## NCERT Solutions Class 6 Maths## Chapter - 3: Playing with Numbers## Exercise 3.1
1 × 24 = 24 2 × 12 = 24 3 × 8 = 24 6 × 4 = 24 Or 24 × 1 = 24 12 × 2 = 24 8 × 3 = 24 4 × 6 = 24
1 × 15 = 15 3 × 5 = 15 Or 15 × 1 = 15 5 × 3 = 15
1 × 21 = 21 3 × 7 = 21 Or 21 × 1 = 21 7 × 3 = 21
1 × 27 = 27 3 × 9 = 27 Or 27 × 1 = 27 9 × 3 = 27
1 × 12 = 12 2 × 6 = 12 3 × 4 = 12 Or 12 × 1 = 12 6 × 2 = 12 4 × 3 = 12
1 × 20 = 20 2 × 10 = 20 5 × 4 = 20 Or 20 × 1 = 20 10 × 2 = 20 4 × 5 = 20
1 × 18 = 18 2 × 9 = 18 3 × 6 = 18 Or 18 × 1 = 18 9 × 2 = 18 6 × 3 = 18
1 × 23 = 23 Or 23 × 1 = 23
1 × 36 = 36 2 × 18 = 36 4 × 9 = 36 3 × 12 = 36 Or 36 × 1 = 36 18 × 2 = 36 9 × 4 = 36 12 × 3 = 36
5 × 1 = 5 5 × 2 = 10 5 × 3 = 15 5 × 4 = 20 5 × 5 = 25
8 × 1 = 8 8 × 2 = 16 8 × 3 = 24 8 × 4 = 32 8 × 5 = 40
9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45
9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45 9 × 6 = 54 9 × 7 = 63 9 × 8 = 72 9 × 9 = 81 9 × 10 = 90 9 × 11 = 99 ## Exercise 3.2
For example, 3 + 3 = 6 5 + 7 = 12 7 + 9 = 16
For example, 4 + 6 = 10 2 + 6 = 8 10 + 14 = 24
(a) The sum of three odd numbers is even.
For example, 3 + 3 + 3 = 9 1 + 1 + 1 = 3 (b) The sum of two odd numbers and one even number is even.
For example, 3 (odd) + 5 (odd) + 2 (even) = 10 (even) 1 (odd) + 7 (odd) + 4 (even) = 12 (even) (c) The product of three odd numbers is odd.
For example, 3 × 1 × 3 = 9 5 × 1 × 3 = 15 (d) If an even number is divided by 2, the quotient is always odd.
For example, 10/2 = 5 (odd) 20/2 = 10 (even) (e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
For example, The factors of prime number 7 are 1 and 7. (g) Sum of two prime numbers is always even.
For example, 2 (prime) + 3 (prime) = 5 (odd) (h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
For example, 4 × 6 = 24 (even) 10 × 4 = 40 (even)
Thus, there ar three such pairs of prime numbers upto 100. 1. 17 and 71 2. 37 and 73 3. 79 and 97
The prime and composite numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19.
7 is the greatest prime number between 1 and 10.
(a) 44
(b) 36
(c) 24
(d) 18
[Remark : Two prime numbers whose difference is 2 are called twin primes].
We can give any three pairs of prime numbers whose difference is 2. Other pairs include 5 and 7, 29 and 31
(a) 23
(b) 51
(c) 37
(d) 26
The seven consecutive composite numbers less than 100 with no prime number between them are 90, 91, 92, 93, 94, 95, and 96. Consecutive numbers are the numbers that follow each other continuously in the order of small to large.
21 = 3 + 5 + 13 Where, 3, 5, and 13 are the odd prime numbers
31 = 3 + 5 + 23 Where, 3, 5, and 13 are the odd prime numbers
3 + 7 + 43 = 53 Where 3, 7, and 43 are the odd prime numbers Or
13 + 17 + 23 = 43 Where, 13, 17, and 23 are the odd prime numbers.
13 + 7 + 41 = 61 Where, 13, 7, and 41 are the odd prime numbers.
(Hint : 3+ 7 = 10)
The five pairs of prime numbers less than 20 are: 1. 2 + 3 = 5 2. 13 + 2 = 15 3. 17 + 3 = 20 4. 3 + 7 = 10 5. 13 + 17 = 30 Other pair is: 6. 11 + 19 = 30
(a) A number which has only two factors is called a (b) A number which has more than two factors is called a (c) 1 is neither (d) The smallest prime number is (e) The smallest composite number is (f) The smallest even number is ## Exercise 3.3
If the last two digits of a number are divisible by 4e number is said to be divisible by 4. If the last three digits of a number are divisible by 8,the number is said to be divisible by 8.
The last three digits of 572, i.e., 572 is not divisible by 8. Hence, the number is also not divisible by 8.
The last three digits of 726352, i.e., 352 is divisible by 8. Hence, the number is also divisible by 8.
The last three digits of 5500, i.e., 500 is not divisible by 8. Hence, the number is also not divisible by 8.
The last three digits of 6000, i.e., 000 is divisible by 8. Hence, the number is also divisible by 8.
The last three digits of 12159, i.e., 159 is not divisible by 8. Hence, the number is also not divisible by 8.
The last three digits of 14560, i.e., 560 is divisible by 8. Hence, the number is also divisible by 8.
The last three digits of 21084, i.e., 084 is not divisible by 8. Hence, the number is also not divisible by 8.
The last three digits of 31795072, i.e., 072 is divisible by 8. Hence, the number is also divisible by 8.
The last three digits of 1700, i.e., 700 is not divisible by 8. Hence, the number is also not divisible by 8.
The last three digits of 2150, i.e., 150 is not divisible by 8. Hence, the number is also not divisible by 8. Thus,
If a number is divisible by both 2 and 3, it is also divisible by 6. A number with 0, 2, 4, 6, and 8 at the unit's place is divisible by 2. If the sum of digits is a multiple of 3, the number is divisible by 3.
297144 has number 4 at the unit place. Hence, it is divisible by 2.
2 + 9 + 7 + 1 + 4 + 4 = 27 27 is a multiple of 3. 9 × 3 = 27 Hence, it is divisible by 3. 297144 is divisible by both 2 and 3. Thus, it is also divisible by 6.
1258 has number 8 at the unit place. Hence, it is divisible by 2.
1 + 2 + 5 + 8 = 16 16 is not a multiple of 3. Hence, it is not divisible by 3. 1258 is divisible by 2 but not by 3. Thus, it is not divisible by 6.
4335 has number 5 at the unit place. Hence, it is not divisible by 2.
4 + 3 + 3 + 5 = 15 15 is a multiple of 3. 3 x 5 = 15 Hence, it is divisible by 3. 1258 is divisible by 3 but not by 2. Thus, it is not divisible by 6.
It is not divisible by 6.
61233 has number 3 at the unit place. Hence, it is not divisible by 2.
6 + 1 + 2 + 3 + 3 = 15 15 is a multiple of 3. 3 x 5 = 15 Hence, it is divisible by 3. 61233 is divisible by 3 but not by 2. Thus, it is not divisible by 6.
901352 has number 2 at the unit place. Hence, it is divisible by 2.
9 + 0 + 1 + 3 + 5 + 2 = 20 20 is not a multiple of 3. Hence, it is not divisible by 3. 901352 is divisible by 2 but not by 3. Thus, it is not divisible by 6.
438750 has number 0 at the unit place. Hence, it is divisible by 2.
4 + 3 + 8 + 7 + 5 + 0 = 27 27 is a multiple of 3. 9 × 3 = 27 Hence, it is divisible by 3. 438750 is divisible by both 2 and 3. Thus, it is also divisible by 6.
1790184 has number 4 at the unit place. Hence, it is divisible by 2.
1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 30 is a multiple of 3. 10 × 3 = 30 Hence, it is divisible by 3. 1790184 is divisible by both 2 and 3. Thus, it is also divisible by 6.
12583 has number 3 at the unit place. Hence, it is not divisible by 2.
1 + 2 + 5 + 8 + 3 = 19 19 is not a multiple of 3. Hence, it is not divisible by 3. 12583 is neither divisible by 2 nor 3. Thus, it is not divisible by 6.
639210 has number 0 at the unit place. Hence, it is divisible by 2.
6 + 3 + 9 + 2 + 1 + 0 = 21 21 is a multiple of 3. 7 × 3 = 21 Hence, it is divisible by 3. 639210 is divisible by both 2 and 3. Thus, it is also divisible by 6.
17852 has number 2 at the unit place. Hence, it is divisible by 2.
1 + 7 + 8 + 5 + 2 = 23 23 is not a multiple of 3. Hence, it is not divisible by 3. 17852 is divisible by 2 but not by 3. Thus, it is not divisible by 6.
A number is divisible by 11 if the
Sum of digits at odd places: 5 + 4 = 9 Sum of digit at even places: 4 + 5 = 9 Difference: 9 - 9 = 0 Hence, 5445 is divisible by 11.
Sum of digits at odd places: 4 + 8 + 1 = 13 Sum of digit at even places: 2 + 0 = 2 Difference: 13 - 2 = 11 11 is divisible by 11. 11 × 1 = 11 Hence, 10824 is divisible by 11.
Sum of digits at odd places: 5 + 9 + 3 + 7 = 24 Sum of digit at even places: 6 + 8 + 1 = 15 Difference: 24 - 15 = 9 9 is not divisible by 11. Hence, 7138965 is not divisible by 11.
Sum of digits at odd places: 8 + 3 + 6 + 0 = 17 Sum of digit at even places: 0 + 9 + 1 + 7 = 17 Difference: 17 - 17 = 0 Hence, 70169308 is divisible by 11.
Sum of digits at odd places: 1 + 0 + 0 + 0 = 1 Sum of digit at even places: 0 + 0 + 0 + 1 = 1 Difference: 1 - 1 = 0 Hence, 10000001 is divisible by 11.
Sum of digits at odd places: 3 + 1 + 0 = 4 Sum of digit at even places: 5 + 1 + 9 = 15 Difference: 15 - 4 = 11 11 is divisible by 11. 11 × 1 = 11 Hence, 901153 is divisible by 11.
If the sum of digits is a multiple of 3, the number is divisible by 3.
a + 6 + 7 + 2 + 4 = a + 19
In case of smallest, we need to consider the nearest number. The nearest number to 19, which is divisible by 3 is 21. a + 19 = 21 a = 21 - 19 a = 2 Thus, 2 is the smallest digit that makes the number divisible by 3.
The farthest number to 19, which is divisible by 3 is 27. If we will consider 30, the gap will be more than 10. We need to consider the gap between the numbers to be less than 9. a + 19 = 27 a = 27 - 19 a = 8 Thus, 8 is the greatest digit that makes the number divisible by 3. (b) 4765 __ 2
4 + 7 + 6 + 5 + a + 2 = a + 24
In case of smallest, we need to consider the nearest number. The nearest number to 24 is 24 only. It is because 24 is also divisible by 3. a + 24 = 24 a = 24 - 24 a = 0 Thus, 0 is the smallest digit that makes the number divisible by 3.
The farthest number to 24, which is divisible by 3 is 33. If we will consider 36, the gap will be more than 10. We need to consider the gap between the numbers to be less or equal to 9. a + 24 = 33 a = 33 - 24 a = 9 Thus, 9 is the greatest digit that makes the number divisible by 3.
A number is divisible by 11 if the
Let the blank space digit be a. Sum of digits at odd places: 9 + 3 + 2 = 14 Sum of digits at even places: 8 + a + 9 = 17 + a Difference = (17 + a) - 14 = 3 + a The nearest number to 3 that is divisible by 11 is 11. 3 + a = 11
We cannot take 0 because the digit may become negative. 3 + a = 0, a = -3 It is not a required digit. We only require positive digits.
Let the blank space digit be a. Sum of digits at odd places: 4 + 4 + a = 8 + a Sum of digits at even places: 8 + 9 + 8 = 25 Difference = 25 - (8 + a) = 25 - 8 - a = 17 - a The nearest number that is divisible by 11 is 11. 17 - a = 11 a = 17 - 11 a = 6 We cannot take 22 because the number (a) is in negative. By taking 22, we may get a negative number. 17 - a = 22 a = -5 It is not a required digit. We only require positive digits. If we take 0, we may get a two digit number. We only require a single digit number in the blank space. 17 - a = 0, a = 17 It is also not required. ## Exercise 3.41. Find the common factors of :
Factors of 20: 1, 2, 4, 5, 10, 20 Factors of 28: 1, 2, 4, 7, 14, 28 These two number has 1, 2, and 4 in common.
Factors of 15: 1, 3, 5, 15 Factors of 25: 1, 5, 25 These two numbers has 1 and 5 in common.
Factors of 35: 1, 5, 7, 35 Factors of 50: 1, 2, 5, 10, 25, 50 These two numbers has 1 and 5 in common.
Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56 Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120 These two numbers has 1, 2, 4, and 8 in common.
Factors of 4: 1, 2, 4 Factors of 8: 1, 2, 4, 8 Factors of 12: 1, 2, 3, 4, 6, 12 These three numbers has 1, 2, and 4 in common.
Factors of 5: 1, 5 Factors of 15: 1, 3, 5 Factors of 25: 1, 5, 25 These three numbers has 1 and 5 in common.
Common multiple refers to the numbers that are divisible by both 6 and 8. Multiples of 6: 6, 12, 18, Multiples of 8: 8, 16, Thus, the first three common multiples are 24, 48, and 72.
Common multiple refers to the numbers that are divisible by both 12 and 18. Multiples of 12: 12, 24, Multiples of 18: 18, Thus, the first three common multiples are 36, 72, and 108.
Multiples of 3: 3, 6, 9, Multiples of 4: 4, 8, Thus, the common multiples of 3 and 4 less than 100 are 12, 24, 36, 48, 60, 72, 84, and 96.
Two numbers having only 1 as a common factor are called co-prime numbers.
18 and 35 have only 1 as a common factor. Factors of 18: 1, 2, 3, 6, 9, 18 Factors of 35: 1, 5, 7, 35
15 and 37 have only 1 as a common factor. Factors of 15: 1, 3, 5, 15 Factors of 37: 1, 37
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30 Factors of 415: 1, 5, 83, 415 30 and 415 have factors 1 and 5 in common. Hence, it is not a co-prime number.
Factors of 17: 1, 17 Factors of 68: 1, 2, 4, 17, 34, 68 17 and 68 have factors 1 and 17 in common. Hence, it is not a co-prime number.
Factors of 216: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216. Factors of 215: 1, 5, 45, 215
Factors of 81: 1, 3, 9, 27, 81 Factors of 16: 1, 2, 4, 8, 16
12 and 5 are co-prime numbers because it has only 1 as a common factor. Thus, if a number is divisible by both 5 and 12, it will be always divisible by its product, i.e. 12 × 5 = 60.
If a number is divisible by 12, it is also divisible by its factors. Factors of 12: 1, 2, 3, 4, 6, and 12 ## Exercise 3.5
For example, Numbers: 6, 12, 15, 21, 24 These numbers are divisible by 3, but not by 9.
For example, Numbers: 9, 18, 27, 36, 45 These numbers are divisible by 9. Hence, also divisible by 3.
For example, Numbers: 12, 24, 30, 42 The above numbers are divisible by both 3 and 6, but not by 18. In some cases, it might be true. For example, Numbers: 18, 36, 54, 72 These numbers are divisible by 3, 6, as well as 18.
For example, 90, 180, 270, 360
For example, 9 and 20 Factors of 9: 1, 3, 9 Factors of 20: 1, 2, 5, 10, 20 Both have only 1 as a common factor and neither of them is a prime number.
For example, Numbers: 12, 16, 20, 24, 28, 32, 36 All these numbers are divisible by 4. But among these, only 16, 24, and 32 are divisible by 8.
Factors of 8: 1, 2, 4, and 8 Thus, if a number is divisible by 8, it is always divible by its factors. For example, Numbers: 8, 16, 24, 32, 40, 48, 56 These numbers are divisible both by 4 and 8.
For example, Number: 4 4 divides 8 and 16. Sum: 8 + 16 = 24 Thus, number 4 exactly divides 8 and 16 separately, and also divides their sum (8 + 16 = 24).
For example, Number: 6 6 divides 24. But, it does not divide their sum (10 + 14). It is possible only in some cases. For example, 6 divides 24 and also divide the numbers separately (18 + 6). Thus, the above condition is true only in some cases.
(a) 2 × 3 = 6 5 × 2 = 10 (b) 30 × 2 = 60 10 × 3 = 30 2 × 5 = 10
9999 is the greatest 4-digit number.
101 × 11 × 3 × 3 = 9999 9999 = 9 × 1111 9999 = (3 × 3) × (101 × 11)
10000 is the smallest 5-digit number.
2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 = 10000 10000 = 4 × 2500 10000 = (2 × 2) × (25 × 100) 10000 = (2 × 2) × (5 × 5 × 4 × 25) 10000 = (2 × 2) × (5 × 5 × 2 × 2 × 5 × 5) 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
The prime factors of 1729 are 7, 13, and 19. 7 × 13 × 19 = 1729 There are three prime factors. The relation between them is: 13 - 7 = 6 19 - 13 = 6 Thus, the three consective prime factors in ascending order differs by 6. ## Note: Factors and prime factors are the different terms. Prime factors does not include factor 1 and the number itself.
1 × 2 × 3 = 6
3 × 4 × 5 = 60
6 × 7 × 8 = 336
8 × 9 × 10 = 720
For example,
1 + 3 = 4 4 is divisible by 4.
3 + 5 = 8 8 is divisible by 4.
5 + 7 = 12 12 is divisible by 4.
7 + 9 = 16 16 is divisible by 4.
9 + 11 = 20 20 is divisible by 4.
Prime factors are the factors that divides the number repeatedly so long as the quotient is divisible by that number.
Correct
Correct
45 = 9 × 5 Since, 5 and 9 are the co-prime numbers. So, if a number is divisible by 5 and 9, it is also divisible by their product (9 × 5 = 45).
If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9. 2 + 5 + 1 + 1 + 0 = 9 9 × 1 = 9 9 is divisible by 9. Hence, 25110 is divisible by 9.
If the unit digit of a number is either 0 or 5, the number is divisible by 5. The unit digit of 25110 is 0. Hence, it is divisible by 5. 25110 is divisible by both 5 and 9. Hence, it is also divisible by their product (9 × 5 = 45).
If a number is divisible by both 4 and 6, it id not necessary that is also divisible by 24. For example, 12, 36, 60 The above numbers are divisible by both 4 and 6, but not by 24.
Number: 2 × 3 × 5 × 7 Number ## Exercise 3.6
The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors.
Factors of 18: Factors of 48: 2 × 2 × 2 × Common factors of 18 and 48 are: 2 × 3 Thus, HCF = 6
Factors of 30: Factors of 42: Common factors of 30 and 42 are: 2 × 3 Thus, HCF = 6
Factors of 18: Factors of 60: Common factors of 18 and 60 are: 2 × 3 Thus, HCF = 6
Factors of 27: Factors of 63: Common factors of 27 and 63 are: 3 × 3 Thus, HCF = 9
Factors of 36: Factors of 84: Common factors of 36 and 84 are: 2 × 2 × 3 Thus, HCF = 12
Factors of 34: Factors of 102: Common factors of 34 and 102 are: Thus, HCF = 34
Factors of 70: 2 × Factors of 105: Factors of 175: Common factors of 70, 105, and 175 are: 7 × 5 Thus, HCF = 35
Factors of 91: Factors of 112: 2 × 2 × 2 × 2 × Factors of 49: Common factors of 91, 112, and 49 are: 7 Thus, HCF = 7
Factors of 18: 2 × Factors of 54: 2 × Factors of 81: 3 × Common factors of 18, 54, and 81 are: Thus, HCF = 9
Factors of 18: 2 × 2 × Factors of 45: 5 × 3 × Factors of 75: 5 × 5 × Common factors of 12, 45, and 75 are: Thus, HCF = 3
There is only one common factor between any two consecutive numbers, i.e. 1. Hence, HCF of two consecutive numbers is 1. For example, 1 and 2 3 and 4 5 and 6 6 and 7
There is only one common factor between any two consecutive even numbers, i.e. 2. Hence, HCF of two consecutive even numbers is 2. For example, 2 and 4 4 and 6 6 and 8 8 and 10
There is only one common factor between any two consecutive odd numbers, i.e. 1. Hence, HCF of two consecutive odd numbers is 1. For example, 1 and 3 5 and 7 7 and 9 11 and 13
The correct answer is HCF of any two numbers can never be 0. The common factor of any two co-prime numbers is 1. Hence, HCF of such numbers with no common prime factors is always 1. ## Exercise 3.7
Weight of two bags of fertilizer = 75 kg and 69 kg Maximum value of weight = HCF (75, 69) Factors of 75: 5 × 5 × 3 Factors of 69: 3 × 23 Common factors of 75 and 69 are: 3 Thus, HCF = 3 The maximum value of weight 75 kg and 69 kg which can measure the weight of the fertiliser exact number of times is
The steps measure of the first boy = 63 cm The steps measure of the second boy = 70 cm The steps measure of the third boy = 77 cm Minimum distance = LCM (63, 70, 77) LCM = 3 × 7 × 3 × 10 × 11 LCM = 6930 Thus, the minimum distance of
The length of a room = 825 cm The breadth of a room = 675 cm The height of a room = 450 cm Longest tape = HCF (825, 675, 450) Factors of 825: 5 × 5 × 3 × 11 Factors of 675: 5 × 5 × 3 × 3 × 3 Factors of 450: 5 × 5 × 3 × 3 × 2 Common factors of 825, 675, 450 are: 5 × 5 × 3 Thus, HCF = 75 The longest tape which can measure the three dimensions of the room exactly is of length 75 cm.
To find the smallest 3-digit number, we need to find the LCM. LCM (Lowest Common Factor) of 6, 8 and 12 is: LCM = 3 × 2 × 2 × 2 LCM = 24 The smallest three digit number should be greater than 100. The nearest multiple of 24 greater than 100 is 24 × 5 = 120. Thus, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is
LCM (8, 10, 12) LCM = 2 × 2 × 2 × 5 × 3 = 120 The nearest 3-digit multiple of 120 near 999 is 120 × 8 = 960. Thus, the greatest 3-digit number exactly divisible by 8, 10 and 12 is
LCM (48, 72, 108) LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 = 60 × 7 + 12 = 7 minutes and 12 seconds Thus, the traffic lights at three different road crossings will change simultaneously at
Maximum capacity = HCF (403, 434, 465) Factors of 403: 13 × 31 Factors of 434: 14 × 31 Factors of 465: 15 × 31 Common factor: 31 Thus, the maximum capacity of a container that can measure the diesel of the three containers exact number of times is
= 90 Least number = 90 Least number leaving remainder 5 = 90 + 5 = 95 Thus,
LCM (18, 24, 32) LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 LCM = 288 The nearest multiple of 288 to 1000 is 288 × 4 = 1152 Thus, the smallest 4-digit number which is divisible by 18, 24 and 32 is
= 9 × 4 = 36
= 60
= 30
= 60
The above two numbers are co-prime numbers, i.e., has only 1 as the common factor. LCM of such numbers is their product.
LCM = 5 × 2 × 2 LCM = 20
LCM = 2 × 3 × 3 LCM = 18
LCM = 2 × 3 × 2 × 2 × 2 LCM = 48
LCM = 3 × 3 × 5 LCM = 45 What do you observe in the results obtained?
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