How to quickly solve the following expression

Question

How do I solve the following equation: ABB (base 16) + 101 (base 16)?

I know how to do it the long way (converting the bases to base 10 and then adding them). But is there a more efficient way without converting the numbers to base 10?

Answer 1

The addition algorithm works in any base. What you need is just the table of elementary additions (or counting with your fingers).

\begin{array}{c|cc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F \\ \hline 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F \\ 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F & 10 \\ 2 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F & 10 & 11 \\ 3 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F & 10 & 11 & 12\\ 4 & 4 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F & 10 & 11 & 12 & 13\\ 5 & 5 & 6 & 7 & 8 & 9 & A & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 \\ 6 & 6 & 7 & 8 & 9 & A & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 \\ 7 & 7 & 8 & 9 & A & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ 8 & 8 & 9 & A & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ 9 & 9 & A & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ A & A & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ B & B & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A \\ C & C & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A & 1B \\ D & D & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A & 1B & 1C \\ E & E & F & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A & 1B & 1C & 1D \\ F & F& 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 1A & 1B & 1C & 1D & 1E \\ \end{array}

Just remember to carry. So

\begin{array}{rrrl} A&B&B & + \\ 1&0&1 & = \\ \hline B&B&C \end{array}

More complicated:

\begin{array}{rrrl} \scriptstyle1&\scriptstyle1& \\ A&B&B & + \\ 2&C&E & = \\ \hline D&8&9 \end{array} where the small ones mean “carry”.

You can note that the sum of two digits plus the possible carry cannot exceed $1F$.

The second sum in base 7: \begin{array}{rrrrl} & & & \scriptstyle1 & \\ 1 & 1 & 0 & 0 & 3 & + \\ & 2 & 0 & 4 & 4 & = \\ \hline 1 & 3 & 0 & 5 & 0 \end{array}