# Class-6 Math Chapter-3: Excercise 3.5 Solutions

## Chapter 3: Playing With Numbers

## Exercise 3.5

**Q1. 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) If a number is divisible by 18, it must be 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 by divisible by 8.

(g) All numbers which are divisible by 8 must also by divisible by 4.

(h) If a number is exactly divides two numbers separately, it must exactly divide their sum.

(i) If a number is exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

**Answer.**

Statements (b), (c), (d), (g) and (h) are true.

**Q2. Here are two different factor trees for 60. Write the missing numbers.**

**(a)**

**(b)**

**Answer.**

(a)

(b)

**Q3. Which factors are not included in the prime factorization of a composite number?**

**Answer.** 1 is the factor which is not included in the prime factorization of a composite number.

**Q4. Write the greatest 4-digit number and express it in terms of its prime factors.**

**Answer.** The greatest 4-digit number = 9999

The prime factors of 9999 are 3 × 3 × 11 × 101.

**Q5. Write the smallest 5-digit number and express it in terms of its prime factors.**

**Answer.** The smallest five digit number is 10000.

The prime factors of 10000 are 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.

**Q6. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any, between, two consecutive prime numbers.**

**Answer.** Prime factors of 1729 are 7 × 13 × 19.

The difference of two consecutive prime factors is 6.

**Q7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.**

**Answer.** 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 x 3 x 4 = 24

(ii) 4 x 5 x 6 = 120

**Q8. The sum of two consecutive odd numbers is always divisible by 4. Verify this statement with the help of some examples.**

**Answer.** 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.

**Q9. In which of the following expressions, prime factorization has been done:**

**(a) 24 = 2 x 3 x 4**

**(b) 56 = 7 x 2 x 2 x 2**

**(c) 70 = 2 x 5 x 7**

**(d) 54 = 2 x 3 x 9**

**Answer.** In expressions (b) and (c), prime factorization has been done.

**Q10. 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.]**

**Answer.** The prime factorization of 45 = 5 x 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 x 9 = 45

**Q11. 18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by 4 and 6. Can we say that the number must be divisible by 4 x 6 = 24? If not, give an example to justify your answer.**

**Answer.** No. Number 12 is divisible by both 6 and 4 but 12 is not divisible by 24.

**Q12. I am the smallest number, having four different prime factors. Can you find me?**

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

Hence, the required number is 2 x 3 x 5 x 7 = 210