Question

The HCF of 2472,1284 and a third number N is 12. If their LCM is $2^3\times 3^2\times 5\times 103\times 107,$ then the number N is:

Answer 1

Hint: We are given LCM,HCF and two numbers so we’ll create the conditions for N to have its value using the HCF and the LCM. Using HCF we’ll get a factor of N in the same way LCM will also provide us a factor of N, using those factors we’ll have the value for N.

Complete step-by-step answer:
Given data: HCF (2472,1284,N)=12
LCM(2472,1284,N)=$2^3\times 3^2\times 5\times 103\times 107$
Using prime factorization we can say that,
$HCF=2^2\times 3$
$LCM=2^3\times 3^2\times 5\times 103\times 107$
$2472=2^3\times 3\times 103$
$1284=2^2\times 3\times 107$
It is well known that HCF of n numbers is the highest common factor that means the highest number that divides all those numbers
Therefore $2^2\times 3$ should also divide N
i.e. N must be a multiple is $2^2\times 3$
Similarly, LCM of any n numbers is that it can be divided by all those numbers.
LCM is having a factor of $3^2\times 5$ but neither of 2472 and 1248 can be divided by $3^2\times 5$
Therefore we can say that N is also a multiple of $3^2\times 5$
From the above statements we can say that the minimum value of N can be $2^2\times 3^2\times 5$
$\therefore N=2^2\times 3^2\times 5$
$=4\times 9\times 5$
$=180$
Hence the number N is 180.

Note: From the above solution we got the value of N=180, but N can also have other values.
We got that minimum value of N as we got the condition that it should be a factor of $2^2\times 3^2\times 5$
But N cannot be greater than its LCM so N can also have the values
$$\begin{gathered} 2^2\times 3^2\times 5\times 103,2^2\times 3^2\times 5\times 107,2^2\times 3^2\times 5\times 103\times 107 \\ i.e.18540,19260,1983780 \end{gathered}$$