2.6: Converting Between (our) Base 10 and Any Other Base (and vice versa)
- Page ID
- 50991
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)To convert any number in (our base) Base 10 to any other base, we must use basic division with remainders. Do not divide using decimals; the algorithm will not work.
Example \(\PageIndex{1}\)
Convert from (our) Base 10 to (weird) Base _____
Change \(236_{\text {ten}}\) to ______ \(_{\text {five}}\)
Solution
Keep dividing by 5, until your quotient is zero.
\[\begin{aligned}
236 \div 5 &=47 \; r \; \mathbf{1} \\
47 \div 5 &=9 \; r \; \mathbf{2} \\
9 \div 5 &=1 \; r \; \mathbf{4} \\
1 \div 5 &=0 \; r \; \mathbf{1}
\end{aligned} \nonumber \]
Now write your remainders backwards!
Answer: \(1421_{\text {five}}\)
Example \(\PageIndex{2}\)
Convert from (weird) Base ____ to (our) Base 10.
Solution
First, notice how to break down \(602_{\text {ten}}\):
\[602_{\text {ten }}: 602=6\left(10^{2}\right)+0\left(10^{1}\right)+2\left(10^{0}\right) \nonumber \]
Now, use the same approach to change \(602_{\text {eight}}\) into Base 10
\[6\left(8^{2}\right)+0\left(8^{1}\right)+2\left(8^{0}\right)=386_{\text {ten}} \nonumber \]
Example \(\PageIndex{3}\)
Convert \(5361_{\text {seven}}\) into Base 10.
Solution
\[5\left(7^{3}\right)+3\left(7^{2}\right)+6\left(7^{1}\right)+1\left(7^{0}\right)=1905_{\text {ten }} \nonumber \]
Partner Activity 1
- Convert the base 10 numbers into base 4
- \(30_{\text {ten}}\) = _____ \(_{\text {four}}\)
- \(2103_{\text {ten}}\) = _____ \(_{\text {four}}\)
- \(16_{\text {ten}}\) = _____ \(_{\text {four}}\)
- Convert the base 5 numbers into base 10
- \(30_{\text {five}}\) = ______ \(_{\text {ten}}\)
- \(2103_{\text {five}}\) = ______\(_{\text {ten}}\)
- \(16_{\text {five}}\) = ______ \(_{\text {ten}}\)
Think carefully about 2c!
***For extra practice, click here.
Practice Problems
- Write the following Base 10 numbers into the new Base.
- 5567 into Base 9
- 12 into Base 4
- 100 into Base 3
- 73 into Base 2
- Write the following numbers into Base 10.
- \(64_{\text {seven}}\)
- \(157_{\text {eight}}\)
- \(1001001_{\text {two}}\)
- \(84671_{\text {eleven}}\)