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UPSC Prelims 2025 CSAT, question 6

Topic: Number system

Q6

Three prime numbers p,qp,q and rr, each less than 20, are such that p−q=q−rp-q=q-r. How many distinct possible values can we get for (p+q+r)(p+q+r) ?

  1. (a) 4
  2. (b) 5
  3. (c) 6
  4. (d) More than 6Correct answer

Solution.

Answer: (d) More than 6

  1. p−q=q−rp - q = q - r means p,q,rp, q, r form an arithmetic progression with middle term qq, so p+q+r=3qp + q + r = 3q.
  2. The primes need not be distinct: p=q=rp = q = r satisfies p−q=q−r=0p - q = q - r = 0 for every prime below 20, giving sums 6,9,15,21,33,39,51,576, 9, 15, 21, 33, 39, 51, 57.
  3. The progressions of distinct primes, such as (3,5,7)(3, 5, 7) and (7,13,19)(7, 13, 19), only repeat sums already in that list, so there are 8 distinct values, more than 6.

Trap: Counting only progressions of three different primes gives sums 15,21,33,3915, 21, 33, 39, which is option (a); nothing in the question says the primes are different.