CBSE Class 6 Maths Chapter 5: Prime Time – Important Questions with Answers (Free PDF Download)

1. Find the prime factorisation of these numbers without multiplying first.
(a) 56 × 25
(b) 108 × 75
(c) 1000 × 81
Solution:
(a) Prime factors of 56 = 2 × 2 × 2 × 7
Prime factors of 25 = 5 × 5
Combined prime factorization of 56 × 25 = 2 × 2 × 2 × 7 × 5 × 5

(b) Prime factors of 108 = 2 × 2× 2 × 3 × 3
Prime factors of 75 = 3 × 5 × 5
Combined prime factorization of
108 × 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

(c) Prime factors of 1000 = 2 × 2 × 2 × 5 × 5 × 5
Prime factors of 81 = 3 × 3 × 3 × 3
Combined prime factorization of 1000 × 81 = 2 × 2 × 2 × 5 × 5 × 5 × 3 ×3 × 3 × 3

2. The prime factorization of a number has one 2, two 3s, and one 11. What is the number?

Solution:

To find the number, we multiply these prime factors together:

2 × 3 × 3 × 11 = 198

Thus, the number is 198.

3. Find the prime factorization of these numbers without multiplying first.
(a) 56 × 25
(b) 108 × 75
(c) 1000 × 81
Solution:
(a) Prime factors of 56 = 2 × 2 × 2 × 7
Prime factors of 25 = 5 × 5
Combined prime factorization of 56 × 25 = 2 × 2 × 2 × 7 × 5 × 5

(b) Prime factors of 108 = 2 × 2 × 2 × 3 × 3 × 3
Prime factors of 75 = 3 × 5 × 5
Combined prime factorization of
108 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

(c) Prime factors of 1000 = 2 × 2 × 2 × 5 × 5 × 5
Prime factors of 81 = 3 × 3 × 3 × 3
Combined prime factorization of 1000 × 81 = 2 × 2 × 2 × 5 × 5 × 5 × 3 × 3 × 3 × 3

4. The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Solution:
If a number is divisible by 8, it will automatically be divisible by 4.
If a number is divisible by 10, it is also divisible by 2 and 5. Therefore, checking divisibility by 8 and 10 confirms divisibility by all other numbers (2, 4, 5).
Thus, the pair of numbers that Guna could check to determine that 14560 is divisible by all of 2, 4, 5, 8, and 10 is: 8 and 5.

5. What is the smallest number whose prime factorization has:
(a) three different prime numbers?
(b) four different prime numbers?
Solution:
(a) The smallest prime numbers are 2, 3, and 5. To find the smallest number with these primes as factors, multiply them together:
2 × 3 × 5 = 30
So, the smallest number whose prime factorization has three different prime numbers is 30.

(b) The smallest four prime numbers are 2, 3, 5, and 7. To find the smallest number with these primes as factors, multiply them together:
2 × 3 × 5 × 7 = 210

Thus, the smallest number whose prime factorization has four different prime numbers is 210.

6. Which of the following pairs of numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
Answer:
(a) Here factors of 18 = 1 × 2 × 3 × 3 and factors of 35 = 1 × 5 × 7
No common factor other than 1.
Hence 18 and 35 are co-prime numbers.

(b) We have factors of 15 = 1 × 3 × 5 and factors of 37 = 1 × 37
No common factor other than 1.
Hence 15 and 37 are co-prime numbers.

(c) Given numbers are 30 and 415 Here factors of 30 = 1 × 2 × 3 × 5 and factors of 415 = 5 × 83
Clearly, 5 is a common factor of 30 and 415.
Hence 30 and 415 are not co-prime numbers.

7. Who am I?

(a) I am a number less than 40. One of my factors is 7. The sum of my digits is 8.

(b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is l more than the other.

Solution:

(a) 7 is the common factor of 7, 14, 21,28, 35, which are less than 40. And there is one number which has a digit sum of 8, is 35 = (3 + 5) = 8.

So, I am 35.

(b) Common factors of 3 and 5 are 15, 30, 45, 60, and 75,90, (which are less than 100). And there is one number one of the digits is 1 more than the other is 45. So, I am 45.

8. The prime factorization of a number has one 2, two 3s, and one 11. What is the number?

Solution:

To find the number, we multiply these prime factors together:

2 × 3 × 3 × 11 = 198

Thus, the number is 198.

9. Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.

Solution:

Largest 4-digit number divisible by 4 and is also palindrome- 8888

The smallest 4-digit number is divisible by 4 and is also palindrome- 2112.

10. Consider these statements:

(a) Only the last two digits matter when deciding if a given number is divisible by 4.

(b) If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4.

(c) If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4.

Do you agree? Why or why not?

Solution:

(a) Yes, that’s correct. When determining if a number is divisible by 4, only the last two digits of the number matter. This is because 100 is divisible by 4, so the divisibility rule for 4 focuses on whether the number formed by the last two digits is divisible by 4.

(b) Yes, that’s correct. If the number formed by the last two digits of a given number is divisible by 4, then the original number is also divisible by 4.

(c) Yes, that’s correct. If the original number is divisible by 4, the last two digits of the number will indeed be divisible by 4.

Check Out Some Extra Questions for More Practise

1. Find the prime factorization of these numbers without multiplying first.

(a) 84 × 45

Prime factors of 84 = 2 × 2 × 3 × 7

Prime factors of 45 = 3 × 3 × 5

Combined prime factorization: 2 × 2 × 3 × 3 × 3 × 5 × 7

(b) 126 × 64

Prime factors of 126 = 2 × 3 × 3 × 7

Prime factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

Combined prime factorization: 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7

2. A number has three 2s, two 5s, and one 7 in its prime factorization. What is the number?

Solution: Multiply the given prime factors together:

2 × 2 × 2 × 5 × 5 × 7 = 1400

Thus, the number is 1400.

3. Who am I?

(a) I am a number less than 50. One of my factors is 5, and the sum of my digits is 6.

Solution: Numbers less than 50 with 5 as a factor: 5, 10, 15, 20, 25, 30, 35, 40, 45

Checking digit sum:

15 → 1 + 5 = 6 

So, I am 15.

(b) I am a number less than 200. Two of my factors are 2 and 3. One of my digits is twice the other.

Solution: Numbers divisible by 2 and 3 (i.e., divisible by 6): 12, 18, 24, 30, 36, 42, 48, 54, 60, 72, 84, 96, 102, 108, 120, 132, 144, 150, 162, 180, 192

Checking for a number where one digit is twice the other:

18 → 8 is not twice 1 

24 → 4 is not twice 2 

36 → 6 is twice 3

So, I am 36.

4. Which of the following pairs of numbers are co-prime?

(a) 22 and 35

(b) 42 and 65

Solution:

(a) Factors of 22 = 1, 2, 11, 22

Factors of 35 = 1, 5, 7, 35

No common factor other than 1 → Co-prime 

(b) Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42

Factors of 65 = 1, 5, 13, 65

No common factor other than 1 → Co-prime 

5. How many prime numbers are there from 21 to 30? 

Solution: In total, there are 2 prime numbers between 21 and 30.

They are 23 and 29.

6. How many composite numbers are there from 21 to 30?

Solution: The total number of composite numbers from 21 to 30 is 8.

They are 21,22,24,25, 26, 27, 28, 30.

This page consists of a different set of short question answers to help students practice effectively. These short question answers will help to solve any kind of problem based on numbers. Use them to improve your understanding and score better on your test papers.

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