Question

The HCF of $2472,1284$and a third number$\text{'}n\text{'}$is$12$. If their LCM is$2^3\times 3^2\times 5\times 103\times 107$. Find the number $\text{'}n\text{'}$.

Answer 1

Step 1

We know that the numbers $2472,1284$ can be expressed as

$$\begin{gathered} 2472=2\times 2\times 2\times 3\times 103=(2^3\times 3\times 103) \\ 1284=2\times 2\times 3\times 107=(2^2\times 3\times 107) \\ HenceHCF(2472,1284)=2\times 2\times 3=(2^2\times 3) \end{gathered}$$

Step 2

Since the HCF is $12$ the number must have a component of $(2^2\times 3)$ at the very least in it.

Since the LCM is $2^3\times 3^2\times 5\times 103\times 107.$

The minimum requirement in the required number has to be $3^2\times 5^1$

Therefore,

$$\begin{gathered} n=2^2\times 3\times (3\times 5) \\ n=2^2\times 3^2\times 5 \\ n=4\times 9\times 5 \\ n=180 \end{gathered}$$

Hence, the other number $n$is $180$