(2022) Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers is:(a) 2 (b) 3 (c) 4 (d) 1

Q1. (2022) Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers is:
(a) 2 (b) 3 (c) 4 (d) 1

Q2. (2022) If n is any natural number, then 12^n cannot end with the digit:
(a) 2 (b) 4 (c) 8 (d) 0

1. Let the two numbers be 12x and 12y where x, y are positive integers and HCF(x, y) = 1.
2. Product = (12x)(12y) = 144xy. Given 144xy = 6336.
3. So xy = 6336 ÷ 144 = 44.
4. Factorise 44 = 2^2 × 11. Divisor pairs (ordered): (1,44), (2,22), (4,11), (11,4), (22,2), (44,1).
5. Check which pairs are coprime (HCF = 1): (1,44) and (4,11) are coprime; (2,22) and (22,2) are not.
6. Counting unordered pairs, valid pairs are {1,44} and {4,11} → 2 pairs.

Answer: 2 (Option a)

Q2. (2022) Solution:

1. Last digit of 12^n depends on last digit of 2^n (since 12 ≡ 2 (mod 10)).
2. Last-digit cycle of powers of 2: 2, 4, 8, 6, … (repeats every 4).
3. So possible last digits of 12^n are 2, 4, 8, 6. Digit 0 never appears.

Answer: 0 (Option d)

1. Divide 385 by 5: 385 ÷ 5 = 77.
2. Factor 77 = 7 × 11.
3. Therefore 385 = 5 × 7 × 11.

Answer: 5 × 7 × 11 (Option b)