Playing With Numbers

## EXERCISE 3.5

CBSE Class 6 Maths

**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) A number is divisible by 18, if it is 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 be divisible by 8.

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

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

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

Answer:

(a) ✗ (e.g. 15 is divisible by 3 but not by 9)

(b) ✔

(c) ✔

(d) ✔

(e) ✗ (e.g. 8 and 9 are co-primes and none of these are prime. )

(f) ✗

(g) ✔

(h) ✔

(i) ✗

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

Answer:

**Q3: Which factors are not included in the prime factorisation of a composite number?**

Answer: 1 is not included in the prime factorisation of a composite number. (∵ 1 is neither prime nor composite)

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

Answer: The greatest 4-digit number = 9999

9999

↙ ↘

3 3333

↙ ↘

3 1111

↙ ↘

11 101

∴ The prime factors of 9999 = 3 × 3 × 11 × 101

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

Answer: The smallest five digit number is 10000.

10000

↙ ↘

2 5000

↙ ↘

2 2500

↙ ↘

5 500

↙ ↘

5 100

↙ ↘

5 20

↙ ↘

5 4

↙ ↘

2 2

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 factors.**

Answer: Prime factors of 1729 = 7 × 13 × 19

The difference of any two consecutive prime numbers = 6

i.e. 13 - 7 = 6, 19 - 13 = 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, at least one number is even and one is multiple of 3. Thus, the product must be multiple of 6.

e.g.

2 × 3 × 4 = 24

6 × 7 × 8 = 336

**Q8: The sum of two consecutive odd numbers is 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 factorisation has been done?**

(a) 24 = 2 × 3 × 4

(b) 56 = 7 × 2 × 2 × 2

(c) 70 = 2 × 5 × 7

(d) 54 = 2 × 3 × 9

Answer:

(a) ✗ ∵ 4 is not a prime factor.

(b) ✔

(c) ✔

(d) ✗ ∵ 9 is not a prime factor.

**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:

Prime factors 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 (2+5+1+1+0 = 9) is divisible by 9.

∴ the number must be divisible by 5 × 9 = 45

**Q11: 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.**

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 × 3 × 5 × 7 = 210

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