1. Stelle die Dezimalzahl ´47 871_10´ in den Positionssystemen zur Basis ´2´, zur Basis ´9´ und zur Basis ´16´ dar. (Ziffern im Hexadezimalsystem: ´0, 1, 2‚ …, 9, A, B, …‚ F´)
  2. Gib die Zahlen ´1100 1110 0010_2´ und ´210 012_3´ im Dezimalsystem an. (Tipp: Horner-Schema)
  3. Gib die Dezimalzahlen ´42´ und ´-42´ in 8-Bit Zweierkomplementdarstellung an.
Approach

"Teil 1\n\ntxt\n47871 / 2 = 23935 Rest: 1\n23935 / 2 = 11967 Rest: 1\n11967 / 2 = 5983 Rest: 1\n 5983 / 2 = 2991 Rest: 1\n 2991 / 2 = 1495 Rest: 1\n 1495 / 2 = 747 Rest: 1\n 747 / 2 = 373 Rest: 1\n 373 / 2 = 186 Rest: 1\n 186 / 2 = 93 Rest: 0\n 93 / 2 = 46 Rest: 1\n 46 / 2 = 23 Rest: 0\n 23 / 2 = 11 Rest: 1\n 11 / 2 = 5 Rest: 1\n 5 / 2 = 2 Rest: 1\n 2 / 2 = 1 Rest: 0\n 1 / 2 = 0 Rest: 1\n\n\n´47 871\_10 = 1011 1010 1111 1111\_2´\n\n\ntxt\n47871 : 9 = 5319 Rest: 0\n 5319 : 9 = 591 Rest: 0\n 591 : 9 = 65 Rest: 6\n 65 : 9 = 7 Rest: 2\n 7 : 9 = 0 Rest: 7\n\n\n´47 871\_10 = 72 600\_9´\n\n\ntxt\n 1011 1010 1111 1111\n= B A F F\n\n\n´47 871\_10 = BAFF\_16´\n\n\nTeil 2\n\ntxt\n 1 = 1\n* 2 + 1 = 3\n* 2 6\n* 2 12\n* 2 + 1 = 25\n* 2 + 1 = 51\n* 2 + 1 = 103\n* 2 206\n* 2 412\n* 2 824\n* 2 + 1 = 1649\n* 2 = 3298\n\n\n´1100 1110 0010\_2 = 3 298\_10´\n\n\ntxt\n 2 = 2\n* 3 + 1 = 7\n* 3 21\n* 3 63\n* 3 + 1 = 190\n* 3 + 2 = 572\n\n\n´210 012_3 = 572_10´\n\n\nTeil 3\n\n´42´:\nBinär: ´101010´\nZweierkomplement: ´00101010´\n\n´-42´:\nZweierkomplement: ´11010101 + 1 = 11010110´"


Solution

  • Teil 1

    ´47 871_10 = 1011 1010 1111 1111_2´

    ´47 871_10 = 72 600_9´

    ´47 871_10 = BAFF_16´

    Teil 2

    ´1100 1110 0010_2 = 3 298_10´

    ´210 012_3 = 572_10´

    Teil 3

    ´42´: ´00101010´

    ´-42´: ´11010110´

  • URL:
  • Language: Deutsch
  • Subjects: math
  • Type: Transform
  • Duration: 30min
  • Credits: 3
  • Difficulty: 0.3
  • Tags: hpi horner's method
  • Note:
    HPI, 2014-04-07, Mathe 2, Blatt 1, Aufgabe 3
  • Created By: ad-si
  • Created At:
    2014-07-25 20:17:37 UTC
  • Last Modified:
    2014-07-25 20:17:37 UTC