NCERT Solutions Class 6th Maths Chapter – 3 Playing With Numbers Exercise – 3.5

NCERT Solutions Class 6th Math Chapter – 3 Playing With Numbers

TextbookNCERT
Class 6th
Subject Mathematics
Chapter3rd
Chapter NamePlaying With Numbers
CategoryClass 6th Mathematics
Medium English
SourceLast Doubt

NCERT Solutions Class 6th Maths Chapter – 3 Playing With Numbers Exercise – 3.5 are helpful for the exam preparation as it covers all types of questions present across the seven exercises of this chapter. The subject experts have created these NCERT Solutions at Last Doubt to provide the best possible methods to solve the problems. These are the ultimate study material a CBSE Class 6th student can get hold of to score good marks in the Maths Examination.

NCERT Solutions Class 6th Math Chapter – 3 Playing With Numbers

Chapter – 3

Playing With Numbers

Exercise – 3.5

Question 1. Which of the following statements are true?

(a) If a number is divisible by 3, it must be divisible by 9.
Solution – False

(b) If a number is divisible by 9, it must be divisible by 3.
Solution – True

(c) A number is divisible by 18, if it is divisible by both 3 and 6.
Solution – False

(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
Solution – True

(e) If two numbers are co-primes, at least one of them must be prime.
Solution – False

(f) All numbers which are divisible by 4 must also be divisible by 8.
Solution – False

(g) All numbers which are divisible by 8 must also be divisible by 4.
Solution – True

(h) If a number exactly divides two numbers separately,- it must exactly divide their sum.
Solution – True

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution – False

Question 2. Here are two different factor trees for 60. Write the missing numbers.

(a)

Playing With Numbers

Solution –

Given that-
Playing With Numbers 2
Here, 6 = 2 x missing number
∴ Missing number = 6 ÷ 2 = 3
Similarly, 10 = 5 x missing number
∴ Missing number = 10 ÷ 5 = 2
Hence, the missing numbers are 3 and 2.

(b)

Playing With Numbers 1

Solution –

Given that-

Let the missing numbers be m1, m2, m3 and m4.
60 = 30 x m1
⇒ m1 = 60 ÷ 30 = 2
30 = 10 x m2
⇒ m2 = 30 ÷ 10 = 3
10 = m3 x m4
⇒ m3 = 2 or 5 and m4 = 5 or 2
Hence, the missing numbers are 2, 3, 2, 5.

Question 3. Which factors are not included in the prime factorisation of a composite number?

Solution –

1 and the number itself are not included in the prime factorisation of a composite number.

Question 4. Write the greatest 4-digit number and express it in terms of its prime factors.

Solution –

The greatest 4-digit number = 9999

Hence, the prime factors of 9999 = 3 x 3 x 11 x 101

Question 5. Write the smallest 5-digit number and express it in the form of its prime factors.

Solution –

The smallest 5-digit number = 10000
Playing With Numbers 5
Hence, the required prime factors: 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5

Question 6. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relations, if any, between the two consecutive prime factors.

Solution –

Given number = 1729

Hence, the prime factors of 1729 = 7 x 13 x 19
Here, 13 – 7 = 6 and 19 – 13 = 6
We see that the difference between two consecutive prime factors is 6

Question 7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Solution –

Among the three consecutive numbers, there must be one even number and one multiple of 3.
As 6 = 2 x 3, thus the product must be multiple of 6.
Example:
(i) 2 × 3 × 4 = 24
(ii) 4 × 5 × 6 = 120

Question 8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Solution –

The sum of any two consecutive numbers is always odd, and the sum of two consecutive odd numbers is always divisible by 4.

= 13 + 15 = 28 and 28 is divisible by 4
= 5 + 7 = 12 and 12 is divisible by 4
= 17 + 19 = 36 and 36 is divisible by 4
= 9 + 11 = 20 and 20 is divisible by 4

Question 9. In which of the following expressions, prime factorization has been done?

(a) 24 = 2 x 3 x 4

Solution –

24 = 2 x 3 x 4
Here, 4 is not a prime number.
Hence, 24 = 2 x 3 x 4 is not a prime factorization.

(b) 56 = 7 x 2 x 2 x 2

Solution –

56 = 7 x 2 x 2 x 2
Here, all factors are prime numbers
Hence, 56 = 7 x 2 x 2 x 2 is a prime factorization.

(c) 70 = 2 x 5 x 7

Solution –

70 = 2 x 5 x 7
Here, all factors are prime numbers.
Hence, 70 = 2 x 5 x 7 is a prime factorization.

(d) 54 = 2 x 3 x 9

Solution –

54 = 2 x 3 x 9
Here, 9 is not a prime number.
Hence, 54 = 2 x 3 x 9 is not a prime factorization.

Question 10. Determine if 25110 is divisible by 45.

Solution –

45 = 5 x 9
Here, 5 and 9 are co-prime numbers.
Test of divisibility by 5- unit place of the given number 25110 is 0. So, it is divisible by 5.
Test of divisibility by 9-
Sum of the digits = 2 + 5 + l + l + 0 = 9 which is divisible by 9.
So, yes the given number is divisible by 5 and 9 both. Hence, the number 25110 is divisible by 45.

Question 11. 18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24? If not, give an example to justify your answer.

Solution –

No, Number 12 is divisible by both 4 and 6; but 12 is not divisible by 24.

Question 12. I am the smallest number, having four different prime factors. Can you find me?

Solution –

We know that the smallest 4 prime numbers are 2, 3, 5 and 7
Hence, the required number = 2 x 3 x 5 x 7 = 210.

NCERT Solution Class 6th Maths All Chapters With Answer

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