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• Pythagorean triple
  <math>(3, 4, 5)</math><nowiki>*</nowiki> ...m^2 + n^2)</math> or <math>(2mn, m^2-n^2, m^2+n^2)</math>, where <math>m > n</math> are relatively prime positive integers of different [[parity]].
  4 KB (684 words) - 16:45, 1 August 2020
• 2003 AMC 12A Problems/Problem 1
  The first <math>2003</math> even counting numbers are <math>2,4,6,...,4006</math>. Thus, the problem is asking for the value of <math>(2+4+6+...+4006)-(1+3+5+...+4005)</math>.
  3 KB (436 words) - 20:31, 28 December 2021
• 2003 AMC 12A Problems/Problem 8
  ...ac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math ...square, exactly half of the positive factors will be less than <math>\sqrt{n}</math>.
  2 KB (326 words) - 15:40, 19 August 2023
• Perfect power
  ...to be a ''perfect <math>k</math>th power''. For example, <math>64 = 8^2 = 4^3 = 2^6</math>, so <math>64</math> is a perfect <math>2</math>nd, <math>3</
  870 bytes (148 words) - 16:52, 18 August 2013
• 1987 IMO Problems/Problem 3
  |a_1x_1 + a_2x_2 + \cdots + a_nx_n| \le \frac{ (k-1) \sqrt{n} }{ k^n - 1 } ...ath> \frac{ (k-1)\sqrt{n} }{k^n - 1} </math>. But since there are <math>k^n </math> such sums, by the [[pigeonhole principle]], two must fall into the
  2 KB (349 words) - 04:36, 28 May 2023
• 1987 IMO Problems
  ...h> be the number of permutations of the set <math>\{ 1, \ldots , n \} , \; n \ge 1 </math>, which have exactly <math>k </math> fixed points. Prove that \sum_{k=0}^{n} k \cdot p_n (k) = n!
  3 KB (459 words) - 14:24, 17 September 2023
• 2002 IMO Shortlist Problems/N3
  ...w that <math> 2^{p_{1}p_{2} \cdots p_{n}} + 1 </math> has at least <math>4^n </math> divisors. ...ime; hence <math>(2^u +1)(2^v + 1)/3 </math> has at least <math>2 \cdot 4^{n-1} </math> factors, by the inductive hypothesis.
  10 KB (1,739 words) - 06:38, 12 November 2019
• Mock AIME 4 2006-2007 Problems/Problem 1
  ...accomplish this by writing 169 four-digit numbers for a total of <math>321+4(169)=997</math> digits. The last of these 169 four-digit numbers is 1168, {{Mock AIME box|year=2006-2007|n=4|before=First question|num-a=2|source=125025}}
  1 KB (149 words) - 23:41, 22 April 2010
• Mock AIME 4 2006-2007 Problems/Problem 7
  ...math> such that <math>N\equiv n\pmod{100}</math> so that <math>3^N\equiv 3^n\pmod{1000}</math>. ...th>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we have the following:
  1 KB (127 words) - 00:15, 5 January 2010
• Mock AIME 4 2006-2007 Problems/Problem 10
  <math>\sum_{k=0}^{n} {n \choose k} =2^n</math>, and <math>\sum_{k=0}^{3n} {3n \choose k} =2^{3n}</math> ...binomial. So we divide the whole sum by 3 and we add or subtract <math>q(n)</math> to correct for the integer based on the modularity of the sum with
  4 KB (595 words) - 12:14, 25 November 2023
• Mock AIME 4 2006-2007 Problems/Problem 15
  ...ime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. ...</math>. By the [[quadratic formula]], <math>MD = \frac{2R + \sqrt{4R^2 - 4\cdot24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>.
  3 KB (532 words) - 20:29, 31 August 2020
• 2020 AMC 10A Problems
  <cmath>x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?</cmath> <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Asymptote: Reference
  ...math>t</math> ranges from <math>0</math> to the number of nodes (say <math>n</math>) that determine the path. The most basic way to make paths is by jo ...tells Asymptote to form a cyclic path by joining the endpoint with <math>t=n</math> to that with <math>t=0</math>.
  7 KB (1,205 words) - 21:38, 26 March 2024
• Asymptote: Useful commands and their Output
  draw((-1,0)--(4,0),EndArrow(5)); draw((0,-1)--(0,4),EndArrow(5)); label("Interest rate", (4,0), E); label("Money supply", (0,4), N);
  6 KB (871 words) - 21:14, 12 June 2023
• Power's of 2 in pascal's triangle
  1 4 6 4 1 Remember that <math>\binom{n}{r}=\frac{n!}{k!(n-k)!}</math> where <math>n \ge r</math>.
  2 KB (341 words) - 16:57, 16 June 2019
• Asymptote (geometry)
  ...n</math>; thus the functions is undefined at <math>x=\frac{\pi}{6} + \frac{n\pi}{3}</math>. [[File:Slantasymptote.png|thumb|500px|The function <math>y=\tfrac{x^2+2x+4} {x+1}</math> has a slant asymptote at <math>y=x+1</math> ]]
  4 KB (664 words) - 11:44, 8 May 2020
• Pigeonhole Principle/Solutions
  Also, note that it is possible to pull out 4 socks without obtaining a pair. ...re exist distinct <math>a, b \in S</math> such that <math>a \equiv b \pmod n</math>, as desired.
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2005 IMO Shortlist Problems/A1
  ...t for <math>|z| \ge 2 </math>, <math>n \ge 2 </math> (<math> n \in \mathbb{N} </math>), we have ...\frac{|z|^n -1}{|z|-1} = \sum_{i=0}^{n-1}|\pm z|^i \ge \left| \sum_{i=0}^{n-1} \pm z^i \right|
  2 KB (346 words) - 08:56, 14 May 2024
• 2006 Canadian MO Problems/Problem 1
  ...(3,6)=1</math>, and <math>f(3,4)=6</math>. Evaluate <math>f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)</math>.
  447 bytes (64 words) - 11:32, 30 November 2023
• 2000 AMC 12 Problems/Problem 7
  ...1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math> If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3<
  616 bytes (86 words) - 23:49, 29 December 2023

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