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====== Weekly Contest for September 27, 2011 ======

===== Join the Contest =====

   % cd your-pc2-directory
   % /homes/cs390cp/pc2/bin/pc2team &

Log in using your individually assigned "teamX" account and password.  You can work in the current window/directory, creating a subdirectory for each problem.

===== Problems =====

  * A: [[http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=35&page=show_problem&problem=1051|Light, More Light]]
  * B: [[Prelude One to Carmichael Numbers -- Primality Testing]]
  * C: [[Prelude Two to Carmichael Numbers -- Efficient Power Mod]]
  * D: [[http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=35&page=show_problem&problem=947|Carmichael Numbers]]
  * E: [[http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=35&page=show_problem&problem=1045|Euclid Problem]]
  * F: [[Prelude to Factovisors]]
  * G: [[http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=35&page=show_problem&problem=1080|Factovisors]]
  * H: [[http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=35&page=show_problem&problem=1030|Repackaging]]

//Remember:// If you've already solved one or more of these problems, try (1) solving again without referring to your old solution, and/or (2) using a different language (Java or C++).  If you want to work on an additional problem from the book, let [[jtk@purdue.edu|me]] know.

**Hints:**

  * A: Find a pattern.  The solution is extremely short and efficient.
  * B: Repeated divisions is fine (be sure to skip even numbers after 2 and only test as large as necessary).
  * 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.
  * E: The Extended Euclid Algorithm for GCD (given in the text), given in the text, solves this problem directly.
  * 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.
  * 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.