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

Question

Accepted Answer

The correct answer is option A. The condition p – q = q – r implies 2q = p + r, which means p, q, and r are in an arithmetic progression. We need to find prime numbers less than 20. These are: 2, 3, 5, 7, 11, 13, 17, 19.
We need to find triplets (p, q, r) from this list such that they form an AP:
1.  (3, 5, 7): p=3, q=5, r=7. Sum = 3+5+7 = 15.
2.  (3, 7, 11): p=3, q=7, r=11. Sum = 3+7+11 = 21.
3.  (5, 11, 17): p=5, q=11, r=17. Sum = 5+11+17 = 33.
4.  (7, 13, 19): p=7, q=13, r=19. Sum = 7+13+19 = 39.
5.  (3, 11, 19): p=3, q=11, r=19. Sum = 3+11+19 = 33. (This gives the same sum as case 3, so not a distinct value for p+q+r).
The distinct possible values for (p + q + r) are 15, 21, 33, 39. There are 4 distinct values. This question combines number theory (prime numbers) with arithmetic progression concepts, a typical CSAT quantitative aptitude problem.

Q36. 36. Three prime numbers p, q, and r, each less than 20, are such that p – q = q – r. How many distinct possible values can we get for (p + q + r)?

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