# What is ##7 \times 7## in base ##7##?

• RChristenk
RChristenk
Homework Statement
What is ##7 \times 7## in base ##7##?
Relevant Equations
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.

RChristenk said:
My first question is how can 7×7 be a valid question if everything is to be in base 7 (only zero to six can be used)?
It isn’t.
RChristenk said:
Or is this actually asking 7×7 in base 10 and for the answer to be converted into base 7?
We cannot know this. Only you have access to the source of the question.

RChristenk said:
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.
Such a representation of a number ##a## to a base ##b## is defined as
$$a=a_n\cdot b_n^n + \ldots + a_2\cdot 7^2+a_1\cdot 7^1+a_0\cdot 7^0$$
with ##a_n,\ldots ,a_2,a_1,a_0 \in \{0,1,2,\ldots, b-1\}.##

This means that seven in base seven is written ##10=1\cdot 7 + 0\cdot 1.## Therefore seven times seven is
##
\begin{array}{c}
1\,0\,\;& \cdot & 1\,0\,\; \\
\hline
1\,0\,\;& & \\
\,0\,0\,0 & &\\
\hline
\,1\,0\,0&&
\end{array}
##
and ##100=1\cdot 7^2 +0\cdot 7^1=0\cdot 7^0=1\cdot 49+0\cdot 7 +0\cdot 1.##

The multiplication is the same, only with fewer numerals.

mathwonk
RChristenk said:
My first question is how can ##7 \times 7## be a valid question if everything is to be in base ##7##
The question is asking what is 710x710 in base 7?
Because if it were asking it in base 7 then the question would be what is 107 x 107? ("10" being 7 - in base 7)
And the answer would be 1007 - which is 4910.

hutchphd, scottdave, e_jane and 3 others
What is ##7 \times 7## in base 7?
It's reasonable to assume that the 7s above are decimal numbers (i.e., in base-10), so I agree with others that the answer is ##100_7##.

Perhaps I am too literal, but to me "##7 \times 7## in base ##7##" is ##10 \times 10##.

hutchphd and e_jane
Hill said:
Perhaps I am too literal, but to me "##7 \times 7## in base ##7##" is ##10 \times 10##.
See my post #3.

Mark44 said:
It's reasonable to assume that the 7s above are decimal numbers (i.e., in base-10), so I agree with others that the answer is ##100_7##.
Me too.

Perhaps a better wording of the problem is to use only words.

What is seven times seven ? Write the result in base seven.

scottdave, PeroK and e_jane
SammyS said:
Me too.

Perhaps a better wording of the problem is to use only words.

What is seven times seven ? Write the result in base seven.
7n x 7n = 4910
for any n larger than 7.

And we know n can't be 7 or less, since that is invalid.

Therefore, 7x7 is actually unambiguous.

75 x 75 = n/a
77 x 77 = n/a
79 x 79 = 4910
717 x 717 = 4910

In fact:

723 x 747 = 49 10

hutchphd and e_jane
One could write the numbers as ##[a]_b## but it is pretty obvious from the context what is meant. If we are talking about ##7## in the base of ##7## then it is clear that ##[7]_{10}=[10]_7## is meant and that ##10\cdot 10 =49## is short for ##[10]_7\cdot [10]_7=[100]_7=[49]_{10}.##

There's no need to search through the crumbs.

I think "b times b, in base b" is always "100" that's how I solved it, anyway.

fresh_42
fresh_42 said:
One could write the numbers as ##[a]_b## but it is pretty obvious from the context what is meant. If we are talking about ##7## in the base of ##7## then it is clear that ##[7]_{10}=[10]_7## is meant and that ##10\cdot 10 =49## is short for ##[10]_7\cdot [10]_7=[100]_7=[49]_{10}.##

There's no need to search through the crumbs.
Still, if we work with classes, rather than numbers, which representative do we choose? [10]_7=[17]_7=...
Maybe I'm nitpicking?

WWGD said:
Still, if we work with classes, rather than numbers, which representative do we choose? [10]_7=[17]_7=...
Maybe I'm nitpicking?
These weren't meant to be classes, only a notation that allows one to interpret the digits of ##a## according to different bases ##b##. We could write ##10_7=7_{10}## but I thought ##[10]_7=[7]_{10}## is easier to read. Rectangular brackets are not automatically equivalence classes.

SammyS and WWGD
fresh_42 said:
These weren't meant to be classes, only a notation that allows one to interpret the digits of ##a## according to different bases ##b##. We could write ##10_7=7_{10}## but I thought ##[10]_7=[7]_{10}## is easier to read. Rectangular brackets are not automatically equivalence classes.
Guess a classic/clear example of notation overload. Notation can only do, help, so much.

scottdave

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