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

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5 are the part of NCERT Solutions for Class 6 Maths (Rationalised Contents). 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 3.5 (Rationalised Contents)

### Ex 3.4 Class 6 Maths Question 1.

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

### Ex 3.4 Class 6 Maths Question 2.

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 3.

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 4.

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 5.

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 6.

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 7.

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 8.

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 9.

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 10.

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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