Chapter - 3 Playing with Numbers

Q&A -Ask Doubts and Get Answers

Q
Playing With Numbers

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: 

(a) False, 15 is divisible by 3 but is not divisible by 9.

(b) True, as 9 = 3 x 3. Hence, if a number is divisible by 9, it will also be divisible by 3.

(c) False, 30 is divisible by both 3 and 6 but is not divisible by 18.

(d) True, 9 x 10 = 90. Hence, if a number is divisible by both 9 and 10 then it is must be divisible by 90.

(e) False, Since 15 and 28 are co-primes and also composite numbers.

(f) False, 12 is divisible by 4 but is not divisible by 8.

(g) True, 2 x 4 = 8. Hence, if a number is divisible by 8, it will also be divisible by 4.

(h) True.

(i) False, 2 divides 8 but it does not divide 5 and 3.

 

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

(a)

 

(b) 

Solution: (a) We know, prime factorisation for 6 and 10 are:

6 = 2 x 3   and   10 = 5 x 2

So, the missing numbers are these:

(b) Prime factorisation for 60 is:

60 = 30 x 2

30 = 10 x 3

10 = 2 x 5

So, the missing numbers are these:

 

3. Which factors are not included in the prime factorisation of a composite number?

Solution: 1 and the number itself 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.

Solution: The greatest 4 digit number is 9999.

The prime factorisation of 9999 is:

Therefore, 9999 = 3 x 3 x 11 x 101

 

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

Solution: The smallest 5 digit number is 10000.

The prime factorisation of 10000 is:

Therefore, 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 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.

Solution: Prime factorisation of 1729 is:

1729 = 7 x 13 x 19

The ascending order is 7, 13 and 19

Here, 13 – 7 = 6 and 19 – 13 = 6
Hence, the difference between 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.

Solution:  Let's take some examples to verify this statement.

(i) 4 x 5 x 6 = 120, 12 is divisible by 6.

(ii) 6 x 7 x 8 = 336

Here, the end digit of the number 336 is 6 so it is divisible by 2.

The sum of digits 3 + 3 + 6 = 12, which is divisible by 3. So, the number is divisible by 3.

The number is divisible by both 2 and 3, hence it is divisible by 6.

 

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

Solution: Let's take some examples to verify this statement.

(i) 3 + 5 = 8, 8 is divisible by 4.

(ii) 11 + 13 = 24, 24 is divisible by 4.

(iii) 19 + 21 = 40, 40 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

Solution: 

(a) 24 = 2 x 3 x 4
Here, 4 is not a prime number.
Hence, 24 = 2 x 3 x 4 is not a complete prime factorisation.

(b) 56 = 7 x 2 x 2 x 2
Here, all factors are prime numbers
Hence, 56 = 7 x 2 x 2 x 2 is a prime factorisation.

(c) 70 = 2 x 5 x 7
Here, all factors are prime numbers.
Hence, 70 = 2 x 5 x 7 is a prime factorisation.

(d) 54 = 2 x 3 x 9
Here, 9 is not a prime number.
Hence, 54 = 2 x 3 x 9 is not a complete prime factorisation.

 

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: 45 = 5 x 9

5 and 9 are the factors of 45.

5 = 1 x 5 and 9 = 3 x 3

Here, 5 and 9 are co-prime numbers.
Test of divisibility by 5: a unit place of the given number 25110 is 0. So, it is divisible by 5.
Test of divisibility by 9:
Sum of the digits = 2 + 5 + l + l + 0 = 9 which is divisible by 9.
So, the given number is divisible by 5 and 9 both. Hence, the number 25110 is divisible by 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.

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 x 6 = 24.
Example: 12 and 36 are divisible by both 4 and 6 but not by 24.

 

12. I am the smallest number, having four different prime factors. Can you find me?

Solution: It is the smallest number, having four different prime factors. So, this number has to be the product of the smallest prime numbers.

The four smallest prime numbers are 2, 3, 5 and 7.

So, the number is 2 x 3 x 5 x 7 = 210.

Answer:


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