CBSE Class 6th ( VI) Mathematics Chapter 3. Playing with Numbers: Exercise 3.5 |Chapter |CBSE Class 6th, 7th, 8th, 9th, 10th|Worksheet |Wikipedia|PPT|PDF|Video|Download|MCQs

CBSE Class 6th ( VI) Mathematics Chapter 3. Playing with Numbers: Exercise 3.5

Remember...

if a number is divisible by another number then it is divisible by each of the factors of that number.
If a number is divisible by two co-prime numbers then it is divisible by their product also.
If two given numbers are divisible by a number, then their sum is also divisible by that number.
If two given numbers are divisible by a number, then their difference is also divisible by that number.

1. Which of the following statements are true?

If a number is divisible by 3, it must be divisible by 9.
If a number is divisible by 9, it must be divisible by 3.
A number is divisible by 18, if it is divisible by both 3 and 6.
If a number is divisible by 9 and 10 both, then it must be divisible by 90.
If two numbers are co-primes, at least one of them must be prime.
All numbers which are divisible by 4 must also be divisible by 8.
All numbers which are divisible by 8 must also be divisible by 4.
If a number exactly divides two numbers separately, it must exactly divide their
sum.
If a number exactly divides the sum of two numbers, it must exactly divide the two
numbers separately.

Answer:

If a number is divisible by 3, it must be divisible by 9. [False]
If a number is divisible by 9, it must be divisible by 3.[True]
A number is divisible by 18, if it is divisible by both 3 and 6.[True]
If a number is divisible by 9 and 10 both, then it must be divisible by 90.[True]
If two numbers are co-primes, at least one of them must be prime.[False]
All numbers which are divisible by 4 must also be divisible by 8.[False]
All numbers which are divisible by 8 must also be divisible by 4.[True]
If a number exactly divides two numbers separately, it must exactly divide their
sum.[True]
If a number exactly divides the sum of two numbers, it must exactly divide the two
numbers separately.[False]

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

Answer:

3. Which factors are not included in the prime factorisation of a composite number?
Answer: We know that :
1. The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers.
2. Numbers having more than two factors are called Composite numbers.1 and the number itself
3. 1 is neither a prime nor a composite number.
Going by the above statement it is clear 1 and the number itself, are the factors which are not included in the prime factorisation of a composite number.

4. Write the greatest 4-digit number and express it in terms of its prime factors.
Answer:

The greatest 4-digit number is 9999 and its factors can be found as described below:

3	9999
3	3333
11	1111
	101

Hence greatest 4-digit number

number 9999 can be expressed in the form of its prime factors as 3 × 3 ×11 × 101

5. Write the smallest 5-digit number and express it in the form of its prime factors.
Answer:

The smallest 5-digit number is 10000 and its factors can be found as described below:

2	10000
2	5000
2	2500
2	1250
5	625
5	125
5	25
	5

Hence the smallest 5-digit number 10000 can be expressed in the form of its prime factors
as 2 ×2 ×2 ×2 ×5 ×5 ×5×5

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.
Answer:

The the prime factors of 1729 can be found as described below :

7	1729
13	247
	19

Hence number 1729 can be expressed in the form of its prime factors as 7 ×13 ×19.
Relation between its two consecutive prime factors. The consecutive prime factors are 7, 13 and 13 , 19. Clearly 13-7=6 and 19-13= 6. Here difference of two consecutive prime factors is 6

7. The product of three consecutive numbers is always divisible by 6. Verify this statement
with the help of some examples.
Answer:

Case 1. Let us take three consecutive numbers 6, 7, 8
Product of these numbers : 6 × 7 × 8 = 336. Also 336 ÷ 6=56
Case 2. Let us take another three consecutive numbers 4, 5, 6
Product of these numbers : 4 × 5 × 6 = 120. Also 120 ÷ 6=20
Case 3. Let us take one more set of three consecutive numbers 9, 10, 11
Product of these numbers : 9 × 10× 11 = 990. Also 990 ÷ 6=165
From the above example it is clear that the product of three consecutive numbers is always divisible by 6

8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with
the help of some examples.
Answer:

Example 1. Let us take two consecutive odd numbers 3 and 5
Sum of these numbers : 3 + 5 = 8.Also 8 ÷ 4=2
Example 2. Let us take two consecutive odd numbers 11 and 13
Sum of these numbers : 11 + 13 = 24. Also 24 ÷ 4=6
Example 3. Let us take others two consecutive odd numbers 21 and 23
Sum of these numbers : 21 + 23 = 44. Also 44 ÷ 4=11
From the above examples it is clear The sum of two consecutive odd numbers is divisible by 4

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
Answer:

(a) 24 = 2 × 3 × 4

The above can be verified after finding the correct prime factorisation for 24 as described below:

2	24
2	12
2	6
	3

The prime factorisation for 24 = 2× 2 ×2 × 3 which is not as given so prime factorisation for 24 has not been done

(b) 56 = 7 × 2 × 2 × 2

The above can be verified after finding the correct prime factorisation for 56 as described below:

2	56
2	28
2	14
	7

The prime factorisation for 56 = 2× 2 ×2 × 7 which is same as given above so prime factorisation for 56 has been done

(c) 70 = 2 × 5 × 7

The above can be verified after finding the correct prime factorisation for 70 as described below:

2	70
5	35
	7

The prime factorisation for 70 = 2× 5 ×7 which is same as given above so prime factorisation for 70 has been done

(d) 54 = 2 × 3 × 9

The above can be verified after finding the correct prime factorisation for 54 as described below:

2	54
3	27
3	9
	3

The prime factorisation for 56 = 2 × 3 ×3 × 3 which is not the same as given above so prime factorisation for 56 has not been done

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].
Answer: We know that If a number is divisible by two co-prime numbers then it is divisible
by their product also.

Here 45 = 5 × 9 and also 5 and 9 are co-prime numbes.

Also 25110 ÷ 5 = 5022 and 25110 ÷ 9 = 2790

As 25110 is divisible by co-prime numbers 4 and 5 there for 25110 will also be divisible by their product i.e. 45

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.
Answer:
No. Number 12 is divisible by both 4 and 6; but 12 is not divisible by 4×6=24.
Similarly Number 36 is divisible by both 4 and 6; but 36 is not divisible by 4×6=24.
Also Number 60 is divisible by both 4 and 6; but 60 is not divisible by 4×6=24.

12. I am the smallest number, having four different prime factors. Can you find me?
Answer: We know that starting from the beginning we have 4 different prime numbers as 2, 3, 5, 7 and their product will be the required number i.e. 2×3×5×7 = 210