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Questions and Solutions for NCERT Class 6th Mathematics Chapter 3 – Playing with Numbers – Exercise 3.5
1. Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
Ans: False
For Example; 15 is divisible by 3 but not by 9
(b) If a number is divisible by 9, it must be divisible by 3.
Ans: True
Example: 27 is divisible by both 9 and 3
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
Ans: True
Example: 36 is divisible by 18 and by both 3 and 6, because 3 and 6 are factors of 18
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
Ans: True
Example: 900 is divisible by 9, 10 and 90
(e) If two numbers are co-primes, at least one of them must be prime.
Ans: False
Example: 35 and 18 are co-prime and both are composite numbers
(f) All numbers which are divisible by 4 must also be divisible by 8.
Ans: False
Example: 12 is divisible by 4 but not by 8
(g) All numbers which are divisible by 8 must also be divisible by 4.
Ans: True
For Example: 16, 24, 32, 88 are divisible by both 8 and 4
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
Ans: True
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two
numbers separately.
Ans: False
2. Here are two different factor trees for 60. Write the missing numbers.
3. Which factors are not included in the prime factorization of a composite number?
Ans: 1
4. Write the greatest 4-digit number and express it in terms of its prime factors.
Sol: Greatest 4-digit number is 9999
Prime factorisation of 9999 = 3 × 3 × 11 × 101
5. Write the smallest 5-digit number and express it in the form of its prime factors.
Sol: Smallest 5-digit number is 10000
Prime factorisation of 10000 = 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.
Sol: Prime Factors of 1729 = 7 × 13 × 19
Difference of two prime factors is 6
13 – 7 = 6 and 1 9 – 13 =6
7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Sol:Among the three consecutive numbers there must be an even number and a multiple of 3.
Thus, the product is multiple of 6.
Examples:
Let us take three consecutive numbers 7, 8 and 9
Product = 7 × 8 × 9 =504
504/6 = 84
Let us take three consecutive numbers 3, 4 and 5
Product = 3× 4 × 5 = 60
60/6 = 10
8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Sol: Let us take some examples
1 + 3 = 4, which is divisible by 4
5 + 7 = 12, which is divisible by 4
7 + 9 = 16, which is divisible by 4
21 + 23 = 44, which is divisible by 4
9. In which of the following expressions, prime factorization has been done?
(a) 24 = 2 × 3 × 4
Ans: Here prime factorisation is not done because 4 is not a prime number.
(b) 56 = 7 × 2 × 2 × 2
Ans: Here prime factorisation is done because all the factors are prime and their product equals to 56.
(c) 70 = 2 × 5 × 7
Ans: Here prime factorisation is done because all the factors are prime and their product equals to 70.
(d) 54 = 2 × 3 × 9
Ans: Here prime factorisation is not done because 9 is not a prime number.
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].
Ans: 45 = 5 × 9
25110 is divisible by 5 as its unit place is 0
25110 is divisible by 9 as the sum of its digits (9) is divisible by 9
Therefore 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.
Ans: No, 36 is divisible by both 4 and 6 but it is not divisible by 24.
12. I am the smallest number, having four different prime factors. Can you find me?
Ans: 2 × 3 × 5 × 7 = 210
210 is the smallest number having four different prime factors.
You can find the solutions for Class 6 Mathematics previous exercises from here
- Playing with Numbers – Exercise 3.4 Solutions
- Playing with Numbers – Exercise 3.3 Solutions
- Playing with Numbers – Exercise 3.2 Solutions
- Playing with Numbers – Exercise 3.1 Solutions
- Whole Numbers – Exercise 2.3 Solutions
- Whole Numbers – Exercise 2.2 Solutions
- Whole Numbers – Exercise 2.1 Solutions
- Knowing Our Numbers Solutions Exercise 1.3 Solutions
- Knowing Our Numbers Solutions Exercise 1.2 Solutions
- Knowing Our Numbers Solutions Exercise 1.1 Solutions
