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  • ...have an inverse, then for some distinct <math>b</math> and <math>c</math> modulo <math>p</math>, <math>ab \equiv ac \pmod{p}</math>, so that <math>a(b-c)</m Let <math>p</math> be a prime. Consider the [[field]] of integers modulo <math>p</math>. By [[Fermat's Little Theorem]], every nonzero element of t
    4 KB (639 words) - 00:53, 2 February 2023
  • ''Let <math>S = {1, 2, 3, \ldots, 2n}</math>. Show that if we choose <math>n+1</math> numbers from < === Example 3 ===
    11 KB (1,986 words) - 18:13, 19 June 2024
  • ...f the product of the elements of <math>S</math> with <math>a</math>, taken modulo <math>p</math>, is simply a permutation of <math>S</math>. In other words, Cancelling the factors <math>1, 2, 3, \ldots, p-1</math> from both sides, we are left with the statement <math>a
    16 KB (2,660 words) - 22:42, 28 August 2024
  • === Divisibility Rule for 3 and 9=== ...or 9, respectively. Note that this does ''not'' work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899
    10 KB (1,572 words) - 21:11, 22 September 2024
  • <math>6 = 3 + 3</math> <math>8 = 3 + 5</math>
    7 KB (1,206 words) - 20:17, 28 December 2024
  • ...>m\neq 0</math>. We say that <math>a</math> is a '''quadratic residue''' [[modulo]] <math>m</math> if there is some integer <math>n</math> so that <math>n^2- ...if }\ p\nmid a\ \mathrm{ and }\ a\ \mathrm{\ is\ a\ quadratic\ nonresidue\ modulo\ }\ p. \end{cases}</math>
    5 KB (778 words) - 12:10, 29 November 2017
  • <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math> ...The same is true in any other [[modulus]] (modular arithmetic system). In modulo <math>5</math>, we [[counting | count]]
    16 KB (2,410 words) - 13:05, 3 January 2025
  • ...<math>n</math> is a perfect square [[iff]] it is a [[quadratic residue]] [[modulo]] all but finitely [[prime]]s. Perfect square trinomials are a type of quadratic equation that have <math>3</math> terms and contain <math>1</math> unique root.
    954 bytes (155 words) - 00:14, 29 November 2023
  • ...0</math>, we say that <math>a</math> is ''congruent to'' <math>b</math> ''modulo'' <math>n</math>, or <math>a \equiv b</math> (mod <math>n</math>), if the d ...This relation gives rise to an algebraic structure called '''the integers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <
    14 KB (2,317 words) - 18:01, 29 October 2021
  • ...at <math>i</math> for some <math>i \leq 2006</math>. We will do this with 3 lemmas. ...s <math>0</math>, then <math>\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})</math>.
    6 KB (1,037 words) - 22:32, 9 November 2024
  • The value of <math>\theta_0</math> only matters [[modulo]] <math>2\pi</math>. Since <math>\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{ == Solution 3 ==
    5 KB (822 words) - 04:55, 3 January 2025
  • <cmath>m | a+(n+2)d-gr^{n+2} \; (3),</cmath> ...om (1) and (2), we have <math>m | d-gr^{n+1}+gr^n</math> and from (2) and (3), we have <math>m | d-gr^{n+2}+gr^{n+1}</math>. Reinterpreting both equatio
    5 KB (883 words) - 00:05, 2 June 2024
  • ...ith the binary string 0001001000 <math>y</math> is 6 and <math>x</math> is 3 (note that it is zero indexed). ...Thus, <math>2^{3}, 2^{9}, 2^{15},</math> etc. work; or, <cmath>y-x \equiv 3 \pmod 6</cmath>
    8 KB (1,283 words) - 18:19, 8 May 2024
  • === Solution 3 (cheap and quick) === === Solution 3===
    3 KB (361 words) - 19:20, 14 January 2023
  • ...+ 6n</math> for nonnegative <math>n</math> are odd composites. We now have 3 cases: ...can also be expressed using case 1, and if <math>x = 40</math>, using case 3. <math>38</math> is the largest even integer that our cases do not cover. I
    8 KB (1,365 words) - 14:38, 10 December 2024
  • P(3)&=\frac13(1-P(2))&&=\frac29, \ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \
    19 KB (3,128 words) - 20:38, 23 July 2024
  • We can also take this equation modulo <math>9</math>; note that <math>m \equiv a+b+c \pmod{9}</math>, so ...h>m \equiv 358</math> mod <math>666</math>. We see that there are no other 3-digit integers that are <math>358</math> mod <math>666</math>, so <math>m =
    3 KB (565 words) - 15:51, 1 October 2023
  • Taking the given equation modulo <math>2,3,</math> and <math>5,</math> respectively, we have n^5&\equiv0\pmod{3}, \
    6 KB (874 words) - 14:50, 20 January 2024
  • == Solution 3 (Only sine and cosine sum formulas) == ...9 x \equiv 141 \pmod{180}</math>. Note that the inverse of <math>19</math> modulo <math>180</math> is itself as <math>19^2 \equiv 361 \equiv 1 \pmod {180}</m
    5 KB (757 words) - 20:59, 23 December 2024
  • <cmath>\begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\
    3 KB (493 words) - 12:51, 22 July 2020

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