Find the HCF of the following numbers: (a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75 Solution: (a) Given numbers are 18 and 48.
Prime factorisations of 18 and 48 are:
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6. (b) The given numbers are 30 and 42.
Prime factorisations of 30 and 42, are:
Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6. (c) Given numbers are 18 and 60.
Prime factorisations of 18 and 60 are: 
Here, the common factors are 2 and 3.
Hence, the HCF of 18 and 60 = 2 x 3 = 6. (d) Given numbers are 27 and 63.
Prime factorisations of 27 and 63 are: 
Here, the common factor is 3 (occurring twice).
Hence, the HCF = 3 x 3 = 9. (e) Given numbers are 36 and 84.
Prime factorisations of 36 and 84 are:
Here, the common factors are 2, 2 and 3.
Hence, the HCF = 2 x 2 x 3 = 12. (f) Given numbers are 34 and 102.
Prime factorisations of 34 and 102 are:
Here, the common factors are 2 and 17.
Thus, HCF is 2 x 17 = 34. (g) The given numbers are 70, 105 and 175.
Prime factorisation of 70, 105 and 175 are:
Here, common factors are 5 and 7.
Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35. (h) Given numbers are 91, 112 and 49.
Prime factorisations of 91, 112 and 49 are:
Here, the common factor is 7.
Hence, the HCF = 7. (i) Given numbers are 18, 54 and 81.
Prime factorisations of 18, 54 and 81 are:
Here, the common factor is 3 (occurring twice).
Thus, the HCF = 3 x 3 = 9. (j) Given numbers are 12, 45 and 75.
Prime factorisations of 12, 45 and 75 are:
Here, the common factor is 3.
Hence, the HCF = 3. |
