Class 6 maths Exercise 3.5 solutions

NCERT mathematics class 6 chapter 3 Playing with Numbers exercise 3.5 solutions are given. You should study the textbook lesson Playing with Numbers very well. You should also observe and practice all example problems and solutions given in the textbook. Observe the solutions given below and try them in your own method. NCERT maths class 6 solutions Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 Exercise 3.6 Exercise 3.7 CA foundation maths solutions 

Chapter 3 exercise 3.5 solutions Playing with Numbers class 6  maths NCERT

6th CBSE maths solutions Chapter 3 Playing with Numbers Exercise 3.5

NCERT chapter 3 class 6 maths exercise 3.5 Playing with Numbers

Std 6 maths textbook solutions  Problem 1 1. Here are two different factor trees for 60. Write the missing numbers. Solution: Problem 2 2. Which factors are not included in the prime factorisation of a composite number? Solution: 1 and itself are not included in the prime factorisation of a composite number. Problem 3 3. Write the greatest 4 – digit number and express it in terms of its prime factors. Solution: The greatest four-digit number is 9999. 9999 = 3 × 333 = 3 × 1111 = 3 × 11 × 101 The prime factors of 9999 = 3 × 3 × 11 × 101 Problem 4 4. Write the smallest 5-digit number and express it in the form of its prime factors. Solution: The smallest 5-digit number is 10000. 10000 = 2 × 5000 10000 = 2 × 2 × 2500 10000 = 2 × 2 × 2 × 1250 10000 = 2 × 2 × 2 × 2 × 625 10000 = 2 × 2 × 2 × 2 × 5 × 125 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 25 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5× 5 Therefore, prime factors of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 Problem 5 5. 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: 1729 = 7 × 247 = 7 × 13 × 19 Therefore, the prime factors of 1729 = 7 × 13 × 19 Relation = The difference of two consecutive prime factors is 6. Problem 6 6. 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 or odd number and one multiple of 3. Therefore, the product must be multiple of 6. Examples: 2 × 3 × 4 =24, 5 × 6 × 7 = 210 Problem 7 7. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples. Solution: The sum of two consecutive odd numbers is divisible by 4. Examples: 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. Problem 8 8. In which of the following expressions, prime factorization 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: In the expressions b and c the prime factorization has been done. Problem 9 9. 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: We cannot say that 12 is divisible by both 6 and 4 but 12 is not divisible by 24. Problem 10 10. I am the smallest number, having four different prime factors. Can you find me? Solution: 2 × 3 × 5 × 7 = 210 Note: Observe the solutions and try them in your own method. Inter maths 1A solutions SSC maths class 10 solutions NCERT maths class 7 solutions NIOS maths 311 book 2 solutions for some chapters