- #1

RChristenk

- 49

- 4

- Homework Statement
- What is ##7 \times 7## in base ##7##?

- Relevant Equations
- Change radix to ##7##

In base ##7##, only the digits ##0,1,2,3,4,5,6## can be used. So ##7 \times 7 = 49## in base ##10##, but ##49## is a multiple of ##7## seven times, so at the right-most digit position of the answer, I can put down a ##0## and carry a ##7 ## to the next-left digit position. I can't put down ##7## in this digit position because it is bigger than the largest allowable digit, ##6##, so I put down a ##0## again, giving me a partial answer of _##00## at this step. Seven goes into seven ##1## time, so now I can put down ##1## in the next-left position and complete this question. Hence the answer is ##100##, or that in base ##7##, ##7 \times 7 = 100##.

My first question is how can ##7 \times 7## be a valid question if everything is to be in base ##7## (only zero to six can be used)? Or is this actually asking ##7 \times 7## in base ##10## and for the answer to be converted into base ##7##? I'm very confused because in the intermediary calculations I'm using the digit seven but calculating according to base ##7## rules.

My second question is how do I write down base ##7## multiplication in long hand on paper? Would something like this be correct:

##

\begin{array}{c}

1&7& \\

\hline

& &7 \\

& &7 \\

\hline

1& 0&0

\end{array}

##

I'm mostly confused here by how to treat the initial ##49##, but I couldn't see a logical place to write it down so I just went ahead and assumed seven goes into forty-nine seven times and wrote down ##7## at the very top.

My first question is how can ##7 \times 7## be a valid question if everything is to be in base ##7## (only zero to six can be used)? Or is this actually asking ##7 \times 7## in base ##10## and for the answer to be converted into base ##7##? I'm very confused because in the intermediary calculations I'm using the digit seven but calculating according to base ##7## rules.

My second question is how do I write down base ##7## multiplication in long hand on paper? Would something like this be correct:

##

\begin{array}{c}

1&7& \\

\hline

& &7 \\

& &7 \\

\hline

1& 0&0

\end{array}

##

I'm mostly confused here by how to treat the initial ##49##, but I couldn't see a logical place to write it down so I just went ahead and assumed seven goes into forty-nine seven times and wrote down ##7## at the very top.