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Chapter 8

Playing With Numbers — Exercise 8(C)

Class - 6 Concise Mathematics Selina

Exercise 8(C)

Question 1(i)

Using the common factor method, find the H.C.F. of:

25 and 20

Answer

Factors of 25, i.e. F25 = 1, 5 and 25

F20 = 1, 2, 4, 5, 10 and 20

Now, Factors that are common to F25 and F20,

= 1 and 5

From the common factors, the highest common factor = 5

⇒ H.C.F. of 25 and 20 = 5

Question 1(ii)

Using the common factor method, find the H.C.F. of:

8, 12 and 18

Answer

F8 = 1, 2, 4 and 8

F12 = 1, 2, 3, 4, 6 and 12

F18 = 1, 2, 3, 6, 9 and 18

Now factors common to F8, F12 and F18,

= 1 and 2

From the common factors, the highest common factor = 2

⇒ H.C.F. of 8, 12 and 18 = 2

Question 1(iii)

Using the common factor method, find the H.C.F. of:

24, 36, 45 and 60

Answer

F24 = 1, 2, 3, 4, 6, 8, 12 and 24

F36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36

F45 = 1, 3, 5, 9, 15 and 45

F60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60

Factors common to F24, F36, F45 and F60,

= 1 and 3

From the common factors, the highest common factor = 3

⇒ H.C.F. of 24, 36, 45 and 60 = 3

Question 2(i)

Using the prime factor method, find the H.C.F. of:

40, 60 and 80

Answer

Splitting each number into its prime factors :

40 = 2 × 2 × 2 × 5

60 = 2 × 2 × 3 × 5

80 = 2 × 2 × 2 × 2 × 5

The prime factors common to the given numbers are 2, 2 and 5.

⇒ H.C.F. of 40, 60 and 80 = 2 × 2 × 5 = 20

Question 2(ii)

Using the prime factor method, find the H.C.F. of:

48, 84 and 88

Answer

Splitting each number into its prime factors :

48 = 2 × 2 × 2 × 2 × 3

84 = 2 × 2 × 3 × 7

88 = 2 × 2 × 2 × 11

The prime factors common to the given numbers are 2 and 2.

⇒ H.C.F. of 48, 84 and 88 = 2 × 2 = 4

Question 2(iii)

Using the prime factor method, find the H.C.F. of:

12, 16 and 28

Answer

Splitting each number into its prime factors :

12 = 2 × 2 × 3

16 = 2 × 2 × 2 × 2

28 = 2 × 2 × 7

The prime factors common to the given numbers are 2 and 2.

⇒ H.C.F. of 12, 16 and 28 = 2 × 2 = 4

Question 3(i)

Using the division method, find the H.C.F. of the following:

16 and 24

Answer

By division method,

16)24(‾1[Dividing the bigger number 24 by the smaller number 16]))−16‾++8)16(‾2[Dividing the divisor 16 by the remainder 8]++))−16‾+++(×[No remainder is left, so the last divisor 8 is the H.C.F.]\begin{array}{ll} 16\overline{\smash{\big)}24\smash{\big(}} 1 & \text{[Dividing the bigger number 24 by the smaller number 16]} \\ \phantom{)}\phantom{)}\underline{-16} & \\ \phantom{++} 8 \overline{\smash{\big)} 16\smash{\big(}} 2 & \text{[Dividing the divisor 16 by the remainder 8]} \\ \phantom{++)}\phantom{)}\underline{-16} & \\ \phantom{+++(} \phantom{} \times & \text{[No remainder is left, so the last divisor 8 is the H.C.F.]} \\ \end{array}

Hence, H.C.F. of 16 and 24 = 8

Question 3(ii)

Using the division method, find the H.C.F. of the following:

7, 14 and 24

Answer

Let us first find the H.C.F. of 7 and 14.

7)14(‾2[Dividing the bigger number 14 by the smaller number 7].−14‾+.×[No remainder is left, so the last divisor 7 is the H.C.F.]\begin{array}{ll} 7\overline{\smash{\big)}14\smash{\big(}} 2 & \text{[Dividing the bigger number 14 by the smaller number 7]} \\ \phantom{}\phantom{.} \underline{-14} & \\ \phantom{+} \phantom{}\phantom{.} \times & \text{[No remainder is left, so the last divisor 7 is the H.C.F.]} \\ \end{array}

H.C.F. of 7 and 14 = 7

Since, the third number is 24 and the H.C.F. obtained above is 7, find the H.C.F. of 24 and 7.

7)24(‾3[Dividing 24 by the H.C.F. 7 obtained above].−21‾+)03)7(‾2[Dividing the divisor 7 by the remainder 3]+).0−6‾++.01)3(‾3[Dividing the divisor 3 by the remainder 1]++)).0−3‾+++(0×[No remainder is left, so the last divisor 1 is the H.C.F.]\begin{array}{ll} 7\overline{\smash{\big)}24\smash{\big(}} 3 & \text{[Dividing 24 by the H.C.F. 7 obtained above]} \\ \phantom{}\phantom{.}\underline{-21} & \\ \phantom{+)} \phantom{0} 3 \overline{\smash{\big)} 7\smash{\big(}} 2 & \text{[Dividing the divisor 7 by the remainder 3]} \\ \phantom{+)}\phantom{.0}\underline{-6} & \\ \phantom{++} \phantom{.0} 1 \overline{\smash{\big)} 3\smash{\big(}} 3 & \text{[Dividing the divisor 3 by the remainder 1]} \\ \phantom{++)}\phantom{)}\phantom{.0}\underline{-3} & \\ \phantom{+++(} \phantom{0} \times & \text{[No remainder is left, so the last divisor 1 is the H.C.F.]} \\ \end{array}

Hence, H.C.F. of 7, 14 and 24 = 1

Question 3(iii)

Using the division method, find the H.C.F. of the following:

32, 56 and 46

Answer

Let us first find the H.C.F. of 32 and 56.

32)56(‾1[Dividing the bigger number 56 by the smaller number 32]))−32‾+()24)32(‾1[Dividing the divisor 32 by the remainder 24]++))−24‾+++()8)24(‾3[Dividing the divisor 24 by the remainder 8]+++))−24‾++++(×[No remainder is left, so the last divisor 8 is the H.C.F.]\begin{array}{ll} 32\overline{\smash{\big)}56\smash{\big(}} 1 & \text{[Dividing the bigger number 56 by the smaller number 32]} \\ \phantom{)}\phantom{)}\underline{-32} & \\ \phantom{+()} 24 \overline{\smash{\big)} 32\smash{\big(}} 1 & \text{[Dividing the divisor 32 by the remainder 24]} \\ \phantom{++)}\phantom{)}\underline{-24} & \\ \phantom{+++()} 8 \overline{\smash{\big)} 24\smash{\big(}} 3 & \text{[Dividing the divisor 24 by the remainder 8]} \\ \phantom{+++)}\phantom{)}\underline{-24} & \\ \phantom{++++(} \phantom{} \times & \text{[No remainder is left, so the last divisor 8 is the H.C.F.]} \\ \end{array}

H.C.F. of 32 and 56 = 8

Since, the third number is 46 and the H.C.F. obtained above is 8, find the H.C.F. of 46 and 8.

8)46(‾5[Dividing 46 by the H.C.F. 8 obtained above])−40‾++6)8(‾1[Dividing the divisor 8 by the remainder 6]++))−6‾+++)2)6(‾3[Dividing the divisor 6 by the remainder 2]+++))−6‾++++(×[No remainder is left, so the last divisor 2 is the H.C.F.]\begin{array}{ll} 8\overline{\smash{\big)}46\smash{\big(}} 5 & \text{[Dividing 46 by the H.C.F. 8 obtained above]} \\ \phantom{}\phantom{)}\underline{-40} & \\ \phantom{++} 6 \overline{\smash{\big)} 8\smash{\big(}} 1 & \text{[Dividing the divisor 8 by the remainder 6]} \\ \phantom{++)}\phantom{)}\underline{-6} & \\ \phantom{+++)} 2 \overline{\smash{\big)} 6\smash{\big(}} 3 & \text{[Dividing the divisor 6 by the remainder 2]} \\ \phantom{+++)}\phantom{)}\underline{-6} & \\ \phantom{++++(} \phantom{} \times & \text{[No remainder is left, so the last divisor 2 is the H.C.F.]} \\ \end{array}

Hence, H.C.F. of 32, 56 and 46 = 2

Question 4(i)

Using any suitable method, find the H.C.F. of:

45, 75 and 135

Answer

Splitting each number into its prime factors :

45 = 3 × 3 × 5

75 = 3 × 5 × 5

135 = 3 × 3 × 3 × 5

The prime factors common to the given numbers are 3 and 5.

⇒ H.C.F. of 45, 75 and 135 = 3 × 5 = 15

Question 4(ii)

Using any suitable method, find the H.C.F. of:

66, 33 and 132

Answer

Splitting each number into its prime factors :

66 = 2 × 3 × 11

33 = 3 × 11

132 = 2 × 2 × 3 × 11

The prime factors common to the given numbers are 3 and 11.

⇒ H.C.F. of 66, 33 and 132 = 3 × 11 = 33

Question 4(iii)

Using any suitable method, find the H.C.F. of:

24, 36, 60 and 132

Answer

Splitting each number into its prime factors :

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

60 = 2 × 2 × 3 × 5

132 = 2 × 2 × 3 × 11

The prime factors common to the given numbers are 2, 2 and 3.

⇒ H.C.F. of 24, 36, 60 and 132 = 2 × 2 × 3 = 12

Question 5

Find the greatest number that divides each of 180, 225 and 315 completely.

Answer

The greatest number that divides each of the given numbers completely is their H.C.F.

Splitting each number into its prime factors :

180 = 2 × 2 × 3 × 3 × 5

225 = 3 × 3 × 5 × 5

315 = 3 × 3 × 5 × 7

The prime factors common to the given numbers are 3, 3 and 5.

⇒ H.C.F. = 3 × 3 × 5 = 45

Hence, the greatest number that divides 180, 225 and 315 completely is 45.

Question 6

Show that 45 and 56 are co-prime numbers.

Answer

Two numbers are said to be co-prime if their H.C.F. is 1.

Splitting each number into its prime factors :

45 = 3 × 3 × 5

56 = 2 × 2 × 2 × 7

Since, there is no prime factor common to 45 and 56,

⇒ H.C.F. of 45 and 56 = 1

Hence, 45 and 56 are co-prime numbers.

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