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• Reader's Digest National Word Power Challenge
  The '''Reader's Digest National Word Power Challenge''' is the first nationwide vocabulary competition for middle scho ...dependencies thereof, who have not previously won a scholarship from Word Power, can participate.
  2 KB (293 words) - 21:13, 20 March 2008
• Bertrand's Postulate
  Note that the power with which a prime <math>p</math> satisfying <math>\frac{2n}3<p\le n</math>
  2 KB (309 words) - 21:43, 11 January 2010
• Perfect square
  * [[Perfect power]]
  954 bytes (155 words) - 01:14, 29 November 2023
• Uncountable
  ...th>|S|<|\mathcal{P}(S)|</math>, where <math>\mathcal{P}(S)</math> is the [[power set]] of <math>S</math>. First, we note that the [[Cantor set]] <math>\math
  2 KB (403 words) - 20:53, 13 October 2019
• Subset
  The set of all subsets of a given set <math>S</math> is called the [[power set]] of <math>S</math> and is denoted <math>\mathcal{P}(S)</math> or <math
  1 KB (217 words) - 09:32, 13 August 2011
• 2006 AIME I Problems
  ...tive integer <math> x, </math> let <math> g(x) </math> denote the greatest power of 2 that divides <math> x. </math> For example, <math> g(20)=4 </math> and
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 13
  ...ve integer]] <math> x </math>, let <math> g(x) </math> denote the greatest power of 2 that [[divisor | divides]] <math> x. </math> For example, <math> g(20) ...ath> that divides <math>n+1</math>. Thus by the above formula, the highest power of <math>2</math> that divides <math>S_n</math> is <math>2^{k+n-1}</math>.
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 9
  The product of <math>a^2</math> and <math>r^{11}</math> is a power of 2. Since both numbers have to be integers, this means that <math>a</mat ...ly not possible. Thus the only restriction r has is that it must be an odd power of 2, so <math>2^{1}</math>, <math>2^{3}</math>, <math>2^{5}</math> .... al
  4 KB (651 words) - 18:27, 22 May 2021
• 2006 AIME I Problems/Problem 4
  A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and ...h> - every <math>n!</math> term for <math>n\geq25</math> has an additional power of <math>5</math> dividing it, for <math>76</math> extra; every n! for <mat
  2 KB (278 words) - 08:33, 4 November 2022
• 2006 AMC 12B Problems/Problem 20
  ...eq 0</math> is the lowest value such that <math>4x</math> becomes a higher power of 10.
  3 KB (485 words) - 14:09, 21 May 2021
• 2006 AMC 12B Problems/Problem 22
  The power of <math>10</math> for any factorial is given by the well-known algorithm ...of <math>2</math> that divides <math>n!</math> is larger or equal than the power of <math>5</math> which divides
  5 KB (881 words) - 15:52, 23 June 2021
• 2006 AMC 12A Problems/Problem 17
  ...> is an isosceles right triangle. Thus <math>DG = r\sqrt{2}</math>. By the Power of a Point Theorem,
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 12A Problems/Problem 24
  ...rent expression is irreducible as each term has a different <math>x</math> power. Thus, when we write <math>a</math> and <math>b</math> back to their origin
  8 KB (1,332 words) - 17:37, 17 September 2023
• 2005 AMC 12B Problems/Problem 23
  We can then put <math>x+y</math> to the third power or <math>(x+y)^{3}=10^{3z}</math>. Basic polynomial multiplication shows us
  5 KB (786 words) - 11:36, 19 May 2024
• Lassiter Invitational
  ...consists of a multiple-choice test, ten ciphering questions, and a pair of power questions, i.e., more in-depth questions on which members of teams collabor
  1 KB (161 words) - 18:35, 25 November 2007
• Power of a Point Theorem/Introductory Problem 1
  Applying the [[Power of a Point Theorem]], we get <math> 3\cdot(3+5) = x (x+10) \rightarrow x^2 ''Back to the [[Power of a Point Theorem]].''
  448 bytes (67 words) - 15:15, 23 March 2020
• Power of a Point Theorem/Introductory Problem 2
  Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = ''Back to the [[Power of a Point Theorem]].''
  289 bytes (45 words) - 13:14, 16 July 2017
• Power of a Point Theorem/Introductory Problem 3
  From the Power of a Point Theorem, we have that ''Back to the [[Power of a Point Theorem]].''
  969 bytes (154 words) - 14:40, 3 July 2006
• Power of a Point Theorem/Introductory Problem 4
  ...3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math>, or <math> ''Back to the [[Power of a Point Theorem]].''
  1 KB (177 words) - 02:14, 26 November 2020
• Power of a Point Introductory Problem 3
  #REDIRECT [[Power of a Point Theorem/Introductory Problem 3]]
  61 bytes (8 words) - 12:07, 10 July 2006

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