Finde alle Paare ´(x,y)´ von ganzen Zahlen, für die ´x >= 0´, ´y >= 0´ sowie ´17x + 13y = 405´ gilt.

Approach

´17x + 13y = 405´ ´17x + 13y = 405 (mod 13)´ ´17x = 405 (mod 13)´ ´4x = 2 (mod 13)´ ´x = 7 (mod 13)´

´x = 13z + 7 (z in ZZ)´

´17(13z + 7) + 13y = 405´ ´17 * 13z + 119 + 13y = 405´ (´-119´) ´17 * 13z + 13y = 286´ (´/13´) ´17z + y = 22´ ´y = 22 - 17z´

Lösung: Alle Paare ´(13z + 7, 22 - 17z)´ mit ´z in ZZ´

´x >= 0´, ´y >= 0´

´13z + 7 >= 0 <=> z >= -7/15´ ´22 - 17z >= 0 <=> z <= 22/17´

´=> -7/13 <= z <= 22/17´ mit ´z in ZZ´ ´=> z = 0, z = 1´

Lösungen: ´(7, 22)´, ´(20, 5)´


Solution

  • ´(7, 22)´, ´(20, 5)´

  • URL:
  • Language:
  • Subjects: math
  • Type: Calculate
  • Duration: 25min
  • Credits: 4
  • Difficulty: 0.5
  • Tags: hpi congruence modulo
  • Note:
    HPI, 2014-06-10, Mathe 2, Aufgabe 38
  • Created By: adius
  • Created At:
    2014-07-26 16:54:48 UTC
  • Last Modified:
    2014-07-26 16:54:48 UTC