Question

A number when divided successively by $6,4$ and $2$ leaves remainders $4,2$ and $1$ respectively. Find the least such number.

Answer 1

Let the number (dividend ) be $D$, and the quotients obtained after the successive divisions be $q,r$ & $s$ respectively.

Writing the given data mathematically using Dividend $=Diviso{r}^{\ast }Quotient+Remainder$, we get

$D=6q+4$

$q=4r+2$

$r=2s+1$

Passing value of $r$ in $q$, & then value of $q$ in $D$, we get

$D=6\left[4\right(2s+1)+2]+4$

$D=6(8s+6)+4$

$\Rightarrow D=48s+40$, where $s$ can be any whole number.

For the smallest possible value of $D$, we put $s=0$

$D=40$