Exercise 3.5 - Chapter 3 - Playing with numbers - Ncert Solutions class 6 - Maths

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Question 1:

Which of the following statements are true?

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

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

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

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

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

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

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

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

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

Answer:

(a) False

6 is divisible by 3, but not by 9.

(b) True, as 9 = 3 × 3

Therefore, if a number is divisible by 9, then it will also be divisible by

3.

(c) False

30 is divisible by 3 and 6 both, but it is not divisible by 18.

(d) True, as 9 × 10 = 90

Therefore, if a number is divisible by 9 and 10 both, then it will also be divisible by 90.

(e) False

15 and 32 are co-primes and also composite.

(f) False

12 is divisible by 4, but not by 8.

(g) True, as 8 = 2 × 4

Therefore, if a number is divisible by 8, then it will also be divisible by 2 and 4.

(h) True

2 divides 4 and 8 as well as 12. (4 + 8 = 12)

(i) False

2 divides 12, but does not divide 7 and 5.

Question 2:

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

 

Question 3:

Which factors are not included in the prime factorization of a composite number?

Answer:

1 and the number itself

Question 4:

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

Question 5:

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

Answer:

Question 6:

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

Answer:

7	1729
13	247
19	19
	1

1729 = 7 × 13 × 19

13 − 7 = 6, 19 − 13 = 6

The difference of 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.

Answer:

2 × 3 × 4 = 24, which is divisible by 6

9 × 10 × 11 = 990, which is divisible by 6

20 × 21 × 22 = 9240, which is divisible by 6

Question 8:

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

Answer:

3 + 5 = 8, which is divisible by 4

15 + 17 = 32, which is divisible by 4

19 + 21 = 40, which is divisible by 4

Question 9:

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

(a) 24 = 2 × 3 × 4 (b) 56 = 7 × 2 × 2 × 2

(c) 70 = 2 × 5 × 7 (d) 54 = 2 × 3 × 9

Answer:

(a) 24 = 2 × 3 × 4

Since 4 is composite, prime factorisation has not been done.

(b) 56 = 7 × 2 × 2 × 2

Since all the factors are prime, prime factorisation has been done.

(c) 70 = 2 × 5 × 7

Since all the factors are prime, prime factorisation has been done.

(d) 54 = 2 × 3 × 9

Since 9 is composite, prime factorisation has not been done.

Question 10:

Determine if 25110 is divisible by 45.

[Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].

Answer:

45 = 5 × 9

Factors of 5 = 1, 5

Factors of 9 = 1, 3, 9

Therefore, 5 and 9 are co-prime numbers.

Since the last digit of 25110 is 0, it is divisible by 5.

Sum of the digits of 25110 = 2 + 5 + 1 + 1 + 0 = 9

As the sum of the digits of 25110 is divisible by 9, therefore, 25110 is divisible by 9.

Since the number is divisible by 5 and 9 both, it is divisible by 45.

Question 11:

18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify our Answer:

Answer:

No. It is not necessary because 12 and 36 are divisible by 4 and 6 both, but are not divisible by 24.

Question 12:

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

Answer:

Since it is the smallest number of such type, it will be the product of 4 smallest prime numbers.

2 × 3 × 5 × 7 = 210