Answer
The capital letters A through Z are assigned 65 through 90.
The lower case letters a through z are assigned 97 through 122.
Accordingly,
\[D=68\]\[a=97\]\[d=100\]
Convert each base ten number to binary by repeated division by 2 with the remainder obtained or mentally convert by forming groups of place-value.
\[\begin{align}
& 68=\left( 1\times {{2}^{6}} \right)+\left( 0\times {{2}^{5}} \right)+\left( 0\times {{2}^{4}} \right)+\left( 0\times {{2}^{3}} \right)+\left( 1\times {{2}^{2}} \right)+\left( 0\times {{2}^{1}} \right)+\left( 0\times {{2}^{0}} \right) \\
& ={{1000100}_{\operatorname{two}}}
\end{align}\]
\[\begin{align}
& 97=\left( 1\times {{2}^{6}} \right)+\left( 1\times {{2}^{5}} \right)+\left( 0\times {{2}^{4}} \right)+\left( 0\times {{2}^{3}} \right)+\left( 0\times {{2}^{2}} \right)+\left( 0\times {{2}^{1}} \right)+\left( 1\times {{2}^{0}} \right) \\
& ={{1100001}_{\operatorname{two}}}
\end{align}\]
\[\begin{align}
& 100=\left( 1\times {{2}^{6}} \right)+\left( 1\times {{2}^{5}} \right)+\left( 0\times {{2}^{4}} \right)+\left( 0\times {{2}^{3}} \right)+\left( 1\times {{2}^{2}} \right)+\left( 0\times {{2}^{1}} \right)+\left( 0\times {{2}^{0}} \right) \\
& ={{1100100}_{\operatorname{two}}}
\end{align}\]
The binary sequence is \[\underline{1000100}\underline{1100001}\underline{1100100}\].
Binary representation for wordDad is\[100010011000011100100\].
