Playing with Numbers- 2 Class 6 Worksheet Maths

Ques 1: Fill up the blanks: 
(i) Write all even prime numbers __________.

Ans: 2

Explanation: 2 is the only even prime number. A prime number has exactly two distinct positive divisors: 1 and itself. Any even number greater than 2 is divisible by 2 and so has at least three divisors, so it is not prime.

(ii) Number of factors of 2 are __________.

Ans: 2 (1 and 2)

Explanation: Factors are numbers that divide the given number exactly. For 2 the only positive factors are 1 and 2, so there are two factors.

(iii) Smallest pair of co-prime numbers is __________.

Ans: (1, 2)

Explanation: Co-prime numbers (also called relatively prime) are pairs whose greatest common divisor is 1. The smallest positive integers that satisfy this are 1 and 2, since gcd(1,2) = 1.

(iv) Smallest twin prime numbers are __________.

Ans: 3, 5

Explanation: Twin primes are a pair of primes that differ by 2. The smallest such pair is 3 and 5 because both are prime and 5 − 3 = 2.

(v) The largest 2-digit number which is odd and prime is __________.

Ans: 97

Explanation: The largest two-digit number is 99, but 99 is divisible by 3. 98 is even, so not prime. 97 is odd and has no divisors other than 1 and 97, so it is the largest two-digit prime.

Ques 2: Find the smallest number which is divisible by 2, 3, 4, and 5. Can we find the largest number which is divisible by 2, 3, 4 and 5?
Ans: 60, No

Explanation:

To find the smallest number divisible by 2, 3, 4 and 5, we find their least common multiple (LCM).

Prime factors:

2 = 2

3 = 3

4 = 2²

5 = 5

Take the highest power of each prime: 2², 3, 5. So LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.

Therefore, the smallest such number is 60.

There is no largest number divisible by these because adding any multiple of 60 (for example 60 itself) gives another larger number divisible by them. Hence, no, a largest number does not exist.

Ques 3: Write the smallest 3-digit number which is exactly divisible by 6.
Ans: 102

Explanation:

A number divisible by 6 must be divisible by both 2 and 3.

The smallest 3-digit number is 100. Check 100: it is even (divisible by 2) but 1 + 0 + 0 = 1, not divisible by 3, so 100 is not divisible by 6.

Next even numbers: 102 → sum of digits 1 + 0 + 2 = 3, which is divisible by 3, so 102 is divisible by both 2 and 3.

Therefore, the smallest 3-digit number divisible by 6 is 102.

Ques 4: Replace '*' by the smallest possible digit so that the following numbers are divisible by 3 and 9: 
(i) *4129 

Ans: 2

Explanation:

For divisibility by 9 (and hence by 3), the sum of the digits must be a multiple of 9.

Sum of known digits: 4 + 1 + 2 + 9 = 16. We need * + 16 to be a multiple of 9. The next multiple of 9 after 16 is 18, so * = 18 − 16 = 2. Thus the smallest digit is 2.

(ii) 2*985 

Ans: 3

Explanation:

Sum of known digits: 2 + 9 + 8 + 5 = 24. We need 24 + * to be a multiple of 9. The next multiple of 9 after 24 is 27, so * = 27 − 24 = 3. Thus the smallest digit is 3.

(iii) 987*32 
Ans: 7

Explanation:

Sum of known digits: 9 + 8 + 7 + 3 + 2 = 29. We need 29 + * to be a multiple of 9. The next multiple of 9 after 29 is 36, so * = 36 − 29 = 7. Thus the smallest digit is 7.

Ques 5: Write the largest 3-digit number which is divisible by 11.
Ans: 990

Explanation:

The largest 3-digit number is 999. Divide 999 by 11: 999 ÷ 11 = 90 remainder 9, so 999 is not divisible by 11.

To find the largest 3-digit number divisible by 11, find the largest multiple of 11 less than or equal to 999. 11 × 90 = 990, and 11 × 91 = 1001 (which is 4-digit).

Therefore, the largest 3-digit number divisible by 11 is 990.

Ques 6: Find HCF of 15 m 60 cm and 20 m 16 cm.
Ans: 24 cm

Explanation:

Convert both lengths to centimetres.

15 m 60 cm = 15 × 100 + 60 = 1560 cm.

20 m 16 cm = 20 × 100 + 16 = 2016 cm.

Find HCF(1560, 2016) using the Euclidean algorithm:

2016 − 1560 = 456

1560 − 3 × 456 = 1560 − 1368 = 192

456 − 2 × 192 = 456 − 384 = 72

192 − 2 × 72 = 192 − 144 = 48

72 − 1 × 48 = 24

48 − 2 × 24 = 0, so HCF = 24.

Thus the highest common factor of the two lengths is 24 cm.

Ques 7: Find the least number which when divided by 25, 40 and 60 leave the remainder 7 in each case.
Ans: 607

Explanation:

If a number N leaves remainder 7 on division by each of 25, 40 and 60, then N − 7 is exactly divisible by each of these numbers. So N − 7 must be a common multiple of 25, 40 and 60.

Find LCM of 25, 40 and 60.

25 = 5²

40 = 2³ × 5

60 = 2² × 3 × 5

LCM = 2³ × 3 × 5² = 8 × 3 × 25 = 600.

The least such number is N = 600 + 7 = 607.

Ques 8: Find HCF of 90 and 243. Check if it divides 90 and 243 both.
Ans: 9

Explanation:

Use the Euclidean algorithm.

243 ÷ 90 = 2 remainder 63 (since 90 × 2 = 180, 243 − 180 = 63).

90 ÷ 63 = 1 remainder 27.

63 ÷ 27 = 2 remainder 9.

27 ÷ 9 = 3 remainder 0, so HCF = 9.

Check: 90 ÷ 9 = 10 and 243 ÷ 9 = 27, so 9 divides both numbers.

Ques 9: Find LCM of 18 and 15. Check if the LCM so obtained is divisible by 15 as well as 18.
Ans: 90

Explanation:

Prime factors:

18 = 2 × 3²

15 = 3 × 5

LCM = 2 × 3² × 5 = 2 × 9 × 5 = 90.

Check divisibility: 90 ÷ 15 = 6 and 90 ÷ 18 = 5, so 90 is divisible by both 15 and 18. Hence LCM = 90.

Ques 10: Match the following:

Ans. (i) ↔ (c)
(ii) ↔ (e)
(iii) ↔ (d)
(iv) ↔ (a)
(v) ↔ (b)

Explanation:

(i) Two consecutive numbers have no common factor other than 1, so HCF = 1 → (c).

(ii) If two numbers are co-prime, their LCM equals their product → product of two numbers → (e).

(iii) Two consecutive even numbers have a common factor 2 (for example, 6 and 8 have gcd 2) and the smallest such HCF is 2 → (d).

(iv) The smallest composite number (a number with a divisor other than 1 and itself) is 4 → (a).

(v) The largest two-digit prime is 97 → (b).