**Concise Mathematics, Middle School, Class 6 ICSE**

**Concise Mathematics, Middle School, Class 6 ICSE**

__Ch – 9__

__Playing with Numbers__:

**Ex – 9A:**

**1.** 19 - (1 + 5) - 3

= 19 – 6 – 3

= 10

**2**. 30 x 6 ÷ (5
- 2)

= 30 x 6÷ 3

= 30 x 2 = 60

**3**. 28 - (3 x
8) ÷ 6

= 28 – 24 ÷ 6

= 28 – 4 = 24

**4**. 9 - [(4 -
3) + 2 x 5]

= 9 - [1 + 10]

= 9 - 11 = -2

**5.** [18 - (15
÷ 5) + 6]

= [18 - 3 + 6]

= 21

**6**. [(4 x 2) - (4 ÷ 2)] +
8

= [8 - 2] + 8 = 6 + 8
= 14

**7**. 48 + 96 ÷ 24
- 6 x 18

= 48 + 4 – 108

= 52 - 108 = -56

**8**. 22 - [3 - {8 - (4 + 6)}]

= 22 - [3 - {8 - 10}]

= 22 - [3 - (-2)]

= 22 - [3 + 2]

= 22 - 5 = 17

**9**. 34 - [29 -
{30 + 66 ÷(24 – bar 28 - 26)}]

= 34 - [29 - {30 + 66 ÷ 22}]

= 34 - [29 - {30 + 3}] =

34 - [29 - 33]

= 34 - (-4)

= 34 + 4 = 38

**10**. 60 - {16 ÷ (4 x 6 - 8)}

= 60 - {16 ÷(24 - 8)} =

60 - \{16 ÷ 16}

= 60 - 1 = 59

**11**. 25 - [12 - {5 + 18 ÷ (4 - bar 5-3)}]

= 25 - [12 - {5 + 18 ÷ (4 - 2)}]

= 25 - [12 - {5 + 18 ÷
2 }]

= 25 - [12 - {5 + 9}]

= 25 - [12 - 14]

= 25 - (-2)

= 25 + 2 = 27

**12**. 15 - [16 - {12 + 21 ÷ (9 - 2)}]

= 15 - [16 - {12 + 21
÷ 7}]

= 15 - [16 - {12 + 3}]

= 15 - [16 - 15]

= 15 - 1 = 14

__Ex – 9B__

1.

Find which of the following numbers are
divisible by 2:

(i) 352: The last digit is 2, which is even.

**Divisible by 2**

**Not divisible by 2.**

**Divisible by 2.**

**Not divisible by 2**

**2**. Find which of the following numbers are divisible
by 4:

(A number is divisible by 4 if the last two digits form a number divisible by 4.)

(i)
222:

The
last two digits are 22.

22 is not divisible by 4. 222 is **not** **Divisible by
4**

(ii) 532:

The last two digits are 32.

32 is divisible by 4.

532is **Divisible by 4**

(iii) 678:

The
last two digits are 78.

78 is not divisible by 4.

**678 Not divisible by 4**.

(iv) 9232:

The last two digits are 32.

32 is divisible by 4.

**9232 Divisible by 4**.

3** .** Find which
of the following numbers are divisible by8 Divisibility by 8

(A number is divisible by 8 if the last three digits form a number divisible by 8.)

(i) 324:

The
last three digits are 324.

324 is not divisible by 8 (324 ÷ 8 = 40.5).

**234 Not divisible by 8**.

(ii) 2536:

The
last three digits are 536.

536 is divisible by 8 (536 ÷ 8 = 67).

**2536 Divisible by 8**

(iii) 92760:

The last three digits are 760.

760 is divisible by 8 (760 ÷ 8 = 95).

**92760 Divisible by 8**.

(iv) 444320:

The last
three digits are 320.

320 is divisible by 8 (320 ÷ 8 = 40).

**444320 Divisible by 8**.

**4** Find which of the following numbers are divisible
by 3:-

(A number is
divisible by 3 if the sum of its digits is divisible by 3.)

(1)
221:

Sum of digits = 2 + 2 + 1 = 5, which is not
divisible by 3.

**221 Not divisible by 3**

(ii) 543:

Sum of digits = 5 + 4 + 3 = 12, Find which of the following numbers are divisible by8 Divisibility by 8

(iii) 28492:

Sum of digits = 2 + 8 + 4 + 9 + 2 = 25, which is not
divisible by 3.

**28492 Not divisible by 3.**

(iv) 92349:

Sum
of digits = 9 + 2 + 3 + 4 + 9 = 27, which is divisible by 3.

**92349 Divisible by 3.**

**5**. Find which
of the following numbers are divisible by 9:

(A number is divisible by 9 if the sum of its digits is divisible by 9).

(i)
1332:

Sum of digits = 1 + 3 + 3 + 2 =
9, which is divisible by 9.

**1332 Divisible by 9**

(ii) 53247:

Sum
of digits = 5 + 3 + 2 + 4 + 7 = 21, which is not divisible by 9.

**53247 Not divisible by 9.**

(iii) 4968:

Sum
of digits = 4 + 9 + 6 + 8 = 27, which is divisible by 9.

**4968 Divisible by 9**

(iv) 200314:

Sum
of digits = 2 + 0 + 0 + 3 + 1 + 4 = 10, which is not divisible by 9.

**200314 Not divisible by 9**

**6**. Find which of the following numbers are divisible
by 6:

(A number is divisible by 6 if it is divisible by both 2 and 3.)

(i)
324:

Last
digit is 4 (even) and sum of digits = 3 + 2 + 4 = 9 (divisible by 3).

**324 Divisible by 6**

(ii)
2010:

Last digit is 0 (even) and sum of digits = 2 +
0 + 1 + 0 = 3 (divisible by 3).

**2010 Divisible by 6**

(iii)
33278:

Last digit is 8 (even) and sum of digits = 3 +
3 + 2 + 7 + 8 = 23 (not divisible by 3).

**33278 Not divisible by 6**

(iv)
15505:

Last
digit is 5 (odd).

**15595 Not divisible by 6.**

** **

**7**. Find which
of the following numbers are divisible by 5:

(A number is divisible by 5 if its last digit is 0 or 5)

(i)
5080:

Last
digit is 0.

**5080 Divisible by
5**

(ii)
66666:

Last digit is 6.

**66666 Not divisible by 5.**

(iii) 755:

Last digit is 5.

**755 Divisible by 5**

(iii) 9207:

Last digit is 7.

**9207Not divisible by 5.**

**8**. Find which
of the following numbers are divisible by 10:

(A number is divisible by 10 if its last digit is 0.)

(i)
9990:

Last digit is 0.

**9990 Divisible by 10**

(ii)
0:

Last
digit is 0.

**0 Divisible by 10**

(iii) 847:

Last
digit is 7.

**847 Not divisible by 10**

iv) 8976

Last
digit is 6.

**8976 Not divisible by 10**

**9**. Find which of the following numbers are divisible
by 11:

(A number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is a multiple of 11)

(i)
5918:

Sum of odd-position digits = 5 +
1 = 6. Sum of even-position digits = 9 + 8 = 17. Difference = |6 - 17| =
11.

**5918 Divisible by 11**

(ii)
68717:

Sum
of odd-position digits = 6 + 7 + 7 = 20. Sum of even-position digits = 8 + 1 =
9. Difference = |20 - 9| = 11.

**68717 Divisible by 11**

(iii)
3882:

Sum
of odd-position digits = 3 + 8 = 11. Sum of even-position digits = 8 + 2 = 10.
Difference = |11 - 10| = 1.

**3882 Not divisible by 11**

(iv)
10857:

Sum
of odd-position digits = 1 + 8 + 7 = 16. Sum of even-position digits = 0 + 5 =
5. Difference = |16 - 5| = 11.

**10857 Divisible by 11**

**10**. Find which of the following numbers are divisible
by 15:

(A number is divisible by 15 if it is divisible by both 3 and 5.)

(i)
960:

Sum
of digits = 9 + 6 + 0 = 15 (divisible by 3) and last digit is 0.

**960 Divisible by 15**

(ii)
8295:

Sum of digits = 8 + 2 + 9 + 5 = 24
(divisible by 3) and last digit is 5.

**8295 Divisible by 15**

iii) 10243:

Sum of digits = 1 + 0 +2 +4+3 = 10( not divisible by 3.)

**10243 not Divisible by 15**

iv) 5013:

Sum of digits = 5 + 1 + 3= 9 (divisible by 3) and last digit is 3 ( not
Divisible by 5).

**5013
not Divisible by 15**

__Exercise 9C__

1. Write all the
factors of:

(i) 15

Factors of 15: 1, 3, 5, 15

(ii) 55

Factors of 55: 1, 5, 11, 55

(iii) 48

Factors of 48: 1, 2,3,4,6,8,12,16,24 and 48

(iv) 36

Factors of 36:

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

(v) 84

Factors of 84: 1,2,3,4,6,7,12,14,21,28,42 and 84

2. Write all the
prime numbers:

(i) Less than 25

- Prime numbers less than 25: 2, 3, 5, 7, 11, 13, 17, 19, 23

ii) between 15 and 35

- Prime numbers between 15 and 35: 17, 19, 23, 29, 31

iii) between 8 and 76

- Prime numbers between 8 and 76: 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73

3. Write the prime
numbers from:

(i) 5 to 45

- Prime numbers from 5 to 45:

5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43

(ii) 2 to 32

- Prime numbers from 2
to 32:

2,3, 5, 7, 11, 13, 17, 19, 23, 29, 31

(iii) 8 to 48

- Prime numbers from 8 to 48:

11, 13, 17, 19, 23,
29, 31, 37, 41, 43, 47.

iv) 9 and 59

- Prime numbers between 9 and 59: 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59

4. Write the prime factors of:

(i) 16

- Prime factors of 16: 2

(ii) 27

- Prime factors of 27: 3

(iii) 35

- Prime factors of 35: 5 and 7

(iv) 49

- Prime factors of 49: 7

5. If P means prime
factors of n, find:

(i) P6

- Prime factors of 6: 2 and 3

ii) P24

- Prime factors of 24: 2and3

(ii) P50

- Prime factors of 50: 2, 5

(iii) P42

- Prime factors of 42: 2, 3 and 7

Ex- 9D.

Let's solve the H.C.F. (Highest Common Factor) problems
using the specified methods:

1. Using the Common
Factor Method , find the HCF of :-

(i) 16 and 35

- Factors of 16: 1, 2, 4, 8, 16

- Factors of 35: 1, 5, 7, 35

- Common factor: 1

- H.C.F. = 1

(ii) 25 and 20

- Factors of 25: 1, 5, 25

- Factors of 20: 1, 2, 4, 5, 10, 20

- Common factor: 1, 5

- H.C.F. = 5

(iii) 27 and 75

- Factors of 27: 1, 3, 9, 27

- Factors of 75: 1, 3, 5, 15, 25, 75

- Common factor: 1, 3

- H.C.F. = 3

(iv) 8, 12, and 18

- 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 factor: 1, 2

- H.C.F. = 2

(v) 24, 36, 45, and 60

- 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, 4, 5, 6, 10, 12, 15, 20, 30, 60

- Common factor: 1, 3

- H.C.F. = 3

2. Using the Prime Factor Method, find the HCF of:-

(i) 5 and 8

- Prime factors of 5: 5

- Prime factors of 8: 2 × 2 × 2

- Common factor: None

- H.C.F. = 1

(ii) 24 and 49

- Prime factors of 24: 2 × 2 × 2 × 3

- Prime factors of 49: 7 × 7

- Common factor: None

- H.C.F. = 1

(iii) 40, 60, and 80

- Prime factors of 40: 2 × 2 × 2 × 5

- Prime factors of 60: 2 × 2 × 3 × 5

- Prime factors of 80: 2 × 2 × 2 × 2 × 5

- Common prime factors: 2 × 2 × 5 = 20

- H.C.F. = 20

(iv) 48, 84, and 88

- Prime factors of 48: 2 × 2 × 2 × 2 × 3

- Prime factors of 84: 2 × 2 × 3 × 7

- Prime factors of 88: 2 × 2 × 2 × 11

- Common prime factors: 2 × 2 = 4

- H.C.F. = 4

(v) 12, 16, and 28

- Prime factors of 12: 2 × 2 × 3

- Prime factors of 16: 2 × 2 × 2 × 2

- Prime factors of 28: 2 × 2 × 7

- Common prime factors: 2 × 2 = 4

- H.C.F. = 4

3. Using the Division
Method, find the HCF of the following

(i) 16 and 24

- Divide 24 by 16: 24 ÷ 16 = 1 remainder 8

- Divide 16 by 8: 16 ÷ 8 = 2 remainder 0

- H.C.F. = 8

(ii) 18 and 30

- Divide 30 by 18: 30 ÷ 18 = 1 remainder 12

- Divide 18 by 12: 18 ÷ 12 = 1 remainder 6

- Divide 12 by 6: 12 ÷ 6 = 2 remainder 0

- H.C.F. = 6

(iii) 7, 14, and 24

- Divide 14 by 7: 14 ÷ 7 = 2 remainder 0

- Divide 24 by 7: 24 ÷ 7 = 3 remainder 3

- H.C.F. of 7 and 3 is 1

- H.C.F. = 1

(iv) 70, 80, 120, and 150

- Divide 80 by 70: 80 ÷ 70 = 1 remainder 10

- Divide 70 by 10: 70 ÷ 10 = 7 remainder 0

- H.C.F. = 10

- Divide 120 by 10: 120 ÷ 10 = 12 remainder 0

- Divide 150 by 10: 150 ÷ 10 = 15 remainder 0

- H.C.F. = 10

(v) 32, 56, and 46

- Divide 56 by 32: 56 ÷ 32 = 1 remainder 24

- Divide 32 by 24: 32 ÷ 24 = 1 remainder 8

- Divide 24 by 8: 24 ÷ 8 = 3 remainder 0

- H.C.F. = 8

- Divide 46 by 8: 46 ÷ 8 = 5 remainder 6

- Divide 8 by 6: 8 ÷ 6 = 1 remainder 2

- Divide 6 by 2: 6 ÷ 2 = 3 remainder 0

- H.C.F. = 2

4. Use a method of your choice to find the HCF of

(i) 45, 75, and 135

- Prime factors of 45: 3 × 3 × 5

- Prime factors of 75: 3 × 5 × 5

- Prime factors of 135: 3 × 3 × 3 × 5

- Common prime factors: 3 × 5 = 15

- H.C.F. = 15

(ii) 48, 36, and 96

- Prime factors of 48: 2 × 2 × 2 × 2 × 3

- Prime factors of 36: 2 × 2 × 3 × 3

- Prime factors of 96: 2 × 2 × 2 × 2 × 2 × 3

- Common prime factors: 2 × 2 × 3 = 12

- H.C.F. = 12

(iii) 66, 33, and 132

- Prime factors of 66: 2 × 3 × 11

- Prime factors of 33: 3 × 11

- Prime factors of 132: 2 × 2 × 3 × 11

- Common prime factors: 3 × 11 = 33

- H.C.F. = 33

(iv) 24, 36, 60, and 132

- Prime factors of 24: 2 × 2 × 2 × 3

- Prime factors of 36: 2 × 2 × 3 × 3

- Prime factors of 60: 2 × 2 × 3 × 5

- Prime factors of 132: 2 × 2 × 3 × 11

- Common prime factors: 2 × 2 × 3 = 12

- H.C.F. = 12

(v) 30, 60, 90, and 105

- Prime factors of 30: 2 × 3 × 5

- Prime factors of 60: 2 × 2 × 3 × 5

- Prime factors of 90: 2 × 3 × 3 × 5

- Prime factors of 105: 3 × 5 × 7

- Common prime factors: 3 × 5 = 15

- H.C.F. = 15

5. Greatest number that divides 180, 225, and 315 completely.

- Prime factors of 180: 2 × 2 × 3 × 3 × 5

- Prime factors of 225: 3 × 3 × 5 × 5

- Prime factors of 315: 3 × 3 × 5 × 7

- Common prime factors: 3 × 3 × 5 = 45

- Greatest number = 45

**6**. Show that
45 and 56 are co prime numbers.

45= 3x3x5

56 = 2x 2x 2x7

45 and 56 have no common factor . They are co prime

**7**. Out of 15, 16 ,21 and 28 ,find out all the pairs
of co prime numbers.

1. 15 and 16:

- Factors of 15: (1,
3, 5, 15)

- Factors of 16: (1,
2, 4, 8, 16)

- Common factor: 1

2. 15 and 21:

- Factors of 15: (1,
3, 5, 15)

- Factors of 21: (1,
3, 7, 21)

- Common factors: (1, 3)

3. 15 and 28:

- Factors of 15: (1,
3, 5, 15)

- Factors of 28: (1,
2, 4, 7, 14, 28)

- Common factor: (1)

4. 16 and 21

- Factors of 16: (1,
2, 4, 8, 16\)

- Factors of 21: (1,
3, 7, 21)

- Common factor: (1)

5. 16 and 28:

- Factors of 16: (1,
2, 4, 8, 16)

- Factors of 28: (1,
2, 4, 7, 14, 28)

- Common factors: (1,
2, 4)

**6**. 21 and 28:

- Factors of 21: (1,
3, 7, 21)

- Factors of 28: (1,
2, 4, 7, 14, 28)

- Common factors: (1,
7)

Thus, the pairs of co-prime numbers are (15, 16), (15, 28),
and (16, 21)

**8**. Find the greatest number that will divide 93, 111, 129, leaving reminder 3 in each case.

Since 93-3 = 90, 111-3= 108, and 129- 3 = 126.

Required number is HCF of 90, 108 and 126

1.

**Calculate the HCF**:

- HCF of 90= ${\mathrm{2x\; 3\; x3x5}}^{}$
- HCF of 126 = $\mathrm{2x\; 3x3x7}$
- HCF of 108 = 2x 2x 3x3x3

HCF=2x3x3 = 18

So, the required number is HCF of 90,
126, and 108 is **18**.

**:**

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