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- ==== Introductory Problem Solvers ==== ...n.artofproblemsolving.com/resources/articles/crt.pdf The Chinese Remainder Theorem] by Evan Chen16 KB (2,192 words) - 22:06, 19 July 2024
- *** [[Chinese Remainder Theorem]] ** [[Euler's Totient Theorem]]734 bytes (79 words) - 17:27, 25 April 2008
- ...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]]1,015 bytes (104 words) - 16:21, 1 September 2024
- ...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 to16 KB (2,660 words) - 22:42, 28 August 2024
- 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) - 18:07, 10 March 2015
- ...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) - 08:45, 18 February 2022
- 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 th19 KB (3,412 words) - 13:57, 21 September 2022
- Additionally, one can use the [[Chinese Remainder Theorem]] to solve polynomial congruences of all moduli. ===Introductory===4 KB (609 words) - 15:17, 8 January 2025
- ...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) - 11:09, 21 August 2022
- ...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 (2,522 words) - 18:33, 31 August 2024