Question

Find the least number which when divided by $6,15$ and $18$ leave remainder $5$ in each case.

Answer 1

Solve for the required number

Given numbers are: $6,15$ and $18$.

Prime factorisation of given numbers is

$\begin{array}{rcl}6& =& 2\times 3\\ 15& =& 3\times 5\\ 18& =& 2\times 3\times 3\end{array}$

$\begin{array}{rcl}\mathrm{LCM}\left(6,15,18\right)& =& 2\times 3\times 3\times 5\\ & =& 90\end{array}$

The number that is $5$ more than their $\mathrm{LCM}$ will leave a remainder of $5$ when divided by the given numbers. Thus, the required number is,

$5+90=95$.

Therefore, when the number $95$ when divided by $6,15$ and $18$, will leave a remainder of $5$.