jmc

Question

Find the absolute value of the difference of single-digit integers A and B such that arrayc@c@;c@c@c@c & & & B& B & A_6 & & & 4 & 1 & B_6& & + & A & 1 & 5_6 2-6 & & A & 1 & 5 & 2_6 array Express your answer in base 6.

Accepted Answer

We start working from the rightmost column. Since

5 > 2

,

A_{6} + B_{6} + 5_{6}

is either equal to

1 2_{6}

or

2 2_{6}

. Therefore,

A_{6} + B_{6}

is either equal to

3_{6}

or

1 3_{6}

.

We then look at the second rightmost digits. If

A_{6} + B_{6} = 1 3_{6}

, then

2_{6} + B_{6} + 1_{6} + 1_{6} = 5_{6}

. This means that

B_{6} = 1

, which makes

A_{6} = 1 2_{6}

. Since

A

has to be a single-digit integer, this is impossible. Therefore, we try

A_{6} + B_{6} = 3_{6}

. This gives us

1_{6} + B_{6} + 1_{6} + 1_{6} = 5_{6}

, which means

B_{6} = 2

, and

A_{6} = 1_{6}

. We plug

B = 2

and

A = 1

into the equation to see if it works.

\begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & &_{1}&\\ & & & 2& 2 & 1_6\\ & & & 4 & 1 & 2_6\\& & + & 1 & 1 & 5_6\\ \cline{2-6} & & 1 & 1 & 5& 2_6\\ \end{array}

Therefore, the difference is

2_{6} - 1_{6} = 1_{6}

.