Answer to Question #250611 in Discrete Mathematics for Alina

Question #250611

1. Convert each of the following to their respective Decimal, Octal, Hexadecimal and binary representation:

(a) (742)8

(b) (1011)2

(c) (47)10

(d) (3EAC)16

1
Expert's answer
2021-10-14T07:47:02-0400

(a) (742)8(742)_8 is already in octal representation. So, we will change it to its Decimal, Hexadecimal and Binary form.

To Decimal

(742)8=7×82+4×81+2×80(742)8=448+32+2=48210.(742)_8=7\times8^2+4\times 8^1+2\times 8^0\\ (742)_8=448+32+2=482_{10}.

To Hexadecimal form

164821630  r  2161  r  E0  r  1\def\arraystretch{1.5} \begin{array}{c|c} 16&482 \\ \hline 16&30~~r~~2\\ 16&1~~r~~E\\ &0~~r~~1 \end{array}

(742)8=1E216(742)_8=1E2_{16}

To Binary form

24822241  r  02120  r  1260  r  0230  r  0215  r  027  r  123  r  121  r  10  r  1\def\arraystretch{1.5} \begin{array}{c|c} 2&482 \\ \hline 2&241~~r~~0\\ 2&120~~r~~1\\ 2&60~~r~~0\\ 2&30~~r~~0\\ 2&15~~r~~0\\ 2&7~~r~~1\\ 2&3~~r~~1\\ 2&1~~r~~1\\ &0~~r~~1 \end{array}

(742)8=1111000102(742)_8=111100010_{2}


(b) (1011)2(1011)_2 is already in binary representation. So, we will change it to its Decimal, Hexadecimal and Octal form.

To Decimal

(1011)2=1×23+0×22+1×21+1×20(1011)2=8+4+1=1310.(1011)_2=1\times2^3+0\times 2^2+1\times 2^1+1\times 2^0\\ (1011)_2=8+4+1=13_{10}.

To Hexadecimal form

1613160  r  D\def\arraystretch{1.5} \begin{array}{c|c} 16&13 \\ \hline 16&0~~r~~D\\ \end{array}

(1011)2=D16(1011)_2=D_{16}

To Octal form

81381  r  50  r  1\def\arraystretch{1.5} \begin{array}{c|c} 8&13 \\ \hline 8&1~~r~~5\\ &0~~r~~1\\ \end{array}

(1011)2=158(1011)_2=15_{8}


(c) (42)10(42)_{10} is already in decimal representation. So, we will change it to its binary, Hexadecimal and Octal form.

To binary

242221  r  0210  r  125  r  022  r  121  r  00  r  1\def\arraystretch{1.5} \begin{array}{c|c} 2&42 \\ \hline 2&21~~r~~0\\ 2&10~~r~~1\\ 2&5~~r~~0\\ 2&2~~r~~1\\ 2&1~~r~~0\\ &0~~r~~1\\ \end{array}

(42)10=1010102(42)_{10}=101010_2

To Hexadecimal form

1642162  r  A0  r  2\def\arraystretch{1.5} \begin{array}{c|c} 16&42 \\ \hline 16&2~~r~~A\\ &0~~r~~2 \end{array}

(42)10=2A16(42)_{10}=2A_{16}

To Octal form

84285  r  20  r  5\def\arraystretch{1.5} \begin{array}{c|c} 8&42\\ \hline 8&5~~r~~2\\ &0~~r~~5\\ \end{array}

(42)10=528(42)_{10}=52_{8}


(d) (3EAC)16(3EAC)_{16} is already in Hexadecimal representation. So, we will change it to its binary, decimal and Octal form.

To decimal

(3EAC)16=3×163+E×162+A×161+C×160(3EAC)16=12288+3584+160+12=1604410.(3EAC)_{16}=3\times16^3+E\times 16^2+A\times 16^1+C\times 16^0\\ (3EAC)_{16}=12288+3584+160+12=16044_{10}.


To Binary form

(3EAC)16=111110101011002(3EAC)_{16}=11111010101100_{2}


To Octal form

(3EAC)16=372548(3EAC)_{16}=37254_{8}




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