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:

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.

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
${2^2} \times {3^2} \times 5 \times 103,{2^2} \times {3^2} \times 5 \times 107,{2^2} \times {3^2} \times 5 \times 103 \times 107 \\ i.e.18540,19260,1983780 \\$