UPSC Prelims 2025 CSAT Paper II · Q17 of 78 Basic Numeracy medium

Let both p and k be prime numbers such that (p² + k) is also a prime number less than 30. What is the number of possible values of k ?

  1. (a) 4
  2. (b) 5 ✓ UPSC's answer
  3. (c) 6
  4. (d) 7

Why the answer is (b)

• Since p is a prime number, p can be 2 or an odd prime. If p is odd, p² is odd, so p² + k is even only if k is odd. The only even prime is 2, so if p² + k = 2, then p² < 2, which is impossible for any prime p. Thus, p² + k must be an odd prime, implying k must be 2 (since odd + even = odd) or p must be 2.

• Case 1: p = 2. Then p² + k = 4 + k. We need 4 + k to be a prime less than 30. So k can be 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 such that 4+k is prime. Testing: 4+2=6 (no), 4+3=7 (yes), 4+5=9 (no), 4+7=11 (yes), 4+11=15 (no), 4+13=17 (yes), 4+17=21 (no), 4+19=23 (yes), 4+23=27 (no), 4+29=33 (no). Valid k values: 3, 7, 13, 19.

• Case 2: p is an odd prime. Then p² is odd. For p² + k to be prime (and > 2), it must be odd. Thus k must be even. The only even prime is k = 2. So we test k = 2 with odd primes p such that p² + 2 < 30. p² < 28. Possible odd primes p: 3, 5. (p=7 gives 49+2=51 > 30). For p=3, 3²+2=11 (prime). For p=5, 5²+2=27 (not prime). So k=2 is a valid value.

• Combining both cases, the possible values for k are {3, 7, 13, 19} from Case 1 and {2} from Case 2. The set of possible k values is {2, 3, 7, 13, 19}.

• The number of distinct possible values of k is 5.

Why the other options are wrong

(a) 4
The count is 5, not 4, because k=2 is also a valid solution when p=3.
(c) 6
The count is 5, not 6, as no other prime k values satisfy the condition for p²+k being a prime less than 30.
(d) 7
The count is 5, not 7, as only five distinct prime values for k yield a prime sum less than 30.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.

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