Q. How many possible values of (p + q + r) are there satisfying 1/p + 1/q + 1/r = 1, where p, q and r are natural numbers (not necessarily distinct)?

Explanation : 

We seek all positive integers p ≤ q ≤ r satisfying
(1) 1/p + 1/q + 1/r = 1.

1. Bound on p.
Since 1/p + 1/q + 1/r ≤ 3/p, we require 3/p ≥ 1 ⇒ p ≤ 3.

2. Case p = 3.
Then 1/3 + 1/q + 1/r = 1 ⇒ 1/q + 1/r = 2/3.
But 1/q + 1/r ≤ 2/q, so 2/q ≥ 2/3 ⇒ q ≤ 3.
With q ≥ p = 3 we get q = 3, and hence 1/r = 2/3 – 1/3 = 1/3 ⇒ r = 3.
Solution: (3, 3, 3), sum = 9.

3. Case p = 2.
Then 1/2 + 1/q + 1/r = 1 ⇒ 1/q + 1/r = 1/2.
Since 1/q + 1/r ≤ 2/q, we need 2/q ≥ 1/2 ⇒ q ≤ 4.
Checking q = 2,3,4:
• q = 2 ⇒ 1/r = 0 (no solution)
• q = 3 ⇒ 1/r = 1/2 – 1/3 = 1/6 ⇒ r = 6 ⇒ (2,3,6), sum = 11
• q = 4 ⇒ 1/r = 1/2 – 1/4 = 1/4 ⇒ r = 4 ⇒ (2,4,4), sum = 10

No other cases arise. Hence the only triples (up to order) are
(3,3,3) with sum 9,
(2,4,4) with sum 10,
(2,3,6) with sum 11.

Therefore there are exactly three distinct values of p + q + r.

Answer: (c) Three.