MP Board Class 6th Maths Solutions 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 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:
Statements (b), (d), (g) and (h) are true.

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5

Question 2.
Here are two different factor trees for 60. Write the missing numbers.
MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 1
Solution:
Factors of 60 are 1, 2, 3, 4, 5, 6,10,12,15, 20, 30, 60.
(a) Since 6 = 2 × 3 and 10 = 5 × 2
∴ The missing numbers are 3 and 2.
MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 2

(b) Since, 60 = 30 × 2, 30 = 10 × 3, and 10 = 5 × 2
MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 3

Question 4.
Write the greatest 4-digit number and express it in terms of its prime factors.
Solution:
The greatest four digit number is 9999.
MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 4

Question 5.
Write the smallest 5-digit number and express it in the form of its prime factors.
Solution:
The smallest five digit number is 10000.
MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 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:
MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 6
∴ 1729 = 7 × 13 × 19.
The difference of two consecutive prime factors is 6. (∵ 13 – 7 = 6 and 19 – 13 = 6)

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5

Question 7.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Solution:
Among the three consecutive numbers, there must be atleast one even number and one multiple of 3. Thus, the product must be divisible by 6.
For example :
(i) 2 × 3 × 4 = 24
(ii) 4 × 5 × 6 = 120,
where both 24 and 120 are divisible by 6.

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5

Question 8.
The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Solution:
The sum of two consecutive odd numbers is divisible by 4.
For example : 3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.

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:
In expressions (b) and (c), prime factorisation has been done.

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5

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:
The prime factorisation of 45 = 5 × 9
25110 is divisible by 5 as ‘0’ is at its unit place.
25110 is divisible by 9 as sum of digits (i.e., 9) is divisible by 9.
Therefore, the number 25110 must be 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. The number 12 is divisible by both 6 and 4, but 12 is not divisible by 24.
∴ A number divisible by both 4 and ( may or may not be divisible by 4 × 6 = 24.

MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.5

Question 12.
I am the smallest number, having four different prime factors. Can you find me?
Solution:
Since, 2 × 3 × 5 × 7 = 210
∴ 210 is the smallest number, having 4
different prime factors i.e., 2, 3, 5 and 7.

MP Board Class 6th Maths Solutions