jmc

Question

Add 175_9 + 714_9 + 61_9. Express your answer in base 9.

Accepted Answer

When summing these three numbers, we notice that

5 + 4 + 1

leaves a residue of

1

when divided by

9

, so it follows that the sum has a rightmost digit of

1

and that carry-over must occur. After carrying over to the next digit, we must find the sum of

1 + 7 + 1 + 6 = 1 6_{9}

, which leaves a residue of

6

when divided by

9

. Thus, we write down

6

as the next digit and carry-over another

1

. Evaluating the next digit, we must find the sum of

1 + 1 + 7 = 1 0_{9}

, which leaves a residue of

0

when divided by

9

. Thus, we must carry-over one more time, yielding that:

\begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & \stackrel{1}{1} & \stackrel{1}{7} & \stackrel{}{5}_{9} \\ & & & 7 & 1 & 4_{9} \\ &+ & & & 6 & 1_{9} \\ \cline{2-6} && 1 & 0 & 6 & 1_{9} \\ \end{array}

Thus, the answer is

106 1_{9}

.

Alternatively, we can notice that

17 5_{9} + 71 4_{9} = 100 0_{9}

, so

100 0_{9} + 6 1_{9} = 106 1_{9}

.