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contest_2011-09-27 [2011/09/25 12:48] – jtkorbcontest_2011-09-27 [2011/09/27 11:00] (current) – jtkorb
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   * C: Observe that, if ''k'' is even, ''n ^ k mod p'' is ''(n ^ (k/2) mod p) ^ 2 mod p''.  Generalize and handle the case for ''k'' odd.   * C: Observe that, if ''k'' is even, ''n ^ k mod p'' is ''(n ^ (k/2) mod p) ^ 2 mod p''.  Generalize and handle the case for ''k'' odd.
   * D: Combine the two preludes.   * D: Combine the two preludes.
-  * E: The GCD algorithm given in the text solves this problem directly. +  * E: The Extended Euclid Algorithm for GCD (given in the text), given in the text, solves this problem directly. 
-  * F:+  * F: Brute force factorization is fine.  You'll want to keep a table of ''(p,e)'' pairs for each prime, ''p'', that divides ''e'' times.
   * G: Theorem: The highest power of ''p'' that divides ''n!'' is ''n/p + n/p^2 + n/p^3 ...''.  So, ''p^e'' divides ''n!'' if this sum is at least ''e'' before ''n/p^k'' reaches 0.   * G: Theorem: The highest power of ''p'' that divides ''n!'' is ''n/p + n/p^2 + n/p^3 ...''.  So, ''p^e'' divides ''n!'' if this sum is at least ''e'' before ''n/p^k'' reaches 0.
-  * H: Use the Diophantine solution method in the text.+  * H: The text suggests using the Diophantine solution method given there.  Note that you don't actually need to compute the solution, just determine whether or not a solution exists.
contest_2011-09-27.1316980139.txt.gz · Last modified: 2011/09/25 12:48 by jtkorb