Question

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

Answer 1

Find the $\mathrm{LCM}$ of given numbers

Given numbers: $6,15,18$ can be written in prime factors as

$$\begin{gathered} 6=2\times 3 \\ 15=3\times 5 \\ and18=2\times 3\times 3 \end{gathered}$$

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

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

So, the required number is $90+5=95$.

Therefore, when $95$ is divided by the given numbers, it will leave a remainder of $5$.