Playing With Numbers Class 6 Ex. 3.6
Ncert Class 6 Math Free Solution.
Exercise 3.6
Question 1 :- 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
Solution1 :-
(a) Given numbers are 18 and 48.
The factors of 18 = 2 x 3 x 3
The factors of 48 = 2 X 2 X 2 X 2 X 3
The common factors are 2 and 3.
Hence, the H.C.F (18, 48) = 2 x 3 = 6.
(b) Given numbers are 30 and 42.
The factors of 30 = 2 X 3 X 5
The factors of 42 = 2 x 3 x 7
The common factors are 2 and 3.
Hence, the H.C.F (30, 42) = 2 x 3 = 6.
(c) Given numbers are 18 and 60.
The factors of 18 = 2 x 3 x 3
The factors of 60 = 2 x 2 x 3 x 5
The common factors are 2 and 3.
Hence, the H.C.F (18, 60) = 2 x 3 = 6.
(d) Given numbers are 27 and 63.
The factors of 27 = 3 x 3 x 3
The factors of 63 = 3 X 3 X 7
The common factors are 2 and 3.
Hence, the H.C.F (27, 63) = 3 x 3 = 9.
(e) Given numbers are 36 and 84.
The factors of 36 = 2 X 2 X 3 X 3
The factors of 84 = 2 x 2 x 3 x 7
The common factors are 2 and 3.
Hence, the H.C.F (36, 84) = 2 x 2 x 3 = 12.
(f) Given numbers are 34 and 102.
The factors of 34 = 2 x 17
The factors of 102 = 2 x 3 x 17
The common factors are 2 and 17.
Hence, the H.C.F (34, 102) = 2 x 17 = 34.
(g) The given numbers are 70, 105 and 175.
The factors of 70 = 2 x 5 x 7
The factors of 105 = 3 x 5 x 7
The factors of 175 = 5 x 5 x 7
The common factors are 5 and 7.
Hence, the H.C.F (70, 105, 175) = 5 x 7 = 35.
(h) The given numbers are 91, 112 and 49.
The factors of 91 = 7 X 13
The factors of 112 = 2 x 2 x 2 x 2 x 7
The factors of 49 = 7 x 7
The common factors are 7.
Hence, the H.C.F (91, 112, 49) = 7
(i) The given numbers are 18, 54 and 81.
The factors of 18 = 2 x 3 x 3
The factors of 54 = 2 X 3 X 3 X 3
The factors of 81 = 3 X 3 X 3 X 3
The common factors are 3 and 3.
Hence, the H.C.F (18, 54, 81) = 3 x 3 = 9.
(j) The given numbers are 12, 45 and 75.
The factors of 12 = 2 X 2 X 3
The factors of 45 = 3 x 3 x 5
The factors of 75 = 3 x 5 x 5
The common factors are 3.
Hence, the H.C.F (12, 45, 75) = 3.
Question 2 :- What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution 2:-
(a) H.C.F of two consecutive numbers is always 1.
(b) H.C.F of two consecutive even numbers are 1 and 2.
So, the H.C.F = 1 x 2 = 2.
(c) H.C.F of two consecutive odd numbers is 1.
So, the H.C.F = 1.
Question 3 :- HCF of co-prime numbers 4 and 15 was found as follows by factorisation: 4 = 2 x 2 and 15 = 3 x 5. Since there is no common prime factors, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?
Solution 3:-
No, answer is not correct.
0 is not a prime factor of any number, for this reason.
The prime factor of a co-prime number is always 1.
Hence, the correct HCF of 4 and 15 is 1.
Playing With Numbers Class 6 Exercise. 3.6 for Free