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- ...h>\{x\} = x - \lfloor x \rfloor</math> denote the fractional part of <math>x </math>. ...ith <math>m_r>n_r</math>, where <math>n_r</math> denotes the row <math>sn \mod p</math> is in. However, since the values of each entry decreases while <m3 KB (506 words) - 17:54, 22 June 2023
- Let <math>a</math> be an element of <math>R</math>. Then the mapping <math>x \mapsto ax</math> of <math>R</math> into <math>R</math> is an [[endomorphis ...isor of <math>x</math>, or <math>y</math> divides <math>x</math>, or <math>x</math> is a multiple of <math>y</math>.6 KB (994 words) - 06:16, 8 April 2015
- ...{(p-1)/2} = 1</math> if and only if <math>-1</math> is a quadratic residue mod <math>p</math>. ...ath>p</math>. Then <math>k^2 = -1</math>, for some residue <math>k</math> mod <math>p</math>, so7 KB (1,182 words) - 16:46, 28 April 2016
- ...> is the [[floor function | greatest integer less than or equal to]] <math>x</math>. If for fixed <math>n</math>, there exists an integer <math>0\le y < ...has a multiplicative [[inverse with respect to an operation | inverse]] [[mod]] <math>p</math>, <math>f_p(n)</math> is the member of the [[set]] <math>\{2 KB (340 words) - 15:52, 3 April 2012
- A '''Linear Congruence''' is a [[congruence]] [[modular arithmetic|mod]] p of the form ...h>b</math>, <math>c</math>, and <math>p</math> are [[constant]]s and <math>x</math> is the [[variable]] to be solved for.1 KB (225 words) - 15:52, 3 August 2022
- ...<math>m</math> with a residue class mod <math>n</math> is a residue class mod <math>mn</math>. Suppose you wish to find the least number <math>x</math> which leaves a remainder of:6 KB (1,022 words) - 14:57, 6 May 2023
- Suppose <math>f(m) = n</math> with <math>m \equiv n \mod k</math>. Then by an easy induction on <math>r</math> we find <math>f(m + k ...1 \pmod k</math>. Hence if <math>m</math> has a different residue <math>r \mod k</math>, then <math>f(m)</math> cannot have residue <math>r_1</math> or <m5 KB (923 words) - 19:51, 21 January 2024
- ...math>u </math> and <math>v </math> are divisible by half the order of 2 in mod <math>t </math>, contradicting our assumption that <math>u </math> and <mat ...}-1}(-x)^{i} </math> has remainder <math>p_k </math> when divided by <math>x+1 </math>. But <math>p_k </math> cannot divide <math>2^q + 1 </math>, as t10 KB (1,739 words) - 06:38, 12 November 2019
- Now, we sort the pigeons <math>\{x\}, \{2x\}, \hdots, \{nx\}</math> into the holes <math>(0, 1/n)), (1/n, 2/n) Then we have that <math>\{(j - i)x\}</math> must fall into the first or last hole, contradiction.10 KB (1,617 words) - 01:34, 26 October 2021
- ...} + (2005)(2004)x^2(x + 1)^{2003} = \sum_{k = 0}^{2005} k^2{2005 \choose k}x^k</math>. Now note that <math>f\left(\dfrac{1}{2}\right) = S</math>. Then w ...) \implies S=2005 \cdot \dfrac{2007}{9}=2005 \cdot 223 \equiv 115 \ (\text{mod} \ 1000)</cmath>3 KB (502 words) - 14:53, 19 July 2020
- In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that t ...and <math>f=2+4</math> where <math>x+y</math> indicates that columns <math>x</math> and <math>y</math> are shaded. From our condition on the columns, we13 KB (2,328 words) - 00:12, 29 November 2023
- ...ghtarrow y = \frac{2007 - 5x}{4}</math>. Substituting, we find that <math>(x - 1)\left(-\frac 54x + \frac{2003}4\right)</math>. ...{-b}{2a}</math> to find the maximum possible value of the quadratic: <math>x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201</math>. However3 KB (399 words) - 21:17, 24 February 2021
- ...od{5} </math>. Since there is an odd digit in each of the residue classes mod 5, <math>k </math> exists and the induction is complete. ...he same residue mod <math> 5^n</math>, and therefore one of them must be 0 mod <math> 5^n</math>.4 KB (736 words) - 22:17, 3 March 2023
- For all integers <math> \displaystyle x </math> we have ...quiv x^3 - (d+e+f)x^2 + (de+ef+fd)x - def </math> <math> \equiv (x-d)(x-e)(x-f) \pmod{S} </math>,2 KB (271 words) - 09:13, 29 August 2011
- | <math>a^{i+1}_3</math>||a^{i+1}_3||<math>x^{3^2}</math>||x^{3^2} ...code> produces <math>x^10</math>, while <code>x^{10}</code> produces <math>x^{10}</math>.12 KB (1,894 words) - 21:24, 7 June 2024
- 0\mod{m}&1\mod{m}&2\mod{m}&...&...&...&(m-1)\mod{m}\\ ...<math>A_{N} = \{(x+kn,y-km) \;:\; k \in \mathbb{Z}\}</math> for any <math>(x,y) \in A_{N}</math>.17 KB (2,748 words) - 19:22, 24 February 2024
- :which has as its only solution satisfying the constraints <math>x = 8</math>, <math>y = 0</math>. Case 3: <math>n < 2000</math>, <math>n = \overline{19xy}</math>, <math>x + y \geq 10</math>15 KB (2,558 words) - 19:33, 4 February 2024
- ...6) \mid 2^6=63</math>, and the only numbers equivalent to <math>-11</math> mod 30 which do not exceed 63 in absolute value are <math>-41, -11, 19, 49</mat ...gers which do not exceed 63 in absolute value and which are congruent to 1 mod 30 are <math>-59, -29, 1, 31, 61</math>. Of these, only 1 is a divisor of5 KB (919 words) - 23:29, 20 January 2016
- ...If <math>3</math> is more efficient than <math>x</math>, then <math>x^3<3^x</math>. We try to prove that all integers greater than 3 are less efficient ...)^3}{x^3}</math>, and we must prove that this is less than 3 for all <math>x</math> greater than 3.4 KB (668 words) - 04:29, 2 January 2023
- Let <math>(x,y)=(a_1,b_1)</math>. Then, we can begin to list out terms as follows: <math>(a_2,b_2)=(x\sqrt{3}-y,y\sqrt{3}+x)</math>5 KB (745 words) - 10:58, 9 December 2022
