## Thinking Mathematically (6th Edition)

Required solution is${{56}_{\text{seven}}}$.
Let's calculate: First, write the multiples of $31$ in base seven. \begin{align} & 31\times 1=31 \\ & 31\times 2=62 \\ & 31\times 3(62+31)=123 \\ & 31\times 4(123+31)=154 \end{align} $31\times 5(154+31)=215$ $31\times 6(215+31)=246...$ ….. (1) Now, divide the given base seven numerals. \begin{align} & {{31}_{\text{seven}}}\overset{{{56}_{\text{seven}}}}{\overline{\left){\begin{align} & {{2426}_{\text{seven}}} \\ & \underline{215} \\ \end{align}}\right.}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,246 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{246} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,0} \\ \end{align} Here, first divide 242 by 31, highest number less than ${{242}_{\text{seven}}}$is ${{215}_{\text{seven}}}$from the table of 31.