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contest_2011-09-27 [2011/09/25 12:48]
jtkorb
contest_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