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• Resources for mathematics competitions
  ==== Introductory Problem Solvers ==== ...n.artofproblemsolving.com/resources/articles/crt.pdf The Chinese Remainder Theorem] by Evan Chen
  24 KB (3,269 words) - 00:43, 24 April 2024
• Number theory/Olympiad
  *** [[Chinese Remainder Theorem]] ** [[Euler's Totient Theorem]]
  734 bytes (79 words) - 18:27, 25 April 2008
• Number theory/Intermediate
  ...ntermediate level study of [[number theory]] extends many of the topics of introductory number theory, but infuses [[mathematical problem solving]] as well as [[al *** [[Chinese Remainder Theorem]]
  1,016 bytes (108 words) - 21:05, 26 January 2016
• Fermat's Little Theorem
  ...y level if they have a hard time following the rest of this article). This theorem is credited to [[Pierre de Fermat]]. ...{p}</math>. As you can see, it is derived by multipling both sides of the theorem by <math>a</math>. The restated form is nice because we no longer need to
  16 KB (2,675 words) - 10:57, 7 March 2024
• 2005 PMWC Problems/Problem I9
  Applying the [[Chinese Remainder Theorem]], we can eventually find that <math>n \equiv 1735 \pmod{3465}</math>. (We [[Category:Introductory Number Theory Problems]]
  1 KB (202 words) - 19:07, 10 March 2015
• 1995 AHSME Problems/Problem 27
  ...n)</math> denote the sum of the numbers in row <math>n</math>. What is the remainder when <math>f(100)</math> is divided by 100? ...20 \cdot 5} - 2 \equiv -1 \pmod{25}</math>, and by the [[Chinese Remainder Theorem]], we have <math>f(100) \equiv 74 \pmod{100} \Longrightarrow \mathrm{(E)}</
  5 KB (682 words) - 09:45, 18 February 2022
• Characteristic polynomial
  By the [[Hamilton-Cayley Theorem]], the characteristic polynomial of a square matrix applied to the square m .../math>s that can be solved for each constant. Refer to the [[#Introductory|introductory problems]] below to see an example of how to do this. In particular, for th
  19 KB (3,412 words) - 14:57, 21 September 2022
• 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1
  ...ath>2017^{2017}\mod 100=17^{2017}\mod 100</math>. By the Chinese remainder theorem, we can find <math>17^{2017}\mod 25</math> and <math>17^{2017}\mod 4</math> [[Category:Introductory Number Theory Problems]]
  1 KB (181 words) - 12:09, 21 August 2022
• 2020 AMC 10A Problems/Problem 24
  ...sible by <math>21</math>, we have to add <math>14</math> to cancel out the remainder. (Note that we don't subtract <math>7</math> to get to <math>53</math>; <ma == Solution 11 (Chinese Remainder Theorem) ==
  17 KB (2,544 words) - 12:09, 1 September 2023