Skip to main content
Mathematics LibreTexts

3.5: Converting Between (our) Base 10 and Any Other Base (and vice versa)

  • Page ID
    51822
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    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.

    clipboard_e2cf9f93b365caf7baf8301833665e427.png
    Figure 2.6.1

    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: \(1421_{\text {five}}\)

    clipboard_eda37c07141625590f75d6c74768c37d2.png
    Figure 2.6.2

    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

    1. Convert the base 10 numbers into base 4
      1. \(30_{\text {ten}}\) = _____ \(_{\text {four}}\)
      2. \(2103_{\text {ten}}\) = _____ \(_{\text {four}}\)
      3. \(16_{\text {ten}}\) = _____ \(_{\text {four}}\)
    2. Convert the base 5 numbers into base 10
      1. \(30_{\text {five}}\) = ______ \(_{\text {ten}}\)
      2. \(2103_{\text {five}}\) = ______\(_{\text {ten}}\)
      3. \(16_{\text {five}}\) = ______ \(_{\text {ten}}\)

    Think carefully about 2c!

    ***For extra practice, click here.

    Practice Problems

    1. Write the following Base 10 numbers into the new Base.
      1. 5567 into Base 9
      2. 12 into Base 4
      3. 100 into Base 3
      4. 73 into Base 2
    2. Write the following numbers into Base 10.
      1. \(64_{\text {seven}}\)
      2. \(157_{\text {eight}}\)
      3. \(1001001_{\text {two}}\)
      4. \(84671_{\text {eleven}}\)

    This page titled 3.5: Converting Between (our) Base 10 and Any Other Base (and vice versa) is shared under a not declared license and was authored, remixed, and/or curated by Amy Lagusker.

    • Was this article helpful?