### Playing with Numbers-Solutions Ex. 3.6

CBSE Class –VI Mathematics

NCERT Solutions

Chaper 3 Playing With Numbers (Ex. 3.6)

NCERT Solutions

Chaper 3 Playing With Numbers (Ex. 3.6)

Question 1. Find the H.C.F. 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

(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

Answer: (a) Factors of 18 = 2 x 3 x 3

Factors of 48 = 2 x 2 x 2 x 2 x 3

H.C.F. (18, 48) = 2 x 3 = 6

Factors of 48 = 2 x 2 x 2 x 2 x 3

H.C.F. (18, 48) = 2 x 3 = 6

(b) Factors of 30 = 2 x 3 x 5

Factors of 42 = 2 x 3 x 7

H.C.F. (30, 42) = 2 x 3 = 6

Factors of 42 = 2 x 3 x 7

H.C.F. (30, 42) = 2 x 3 = 6

(c) Factors of 18 = 2 x 3 x 3

Factors of 60 = 2 x 2 x 3 x 5

H.C.F. (18, 60) = 2 x 3 = 6

Factors of 60 = 2 x 2 x 3 x 5

H.C.F. (18, 60) = 2 x 3 = 6

(d) Factors of 27 = 3 x 3 x 3

Factors of 63 = 3 x 3 x 7

H.C.F. (27, 63) = 3 x 3 = 9

Factors of 63 = 3 x 3 x 7

H.C.F. (27, 63) = 3 x 3 = 9

(e) Factors of 36 = 2 x 2 x 3 x 3

Factors of 84 = 2 x 2 x 3 x 7

H.C.F. (36, 84) = 2 x 2 x 3 = 12

Factors of 84 = 2 x 2 x 3 x 7

H.C.F. (36, 84) = 2 x 2 x 3 = 12

(f) Factors of 34 = 2 x 17

Factors of 102 = 2 x 3 x 17

H.C.F. (34, 102) = 2 x 17 = 34

Factors of 102 = 2 x 3 x 17

H.C.F. (34, 102) = 2 x 17 = 34

(g) Factors of 70 = 2 x 5 x 7

Factors of 105 = 3 x 5 x 7

Factors of 175 = 5 x 5 x 7

H.C.F. (70, 105, 175) = 5 x 7 = 35

Factors of 105 = 3 x 5 x 7

Factors of 175 = 5 x 5 x 7

H.C.F. (70, 105, 175) = 5 x 7 = 35

(h) Factors of 91 = 7 x 13

Factors of 112 = 2 x 2 x 2 x 2 x 7

Factors of 49 = 7 x 7

H.C.F. (91, 112, 49) = 7

Factors of 112 = 2 x 2 x 2 x 2 x 7

Factors of 49 = 7 x 7

H.C.F. (91, 112, 49) = 7

(i) Factors of 18 = 2 x 3 x 3

Factors of 54 = 2 x 3 x 3 x 3

Factors of 81 = 3 x 3 x 3 x 3

H.C.F. (18, 54, 81) = 3 x 3 = 9

Factors of 54 = 2 x 3 x 3 x 3

Factors of 81 = 3 x 3 x 3 x 3

H.C.F. (18, 54, 81) = 3 x 3 = 9

(j) Factors of 12 = 2 x 2 x 3

Factors of 45 = 3 x 3 x 5

Factors of 75 = 3 x 5 x 5

H.C.F. (12, 45, 75) = 3

Factors of 45 = 3 x 3 x 5

Factors of 75 = 3 x 5 x 5

H.C.F. (12, 45, 75) = 3

Question 2. What is the H.C.F. of two consecutive:

(a) numbers?

(b) even numbers?

(c) odd numbers?

Answer:

(a) H.C.F. of two consecutive numbers be 1.

(b) H.C.F. of two consecutive even numbers be 2.

(c) H.C.F. of two consecutive odd numbers be 1.

(a) H.C.F. of two consecutive numbers be 1.

(b) H.C.F. of two consecutive even numbers be 2.

(c) H.C.F. of two consecutive odd numbers be 1.

Question 3. H.C.F. of co-prime numbers 4 and 15 was found as follows by factorization:

4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0.

Is the answer correct? If not, what is the correct H.C.F.?

4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0.

Is the answer correct? If not, what is the correct H.C.F.?

Answer: No. The correct H.C.F. is 1.