## Thinking Mathematically (6th Edition)

${{1102}_{five}}$
To multiply two numbers those are not of base 10, follow the below steps, To multiply such numbers, multiply just as it is done in base ten. That is first multiply the right most column numbers and continue same as it is done for base ten but in base four there are only 0,1,2, 3 and 4, if multiplication exceeds 4, then change it in base five. \begin{align} {{123}_{five}} & \\ \times {{4}_{five}} & \\ \end{align} To solve this, follow the below step, \begin{align} & {{3}_{five}}\times {{4}_{five}}={{12}_{ten}} \\ & {{4}_{five}}\times {{2}_{five}}={{8}_{ten}} \\ & {{4}_{five}}\times {{1}_{five}}={{4}_{five}} \end{align} And here 12 and 8 are not in base five, so convert it. ${{12}_{ten}}=\left( 5\times 2 \right)+\left( 2\times 1 \right)={{22}_{five}}$ As 2 is the carry part, add 2 to 8, that is $8+2=10$ ${{10}_{ten}}=\left( 5\times 2 \right)+\left( 0\times 1 \right)={{20}_{five}}$ As 2 is the carry part, add 2 to 4, that is $4+2=6$ ${{6}_{ten}}=\left( 5\times 1 \right)+\left( 1\times 1 \right)={{11}_{five}}$ \begin{align} & \underline{\begin{align} {{123}_{five}} & \\ \times {{4}_{five}} & \\ \end{align}} \\ & \underline{{{1102}_{five}}} \\ \end{align} So, here is the solution of the multiplication.