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

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.5

Question 1.
Which of the following state¬ments are true ?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a num ber is divisible is 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 that numbers divisible by 4 must also be divisible by 8.
(g) All that numbers divisible by 8 must also be divisible by 4.
(h) The sum of two consecutive odd numbers is divisible by 4.
(i) If a number exactly divides two numbers separately, it must exactly divide their sum.
(j) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution:
(a) F, (b) T, (c) T, (d) T, (e) F, (/) F, (g) T, (h) T, (i) T, (/) F.

Question 2.
Here are two different factor trees for 60. Write the missing numbers.
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 (1)
Solution:
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 (2)

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

Question 3.
Which factors are not included in the prime factorisation of a composite number ?
Solution:
Let us consider any composite number say 12.
12 = 1 x 12
= 2 x 6
= 3 x 4
and
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 (3)
∴ Factors of 12 are : 1, 2, 3, 4, 6, 12.
Prime factorisation of 12 = 2 x 2 x 3
We clearly see that composite factors 4, 6, 12 are not included in the prime factorisation of a composite number.
Hence, composite factors are not included in the prime factorisation of a composite number.

Question 4.
Write the greatest four-digit number and express it into the form of prime factorisation.
Solution:
Greatest 4-digit number is 9999.
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 (4)
∴ Prime factorisation of 9999
= 3 x 3 x 11 x 101

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

Question 5.
Write the smallest five-digit number and express it into the form of prime factorisation.
Solution:
Smallest 5-digit number is 10000
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 (5)
∴ Prime factorisation of 10000
= 2 x 2 x 2 x 2 x 5 x 5 x 5 x 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:
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 6
∴ Prime factors of 1729 are 7, 13, 19
Difference between two consecutive prime factors is 6.

Question 7.
The product of three consecutive numbers is always divisible by 6. Explain this statement with the help of some examples.
Solution:
(i) Let us take three consecutive numbers 5, 6, 7.
Their product = 5 x 6 x 7 = 210 which is divisible by 6.
(ii) Let us take three consecutive numbers 9, 10, 11.
Their product = 9 x 10 x 11 = 990 which is divisible by 6.
(iii) Let us take three consecutive numbers 23, 24, 25.
Their product = 23 x 24 x 25 = 13800 which is divisible by 6.
(iv) Let us take three consecutive numbers 2, 3, 4.
Their product = 2 x 3 x 4 = 24 which indivisible by 6.

Question 8.
In which of the following expressions, prime factorisation has been done :
(a) 24 = 2 x 3 x 4
(b) 56 = 1 x 7 x 2 x 2 x 2
(c) 70 = 2 x 5 x 7
(d) 54 = 2 x 3 x 9.
Solution:
(c) 70 = 2 x 5 x 7.
In this expression prime factorisation has been done.
(b) 56 = 1 x 7 x 2 x 2 x 2 In this expression prime factorisation has been done.

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

Question 9.
Write the prime factorisation of 15470.
HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 7
∴ Prime factorisation of 15470
= 2 x 5 x 7 x 13 x 17.

Question 10.
Determine if 25110 is divisible by 45.
Solution:
∵ The unit’s digit of 25110 is 0,
∴ 25110 is divisible by 5.
Sum of the digits = 2 + 5+ l + l + 0 = 9 which is divisible by 9.
∴ 25110 is divisible by 9.
Now, 5 and 9 are co-prime numbers.
∴ 25110 is divisible by their product 5 x 9 = 45.

Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24 ? If not, give an example to justify your answer.
Solution:
No, because 4 and 6 are not co-primes.
e.g. (i) 36 is divisible by both 4 and 6. But it is not divisible by 24.
(ii) 12 is divisible by both 4 and 6 but 12 is not divisible by 24.

HBSE 6th Class 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:
The smallest four different prime factors are 2, 3, 5, 7.
Hence, the smallest number, having four different prime factors = 2 x 3 x 5 x 7 = 210.

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