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## Here's the Solution to this Question

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

To Decimal

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

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

$(742)_8=1E2_{16}$

To Binary form

$\def\arraystretch{1.5} \begin{array}{c|c} 2&482 \\ \hline 2&241~~r~~0\\ 2&120~~r~~1\\ 2&60~~r~~0\\ 2&30~~r~~0\\ 2&15~~r~~0\\ 2&7~~r~~1\\ 2&3~~r~~1\\ 2&1~~r~~1\\ &0~~r~~1 \end{array}$

$(742)_8=111100010_{2}$

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

To Decimal

$(1011)_2=1\times2^3+0\times 2^2+1\times 2^1+1\times 2^0\\ (1011)_2=8+4+1=13_{10}.$

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

$(1011)_2=D_{16}$

To Octal form

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

$(1011)_2=15_{8}$

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

To binary

$\def\arraystretch{1.5} \begin{array}{c|c} 2&42 \\ \hline 2&21~~r~~0\\ 2&10~~r~~1\\ 2&5~~r~~0\\ 2&2~~r~~1\\ 2&1~~r~~0\\ &0~~r~~1\\ \end{array}$

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

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

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

To Octal form

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

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

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

To decimal

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

To Binary form

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

To Octal form

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