- A. 4
- B. 5
- C. 6
- D. 7
Answer: B
Explanation
We are given that p and k are prime numbers, and (p² + k) is also a prime number less than 30. The prime numbers less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Since p² + k < 30, p² must be less than 30. Possible prime values for p are 2, 3, and 5.
Case 1: p = 2
If p = 2, then p² = 4. We need 4 + k < 30, so k k = 3 is a solution
– 4 + 5 = 9 (not prime)
– 4 + 7 = 11 (prime) -> k = 7 is a solution
– 4 + 11 = 15 (not prime)
– 4 + 13 = 17 (prime) -> k = 13 is a solution
– 4 + 17 = 21 (not prime)
– 4 + 19 = 23 (prime) -> k = 19 is a solution
– 4 + 23 = 27 (not prime)
For p = 2, there are 4 possible values for k: {3, 7, 13, 19}.
Case 2: p = 3
If p = 3, then p² = 9. We need 9 + k < 30, so k k = 2 is a solution
– If k is any other prime (which would be an odd number), then 9 + k will be an even number greater than 2 (odd + odd = even), and thus not prime. So, for p = 3, k = 2 is the only solution.
For p = 3, there is 1 possible value for k: {2}.
Case 3: p = 5
If p = 5, then p² = 25. We need 25 + k < 30, so k 30. So, no further cases are possible.
Total number of possible values for k = 4 (from p=2) + 1 (from p=3) = 5. This question tests the understanding of prime numbers and systematic case analysis, which are important for number theory problems in UPSC CSAT.