Question

(v) The HCF of two numbers
3. Find the HCF of each set of numbers using prime factorisation method.
18,24 (W) 51,85 (1) 61,76 (1) 84,120 (V) 27,45,81 (vi) 45,55,95
of each of numbers using prime factorisation method.​

Answer 1

Answer:

Explanation:

Prime factorisation is the process through which we express the number as the product of its prime factors.The prime factors only consists of prime numbers and do not include composite numbers.

When we use the method of prime factorisation to find the H.C.F. we gind the greatest common factor among the numbersor prime factors. Then we identify the common factor and multiply the common factor.  The resultant is the H.C.F of the numbers.

• H.C.F. of 18 and 24 using prime factorisation are as below:  The factors of 18 are 1, 3 , 3  and 2 and the factors of  24 are 1, 2 , 2, 2 and 3 . The common factor are 3 and 2 so the H.C.F is 1*3*2=6
• H.C.F of 51 and 85 is :  The factor of 51 are 1, 17 and 3 and the factors of 85 are 1, 5 and 17 so the common factor is 17 and 1 therefore the H.C.F is 17*1 =17.

• H.C.F of 61 and 76 is : The factor of 61 are 1 and 61 and the factors of 76 are 1, 2, 2 and 19 is 1 so the H.C.F is 1
• H.C.F of 84 and 120 is : The factor of 84 are 1,2,2,3 and 7 and the factor of 120 are 1,2,2,2,3 and 5. So the common factor are 1,2,2 and 3 H.C.F is 2*2*3= 12
• H.C.F of 27, 45 and 81 is: The factor of 27 are 1,3 ,3 and 3, the factor of 45 are 1, 3,3 and 5 and the factor of  81 are 1,3,3,3 and 3 so the common factor are  3 and 3  therefore the H.C.F is 3*3=9

• H.C.F of 45, 55 and 95 is : the factor of 45 are 1, 3,3 and 5 the factor of, 55 are 1, 5, 11  and the factor of 95 is 1,5 and 19 so the common factor is 5 so the H.C.F is 5

Hope this helps.