• Stellenwertsysteme
  • ttry-Katalog
  • 06.10.2020
  • Mathematik
  • 6, 5
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  • 1
    Übertrage die Zahlen aus dem Dezimalsystem ins Binärsystem.
    • 27=16+8+0+2+1=(11011)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 27 = \cloze{16} + \cloze{8} + \cloze{0} + \cloze{2} + \cloze{1} = ( \cloze{1 1 0 1 1} )_2
    • 121=64+32+16+8+0+0+1=(1111001)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 121 = \cloze{64} + \cloze{32} + \cloze{16} + \cloze{8} + \cloze{0} + \cloze{0} + \cloze{1} = ( \cloze{1 1 1 1 0 0 1} )_2
    • 8=8+0+0+0=(1000)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 8 = \cloze{8} + \cloze{0} + \cloze{0} + \cloze{0} = ( \cloze{1 0 0 0} )_2
    • 65=64+0+0+0+0+0+1=(1000001)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 65 = \cloze{64} + \cloze{0} + \cloze{0} + \cloze{0} + \cloze{0} + \cloze{0} + \cloze{1} = ( \cloze{1 0 0 0 0 0 1} )_2
    • 38=32+0+0+4+2+0=(100110)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 38 = \cloze{32} + \cloze{0} + \cloze{0} + \cloze{4} + \cloze{2} + \cloze{0} = ( \cloze{1 0 0 1 1 0} )_2
    • 15=8+4+2+1=(1111)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 15 = \cloze{8} + \cloze{4} + \cloze{2} + \cloze{1} = ( \cloze{1 1 1 1} )_2

    Natürliche Zahlen im Fünfersystem

    2
    Übertrage die Zahlen ins Dezimalsystem.
    • (33)5=351+350=15+3=18\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (3 3)_5 = \cloze{3} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{15} + \cloze{3} = \cloze{18}
    • (13)5=151+350=5+3=8\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 3)_5 = \cloze{1} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{5} + \cloze{3} = \cloze{8}
    • (20)5=251+050=10+0=10\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (2 0)_5 = \cloze{2} \cdot 5^\cloze{1} + \cloze{0} \cdot 5^\cloze{0} = \cloze{10} + \cloze{0} = \cloze{10}
    • (42)5=451+250=20+2=22\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (4 2)_5 = \cloze{4} \cdot 5^\cloze{1} + \cloze{2} \cdot 5^\cloze{0} = \cloze{20} + \cloze{2} = \cloze{22}
    • (43)5=451+350=20+3=23\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (4 3)_5 = \cloze{4} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{20} + \cloze{3} = \cloze{23}
    • (32)5=351+250=15+2=17\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (3 2)_5 = \cloze{3} \cdot 5^\cloze{1} + \cloze{2} \cdot 5^\cloze{0} = \cloze{15} + \cloze{2} = \cloze{17}

    Bei der ersten Ziffer sollte die 0 ausgeschlossen werden

    3
    Übertrage die Zahlen ins Dezimalsystem.
    • (100)5=152+051+050=25+0+0=25\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 0 0)_5 = \cloze{1} \cdot 5^\cloze{2} + \cloze{0} \cdot 5^\cloze{1} + \cloze{0} \cdot 5^\cloze{0} = \cloze{25} + \cloze{0} + \cloze{0} = \cloze{25}
    • (322)5=352+251+250=75+10+2=87\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (3 2 2)_5 = \cloze{3} \cdot 5^\cloze{2} + \cloze{2} \cdot 5^\cloze{1} + \cloze{2} \cdot 5^\cloze{0} = \cloze{75} + \cloze{10} + \cloze{2} = \cloze{87}
    • (401)5=452+051+150=100+0+1=101\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (4 0 1)_5 = \cloze{4} \cdot 5^\cloze{2} + \cloze{0} \cdot 5^\cloze{1} + \cloze{1} \cdot 5^\cloze{0} = \cloze{100} + \cloze{0} + \cloze{1} = \cloze{101}
    • (233)5=252+351+350=50+15+3=68\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (2 3 3)_5 = \cloze{2} \cdot 5^\cloze{2} + \cloze{3} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{50} + \cloze{15} + \cloze{3} = \cloze{68}
    • (124)5=152+251+450=25+10+4=39\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 2 4)_5 = \cloze{1} \cdot 5^\cloze{2} + \cloze{2} \cdot 5^\cloze{1} + \cloze{4} \cdot 5^\cloze{0} = \cloze{25} + \cloze{10} + \cloze{4} = \cloze{39}
    • (101)5=152+051+150=25+0+1=26\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 0 1)_5 = \cloze{1} \cdot 5^\cloze{2} + \cloze{0} \cdot 5^\cloze{1} + \cloze{1} \cdot 5^\cloze{0} = \cloze{25} + \cloze{0} + \cloze{1} = \cloze{26}

    Wenn man zwei- und dreistellige Fünferzahlen in einer Aufgabe kombinieren möchte (siehe nächste Aufgabe), ist das ein wenig komplizierter als im Dualsytem - die Variable #a1 ist hier für die dreistellige Ausgabe nicht verwendbar, da sonst an zweiter Stelle keine 0 stehen könnte; deshalb werden die Variablen #a10 und #b10 eingeführt.

  • Natürliche Zahlen im Binärsystem (Zweiersystem)

    4
    Übertrage die Zahlen ins Dezimalsystem.
    • (11)2=121+120=2+1=3\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1)_2 = \cloze{1} \cdot 2^\cloze{1} + \cloze{1} \cdot 2^\cloze{0} = \cloze{2} + \cloze{1} = \cloze{3}
    • (101)2=122+021+120=4+0+1=5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 0 1)_2 = \cloze{1} \cdot 2^\cloze{2} + \cloze{0} \cdot 2^\cloze{1} + \cloze{1} \cdot 2^\cloze{0} = \cloze{4} + \cloze{0} + \cloze{1} = \cloze{5}
    • (10)2=121+020=2+0=2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 0)_2 = \cloze{1} \cdot 2^\cloze{1} + \cloze{0} \cdot 2^\cloze{0} = \cloze{2} + \cloze{0} = \cloze{2}
    • (110)2=122+121+020=4+2+0=6\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1 0)_2 = \cloze{1} \cdot 2^\cloze{2} + \cloze{1} \cdot 2^\cloze{1} + \cloze{0} \cdot 2^\cloze{0} = \cloze{4} + \cloze{2} + \cloze{0} = \cloze{6}
    • (111)2=122+121+120=4+2+1=7\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1 1)_2 = \cloze{1} \cdot 2^\cloze{2} + \cloze{1} \cdot 2^\cloze{1} + \cloze{1} \cdot 2^\cloze{0} = \cloze{4} + \cloze{2} + \cloze{1} = \cloze{7}
    • (1000)2=123+022+021+020=8+0+0+0=8\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} \begin{aligned} (1 0 0 0)_2 &= \cloze{1} \cdot 2^\cloze{3} + \cloze{0} \cdot 2^\cloze{2} + \cloze{0} \cdot 2^\cloze{1} + \cloze{0} \cdot 2^\cloze{0} \\ &= \cloze{8} + \cloze{0} + \cloze{0} + \cloze{0} = \cloze{8} \end{aligned}
    5
    Übertrage die Zahlen ins Dezimalsystem. Wandle zunächst jede Stelle einzeln um. Tipp: Beginne bei der letzten Stelle.
    • (11100)2=16+8+4+0+0=28\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1 1 0 0)_2 = \cloze{16} + \cloze{8} + \cloze{4} + \cloze{0} + \cloze{0} = \cloze{28}
    • (101011)2=32+0+8+0+2+1=43\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 0 1 0 1 1)_2 = \cloze{32} + \cloze{0} + \cloze{8} + \cloze{0} + \cloze{2} + \cloze{1} = \cloze{43}
    • (110010)2=32+16+0+0+2+0=50\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1 0 0 1 0)_2 = \cloze{32} + \cloze{16} + \cloze{0} + \cloze{0} + \cloze{2} + \cloze{0} = \cloze{50}
    • (1111)2=8+4+2+1=15\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1 1 1)_2 = \cloze{8} + \cloze{4} + \cloze{2} + \cloze{1} = \cloze{15}
    • (1010011)2=64+0+16+0+0+2+1=83\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 0 1 0 0 1 1)_2 = \cloze{64} + \cloze{0} + \cloze{16} + \cloze{0} + \cloze{0} + \cloze{2} + \cloze{1} = \cloze{83}
    • (1101)2=8+4+0+1=13\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 1 0 1)_2 = \cloze{8} + \cloze{4} + \cloze{0} + \cloze{1} = \cloze{13}

    An der ersten Stelle sollte immer eine 1 stehen

    6
    Übertrage die Zahlen aus dem Dezimalsystem ins Binärsystem.
    • 3=2+1=(11)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 3 = \cloze{2} + \cloze{1} = ( \cloze{1 1} )_2
    • 2=2+0=(10)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 2 = \cloze{2} + \cloze{0} = ( \cloze{1 0} )_2
    • 5=4+0+1=(101)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 5 = \cloze{4} + \cloze{0} + \cloze{1} = ( \cloze{1 0 1} )_2
    • 15=8+4+2+1=(1111)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 15 = \cloze{8} + \cloze{4} + \cloze{2} + \cloze{1} = ( \cloze{1 1 1 1} )_2
    • 6=4+2+0=(110)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 6 = \cloze{4} + \cloze{2} + \cloze{0} = ( \cloze{1 1 0} )_2
    • 4=4+0+0=(100)2\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 4 = \cloze{4} + \cloze{0} + \cloze{0} = ( \cloze{1 0 0} )_2
  • 7
    Übertrage die Zahlen ins Dezimalsystem.
    • (30)5=351+050=15+0=15\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (3 0)_5 = \cloze{3} \cdot 5^\cloze{1} + \cloze{0} \cdot 5^\cloze{0} = \cloze{15} + \cloze{0} = \cloze{15}
    • (13)5=151+350=5+3=8\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (1 3)_5 = \cloze{1} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{5} + \cloze{3} = \cloze{8}
    • (23)5=251+350=10+3=13\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (2 3)_5 = \cloze{2} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{10} + \cloze{3} = \cloze{13}
    • (241)5=252+451+150=50+20+1=71\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (2 4 1)_5 = \cloze{2} \cdot 5^\cloze{2} + \cloze{4} \cdot 5^\cloze{1} + \cloze{1} \cdot 5^\cloze{0} = \cloze{50} + \cloze{20} + \cloze{1} = \cloze{71}
    • (213)5=252+151+350=50+5+3=58\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} (2 1 3)_5 = \cloze{2} \cdot 5^\cloze{2} + \cloze{1} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = \cloze{50} + \cloze{5} + \cloze{3} = \cloze{58}
    8
    Übertrage die Zahlen ins Fünfersystem.
    • 10=10+0=251+050=(20)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 10 = \cloze{10} + \cloze{0} = \cloze{2} \cdot 5^\cloze{1} + \cloze{0} \cdot 5^\cloze{0} = ( \cloze{2 0} )_5
    • 6=5+1=151+150=(11)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 6 = \cloze{5} + \cloze{1} = \cloze{1} \cdot 5^\cloze{1} + \cloze{1} \cdot 5^\cloze{0} = ( \cloze{1 1} )_5
    • 13=10+3=251+350=(23)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 13 = \cloze{10} + \cloze{3} = \cloze{2} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = ( \cloze{2 3} )_5
    • 12=10+2=251+250=(22)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 12 = \cloze{10} + \cloze{2} = \cloze{2} \cdot 5^\cloze{1} + \cloze{2} \cdot 5^\cloze{0} = ( \cloze{2 2} )_5
    • 5=5+0=151+050=(10)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 5 = \cloze{5} + \cloze{0} = \cloze{1} \cdot 5^\cloze{1} + \cloze{0} \cdot 5^\cloze{0} = ( \cloze{1 0} )_5
    9
    Übertrage die Zahlen ins Fünfersystem.
    • 89=75+10+4=352+251+450=(324)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 89 = \cloze{75} + \cloze{10} + \cloze{4} = \cloze{3} \cdot 5^\cloze{2} + \cloze{2} \cdot 5^\cloze{1} + \cloze{4} \cdot 5^\cloze{0} = ( \cloze{3 2 4} )_5
    • 98=75+20+3=352+451+350=(343)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 98 = \cloze{75} + \cloze{20} + \cloze{3} = \cloze{3} \cdot 5^\cloze{2} + \cloze{4} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = ( \cloze{3 4 3} )_5
    • 83=75+5+3=352+151+350=(313)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 83 = \cloze{75} + \cloze{5} + \cloze{3} = \cloze{3} \cdot 5^\cloze{2} + \cloze{1} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = ( \cloze{3 1 3} )_5
    • 107=100+5+2=452+151+250=(412)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 107 = \cloze{100} + \cloze{5} + \cloze{2} = \cloze{4} \cdot 5^\cloze{2} + \cloze{1} \cdot 5^\cloze{1} + \cloze{2} \cdot 5^\cloze{0} = ( \cloze{4 1 2} )_5
    • 73=50+20+3=252+451+350=(243)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 73 = \cloze{50} + \cloze{20} + \cloze{3} = \cloze{2} \cdot 5^\cloze{2} + \cloze{4} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = ( \cloze{2 4 3} )_5
    10
    Übertrage die Zahlen ins Fünfersystem.
    • 63=50+10+3=252+251+350=(223)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 63 = \cloze{50} + \cloze{10} + \cloze{3} = \cloze{2} \cdot 5^\cloze{2} + \cloze{2} \cdot 5^\cloze{1} + \cloze{3} \cdot 5^\cloze{0} = ( \cloze{2 2 3} )_5
    • 16=15+1=351+150=(31)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 16 = \cloze{15} + \cloze{1} = \cloze{3} \cdot 5^\cloze{1} + \cloze{1} \cdot 5^\cloze{0} = ( \cloze{3 1} )_5
    • 57=50+5+2=252+151+250=(212)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 57 = \cloze{50} + \cloze{5} + \cloze{2} = \cloze{2} \cdot 5^\cloze{2} + \cloze{1} \cdot 5^\cloze{1} + \cloze{2} \cdot 5^\cloze{0} = ( \cloze{2 1 2} )_5
    • 15=15+0=351+050=(30)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 15 = \cloze{15} + \cloze{0} = \cloze{3} \cdot 5^\cloze{1} + \cloze{0} \cdot 5^\cloze{0} = ( \cloze{3 0} )_5
    • 106=100+5+1=452+151+150=(411)5\gdef\cloze#1{\colorbox{none}{\color{transparent}{\large{$\displaystyle #1$}}}} 106 = \cloze{100} + \cloze{5} + \cloze{1} = \cloze{4} \cdot 5^\cloze{2} + \cloze{1} \cdot 5^\cloze{1} + \cloze{1} \cdot 5^\cloze{0} = ( \cloze{4 1 1} )_5