**NCERT
Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5**

NCERT
Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5 are the part of NCERT Solutions for
Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter
3 Playing with Numbers Ex 3.5.

**NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7**

**Ex 3.5 Class 6 Maths 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.

**Solution:
**(a) False

(b) True

(c) False

(d) True

(e) False

(f) False

(g) True

(h) True

(i) False

**Ex 3.4 Class 6 Maths Question
2.**

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

(a) Given that

Here, 6 = 2 × missing number

∴ Missing number = 6 ÷ 2 = 3

Similarly, 10 = 5 × missing number

∴ Missing number = 10 ÷ 5 = 2

Hence, the missing numbers are 3 and 2.

(b) Given that

Let the missing numbers be m_{1}, m_{2}, m_{3} and
m_{4}.

60 = 30 × m_{1}

⇒ m_{1} = 60 ÷ 30 = 2

30 = 10 × m_{2}

⇒ m_{2} = 30 ÷ 10 = 3

10 = m_{3} × m_{4}

⇒ m_{3} = 2 or 5 and m_{4} = 5 or 2

Hence, the missing numbers are 2, 3, 2, 5.

**Ex 3.4 Class 6 Maths Question
3.**

Which factors are not included in the prime
factorisation of a composite number?**Solution:
**In the prime factorisation of a composite number, 1
and the number itself are not included.

**Ex 3.4 Class 6 Maths Question
4.**

Write the greatest 4-digit number and express it in
terms of its prime factors.**Solution:
**The greatest 4-digit number is 9999.

Hence, the prime factorisation of 9999 = 3 × 3 × 11 × 101

**Ex 3.4 Class 6 Maths Question
5.**

Write the smallest 5-digit number and express it in
the form of its prime factors.**Solution:
**The smallest 5-digit number is 10000.

Hence, the prime factorisation of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

**Ex 3.4 Class 6 Maths 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:
**The given number is 1729.

Hence, the prime factorisation of 1729 = 7 × 13 × 19

Here, 13 – 7 = 6 and 19 – 13 = 6

We see that the difference between two consecutive prime factors is 6.

**Ex 3.4 Class 6 Maths Question
7.**

The product of three consecutive numbers is always
divisible by 6. Verify this statement with the help of some examples.**Solution:
**Example 1:

Take three consecutive numbers 7, 8 and 9.

Here, 9 is divisible by 3 and 8 is divisible by 2.

Therefore, the product 7 × 8 × 9 = 504 which is divisible by 6.

Example 2:

Take three consecutive numbers 20, 21 and 22.

Here, 20 is divisible by 2 and 21 is divisible by 3.

Therefore, the product 20 × 21 × 22 = 9240 is divisible by 6.

Example 3:

Take three consecutive numbers 30, 31 and 32.

Here, 30 is divisible by 3 and 32 is divisible by 2.

Therefore, the product 30 × 31 × 32 = 29760 is divisible by 6.

**Ex 3.4 Class 6 Maths 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:

Let us take two consecutive odd numbers 15 and 17.

Sum = 15 + 17 = 32

Here, the number 32 is divisible by 4.

Example 2:

Let us take two consecutive odd numbers 97 and 99.

Sum = 97 + 99 = 196

Here, the number formed by last two digits is 96 which is divisible by 4.

Hence, the sum of numbers 97 and 99, i.e. 196 is divisible by 4.

Example 3:

Let us take two consecutive odd numbers 321 and 323.

Sum = 321 + 323 = 344

Here, the number formed by last two digits is 44 which is divisible by 4.

Hence, the sum of numbers 321 and 323, i.e. 344 is
divisible by 4.

**Ex 3.4 Class 6 Maths Question
9.**

In which of the following expressions, prime
factorisation has been done?(a) 24 = 2 × 3 × 4

(b) 56 = 7 × 2 × 2 × 2

(c) 70 = 2 × 5 × 7

(d) 54 = 2 × 3 × 9

**Solution:
**(a) 24 = 2 × 3 × 4

Here, 4 is not a prime number.

Hence, 24 = 2 × 3 × 4 is not a prime factorisation.

(b) 56 = 7 × 2 × 2 × 2

Here, all the factors are prime numbers.

Hence, 56 = 7 × 2 × 2 × 2 is a prime factorisation.

(c) 70 = 2 × 5 × 7

Here, all the factors are prime numbers.

Hence, 70 = 2 × 5 × 7 is a prime factorisation.

(d) 54 = 2 × 3 × 9

Here, 9 is not a prime number.

Hence, 54 = 2 × 3 × 9 is not a prime factorisation.

**Ex 3.4 Class 6 Maths Question
10.**

Determine if 25110 is divisible by 45.**Solution:
**45 = 5 × 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 + 1 + 1 + 0 = 9 which is divisible by 9.

So, the given number is divisible by 5 and 9 both. Hence, the number 25110 is divisible by 45.

**Ex 3.4 Class 6 Maths 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:
**Here, the given two numbers are not co-prime. So, it
is not necessary that a number divisible by both 4 and 6, must also be
divisible by their product 4 × 6 = 24.

Example: 36 and 60 are divisible by both 4 and 6 but not by 24.

**Ex 3.4 Class 6 Maths Question
12.**

I am the smallest number, having four different
prime factors. Can you find me?**Solution:
**We know that the four smallest prime numbers are 2,
3, 5 and 7.

Hence, the required number = 2 × 3 × 5 × 7 = 210

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