Selina Concise Class 6 Math Chapter 8 H.C.F. and L.C.M. Exercise 8(B) Solution:
(1) Using the common factor method, find the H.C.F. of :
(i) 16 and 35
Solution: Factors of 16 = 1, 2, 4, 8, 16
Factors of 35 = 1, 5, 7, 35
Factors that are common = 1.
Required H.C.F. 1
(ii) 25 and 20
Solution: Factors of 25 = 1, 5, 25
Factors of 20 = 1, 2, 4, 5, 10, 20
Factors that are common = 1, 5
Required H.C.F. 5
(iii) 27 and 75
Solution: Factors of 27 = 1, 3, 9, 27
Factors of 75 = 1, 3, 5, 15, 25, 75
Common Factors = 1, 3
Required H.C.F. = 3
(iv) 8, 12 and 18
Solution: Factors of 8 = 1, 2, 4, 8
Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 18 = 1, 2, 3, 6, 9, 18
Common Factors- 1, 2
Required H.C.F. = 2
(v) 24, 36, 45 and 60
Solution: Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 – 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 45 – 1, 3, 5, 9, 15, 45
Factors of 60 – 1, 2, 3, 5, 6, 10, 12, 15, 20, 30, 60
Common factors: 1, 2, 3
Required H.C.F. 3
(2) Using the prime factor method, find the H.C.F. of:
(i) 5 and 8
Prime factors of 5 = 1 x 5
Prime factors of 8 = 2 x 2 x2 x 1
Since the common prime factor is 1 only,
Therefore, `H.C.F. of 5 and 8 = 1
(ii) 24 and 49
Prime factors 24 = 2 x 2 x 2 x 3 x 1
Prime factors of 49 = 7 x 7 x 1
Since the common prime factor is 1 only
Therefore, H.C.F. of 24 and 49 = 1
(iii) 40, 60 , and 80
Prime factors of 40 = 2 x 2 x 2 x5 x1
Prime factors of 60 = 2 x 2 x 3 x 5 x 1
Prime factors of 80 = 2 x 2 x 2 x 2 x 5 x 1
The prime factors common to be given number 2, 2, 5
Therefore, H.C.F. of 40, 60 and 80 are 2 x 2 x 5 = 20
(iv) 48, 84, and 88
Prime factors of 40 = 2 x 2 x 2 x 2 x 3 x 1
Prime factors of 84 = 2 x 2 x 3 x 7 x 1
Factors of 88 = 2 x 2 x 2 x 11 x1
The prime factors common to the given numbers are 2, 2
Therefore H.C.F. = 2 x 2 = 4
(v) 12, 16 and 28
Prime factors of 12 = 2 x 2 x 3 x 1
Prime factors of 16 = 2 x 2 x 2 x 2 x 1
Prime factor of 28 = 2 x 2 x 7
The prime factors common to be given numbers are 2, 2
Therefore H.C.F. = 2 x 2 = 4
(3) Using the division method find the H.C.F. of the following
(4) Use a method of your own choice to find the H.C.F. of :
(i) 45, 75 and 135
Solution: Factors of 45 = 1, 3, 5, 9, 15, 45
Factors of 75 = 1, 3, 5, 15, 25, 75
Factors of 135 = 1, 3, 5, 9, 15, 135
Factors that are common = 1, 3, 5, 15
Required H.C.F. = 15
(ii) 48, 36 and 96
Solution: Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 24, 48
Factors of 36 = 1, 2, 3, 4, 9, 12, 18, 36
Factors of 96 = 1, 2, 3, , 4, 6, 8, 12, 16, 96
Common factor = 1, 2, 3, 4, 12
Required H.C.F. 12
(iii) 66, 33, and 132
Solution: Factors of 66 = 1, 2, 3, 6, 11, 33, 66
Factors of 33 = 1, 3, 11, 33
Factors of 132 = 1, 2, 3, 4, 6, 11, 12, 33
Common factors = 1, 3, 11, 33
Required H.C.F. = 33
(iv) 24, 36, 60, 132
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 = 1, 2, 3, 4, 6, 12, 18, 36
Factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 20, 30, 60
Common Factors = 1, 2, 3, 4, 12
Required H.C.F. = 12
(v) 30, 60, 90, 105
Factors of 30 = 1, 2, 3, 5, 6, 15, 30
Factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12,15, 30, 60
Factors of 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 90
Factors of 105 = 1, 3, 5, 7, 15, 105
Required H.C.F. 15
(5) Find the greatest number that divides each of 180, 225 and 315 completely
(6) Show that 45 and 56 are co-prime numbers.
Solution: 45 = 3 x 3 x 5
56 = 2 x 2 x 2 x 7
Hence 39 56 have no common factor
Therefore, 45 and 56 are co-prime.
(7) Out of 15, 16, 21, and 28, find out all the pairs of co prime number
Solution: HCF of 15 and 16 = 1 so 15 and 16 are co prime.
15 and 28 are co prime
16 and 21 are co prime
(8) Find the greatest number that will divide 93, 111, and 129, leaving reminder 3 in each case.