AP 6th Class Maths 3rd Chapter Playing with Numbers Exercise 3.5 Solutions

Well-designed AP 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise 3.5 offers step-by-step explanations to help students understand problem-solving strategies.

Playing with Numbers Class 6 Exercise 3.5 Solutions – 6th Class Maths 3.5 Exercise Solutions

Question 1.
Here are two different factor trees for 60. Write the missing numbers.

Solution:

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.

Solution:

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.

Question 2.
Which factors are not included in the prime factorisation of a composite number?
Answer:
1 and the number itself are not included in the prime factorisation of a composite number.

Question 3.
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 × 3 × 11 × 101

Question 4.
Write the smallest 5 -digit number and express it in the form of its prime factors.
Solution:
The smallest 5 – digit number = 10000

Hence, the required prime factors of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Question 5.
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:
Given number = 1729
Hence, the prime factors 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.

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 : Let us take three consecutive numbers 2,3 and 4.
The product = 2 × 3 × 4 = 24 is divisible by 6.

Example 2: Let us take three consecutive numbers 10,11 and 12.
The product = 10 × 11 × 12 = 1320
1320 is divisible by 2 [∵ ones place = 0]
1320 is also divisible by 3 [∵ 1 + 3 + 2 + 0 = 6]
Hence, the product 1320 is divisible by 6 .

Example 3 : Let us take three consecutive numbers 31,32 and 33 .
The product = 31 × 32 × 33 = 32736
32736 is divisible by 2 [∵ ones place = 6]
32736 is also divisible by 3
[∵ 3 + 2 + 7 + 3 + 6 = 21]
Hence, the product 32736 is divisible by 6.

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 7 and 9.
Sum = 7 + 9 = 16 which is divisible by 4.
Example 2 : Let us take two consecutive odd numbers 21 and 23.
Sum = 21 + 23 = 44, which is divisible by 4.
Example 3 : Let us take two consecutive odd numbers 101 and 103.
Sum 101 + 103 = 204, which is divisible by 4.
(∵ Here, the number formed by last two digits is 04 which is divisible by 4 )

Question 8.
In which of the following expressions, prime factorisation has been done?
a) 24 = 2 × 3 × 4
Solution:
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
Solution:
56 = 7 × 2 × 2 × 2
Here all factors are prime numbers.
Hence, 56 = 7 × 2 × 2 × 2 is a prime factorisation.

c) 70 = 2 × 5 × 7
Solution:
70 = 2 × 5 × 7
Here all factors are prime numbers.
Hence, 70 = 2 × 5 × 7 is a prime factorisation.

d) 54 = 2 × 3 × 9
Solution:
54 = 2 × 3 × 9
Here 9 is not a prime number.
Hence, 54 = 2 × 3 × 9 is not a prime factorisation.

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.

Question 10.
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 × 3 × 5 × 7 = 210