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• 2005 Alabama ARML TST Problems/Problem 14
  ...lutions we want to the equation. Raising the first equation to the fourth power gives us
  3 KB (478 words) - 23:41, 5 January 2014
• 2008 AIME I Problems
  ...< 1000</math>, and <math>m</math> is not divisible by the <math>n</math>th power of any prime. Find <math>m + n</math>.
  9 KB (1,536 words) - 00:46, 26 August 2023
• 2008 AIME I Problems/Problem 6
  ...ath>r</math> by <math>2^{r-1}</math> (we can do this because dividing by a power of 2 doesn't affect divisibility by <math>67</math>). The second row will b
  3 KB (509 words) - 17:21, 22 March 2018
• 2008 AIME I Problems/Problem 15
  ...m<1000</math>, and <math>m</math> is not divisible by the <math>n</math>th power of any prime. Find <math>m+n</math>.
  6 KB (1,041 words) - 00:54, 1 February 2024
• 2006 IMO Shortlist Problems/A5
  By the [[Power Mean Inequality]],
  3 KB (529 words) - 08:03, 29 March 2008
• 2006 IMO Shortlist Problems/C5
  ...yers in any minimal subtournament of an <math>(n,k)</math>-tournament is a power of 2. ...ple of <math>2^{t+1}</math> players, since each must have a size that is a power of 2, and no two players meet more than once. It follows that <math>2^{t+1
  4 KB (641 words) - 11:09, 30 March 2008
• Mock AIME 1 2007-2008 Problems/Problem 1
  ...e sum of the degrees of the power of each of the two terms must sum to the power being raised to, here <math>17</math>. However, <math>3 + 13 = 16</math>, c
  552 bytes (84 words) - 17:16, 2 April 2008
• Mock AIME 1 2007-2008 Problems/Problem 9
  ...th> upon division by <math>3</math>). The third condition implies that the power of each prime factor of <math>n</math> must be divisible by <math>5</math> ...th>n</math>, we want to just use the prime factors <math>2,3,5</math>. The power of <math>2</math> must be divisible by <math>3,5</math>, and <math>2^{15}</
  2 KB (323 words) - 00:32, 31 August 2015
• 2008 AIME II Problems/Problem 7
  ...> and <math>(t+r)</math> are roots of this polynomial, we know that (using power reduction)
  7 KB (1,251 words) - 19:18, 2 January 2024
• 2005 USAMO Problems/Problem 2
  We now apply the lifting the exponent lemma to examine the power of 3 that divides each side of the equation when <math>m > 0</math> to obta
  7 KB (1,053 words) - 10:38, 12 August 2015
• 2008 USAMO Problems/Problem 1
  Now, for <math>1 \le i \le n</math>, take <math>k_i</math> to be the greatest power of <math>p_i</math> that divides <math>a^2 + a +1</math>, and let <math>k_0
  11 KB (1,964 words) - 03:38, 17 August 2019
• 2008 USAMO Problems/Problem 4
  ...stinct powers of <math>2</math>, where <math>1= 2^0</math> is considered a power of <math>2</math>.
  8 KB (1,318 words) - 12:37, 20 April 2022
• 2008 USAMO Problems/Problem 6
  ...er of ways that the mathematicians may be split between the two rooms is a power of two (i.e., is of the form <math>2^k</math> for some positive integer <ma ...Therefore, the number of good configurations of the mathematicians was a power of two.
  13 KB (2,414 words) - 14:37, 11 July 2016
• 2008 USAMO Problems
  ...er of ways that the mathematicians may be split between the two rooms is a power of two (i.e., is of the form <math>2^k</math> for some positive integer <ma
  4 KB (674 words) - 21:48, 12 August 2014
• Mathleague.org
  ..., which is a proof-based problem set of around 10 problems for a team. The Power Round is scored out of 100, with possible partial credit on most problems. ...hool). The total points possible for a team is 400. At contests where the Power Round is taken, that score is also included in the total sum, making it out
  5 KB (801 words) - 12:47, 23 September 2023
• Mock ARML
  ...rom March to May of the year (some times including the full set with Team, Power, Individual, Relay, Super Relay, and Tiebreaker, other times with only Indi
  1 KB (153 words) - 11:44, 16 June 2019
• Nilpotent group
  ...rder subgroup generated by the <math>P_p</math> must be divisible by every power of a prime that divides <math>G</math>, but it must also divide <math>G</ma
  9 KB (1,768 words) - 17:55, 5 June 2008
• P-group
  ...up''' is a [[finite]] [[group]] whose [[order (group theory) |order]] is a power of a [[prime]] <math>p</math>.
  4 KB (814 words) - 22:50, 3 November 2023
• Sylow Theorems
  ...G</math> modulo <math>H</math>. Since the order of <math>K</math> is some power of <math>p</math>, say <math>p^s</math>, the size of each orbit must divide
  11 KB (2,071 words) - 12:25, 9 April 2019
• Poncelet
  ...wer]] named after [[Jean-Victor Poncelet]]. One poncelet is defined as the power required to raise a hundred-[[kilogram]] [[mass]] (quintal) at a [[velocity
  284 bytes (40 words) - 11:36, 6 June 2008

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