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
= 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
= 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
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=
- HCF of 126 =
- HCF of 108 = 2x 2x 3x3x3
HCF=2x3x3 = 18
So, the required number is HCF of 90, 126, and 108 is 18.
