Question

The HCF of $2472,1284$ and a third number ‘$\mathrm{n}$’ is $12$. If their lcm is $2^3\times 3^2\times 5\times 103\times 107$, then the number ‘$\mathrm{n}$’ is:

Answer 1

Step 1: Find the prime factorisation of $2472$ and $1284$

HCF of $2472,1284$ and a third number ‘$\mathrm{n}$’ is $12$.

Their LCM is $2^3\times 3^2\times 5\times 103\times 107$

Prime factorisation of $\begin{array}{rcl}2472& =& 2\times 2\times 2\times 3\times 103\end{array}$

$\begin{array}{rcl}& =& 2^3\times 3\times 103\end{array}$

Prime factorisation of $1284=2\times 2\times 3\times 107$

$=2^2\times 3\times 107$

$\therefore$ HCF ($2472,1284$) $=2\times 2\times 3$

$=2^2\times 3$

Thus, $\mathrm{n}$ also has the factors $2^2\times 3$

Step 2: Find the value of $\mathrm{n}$

From the LCM of three numbers, another factor is the missing element of another number, i.e.,$3\times 5$

$$\begin{gathered} \mathrm{n}=2^2\times 3\times 3\times 5 \\ \mathrm{n}=180 \end{gathered}$$

Hence, the number $\mathrm{n}$ is $180$