Question

321five × 22five

in base 5

Answer 1

`321_5=(3)5^2+(2)5^1+(1)5^0`

`321_5=\left(\frac34\right)10^2+(1)10^1+(1)10^0`

`321_5=75+10+1`

`321_5=86_(10)`


`22_5=(2)5^1+(2)5^0=(1)10^1+(2)10^0=12_(10)`


`321_5*22_5=86_(10)*12_(10)=1032_(10)`


`\frac{1032}5=206+\frac25`

`\frac{206}5=41+\frac15`

`\frac{41}5=8+\frac15`

`\frac85=1+\frac35`

`\frac15=0+\frac15`

`\implies1032_(10)=13112_5`