# Thread: [SOLVED] Convert from base8 to base7

1. ## [SOLVED] Convert from base8 to base7

How would i convert base8 number of 725 to a base7 number?

2. Originally Posted by Edbaseball17
How would i convert base8 number of 725 to a base7 number?
The easiest way I know is to convert the base 8 number to base 10, then convert the base 10 number to base 7.
$725_8 = 7 \cdot 8^2 + 2 \cdot 8^1 + 5 \cdot 8^0 = 469$
where the 469 is in base 10.

Now to convert 469 to base 7. The largest power of 7 that divides 469 (without being larger than it) is 3:
$469 = 1 \cdot 7^3 + 126$

The next largest power of 7 is 2:
$469 = 1 \cdot 7^3 + 2 \cdot 7^2 + 28$

The next largest power of 7 is 1:
$469 = 1 \cdot 7^3 + 2 \cdot 7^2 + 4 \cdot 7^1 + 0$

And the last power of 7 is 0:
$469 = 1 \cdot 7^3 + 2 \cdot 7^2 + 4 \cdot 7^1 + 0 \cdot 7^0$

So $725_8 = 469_{10} = 1240_7$

I would not be surprised to find that the Euclidean Algorithm provides a way to do this directly from base 8 to base 7, but I don't know how to set it up.

-Dan

3. Hello, Edbaseball17!

How would i convert $725_8$ to a base 7 number?

Convert to base-ten: . $725_8 \:=\:469_{10}$

Then convert to base-seven: . $469_{10} \:=\:1240_7$

I trust you know the necessary procedures.
Otherwise, you shouldn't have been assigned this problem.
.

4. Hey, thanks alot. You made my day, now i understand it perfectly. I was trying to go directly from 8 to 7 instead of working with the base10 instead.

5. Hello, Edbaseball17!

I was trying to go directly from base 8 to base 7.

This can be done IF you can divide in base 8 . . .

$\begin{array}{cccccc} & & 1 & 0 & 3 \\
& & -- & -- & -- \\ 7 & ) & 7 & 2 & 5 \\ & & 7 \\
& & --& -- \\
& & & 2 & 5\\
& & & 2 & 5 \\
& & & --& -- \\
& & & & 0
\end{array}$
. . Remainder: 0

. . $\begin{array}{cccccc}& & & 1 & 1 \\
& & -- & -- & --\\
7 & ) & 1 & 0 & 3 \\
& & & 7 \\
& & & -- \\
& & & 1 & 3 \\
& & & & 7 \\
& & & --& -- \\
& & & & 4 \end{array}$
. . Remainder: 4

. . . . $\begin{array}{cccccc} & & & 1 \\
& & -- & -- \\
7 & ) & 1 & 1 \\
& & & 7 \\
& & -- & -- \\
& & & 2 \end{array}$
. . Remainder: 2

. . . . . . $\begin{array}{cccccc}& & 0 \\
& & -- \\
7 & ) & 1 \\
& & 0 \\
& & -- \\
& & 1 \end{array}$
. . Remainder: 1

Therefore: . $\boxed{\;725_8 \;=\;1240_7\;}$