3. Playing with Numbers Mathematics Exercise - 3.5 class 6 Maths in English - CBSE Study
NCERT Solutions for Class 6 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 3. Playing with Numbers with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 3.5 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 6 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 6 English Medium Mathematics All Chapters:
3. Playing with Numbers
5. Exercise 3.5
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.
Answer : False
(b) If a number is divisible by 9, it must be divisible by 3.
Answer : True
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
Answer : False
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
Answer : True
(e) If two numbers are co-primes, at least one of them must be prime.
Answer : False
(f) All numbers which are divisible by 4 must also be divisible by 8.
Answer : False
(g) All numbers which are divisible by 8 must also be divisible by 4.
Answer : True
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
Answer : True
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Answer : False
2. Here are two different factor trees for 60. Write the missing numbers.
Solution:
There are two different way as follow:
3. Which factors are not included in the prime factorization of a composite number?
Solution: 1 and the composite number itself not included in the prime factorization 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 -
5. Write the smallest 5-digit number and express it in the form of its prime factors.
Solution:
The smallest five diigit number is 10000.
It's tree factor is :
Hence the prime factorisation =
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.
Solution:
Prime factors of 1729 are 7 × 13 × 19.
7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Solution: Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
Example:
(i) 2 × 3 × 4 = 24
(ii) 4 × 5 × 6 = 120
8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.
Solution: 3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.
9. In which of the following expressions, prime factorisation has been done?
Solution: In expressions (b) and (c), prime factorization has 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].
Solution: The prime factorization of 45 = 5 × 9 25110 is divisible by 5 as ‘0’ is at its unit place.
25110 is divisible by 9 as sum of digits is divisible by 9.
Therefore, the number must be divisible by 5 × 9 = 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: No. Number 12 is divisible by both 6 and 4 but 12 is not divisible by 24.
12. I am the smallest number, having four different prime factors. Can you find me?
Solution: The smallest four prime numbers are 2, 3, 5 and 7.
Hence, the required number is 2 × 3 × 5 × 7 = 210
Topic Lists: