.
  870 bytes (148 words) - 16:52, 18 August 2013
• 1987 IMO Problems/Problem 3
  We first note that by the [[Power Mean Inequality]], <math> \sum_{i=1}^{n} x_i \le \sqrt{n} </math>. Therefore al
  2 KB (349 words) - 04:36, 28 May 2023
• Cherokee County Open Math Tournament
  .... Calculators are not permitted on any part of the competition except the power round.
  1 KB (178 words) - 21:59, 21 December 2006
• 2002 IMO Shortlist Problems/N3
  ...<math> 2^{p_1} +1 </math> is clearly greater than 3 and cannot be a larger power of three, since this would require <math> p_1 \equiv 3 \pmod{6} </math>. T ...is not the square of the latter (i.e., that the former is either a higher power of the latter, or some other prime divides the latter, either of which impl
  10 KB (1,739 words) - 06:38, 12 November 2019
• ELMO
  As suggested by the names, the PSC holds all power, relegating the jury to a purely advisory role. This system may appear olig
  4 KB (696 words) - 00:56, 9 June 2024

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