Question

Find the least number which when divided by $10,15,30$ and $60$ leaves remainder $9$ in each case

Answer 1

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

Prime factorisation of given numbers is $10,15,30$ and $60$.

$$\begin{gathered} \Rightarrow 10=2\times 5 \\ \Rightarrow 15=3\times 5 \\ \Rightarrow 30=2\times 3\times 5 \\ \Rightarrow 60=2\times 2\times 3\times 5 \end{gathered}$$

$\therefore \mathrm{LCM}\left(10,15,30,60\right)=60$.

So, the number $9$ more than the $\mathrm{LCM}$ of the given numbers will leave a remainder of $9$

So, the required number is $60+9=69$.

Therefore, when $69$ is divided by $10,15,30$ and $60$, remainder is $9$ in each case.