NCERT Class 6 Mathematics Third Chapter Playing with Numbers Exercise 3.6 Solutions
EXERCISE 3.6
(1) Find the HCF of the following numbers:
(a) 18, 48
Solution: The HCF of 18 and 48 can also be found by prime factorization of these numbers as follows:
Thus, 18 = 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
∴ HCF of 18 and 48 is 2 × 3 = 6.
(b) 30, 42
Solution: The HCF of 30 and 42 can also be found by prime factorization of these numbers as follows:
Thus, 30 = 2 × 3 × 5
42 = 2 × 3 × 7
∴ HCF of 30 and 42 is 2 × 3 = 6.
(c) 18, 60
Solution: The HCF of 18 and 60 can also be found by prime factorization of these numbers as follows:
Thus, 18 = 2 × 3 × 3
60 = 2 × 2 × 3 × 5
∴ HCF of 18 and 60 is 2 × 3 = 6.
(d) 27, 63
Solution: The HCF of 27 and 63 can also be found by prime factorization of these numbers as follows:
Thus, 27 = 3 × 3 × 3
63 = 3 × 3 × 7
∴ HCF of 27 and 63 is 3 × 3 = 9.
(e) 36, 84
Solution: The HCF of 36 and 84 can also be found by prime factorization of these numbers as follows:
Thus, 36 = 2 × 2 × 3 × 3
84 = 2 × 2 × 3 × 7
∴ HCF of 36 and 84 is 2 × 2 × 3 = 12.
(f) 34, 102
Solution: The HCF of 34 and 102 can also be found by prime factorization of these numbers as follows:
Thus, 34 = 2 × 17
102 = 2 × 3 × 17
∴ HCF of 34 and 102 is 2 × 17 = 34.
(g) 70, 105, 175
Solution: The HCF of 70, 105 and 175 can also be found by prime factorization of these numbers as follows:
Thus, 70 = 2 × 5 × 7
105 = 3 × 5 × 7
175 = 5 × 5 × 7
∴ HCF of 70, 102 and 175 is 5 × 7 = 35.
(h) 91, 112, 49
Solution: The HCF of 91, 112 and 49 can also be found by prime factorization of these numbers as follows:
Thus, 91 = 7 × 13
112 = 2 × 2 × 2 × 2 × 7
49 = 7 × 7
∴ HCF of 91, 112 and 49 is 7.
(i) 18, 54, 81
Solution: The HCF of 18, 54 and 81 can also be found by prime factorization of these numbers as follows:
Thus, 18 = 2 × 3 × 3
54 = 2 × 3 × 3 × 3
81 = 3 × 3 × 3 × 3
∴ HCF of 18, 54 and 81 is 3 × 3 = 9.
(j) 12, 45, 75
Solution: The HCF of 18, 54 and 81 can also be found by prime factorization of these numbers as follows:
Thus, 12 = 2 × 2 × 3
45 = 3 × 3 × 5
75 = 3 × 5 × 5
∴ HCF of 12, 45 and 75 is 3.
(2) What is the HCF of two consecutive?
Ans: (a) Numbers = 1
(b) Even numbers = 2
(c) Odd numbers = 1
(3) HCF of co-prime numbers 4 and 15 was found as follows by factorization: 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
Ans: No; the correct HCF is 1.
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