Question

Find the HCF of the following numbers by prime factorisation method ?
$\left(i\right)18,27,36$
$\left(ii\right)106,159,265$
$\left(iii\right)10,35,40$
$\left(iv\right)32,64,96,128$

Answer 1

$\left(i\right)18,27,36$

The three numbers can be represented as the product of their prime factors, as

$18=3^2\times 2$

$27=3^3$

$36=3^2\times 2^2$

Now, HCF is product of the lowest powers of the common factors among the three numbers.

$\Rightarrow HCF=3^2=9$

$\therefore HCF=9$

$\left(ii\right)106,159,165$

The three numbers can be represented as the product of their prime factors, as

$106=2\times 53$

$159=3\times 53$

$265=5\times 53$

Now, HCF is the product of the lowest powers of the common factors among the three numbers.

$\Rightarrow HCF={53}^1$

$\therefore HCF=53$

$\left(iii\right)10,35,40$

The three numbers can be represented as the product of their prime factors, as

$10=5\times 2$

$35=5\times 7$

$40=2^3\times 5$

Now, HCF is the product of the lowest powers of the common factors among the three numbers.

$\Rightarrow HCF=5^1$

$\therefore HCF=5$

$\left(iv\right)32,64,96,128$

The three numbers can be represented as the product of their prime factors, as

$32=2^5$

$64=2^6$

$96=2^5\times 3$

$128=2^7$

Now, HCF is the product of the lowest powers of the common factors among the three numbers.

$\Rightarrow HCF=2^5=32$

$\therefore HCF=32$