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: Relation between the HCF and LCM of three numbers a, b and c is given by the formula, \[abc = \dfrac{{{\text{LCM}}(a,b,c) \times {\text{HCF}}(a,b) \times {\text{HCF}}(b,c) \times {\text{HCF}}(c,a)}}{{{\text{HCF}}(a,b,c)}}\] .So for obtaining the unknown number we will use this property of LCM & HCF; and we equate them.

Formula used:
\[abc = \dfrac{{{\text{LCM}}(a,b,c) \times {\text{HCF}}(a,b) \times {\text{HCF}}(b,c) \times {\text{HCF}}(c,a)}}{{{\text{HCF}}(a,b,c)}}\]

Complete step-by-step answer:
According to the question, two of the numbers from three are known to us along with their HCF.
So, let \[a = 2472\], \[b = 1284\] and \[c = \]N are the three numbers.
First we will factorize the 2472,1284 & N.
Prime factorization of 2472 gives us:
$ \Rightarrow $2472 = \[2 \times 2 \times 2 \times 3 \times 103\]
Prime factorization of 1284 gives us:
$ \Rightarrow $1284 = \[2 \times 2 \times 3 \times 107\]
Now the HCF of these three numbers is given to us in the question:
$ \Rightarrow $HCF = \[2 \times 2 \times 3 = 12\]
Based on HCF value, we can conclude that the prime factorization of N will be of the form:
$ \Rightarrow $N = \[2 \times 2 \times 3 \times X\]
And LCM is also given = \[{2^3} \times {3^2} \times 5 \times 103 \times 107\]
We know that the relation between product of three numbers, their HCF and LCM is given as:
\[ \Rightarrow abc = \dfrac{{{\text{LCM}}(a,b,c) \times {\text{HCF}}(a,b) \times {\text{HCF}}(b,c) \times {\text{HCF}}(c,a)}}{{{\text{HCF}}(a,b,c)}}\]
As \[{\text{HCF}}(a,b,c) = 12\], \[{\text{HCF}}(a,b) = {\text{HCF}}(b,c) = {\text{HCF}}(c,a) = 12\]. Using the formula and putting all these values, we’ll get:
\[ \Rightarrow 1284 \times 2472 \times c = \dfrac{{({2^3} \times {3^2} \times 5 \times 103 \times 107 \times 2 \times 2 \times 3) \times 12 \times 12 \times 12}}{{12}}\]
Now simplifying the equation further, we’ll get:
\[ \Rightarrow c = \dfrac{{({2^3} \times {3^2} \times 5 \times 103 \times 107 \times 2 \times 2 \times 3) \times 12 \times 12 \times 12}}{{12 \times 1284 \times 2472}}\]
\[ \Rightarrow c = 2160\]

Thus the third number i.e. N is 2160.

Note: The LCM of any three numbers is always divisible by all those three numbers while the HCF of any three is always a factor of all those three numbers. This rule is not only limited for two or three numbers. This is true for any collection of natural numbers.