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## NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 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.

**(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 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.

Solution :

**(a)** This statement is false.

**(b)** This statement is true.

**(c)** This statement is false.

**(d)** This statement is true.

**(e)** This statement is false.

**(j)** This statement is false.

**(g)** This statement is true.

**(h)** This statement is true.

**(i)** This statement is false.

Question 2.

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

**(a)
(b)
**

Solution :

**(a)**

(b)

Question 3.

(b)

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

Solution :

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

Question 4.

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

Solution :

The greatest 4 digit number is 9999.

∴ 9999 = 3×3× 11 × 101.

Question 5.

Write the smallest 5-digit number and e×press it into the form of its prime factors.

Solution :

The smallest 5-digit number is 10000.

∴ 10000 = 2×2×2×2×5×5×5×5.

Question 6.

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

Solution :

All the prime factors of 1729 are 7, 13 and 19. When arranged in ascending order, these are 7, 13, 19. We observe that 13 – 7 = 6 19 – 13 = 6

**Relation:** The difference between two consecutive prime factors is 6.

Question 7.

The product of three consecutive numbers is always divisible by 6. E×plain this statement with the help of some examples.

Solution :

**Example 1:** Take three consecutive numbers 21, 22 and 23.

21 is divisible by 3.

22 is divisible by 2.

∴ 21 × 22 is divisible by 3 × 2 ( = 6)

∴ 21 × 22 × 23 is divisible by 6.

**E×ample 2:** Take three consecutive numbers 47, 48 and 49.

48 is divisible by 2 and 3 both.

∴ 48 is divisible by 2 × 3 (= 6)

47 × 48 × 49 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.

Solution :

**Example 1:** Take two consecutive odd numbers 5 and 7.

Sum of these numbers = 5 + 7=12 12 is divisible by 4.

**Example 2:** 13 and 15

Sum of 13 and 15 =13+15 = 28

28 is divisible by 4.

Question 9.

In which of the following expressions, prime ffactorizationhas been done: j

**(a)** 24 = 2 × 3 × 4

**(b)** 56 = 7 × 2 × 2 × 2

**(c)** 70 = 2 × 5 × 7

(d) 54 = 2 × 3 × 9

Solution :

**(a)** Prime factorisation has not been done.

**(b)** Prime factorisation has been done.

**(c)** Prime factorisation has been done.

**(d)** 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]

Solution :

Divisibility of 25110 by 5

Number in the unit’s place of 25110 = 0

∴ 25110 is divisible by 5.

Divisibility of 25110 by 9

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

9 is divisible by 9.

∴ 25110 is divisible by 9

As 25110 is divisible by 5 and 9 both and 5 and 9 are co-prime numbers, so 25110 is divisible by 5 × 9 = 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 your answer.

Solution :

No we cannot say that the number will be divisible by 4 × 6 = 24, if it is divisible by both 4 and 6 because 4 and 6 are not co-prime numbers (they have two common factors 1 and 2).

**Example :** 36 is divisible by both 4 and 6.

But, 36 is not divisible by 24.

Question 12.

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

Solution :

The smallest four different prime numbers are 2. 3. 5 and 7.

∴ The smallest number, having four different prime factors is 2 × 3 × 5 × 7 = 210.