Question

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

Answer 1

Hint: The prime factorization procedure can be used to determine the value of H.C.F. To get our H.C.F., we use the prime factorization method, which involves finding the prime factor of all the numbers and then multiplying all the common factors in them.

Complete step-by-step answer:
We have to obtain all the prime factors of all the numbers and write the numbers as the product of their prime factors. This method is known as the prime factorisation method.

 $ \left( i \right)18,27,36 $
The prime factors of $ 18 $ are: $ 18 = 3 \times 3 \times 2......\left( 1 \right) $
The prime factors of $ 27 $ are: $ 27 = 3 \times 3 \times 3......\left( 2 \right) $
The prime factors of $ 36 $ are: $ 36 = 3 \times 3 \times 2 \times 2......\left( 3 \right) $
From the equations $ \left( 1 \right),\left( 2 \right) $ and $ \left( 3 \right) $ , the common factors are $ 3 \times 3 $ .
Therefore, H.C.F of $ 18,27,36 = 3 \times 3 $
Now, on multiplying the terms, we get,
 $ \therefore $ H.C.F $ \left( {18,27,36} \right) = 9 $

 $ \left( {ii} \right)106,159,165 $
The prime factors of $ 106 $ are: $ 106 = 53 \times 2......\left( 1 \right) $
The prime factors of $ 159 $ are: $ 159 = 53 \times 3......\left( 2 \right) $
The prime factors of $ 165 $ are: $ 165 = 53 \times 5......\left( 3 \right) $
From the equations $ \left( 1 \right),\left( 2 \right) $ and $ \left( 3 \right) $ , the common factor is $ 53 $ .
Therefore, H.C.F of $ 106,159,165 = 53 $
 $ \therefore $ H.C.F $ \left( {106,159,165} \right) = 53 $

 $ \left( {iii} \right)10,35,40 $
The prime factors of $ 10 $ are: $ 10 = 5 \times 2......\left( 1 \right) $
The prime factors of $ 35 $ are: $ 35 = 5 \times 7......\left( 2 \right) $
The prime factors of $ 40 $ are: $ 40 = 5 \times 2 \times 2 \times 2......\left( 3 \right) $
From the equations $ \left( 1 \right),\left( 2 \right) $ and $ \left( 3 \right) $ , the common factor is $ 5 $ .
Therefore, H.C.F of $ 10,35,40 = 5 $
 $ \therefore $ H.C.F $ \left( {10,35,40} \right) = 5 $

 $ \left( {iv} \right)32,64,96,128 $
The prime factors of $ 32 $ are: $ 32 = 2 \times 2 \times 2 \times 2 \times 2......\left( 1 \right) $
The prime factors of $ 64 $ are: $ 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2......\left( 2 \right) $
The prime factors of $ 96 $ are: $ 96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3......\left( 3 \right) $
The prime factors of $ 128 $ are: $ 128 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2......\left( 4 \right) $
From the equations $ \left( 1 \right),\left( 2 \right),\left( 3 \right) $ and $ \left( 4 \right) $ ,the common factors are $ 2 \times 2 \times 2 \times 2 \times 2 $
Therefore, H.C.F of $ 32,64,96,128 = 2 \times 2 \times 2 \times 2 \times 2 $
Now, on multiplying the terms, we get,
 $ \therefore $ H.C.F $ \left( {32,64,96,128} \right) = 32 $

Note: The largest number that absolutely divides any of the given integers is the H.C.F (Highest Common Factor). H.C.F. includes all of the common factors of two numbers, so H.C.F. of a co prime number is always equal to 1 since they have no common divisor. The product of two numbers is equivalent to the product of their H.C.F and L.C.M.