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• Relatively prime
  ...n. The positive integers <math>m</math> and <math>n</math> are relatively prime if and only if <math>\frac{m}{n}</math> is in lowest terms. Relatively prime numbers show up frequently in [[number theory]] formulas and derivations:
  2 KB (245 words) - 15:51, 25 February 2020
• Fermat's Little Theorem
  If <math>{a}</math> is an [[integer]], <math>{p}</math> is a [[prime number]] and <math>{a}</math> is not [[divisibility|divisible]] by <math>{p ...ler's totient function]]. In particular, <math>\varphi(p) = p-1</math> for prime numbers <math>p</math>. In turn, this is a special case of [[Lagrange's The
  16 KB (2,660 words) - 23:42, 28 August 2024
• Goldbach Conjecture
  ...] stating that every [[even integer]] greater than two is the sum of two [[prime number]]s. The conjecture has been tested up to 400,000,000,000,000. He considered 1 to be a prime number, a [[mathematical convention|convention]] subsequently abandoned. So
  7 KB (1,201 words) - 16:59, 19 February 2024
• Quadratic residues
  ...e modulo <math>m</math> is easiest if <math>m=p</math> is a [[prime number|prime]]. In this case we write <math>\left(\frac{a}{p}\right)=\begin{cases} 0 & \ ...e above are properties of the Legendre symbol, they still hold for any odd coprime integers <math>p</math> and <math>q</math> when using the Jacobi symbol def
  5 KB (778 words) - 13:10, 29 November 2017
• Mathematicial notation
  <math>\mathbb{P}</math>: Also an ambiguous notation, use for the positive [[prime]]s or the positive integers. ...ath>: the residues <math>\mod n</math> (a ring; a field for <math>n</math> prime).
  8 KB (1,401 words) - 13:11, 17 June 2008
• 1985 AIME Problems/Problem 13
  Because <math>n</math> and <math>2n+1</math> will be coprime, the only thing stopping the GCD from being <math>1</math> is <math>n-200.< ...hm on <math>x</math> and <math>2x+1</math> yields that they are relatively prime. Thus, the only way the GCD will not be 1 is if the<math> x-200</math> term
  4 KB (671 words) - 20:04, 6 March 2024
• 2000 AIME I Problems/Problem 11
  ...<math>a/b,</math> where <math>a</math> and <math>b</math> are [[relatively prime]] positive [[divisor]]s of <math>1000.</math> What is the [[floor function| ...them in quantities to reach our answer. First, separate the fractions with coprime parts into those that are combinations of powers of 2 and 5, and those that
  4 KB (667 words) - 13:58, 31 July 2020
• 1959 IMO Problems/Problem 1
  So it follows that <math>7n+1</math> and <math>14n+3</math> must be coprime for every natural number <math>n</math> for the fraction to be irreducible. ...ath>7n+1</math> by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that <math> \frac{21n+4}{14n+3}</math> is
  5 KB (767 words) - 10:59, 23 July 2023
• 2007 iTest Problems
  ...math>. What is the arithmetic mean of the two primes in the smallest twin prime pair? ...be the number of positive integers <math>k< n</math> which are relatively prime to <math>n</math>. For how many distinct values of <math>n</math> is <math>
  30 KB (4,794 words) - 23:00, 8 May 2024
• Chinese Remainder Theorem
  Let <math>m</math> be [[relatively prime]] to <math>n</math>. Then each [[residue class]] mod <math>mn</math> is eq ...}</math> , <math>d_{2}</math> , ... <math>d_{n}</math> are all relatively prime.
  6 KB (1,022 words) - 14:57, 6 May 2023
• Mock AIME 3 Pre 2005 Problems
  ...positive integers and <math>r</math> is not divisible by the square of any prime. Determine <math>p + q + r</math>. ...ac{p\pi}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Determine <math>p + q</math>.
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 14
  ...math> are coprime and <math>q</math> is not divisible by the square of any prime. Determine <math>p + q + r</math>.
  3 KB (563 words) - 02:05, 25 November 2023
• Mock AIME 4 2005-2006/Problems
  ...2006}}{V_1}=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find the remainder when <math>m+n</math> is divided by < ...<math>f(x)=\frac{m}{n}</math> where <math>m</math> and <math>n</math> are coprime positive integers, find <math>m^2+n^2</math>.
  7 KB (1,176 words) - 04:44, 26 February 2007
• Chicken McNugget Theorem
  ...m''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <math>m,n</math>, the greatest integer that cannot ..._{2}) = n(y_{2}-y_{1})</math>. Since <math>m</math> and <math>n</math> are coprime and <math>m</math> divides <math>n(y_{2}-y_{1})</math>, <math>m</math> divi
  17 KB (2,748 words) - 15:53, 11 August 2024
• 2008 USAMO Problems/Problem 1
  ...at for each positive integer <math>n</math>, there are pairwise relatively prime integers <math>k_0, k_1 \dotsc, k_n</math>, all strictly greater than 1, su ...ep, suppose that <math>k_0, \dotsc, k_{n-1}</math> are pairwise relatively prime integers such that
  11 KB (1,964 words) - 03:38, 17 August 2019
• 1972 USAMO Problems/Problem 1
  Consider an arbitrary prime <math>p</math>. Let <math>p^\alpha</math>, <math>p^\beta</math>, and <math> ...e is 2\alpha -(<math>\alpha + \beta + \alpha) = -\beta</math>. Since every prime has the same power in both expressions, the expressions are equal. <math>\b
  5 KB (1,018 words) - 11:14, 6 October 2023
• 2010 AMC 12A Problems/Problem 20
  The prime factorization of <math>2010</math> is <math>2\cdot{3}\cdot{5}\cdot{67}</mat ...ximize <math>n-1</math>. Therefore <math>k</math> and <math>i</math> are [[coprime]] and <math>n-1</math> is the [[Greatest common factor|GCF]] of any corresp
  5 KB (864 words) - 17:41, 26 August 2024
• Mobius function
  ...ers less than or equal to <math>n</math> and not divisible by any one of a coprime set of integers <math>a_1,a_2,\ldots</math> is ...of integers <math>a_1,a_2,\ldots</math> to be the different (and distinct) prime factors of <math>n</math> we get
  5 KB (910 words) - 02:58, 1 March 2022
• 1997 USAMO Problems/Problem 1
  Let <math>p_1,p_2,p_3,...</math> be the prime numbers listed in increasing order, and let <math>x_0</math> be a real numb ...e integers. Let <math>x_0 = \frac{m}{n}</math>, where <math>m,n</math> are coprime positive integers. Since <math>0<x_0<1</math>, <math>0<m<n</math>. Now <cma
  2 KB (348 words) - 13:14, 13 February 2015
• 2012 AMC 10A Problems/Problem 17
  Let <math>a</math> and <math>b</math> be relatively prime positive integers with <math>a>b>0</math> and <math>\dfrac{a^3-b^3}{(a-b)^3 Since <math>a</math> and <math>b</math> are relatively prime, <math>a^3-b^3</math> and <math>(a-b)^3</math> are both integers as well. T
  6 KB (1,028 words) - 20:24, 14 June 2024

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